initial commit of ebook 13609HEADmain

35 files changed, 6465 insertions, 0 deletions

diff --git a/.gitattributes b/.gitattributes
new file mode 100644
index 0000000..6833f05
--- /dev/null
+++ b/.gitattributes
@@ -0,0 +1,3 @@
+* text=auto
+*.txt text
+*.md text
diff --git a/13609-pdf.pdf b/13609-pdf.pdf
new file mode 100644
index 0000000..d2b963b
--- /dev/null
+++ b/13609-pdf.pdf
diff --git a/13609-t/13609-t.tex b/13609-t/13609-t.tex
new file mode 100644
index 0000000..4e2515c
--- /dev/null
+++ b/13609-t/13609-t.tex
@@ -0,0 +1,3216 @@
+\documentclass[oneside]{book}
+\usepackage[latin1]{inputenc}
+\usepackage[reqno]{amsmath}
+\usepackage{makeidx,graphicx}
+\makeindex
+\renewcommand{\chaptername}{Article}
+\DeclareMathOperator{\Sin}{Sin}
+\begin{document}
+\thispagestyle{empty}
+\small
+\begin{verbatim}
+Project Gutenberg's Vector Analysis and Quaternions, by Alexander Macfarlane
+
+This eBook is for the use of anyone anywhere in the United States and   
+most other parts of the world at no cost and with almost no restrictions
+whatsoever. You may copy it, give it away or re-use it under the terms  
+of the Project Gutenberg License included with this eBook or online at  
+www.gutenberg.org. If you are not located in the United States, you     
+will have to check the laws of the country where you are located before 
+using this eBook.
+
+Title: Vector Analysis and Quaternions
+
+Author: Alexander Macfarlane
+
+Release Date: October 5, 2004 [EBook #13609]
+
+Language: English
+
+Character set encoding: TeX
+
+Produced by David Starner, Joshua Hutchinson, John Hagerson, and the 
+Project Gutenberg On-line Distributed Proofreaders.
+
+*** START OF THIS PROJECT GUTENBERG EBOOK VECTOR ANALYSIS AND QUATERNIONS ***
+\end{verbatim}
+\normalsize
+\newpage
+
+\frontmatter
+
+\begin{center}
+\noindent \Large MATHEMATICAL MONOGRAPHS.
+
+\bigskip\footnotesize{EDITED BY} \\
+\normalsize \textsc{MANSFIELD MERRIMAN and ROBERT S. WOODWARD.}
+
+\bigskip\bigskip\huge
+No. 8.
+
+\bigskip\bigskip\huge VECTOR ANALYSIS \\
+\bigskip\footnotesize \textsc{and} \\
+\bigskip\huge QUATERNIONS.
+
+\bigskip\bigskip\footnotesize \textsc{by} \\
+\bigskip \large ALEXANDER MACFARLANE, \\
+\footnotesize\textsc{Secretary of International Association for
+Promoting the Study of Quaternions.} \\
+
+\bigskip\bigskip\normalsize NEW YORK: \\
+\medskip JOHN WILEY \& SONS. \\
+\medskip \textsc{London: CHAPMAN \& HALL, Limited.} \\
+\medskip 1906.
+\end{center}
+
+\bigskip\bigskip
+\scriptsize \noindent \textsc{Transcriber's Notes:} \emph{This
+material was originally published in a book by Merriman and Woodward
+titled \emph{Higher Mathematics}. I believe that some of the page
+number cross-references have been retained from that presentation of
+this material.}
+
+\emph{I did my best to recreate the index.} \normalsize
+
+\newpage
+
+\fbox{\parbox{11cm}{
+\begin{center}
+\textbf{MATHEMATICAL MONOGRAPHS.} \\
+\footnotesize\textsc{edited by}\normalsize \\
+\textbf{Mansfield Merriman and Robert S. Woodward.} \\
+\smallskip \footnotesize \textbf{Octavo. Cloth. \$1.00 each.}
+
+\bigskip
+\textbf{No. 1. History of Modern Mathematics.} \\
+By \textsc{David Eugene Smith.}
+
+\smallskip
+\textbf{No. 2. Synthetic Projective Geometry.} \\
+By \textsc{George Bruce Halsted.}
+
+\smallskip
+\textbf{No. 3. Determinants.} \\
+By \textsc{Laenas Gifford Weld.}
+
+\smallskip
+\textbf{No. 4. Hyperbolic Functions.} \\
+By \textsc{James McMahon.}
+
+\smallskip
+\textbf{No. 5. Harmonic Functions.} \\
+By \textsc{William E. Byerly.}
+
+\smallskip
+\textbf{No. 6. Grassmann's Space Analysis.} \\
+By \textsc{Edward W. Hyde.}
+
+\smallskip
+\textbf{No. 7. Probability and Theory of Errors.} \\
+By \textsc{Robert S. Woodward.}
+
+\smallskip
+\textbf{No. 8. Vector Analysis and Quaternions.} \\
+By \textsc{Alexander Macfarlane.}
+
+\smallskip
+\textbf{No. 9. Differential Equations.} \\
+By \textsc{William Woolsey Johnson.}
+
+\smallskip
+\textbf{No. 10. The Solution of Equations.} \\
+By \textsc{Mansfield Merriman.}
+
+\smallskip
+\textbf{No. 11. Functions of a Complex Variable.} \\
+By \textsc{Thomas S. Fiske.}
+
+\normalsize \bigskip PUBLISHED BY \\
+\smallskip \textbf{JOHN WILEY \& SONS, Inc., NEW YORK. \\
+CHAPMAN \& HALL, Limited, LONDON.}
+\end{center}}}
+
+\chapter{Editors' Preface}
+
+The volume called Higher Mathematics, the first edition of which was
+published in 1896, contained eleven chapters by eleven authors, each
+chapter being independent of the others, but all supposing the
+reader to have at least a mathematical training equivalent to that
+given in classical and engineering colleges. The publication of that
+volume is now discontinued and the chapters are issued in separate
+form. In these reissues it will generally be found that the
+monographs are enlarged by additional articles or appendices which
+either amplify the former presentation or record recent advances.
+This plan of publication has been arranged in order to meet the
+demand of teachers and the convenience of classes, but it is also
+thought that it may prove advantageous to readers in special lines
+of mathematical literature.
+
+It is the intention of the publishers and editors to add other
+monographs to the series from time to time, if the call for the same
+seems to warrant it. Among the topics which are under consideration
+are those of elliptic functions, the theory of numbers, the group
+theory, the calculus of variations, and non-Euclidean geometry;
+possibly also monographs on branches of astronomy, mechanics, and
+mathematical physics may be included. It is the hope of the editors
+that this form of publication may tend to promote mathematical study
+and research over a wider field than that which the former volume
+has occupied.
+
+\medskip \footnotesize December, 1905. \normalsize
+
+\chapter{Author's Preface}
+
+Since this Introduction to Vector Analysis and Quaternions was first
+published in 1896, the study of the subject has become much more
+general; and whereas some reviewers then regarded the analysis as a
+luxury, it is now recognized as a necessity for the exact student of
+physics or engineering. In America, Professor Hathaway has published
+a Primer of Quaternions (New York, 1896), and Dr. Wilson has
+amplified and extended Professor Gibbs' lectures on vector analysis
+into a text-book for the use of students of mathematics and physics
+(New York, 1901). In Great Britain, Professor Henrici and Mr. Turner
+have published a manual for students entitled Vectors and Rotors
+(London, 1903); Dr. Knott has prepared a new edition of Kelland and
+Tait's Introduction to Quaternions (London, 1904); and Professor
+Joly has realized Hamilton's idea of a Manual of Quaternions
+(London, 1905). In Germany Dr. Bucherer has published Elemente der
+Vektoranalysis (Leipzig, 1903) which has now reached a second
+edition.\index{Bibliography}
+
+Also the writings of the great masters have been rendered more
+accessible. A new edition of Hamilton's classic, the Elements of
+Quaternions, has been prepared by Professor Joly (London, 1899,
+1901); Tait's Scientific Papers have been reprinted in collected
+form (Cambridge, 1898, 1900); and a complete edition of Grassmann's
+mathematical and physical works has been edited by Friedrich Engel
+with the assistance of several of the eminent mathematicians of
+Germany (Leipzig, 1894--). In the same interval many papers,
+pamphlets, and discussions have appeared. For those who desire
+information on the literature of the subject a Bibliography has been
+published by the Association for the promotion of the study of
+Quaternions and Allied Mathematics (Dublin, 1904).
+
+There is still much variety in the matter of notation, and the
+relation of Vector Analysis to Quaternions is still the subject of
+discussion (see Journal of the Deutsche Mathematiker-Vereinigung for
+1904 and 1905).
+
+\medskip \footnotesize \textsc{Chatham, Ontario, Canada,}
+December, 1905. \normalsize
+
+\tableofcontents
+
+%%  1. INTRODUCTION
+%%  2. ADDITION OF COPLANAR VECTORS
+%%  3. PRODUCTS OF COPLANAR VECTORS
+%%  4. COAXIAL QUATERNIONS
+%%  5. ADDITION OF VECTORS IN SPACE
+%%  6. PRODUCT OF TWO VECTORS
+%%  7. PRODUCT OF THREE VECTORS
+%%  8. COMPOSITION OF LOCATED QUANTITIES
+%%  9. SPHERICAL TRIGONOMETRY
+%% 10. COMPOSITION OF ROTATIONS
+%% INDEX
+
+\mainmatter
+
+\chapter{Introduction.}
+
+By ``Vector Analysis'' is meant a space analysis in which the vector
+is the fundamental idea; by ``Quaternions'' is meant a
+space-analysis in which the quaternion is the fundamental idea.%
+\index{Quaternions!definion of}%
+\index{Quaternions!relation to vector analysis}%
+\index{Space-analysis}%
+\index{Vector analysis!definition of}%
+\index{Vector analysis!relation to Quaternions} They are in truth
+complementary parts of one whole; and in this chapter they will be
+treated as such, and developed so as to harmonize with one another
+and with the Cartesian Analysis\footnote{For a discussion of the
+relation of
+Vector Analysis to Quaternions, see Nature, 1891--1893.}.%
+\index{Cartesian analysis} The subject to be treated is the analysis
+of quantities in space, whether they are vector in nature, or
+quaternion in nature, or of a still different nature, or are of such
+a kind that they can be adequately represented by space quantities.
+
+Every proposition about quantities in space ought to remain true
+when restricted to a plane; just as propositions about quantities in
+a plane remain true when restricted to a straight line. Hence in the
+following articles the ascent to the algebra of space%
+\index{Algebra!of space} is made through the intermediate algebra of
+the plane\index{Algebra!of the plane}. Arts.\ 2--4 treat of the more
+restricted analysis, while Arts.\ 5--10 treat of the general
+analysis.
+
+This space analysis is a universal Cartesian analysis, in the same
+manner as algebra is a universal arithmetic. By providing an
+explicit notation for directed quantities, it enables their general
+properties to be investigated independently of any particular system
+of coordinates, whether rectangular, cylindrical, or polar. It also
+has this advantage that it can express the directed quantity by a
+linear function of the coordinates, instead of in a roundabout way
+by means of a quadratic function.%
+\index{Space-analysis!advantage over Cartesian analysis}
+
+The different views of this extension of analysis which have been
+held by independent writers are briefly indicated by the titles of
+their works:\index{Bibliography}
+
+\small\begin{itemize}
+\item Argand, Essai sur une mani�re de repr�senter les
+quantit�s imaginaires dans les constructions g�om�triques, 1806.
+
+\item Warren, Treatise on the geometrical representation of the
+square roots of negative quantities, 1828.
+
+\item Moebius, Der barycentrische Calcul, 1827.
+
+\item Bellavitis, Calcolo delle Equipollenze, 1835.
+
+\item Grassmann, Die lineale Ausdehnungslehre, 1844.
+
+\item De~Morgan, Trigonometry and Double Algebra, 1849.
+
+\item O'Brien, Symbolic Forms derived from the conception of the
+translation of a directed magnitude. Philosophical Transactions,
+1851.
+
+\item Hamilton, Lectures on Quaternions, 1853, and Elements of
+Quaternions, 1866.
+
+\item Tait, Elementary Treatise on Quaternions, 1867.
+
+\item Hankel, Vorlesungen �ber die complexen Zahlen und ihre
+Functionen, 1867.
+
+\item Schlegel, System der Raumlehre, 1872.
+
+\item Ho�el, Th�orie des quantit�s complexes, 1874.
+
+\item Gibbs, Elements of Vector Analysis, 1881--4.
+
+\item Peano, Calcolo geometrico, 1888.
+
+\item Hyde, The Directional Calculus, 1890.
+
+\item Heaviside, Vector Analysis, in ``Reprint of Electrical
+Papers,'' 1885--92.
+
+\item Macfarlane, Principles of the Algebra of Physics, 1891. Papers
+on Space Analysis, 1891--3.
+\end{itemize}
+
+An excellent synopsis is given by Hagen in the second volume of his
+``Synopsis der h�heren Mathematik.'' \normalsize
+
+\chapter{Addition of Coplanar Vectors.}
+
+By a ``vector'' is meant a quantity which has magnitude and
+direction.\index{Vector!definition of} It is graphically represented
+by a line whose length represents the magnitude on some convenient
+scale, and whose direction coincides with or represents the
+direction of the vector. Though a vector is represented by a line,
+its physical dimensions may be different from that of a line.
+Examples are a linear velocity which is of one dimension in length,
+a directed area which is of two dimensions in length, an axis which
+is of no dimensions in length.
+
+A vector will be denoted by a capital italic letter, as
+$B$,\footnote{This notation is found convenient by electrical
+writers in order to harmonize with the Hospitalier system of symbols
+and abbreviations.\index{Hospitalier system}} its magnitude by a
+small italic letter, as $b$, and its direction by a small Greek
+letter, as $\beta$.\index{Notation for vector}%
+\index{Vector!dimensions of}%
+\index{Vector!notation for} For example, $B = b\beta$, $R = r\rho$.
+Sometimes it is necessary to introduce a dot or a mark $\angle$ to
+separate the specification of the direction from the expression for
+the magnitude;\footnote{The dot was used for this purpose in the
+author's Note on Plane Algebra, 1883; Kennelly has since used
+$\angle$ for the same purpose in his electrical
+papers.\index{Kennelly's notation}}%
+\index{Meaning!of dot}%
+\index{Meaning!of $\angle$} but in such simple expressions as the
+above, the difference is sufficiently indicated by the difference of
+type. A system of three mutually rectangular axes will be indicated,
+as usual, by the letters $i$, $j$, $k$.\index{Unit-vector}
+
+The analysis of a vector here supposed is that into magnitude and
+direction. According to Hamilton and Tait and other writers on
+Quaternions, the vector is analyzed into tensor and unit-vector,
+which means that the tensor is a mere ratio destitute of dimensions,
+while the unit-vector is the physical magnitude.%
+\index{Hamilton's!analysis of vector}%
+\index{Tait's analysis of vector} But it will be found that the
+analysis into magnitude and direction is much more in accord with
+physical ideas, and explains readily many things which are difficult
+to explain by the other analysis.
+
+A vector quantity may be such that its components have a common
+point of application and are applied simultaneously;%
+\index{Simultaneous components}%
+\index{Vector!simultaneous} or it may be such that its components
+are applied in succession, each component
+starting from the end of its predecessor.%
+\index{Successive components}%
+\index{Vector!successive} An example of the former is found in two
+forces applied simultaneously at the same point, and an example of
+the latter in two rectilinear displacements made in succession to
+one another.
+
+\begin{center}
+\includegraphics[width=40mm]{fig01.png}
+\end{center}
+
+\smallskip Composition of Components having a common Point of
+Application.%
+\index{Composition!of two simultaneous components}%
+\index{Parallelogram of simultaneous components}%
+\index{Simultaneous components!composition of}%
+\index{Simultaneous components!parallelogram of}---Let $OA$ and $OB$
+represent two vectors of the same kind simultaneously applied at the
+point $O$. Draw $BC$ parallel to $OA$, and $AC$ parallel to $OB$,
+and join $OC$. The diagonal $OC$ represents in magnitude and
+direction and point of application the resultant of $OA$ and $OB$.
+This principle was discovered with reference to force, but it
+applies to any vector quantity coming under the above conditions.
+
+Take the direction of $OA$ for the initial direction; the direction
+of any other vector will be sufficiently denoted by the angle round
+which the initial direction has to be turned in order to coincide
+with it. Thus $OA$ may be denoted by $f_1\underline{/0}$, $OB$ by
+$f_2\underline{/\theta_2}$, $OC$ by $f\underline{/\theta}$. From the
+geometry of the figure it follows that
+\begin{gather*}
+f^2 = f_1^2 + f_2^2 + 2f_1f_2 \cos \theta_2 \\
+\intertext{and}
+\tan \theta =\frac{f_2\sin\theta_2}{f_1 + f_2\cos\theta_2}; \\
+\intertext{hence}
+OC = \sqrt{f_1^2 + f_2^2 + 2f_1f_2\cos\theta_2}
+  \underline{\left/\tan^{-1}
+    \frac{f_2\sin \theta_2}{f_1 + f_2\cos\theta_2}\right.}.
+\end{gather*}
+
+\smallskip Example.---Let the forces applied at a point be
+$2\underline{/0^{\circ}}$ and $3\underline{/60^{\circ}}$. Then the
+resultant is $\sqrt{4 + 9 + 12 \times \frac{1}{2}}\,
+\underline{\left/\tan^{-1}\frac{3\sqrt{3}}{7}\right.} =
+4.36\underline{/36^{\circ}\,30'}$.
+
+\smallskip If the first component is given as
+$f_1\underline{/\theta_1}$, then we have the more symmetrical
+formula
+\begin{equation*}
+OC = \sqrt{f_1^2 + f_2^2 + 2f_1f_2\cos(\theta_2 - \theta_1)}\,
+     \underline{\left/\tan^{-1} \frac{f_1\sin\theta_1 +
+       f_2\sin\theta_2}{f_1\cos\theta_1 + f_2\cos\theta_2}\right.}.
+\end{equation*}
+
+When the components are equal, the direction of the resultant
+bisects the angle formed by the vectors; and the magnitude of the
+resultant is twice the projection of either component on the
+bisecting line. The above formula reduces to
+\begin{equation*}
+OC = 2f_1\cos\frac{\theta_2}{2}
+  \underline{\left/\frac{\theta_2}{2}\right.}.
+\end{equation*}
+
+\smallskip Example.---The resultant of two equal alternating
+electromotive forces which differ $120^\circ$ in phase is equal in
+magnitude to either and has a phase of $60^\circ$.
+
+\begin{center}
+\includegraphics[width=40mm]{fig02.png}
+\end{center}
+
+\smallskip Given a vector and one component, to find the other
+component.---Let $OC$ represent the resultant, and $OA$ the
+component. Join $AC$ and draw $OB$ equal and parallel to $AC$. The
+line $OB$ represents the component required, for it is the only line
+which combined with $OA$ gives $OC$ as resultant. The line $OB$ is
+identical with the diagonal of the parallelogram formed by $OC$ and
+$OA$ reversed; hence the rule is, ``Reverse the direction of the
+component, then compound it with the given resultant to find the
+required component.'' Let $f\underline{/\theta}$ be the vector and
+$f_1\underline{/0}$ one component; then the other component is
+\begin{equation*}
+f_2\underline{/\theta_2} = \sqrt{f^2 + f_1^2 - 2ff_1\cos\theta}
+  \underline{\left/\tan^{-1}
+             \frac{f\sin\theta}{-f_1 + f\cos\theta}\right.}
+\end{equation*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig03.png}
+\end{center}
+
+\smallskip Given the resultant and the directions of the two
+components, to find the magnitude of the components.%
+\index{Resolution!of a vector}%
+\index{Simultaneous components!resolution of}---The resultant is
+represented by $OC$, and the directions by $OX$ and $OY$. From C
+draw $CA$ parallel to $OY$, and $CB$ parallel to $OX$; the lines
+$OA$ and $OB$ cut off represent the required components. It is
+evident that $OA$ and $OB$ when compounded produce the given
+resultant $OC$, and there is only one set of two components which
+produces a given resultant; hence they are the only pair of
+components having the given directions.
+
+\smallskip Let $f\underline{/\theta}$ be the vector and
+$\underline{/\theta_1}$ and $\underline{/\theta_2}$ the given
+directions. Then
+\begin{align*}
+f_1 + f_2\cos(\theta_2 - \theta_1) &= f\cos(\theta - \theta_1), \\
+f_1\cos(\theta_2 - \theta_1) + f_2 &= f\cos(\theta_2 - \theta),
+\end{align*}
+from which it follows that
+\begin{equation*}
+f_1 = f\frac{ \{\cos(\theta - \theta_1)
+        - \cos(\theta_2 - \theta)\cos(\theta_2 - \theta_1)\} }
+      {1 - \cos^2(\theta_2 - \theta_1)}.
+\end{equation*}
+
+For example, let $100\underline{/60^\circ}$,
+$\underline{/30^\circ}$, and $\underline{/90^\circ}$ be given; then
+\begin{equation*}
+f_1 = 100 \frac{\cos 30^\circ}{1 + \cos 60^\circ} .
+\end{equation*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig04.png}
+\end{center}
+
+\smallskip Composition of any Number of Vectors applied at a common
+Point.%
+\index{Composition!of any number of simultaneous components}%
+\index{Polygon of simultaneous components}%
+\index{Simultaneous components!polygon of}---The resultant may be
+found by the following graphic construction: Take the vectors in any
+order, as $A$, $B$, $C$. From the end of $A$ draw $B'$ equal and
+parallel to $B$, and from the end of $B'$ draw $C'$ equal and
+parallel to $C$; the vector from the beginning of $A$ to the end of
+$C'$ is the resultant of the given vectors. This follows by
+continued application of the parallelogram construction. The
+resultant obtained is the same, whatever the order; and as the order
+is arbitrary, the area enclosed has no physical meaning.
+
+The result may be obtained analytically as follows:
+
+Given
+\begin{gather*}
+f_1\underline{/\theta_1} + f_2\underline{/\theta_2} +
+  f_3\underline{/\theta_3} + \cdots + f_n\underline{/\theta_n}. \\
+\intertext{Now}
+f_1\underline{/\theta_1} = f_1\cos\theta_1\underline{/0} +
+  f_1\sin\theta_1\underline{\left/ \frac{\pi}{2}\right.}. \\
+\intertext{Similarly}
+f_2\underline{/\theta_2} = f_2\cos\theta_2\underline{/0}
+  + f_2\sin\theta_2\underline{\left/ \frac{\pi}{2}\right.}. \\
+\intertext{and}
+f_n\underline{/\theta_n} = f_n\cos\theta_n\underline{/0} +
+  f_n\sin\theta_n\underline{\left/ \frac{\pi}{2}\right.}. \\
+\intertext{Hence}
+\begin{align*}
+\sum\Bigl\lbrace f\underline{/\theta} \Bigr\rbrace & =
+  \Bigl\lbrace \sum f\cos\theta \Bigr\rbrace \underline{/0} +
+  \Bigl\lbrace \sum f\sum\theta \Bigr\rbrace
+    \underline{\left/ \frac{\pi}{2}\right.} \\
+& =\sqrt{\left( \sum f\cos\theta \right)^2 +
+     \left( \sum f\sin\theta \right )^2} \cdot
+  \tan^{-1}\frac{\sum f\sin\theta}{\sum f\cos\theta}.
+\end{align*}
+\end{gather*}
+
+In the case of a sum of simultaneous vectors applied at a common
+point, the ordinary rule about the transposition of a term in an
+equation holds good. For example, if $A + B + C = 0$, then $A + B =
+-C$, and $A + C = -B$, and $B + C = -A$, etc. This is permissible
+because there is no real order of succession among the given
+components.\footnote{ This does not hold true of a sum of vectors
+having a real order of succession. It is a mistake to attempt to
+found space-analysis upon arbitrary formal laws; the fundamental
+rules must be made to express universal properties of the thing
+denoted. In this chapter no attempt is made to apply formal laws to
+directed quantities. What is attempted is an analysis of these
+quantities.}\index{Space-analysis!foundation of}
+
+\begin{center}
+\includegraphics[width=50mm]{fig05.png} \qquad
+\includegraphics[width=20mm]{fig06.png}
+\end{center}
+
+\smallskip Composition of Successive Vectors.%
+  \index{Composition!of successive components}%
+  \index{Successive components!composition of}---The
+composition of successive vectors partakes more of the nature of
+multiplication than of addition. Let $A$ be a vector starting from
+the point $O$, and $B$ a vector starting from the end of $A$. Draw
+the third side $OP$, and from $O$ draw a vector equal to $B$, and
+from its extremity a vector equal to $A$. The line $OP$ is not the
+complete equivalent of $A + B$; if it were so, it would also be the
+complete equivalent of $B + A$. But $A + B$ and $B + A$ determine
+different paths; and as they go oppositely around, the areas they
+determine with $OP$ have different signs. The diagonal $OP$
+represents $A + B$ only so far as it is considered independent of
+path. For any number of successive vectors, the sum so far as it is
+independent of path is the vector from the initial point of the
+first to the final point of the last. This is also true when the
+successive vectors become so small as to form a continuous curve.
+The area between the curve $OPQ$ and the vector $OQ$ depends on the
+path, and has a physical meaning. \bigskip
+
+\small \begin{enumerate}
+
+\item[Prob.~1.] The resultant vector is $123\underline{/45^\circ}$,
+and one component is $100\underline{/0^\circ}$; find the other
+component.
+
+\item[Prob.~2.] The velocity of a body in a given plane is
+$200\underline{/75^\circ}$, and one component is
+$100\underline{/25^\circ}$; find the other component.
+
+\item[Prob.~3.] Three alternating magnetomotive forces are of equal
+virtual value, but each pair differs in phase by $120^\circ$; find
+the resultant. \hfill (Ans.~Zero.)
+
+\item[Prob.~4.] Find the components of the vector
+$100\underline{/70^\circ}$ in the directions $20^\circ$ and
+$100^\circ$.
+
+\item[Prob.~5.] Calculate the resultant vector of
+$1\underline{/10^\circ}$, $2\underline{/20^\circ}$,
+$3\underline{/30^\circ}$, $4\underline{/40^\circ}$.
+
+\item[Prob.~6.] Compound the following magnetic fluxes: $h \sin nt + h
+\sin (nt - 120^\circ)\underline{/120^\circ} + h \sin (nt -
+240^\circ)\underline{/240^\circ}$. \hfill
+(Ans.~$\frac{3}{2}h\underline{/nt}$.)
+
+\item[Prob.~7.] Compound two alternating magnetic fluxes at a point $a
+\cos nt \underline{/0}$ and $a \sin nt \underline{/\frac{\pi}{2}}$.
+(Ans.~$a \underline{/nt}$.)
+
+\item[Prob.~8.] Find the resultant of two simple alternating
+electromotive forces $100\underline{/20^\circ}$ and
+$50\underline{/75^\circ}$.
+
+\item[Prob.~9.] Prove that a uniform circular motion is obtained by
+compounding two equal simple harmonic motions which have the
+space-phase of their angular positions equal to the supplement of
+the time-phase of their motions.
+\end{enumerate} \normalsize
+
+\chapter{Products of Coplanar Vectors.}%
+\index{Coplanar vectors}%
+\index{Product!of two coplanar vectors}%
+\index{Rules!for vectors}%
+\index{Scalar product!of two coplanar vectors}%
+\index{Vector!co-planar}
+
+When all the vectors considered are confined to a common plane, each
+may be expressed as the sum of two rectangular components. Let $i$
+and $j$ denote two directions in the plane at right angles to one
+another; then $A = a_1i + a_2j$, $B = b_1i + b_2j$, $R = xi + yj$.
+Here $i$ and $j$ are not unit-vectors, but rather signs of
+direction.
+
+\smallskip Product of two Vectors.---Let $A = a_1i + a_2j$ and $B =
+b_1i + b_2j$ be any two vectors, not necessarily of the same kind
+physically. We assume that their product is obtained by applying the
+distributive law, but we do not assume that the order of the factors
+is indifferent. Hence
+\begin{equation*}
+AB = (a_1i + a_2j)(b_1i+b_2j) = a_1b_1ii + a_2b_2jj +
+        a_1b_2ij + a_2b_2ji.
+\end{equation*}
+
+If we assume, as suggested by ordinary algebra, that the
+square of a sign of direction is $+$, and further that the product
+of two directions at right angles to one another is the direction
+normal to both, then the above reduces to
+\begin{equation*}
+AB = a_1b_1 + a_2b_2 + (a_1b_2 - a_2b_1)k.
+\end{equation*}
+
+Thus the complete product%
+\index{Complete product!of two vectors}%
+\index{Product!complete} breaks up into two partial products%
+\index{Partial products}%
+\index{Product!partial}, namely, $a_1b_1 + a_2b_2$ which is
+independent of direction, and $(a_1b_2 - a_2b_1)k$ which has the
+axis of the plane for direction.\footnote{A common explanation which
+is given of $ij = k$ is that $i$ is an operator, $j$ an operand, and
+$k$ the result. The kind of operator which $i$ is supposed to denote
+is a quadrant of turning round the axis $i$; it is supposed not to
+be an axis, but a quadrant of rotation round an axis. This explains
+the result $ij = k$, but unfortunately it does not explain $ii = +$;
+for it would give $ii = i$.}
+
+\begin{center}
+\includegraphics[width=40mm]{fig07.png}
+\end{center}
+
+\smallskip Scalar Product of two Vectors.%
+\index{Product!scalar}%
+\index{Scalar product}---By a scalar quantity is meant a quantity
+which has magnitude and may be positive or negative but is destitute
+of direction. The former partial product is so called because it is
+of such a nature. It is denoted by $\mathrm{S}AB$ where the symbol
+S, being in Roman type, denotes, not a vector, but a function of the
+vectors $A$ and $B$.\index{Meaning!of S} The geometrical meaning of
+$\mathrm{S}AB$ is the product of $A$ and the orthogonal projection
+of $B$ upon $A$.\index{Scalar product!geometrical meaning} Let $OP$
+and $OQ$ represent the vectors $A$ and $B$; draw $QM$ and $NL$
+perpendicular to $OP$. Then
+\begin{align*}
+(OP)(OM) &= (OP)(OL) + (OP)(LM),  \\
+         &= a\left\{ b_1\frac{a_1}{a} + b_2\frac{a_2}{a} \right\},  \\
+         &= a_1b_1 + a_2b_2.
+\end{align*}
+
+\smallskip Corollary 1.---$\mathrm{S}BA = \mathrm{S}AB$. For instance,
+let $A$ denote a force and $B$ the velocity of its point of
+application; then $\mathrm{S}AB$ denotes the rate of working of the
+force. The result is the same whether the force is projected on the
+velocity or the velocity on the force.
+
+\medskip Example 1.---A force of $2$ pounds East + $3$ pounds North
+is moved with a velocity of $4$ feet East per second + $5$ feet
+North per second; find the rate at which work is done.
+\begin{equation*}
+  2\times 4 + 3\times 5 = 23 \text{ foot-pounds per second.}
+\end{equation*}
+
+\smallskip Corollary 2.---$A^2 = a_1^2 + a_2^2 = a^2$. The square of
+any vector is independent of direction;\index{Square!of a vector} it
+is an essentially positive or signless quantity; for whatever the
+direction of $A$, the direction of the other $A$ must be the same;
+hence the scalar product cannot be negative.
+
+\medskip Example 2.---A stone of $10$ pounds mass is moving with a
+velocity $64$ feet down per second + $100$ feet horizontal per
+second. Its kinetic energy then is
+\begin{equation*}
+  \frac{10}{2} (64^2 + 100^2) \text{ foot-poundals,}
+\end{equation*}
+a quantity which has no direction. The kinetic energy due to the
+downward velocity is $10\times\dfrac{64^2}{2}$ and that due to the
+horizontal velocity is $\dfrac{10}{2} \times 100^2$; the whole
+kinetic energy is obtained, not by vector, but by simple addition,
+when the components are rectangular.
+
+\begin{center}
+\includegraphics[width=40mm]{fig08.png}
+\end{center}
+
+\smallskip Vector Product of two Vectors.%
+\index{Product!vector}%
+\index{Vector product}%
+\index{Vector product!of two vectors}---The other partial product
+from its nature is called the vector product, and is denoted by
+$\mathrm{V}AB$.\index{Meaning!of V} Its geometrical meaning is the
+product of $A$ and the projection of $B$ which is perpendicular to
+$A$, that is, the area of the parallelogram formed upon $A$ and $B$.
+Let $OP$ and $OQ$ represent the vectors $A$ and $B$, and draw the
+lines indicated by the figure. It is then evident that the area of
+the triangle $OPQ = a_1 b_2 - \frac{1}{2} a_2 a_2 - \frac{1}{2} b_1
+b_2 - \frac{1}{2} (a_1 - b_1)(b_2 - a_2) = \frac{1}{2}(a_1 b_2 - a_2
+b_1)$.
+
+Thus $(a_1 b_2 - a_2 b_1)k$ denotes the magnitude of the
+parallelogram formed by $A$ and $B$ and also the axis of the plane
+in which it lies.
+
+It follows that $\mathrm{V}BA = -\mathrm{V}AB$. It is to be observed
+that the coordinates of $A$ and $B$ are mere component vectors,
+whereas $A$ and $B$ themselves are taken in a real order.
+
+\medskip Example.---Let $A = (10i + 11j)$~inches and $B = (5i +
+12j)$~inches, then $\mathrm{V}AB = (120-55)k$~square inches; that
+is, 65~square inches in the plane which has the direction $k$ for
+axis.
+
+\medskip If $A$ is expressed as $a\alpha$ and $B$ as $b\beta$, then
+$\mathrm{S}AB = ab \cos \alpha\beta$, where $\alpha\beta$ denotes
+the angle between the directions $\alpha$ and $\beta$.
+
+\medskip Example.---The effective electromotive force of $100$~volts
+per inch $\underline{/90^\circ}$ along a conductor $8$~inch
+$\underline{/45^\circ}$ is $\mathrm{S}AB = 8 \times 100\,
+\cos\underline{/45^\circ}\underline{/90^\circ}$~volts, that is, $800
+\cos 45^\circ$ volts. Here $\underline{/45^\circ}$ indicates the
+direction $\alpha$ and $\underline{/90^\circ}$ the direction
+$\beta$, and $\underline{/45^\circ}\underline{/90^\circ}$ means the
+angle between the direction of $45^\circ$ and the direction of
+$90^\circ$.
+
+\smallskip Also $\mathrm{V}AB = ab \sin \alpha\beta \cdot
+\overline{\alpha\beta}$, where $\overline{\alpha\beta}$ denotes the
+direction which is normal to both $\alpha$ and $\beta$, that is,
+their pole.\index{Meaning!of vinculum over two axes}
+
+\smallskip Example.---At a distance of $10$ feet
+$\underline{/30^\circ}$ there is a force of $100$ pounds
+$\underline{/60^\circ}$ The moment is $\mathrm{V}AB$
+\begin{align*}
+&= 10 \times 100 \sin \underline{/30^\circ} \underline{/60^\circ}
+  \text{ pound-feet } \overline{90^\circ/}\underline{/90^\circ} \\
+&= 1000 \sin 30^\circ \text{ pound feet }
+  \overline{90^\circ/}\underline{/90^\circ}
+\end{align*}
+
+Here $\overline{90^\circ/}$ specifies the plane of the angle and
+$\underline{/90^\circ}$ the angle. The two together written as above
+specify the normal $k$.
+
+\medskip Reciprocal of a Vector.%
+\index{Reciprocal!of a vector}%
+\index{Vector!reciprocal of}---By the reciprocal of a vector is
+meant the vector which combined with the original vector produces
+the product $+1$. The reciprocal of $A$ is denoted by $A^{-1}$.
+Since $AB = ab (\cos \alpha\beta + \sin \alpha\beta \cdot
+\overline{\alpha\beta})$, $b$ must equal $a^{-1}$ and $\beta$ must
+be identical with $\alpha$ in order that the product may be $1$. It
+follows that
+\begin{equation*}
+A^{-1} = \frac{1}{a}\alpha = \frac{a\alpha}{a^2} = \frac{a_1i +
+a_2j}{a_1^2 + a_2^2}.
+\end{equation*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig09.png}
+\end{center}
+
+The reciprocal and opposite vector is $-A^{-1}$.%
+\index{Opposite vector}%
+\index{Vector!opposite of} In the figure let $OP = 2\beta$ be the
+given vector; then $OQ = \frac{1}{2}\beta$ is its reciprocal, and
+$OR =\frac{1}{2}(-\beta)$ is its reciprocal and
+opposite.\footnote{Writers who identify a vector with a quadrantal
+versor\index{Quadrantal versor} are logically led to define the
+reciprocal of a vector as being opposite in direction as well as
+reciprocal in magnitude.}
+
+\smallskip Example.---If $A = 10 \text{ feet East} + 5 \text{ feet
+North}$, $A^{-1}= \dfrac{10}{125} \text{ feet East} \ +$ \\
+$\dfrac{5}{125} \text{ feet North}$ and $-A^{-1}=-\dfrac{10}{125}
+\text{ feet East} - \dfrac{5}{125}\text{ feet North}$.
+
+\smallskip Product of the reciprocal of a vector and another
+vector.---
+\begin{align*}
+A^{-1}B &= \frac{1}{a^2}AB, \\
+  &= \frac{1}{a^2}\left\{a_1b_1 + a_2b_2 + (a_1b_2 - a_2b_1)
+       \overline{\alpha\beta}\right\}, \\
+  &= \frac{b}{a}(\cos \alpha\beta + \sin\alpha\beta \cdot
+       \overline{\alpha\beta}).
+\end{align*}
+
+Hence $\mathrm{S}A^{-1}B = \dfrac{b}{a}\cos \alpha\beta$ and
+$\mathrm{V}A^{-1}B = \dfrac{b}{a} \sin \alpha\beta \cdot
+\overline{\alpha\beta}$.
+
+\newpage
+\medskip Product of three Coplanar Vectors.%
+\index{Association of three vectors}%
+\index{Product!of three coplanar vectors}---Let $A = a_1i + a_2j$,
+$B = b_1i + b_2j$, $C = c_1i + c_2j$ denote any three vectors in a
+common plane. Then
+\begin{align*}
+(AB)C &= \bigl\{(a_1b_1 + a_2b_2) + (a_1b_2 - a_2b_1)k \bigr\}
+            (c_1i + c_2j) \\
+      &= (a_1b_1 + a_2b_2)(c_1i + c_2j) +
+            (a_1b_2 - a_2b_1)(-c_2i + c_1j).
+\end{align*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig10.png}
+\end{center}
+
+The former partial product means the vector $C$ multiplied by the
+scalar product of $A$ and $B$; while the latter partial product
+means the complementary vector of $C$ multiplied by the magnitude of
+the vector product of $A$ and $B$. If these partial products
+(represented by $OP$ and $OQ$) unite to form a total product, the
+total product will be represented by $OR$, the resultant of $OP$ and
+$OQ$.
+
+The former product is also expressed by $\mathrm{S}AB \cdot C$,
+where the point separates the vectors to which the $\mathrm{S}$
+refers; and more analytically by $abc \cos \alpha\beta \cdot
+\gamma$.
+
+The latter product is also expressed by $(\mathrm{V}AB)C$, which is
+equivalent to $\mathrm{V}(\mathrm{V}AB)C$, because $\mathrm{V}AB$ is
+at right angles to $C$. It is also expressed by $abc \sin
+\alpha\beta \cdot \overline{\overline{\alpha\beta}\gamma}$, where
+$\overline{\overline{\alpha\beta}\gamma}$ denotes the direction
+which is perpendicular to the perpendicular to $\alpha$ and $\beta$
+and $\gamma$.
+
+If the product is formed after the other mode of association
+we have
+\begin{align*}
+A(BC) &= (a_1i + a_2j)(b_1c_1 + b_2c_2) +
+           (a_1i + a_2j)(b_1c_2 - b_2c_1)k \\
+      &= (b_1c_1 + b_2c_2)(a_1i + a_2j) +
+           (b_1c_2 - b_2c_1)(a_2i - a_1j) \\
+      &= \mathrm{S}BC \cdot A + \mathrm{V}A(\mathrm{V}BC).
+\end{align*}
+
+The vector $a_2i - a_1j$ is the opposite of the complementary
+vector of $a_1i + a_2j$. Hence the latter partial product differs
+with the mode of association.
+
+\smallskip Example.---Let $A = 1\underline{/0^\circ} +
+2\underline{/90^\circ}$, $B = 3\underline{/0^\circ} +
+4\underline{/90^\circ}$, $C = 5\underline{/0^\circ} +
+6\underline{/90^\circ}$. The fourth proportional to $A, B, C$ is
+\begin{align*}
+(A^{-1}B)C &=
+   \frac{1 \times 3 + 2 \times 4}{1^2 + 2^2}
+   \left\{ 5 \underline{/0^\circ}
+         + 6 \underline{/90^\circ}\right\} \\
+&\quad + \frac{1 \times 4 - 2 \times 3}{1^2 + 2^2}
+     \left\{ -6 \underline{/0^\circ} +
+                                  5 \underline{/90^\circ}\right \} \\
+&= 13.4 \underline{/0^\circ} + 11.2 \underline{/90^\circ}.
+\end{align*}
+
+\medskip Square of a Binomial of Vectors.%
+\index{Square!of two simultaneous components}---If $A + B$
+denotes a sum of non-successive vectors, it is entirely equivalent
+to the resultant vector $C$. But the square of any vector is a
+positive scalar, hence the square of $A + B$ must be a positive
+scalar. Since $A$ and $B$ are in reality components of one vector,
+the square must be formed after the rules for the products of
+rectangular components (p.\ 432). Hence
+\begin{align*}
+(A + B)^2 &= (A + B)(A + B), \\
+          &= A^2 + AB + BA + B^2, \\
+          &= A^2 + B^2 + \mathrm{S}AB + \mathrm{S}BA +
+               \mathrm{V}AB + \mathrm{V}BA, \\
+          &= A^2 + B^2 + 2\mathrm{S}AB.
+\end{align*}
+This may also be written in the form
+\begin{equation*}
+a^2 + b^2 + 2ab\cos\alpha\beta.
+\end{equation*}
+
+But when $A + B$ denotes a sum of successive vectors, there is no
+third vector $C$ which is the complete equivalent; and consequently
+we need not expect the square to be a scalar quantity.%
+\index{Square!of two successive components} We observe that there is
+a real order, not of the factors, but of the terms in the binomial;
+this causes both product terms to be $AB$, giving
+\begin{align*}
+(A + B)^2 &=  A^2 + 2AB + B^2 \\
+          &=  A^2+B^2 + 2\mathrm{S}AB + 2\mathrm{V}AB.
+\end{align*}
+
+The scalar part gives the square of the length of the third
+side, while the vector part gives four times the area included
+between the path and the third side.
+
+\smallskip Square of a Trinomial of Coplanar Vectors.%
+\index{Square!of three successive components}---Let $A + B + C$
+denote a sum of successive vectors. The product terms must be formed
+so as to preserve the order of the vectors in the trinomial; that
+is, $A$ is prior to $B$ and $C$, and $B$ is prior to $C$. Hence
+\begin{align}
+(A + B + C)^2 &= A^2 + B^2 + C^2 + 2AB + 2AC + 2BC, \notag \\
+              &= A^2 + B^2 + C^2 + 2(\mathrm{S}AB +
+                  \mathrm{S}AC + \mathrm{S}BC), \tag{1} \\
+              &\qquad + 2(\mathrm{V}AB + \mathrm{V}AC +
+                  \mathrm{V}BC). \tag{2}
+\end{align}
+Hence
+\begin{gather*}
+\mathrm{S}(A+B+C)^2 = (1) \\
+= a^2 + b^2 + c^2 + 2ab\cos \alpha\beta
+  + 2ac\cos \alpha\gamma + 2bc\cos\beta\gamma \\
+\intertext{and}
+\mathrm{V}(A+B+C)^2 = (2) \\
+= \{ 2ab\sin\alpha\beta + 2ac\sin\alpha\gamma + 2bc\sin\beta\gamma\}
+  \cdot \overline{\alpha\beta}
+\end{gather*}
+
+\begin{center}
+\includegraphics[width=35mm]{fig11.png}
+\end{center}
+
+The scalar part gives the square of the vector from the beginning of
+$A$ to the end of $C$ and is all that exists when the vectors are
+non-successive. The vector part is four times the area included
+between the successive sides and the resultant side of the polygon.
+
+Note that it is here assumed that $\mathrm{V}(A + B)C = \mathrm{V}AC
++ \mathrm{V}BC$, which is the theorem of moments. Also that the
+product terms are not formed in cyclical order, but in accordance
+with the order of the vectors in the trinomial.%
+\index{Cyclical and natural order}%
+\index{Natural order}
+
+\smallskip Example.---Let $A = 3\underline{/0^\circ}$,
+$B = 5\underline{/30^\circ}$, $C = 7\underline{/45^\circ}$; find the
+area of the polygon.
+\begin{align*}
+\frac{1}{2}\mathrm{V}(AB + AC + BC) &=
+\frac{1}{2}\{15\sin\underline{/0^\circ}\underline{/30^\circ} +
+        21\sin\underline{/0^\circ}\underline{/45^\circ} +
+        35\underline{/30^\circ}\underline{/45^\circ}\}, \\
+&= 3.75 + 7.42 + 4.53 = 15.7.
+\end{align*}
+
+\small \begin{enumerate}
+\item[Prob.~10.] At a distance of $25$ centimeters
+$\underline{/20^\circ}$ there is a force of 1000~dynes
+$\underline{/80^\circ}$; find the moment.
+
+\item[Prob.~11.] A conductor in an armature has a velocity of
+240~inches per second $\underline{/300^\circ}$ and the magnetic flux
+is 50,000~lines per square inch $\underline{/0^\circ}$; find the
+vector product. (Ans.~$1.04 \times 10^7$~lines per inch per second.)
+
+\item[Prob.~12.] Find the sine and cosine of the angle between the
+directions 0.8141~E.\ + 0.5807~N., and 0.5060~E.\ + 0.8625~N.
+
+\item[Prob.~13.] When a force of 200~pounds $\underline{/270^\circ}$ is
+displaced by 10~feet $\underline{/30^\circ}$, what is the work done
+(scalar product)? What is the meaning of the negative sign in the
+scalar product?
+
+\item[Prob.~14.] A mass of $100$ pounds is moving with a velocity of
+30 feet E.\ per second + 50 feet SE.\ per second; find its kinetic
+energy.
+
+\item[Prob.~15.] A force of $10$ pounds $\underline{/45^\circ}$ is
+acting at the end of $8$ feet $\underline{/200^\circ}$; find the
+torque, or vector product.
+
+\item[Prob.~16.] The radius of curvature of a curve is
+$2\underline{/0^\circ} + 5\underline{/90^\circ}$; find the
+curvature. \\ (Ans.~$.03\underline{/0^\circ} +
+.17\underline{/90^\circ}$.)
+
+\item[Prob.~17.] Find the fourth proportional to
+$10\underline{/0^\circ} + 2\underline{/90^\circ}$,
+$8\underline{/0^\circ} - 3\underline{/90^\circ}$, and
+$6\underline{/0^\circ} + 5\underline{/90^\circ}$.
+
+\item[Prob.~18.] Find the area of the polygon whose successive sides
+are $10\underline{/30^\circ}$, $9\underline{/100^\circ}$,
+$8\underline{/180^\circ}$, $7\underline{/225^\circ}$.
+\end{enumerate} \normalsize
+
+\chapter{Coaxial Quaternions.}%
+\index{Coaxial Quaternions}%
+\index{Quaternions!Coaxial}
+
+By a ``quaternion'' is meant the operator which changes one vector
+into another. It is composed of a magnitude and a turning factor.%
+\index{Components!of versor}%
+\index{Quaternion!definition of}%
+\index{Versor!components of} The magnitude may or may not be a mere
+ratio, that is, a quantity destitute of physical dimensions; for the
+two vectors may or may not be of the same physical kind. The turning
+is in a plane, that is to say, it is not conical. For the present
+all the vectors considered lie in a common plane; hence all the
+quaternions considered have a common axis.\footnote{The idea of the
+``quaternion'' is due to Hamilton.\index{Hamilton's!idea of
+quaternion} Its importance may be judged from the fact that it has
+made solid trigonometrical analysis possible. It is the most
+important key to the extension of analysis to space. Etymologically
+``quaternion'' means defined by four elements; which is true in
+space; in plane analysis it is defined by two.%
+\index{Quaternion!etymology of}}
+
+\begin{center}
+\includegraphics[width=30mm]{fig12.png}
+\end{center}
+
+Let $A$ and $R$ be two coinitial vectors; the direction normal to
+the plane may be denoted by $\beta$. The operator which changes $A$
+into $R$ consists of a scalar multiplier and a turning round the
+axis $\beta$. Let the former be denoted by $r$ and the latter by
+$\beta^\theta$, where $\theta$ denotes the angle in radians. Thus $R
+= r\beta^\theta A$ and reciprocally $A =
+\dfrac{1}{r}\beta^{-\theta}R$. Also $\dfrac{1}{A}R = r\beta^\theta$
+and $\dfrac{1}{R}A = \dfrac{1}{r}\beta^{-\theta}$.
+
+The turning factor $\beta^\theta$ may be expressed as the sum of two
+component operators, one of which has a zero angle and the other an
+angle of a quadrant. Thus
+\begin{equation*}
+\beta^\theta = \cos\theta \cdot \beta^\theta + \sin\theta \cdot
+\beta^\frac{\pi}{2}.
+\end{equation*}
+
+When the angle is naught, the turning-factor may be omitted; but the
+above form shows that the equation is homogeneous, and expresses
+nothing but the equivalence of a given quaternion to two component
+quaternions.\footnote{In the method of complex numbers
+$\beta^\frac{\pi}{2}$ is expressed by $i$, which stands for
+$\sqrt{-1}$.%
+\index{Algebraic imaginary}%
+\index{Imaginary algebraic} The advantages of using the above
+notation are that it is capable of being applied to space, and that
+it also serves to specify the general turning factor $\beta^\theta$
+as well as the quadrantal turning factor
+$\beta^\frac{\pi}{2}$.}\index{Components!of quaternion}
+
+Hence
+\begin{align*}
+r\beta^\theta & = r\cos\theta +
+                             r\sin\theta \cdot \beta^\frac{\pi}{2} \\
+& = p + q \cdot \beta^\frac{\pi}{2} \\
+\intertext{and}
+r\beta^\theta A & = pA + q \beta^\frac{\pi}{2} A \\
+& = pa \cdot \alpha + qa \cdot \beta^\frac{\pi}{2} \alpha.
+\end{align*}
+
+The relations between $r$ and $\theta$, and $p$ and $q$, are given by
+\begin{equation*}
+r = \sqrt{p^2 + q^2}, \quad \theta = \tan^{-1} \frac{p}{q}.
+\end{equation*}
+
+\medskip Example.---Let $E$ denote a sine alternating electromotive
+force in magnitude and phase, and $I$ the alternating current in
+magnitude and phase, then
+\begin{equation*}
+E = \left(r + 2\pi n l \cdot \beta^\frac{\pi}{2} \right) I,
+\end{equation*}
+where $r$ is the resistance, $l$ the self-induction, $n$ the
+alternations per unit of time, and $\beta$ denotes the axis of the
+plane of representation. It follows that $E = rI + 2\pi n l \cdot
+\beta^\frac{\pi}{2} I$; also that
+\begin{equation*}
+I^{-1} E = r + 2\pi n l \cdot \beta^\frac{\pi}{2},
+\end{equation*}
+that is, the operator which changes the current into the
+electromotive force is a quaternion. The resistance is the scalar
+part of the quaternion, and the inductance is the vector part.
+
+\medskip Components of the Reciprocal of a Quaternion.%
+\index{Components!of reciprocal of quaternion}%
+\index{Quaternion!reciprocal of}%
+\index{Reciprocal!of a quaternion}---Given
+\begin{equation*}
+R = \left(p + q \cdot \beta^\frac{\pi}{2} \right) A,
+\end{equation*}
+then
+\begin{align*}
+A & = \frac{1}{p + q \cdot \beta^\frac{\pi}{2}} R \\
+  & = \frac{p - q \cdot \beta^\frac{\pi}{2}}
+                       {\left(p + q \cdot \beta^\frac{\pi}{2} \right)
+      \left(p - q \cdot \beta^\frac{\pi}{2} \right)} R \\
+  & = \frac{p - q \cdot \beta^\frac{\pi}{2}}{p^2 + q^2} R \\
+  & = \left\{ \frac{p}{p^2 + q^2} - \frac{q}{p^2 + q^2} \cdot
+    \beta^\frac{\pi}{2} \right\} R.
+\end{align*}
+
+\smallskip Example.---Take the same application as above. It is
+important to obtain $I$ in terms of $E$. By the above we deduce that
+from $E = (r + 2\pi nl \cdot \beta^\frac{\pi}{2})I$
+\begin{equation*}
+I = \left\{\frac{r}{r^2+(2\pi nl)^2} -
+  \frac{2\pi nl}{r^2+(2\pi nl)^2}\cdot \beta^\frac{\pi}{2}\right\}E.
+\end{equation*}
+
+\medskip Addition of Coaxial Quaternions.%
+\index{Coaxial Quaternions!Addition of}---If the ratio of each of
+several vectors to a constant vector $A$ is given, the ratio of
+their resultant to the same constant vector is obtained by taking
+the sum of the ratios. Thus, if
+\begin{align*}
+R_1 &= (p_1 + q_1 \cdot \beta^\frac{\pi}{2}) A, \\
+R_2 &= (p_2 + q_2 \cdot \beta^\frac{\pi}{2}) A, \\
+\qquad \qquad \vdots & \qquad \vdots \qquad \vdots \qquad \vdots \\
+R_n &= (p_n + q_n \cdot \beta^\frac{\pi}{2}) A, \\
+\intertext{then}
+\sum R &= \left\{\sum p + \left(\sum q\right) \cdot
+                                   \beta^\frac{\pi}{2}\right\}A, \\
+\intertext{and reciprocally}
+A &= \frac{\sum p - \left(\sum q \right) \cdot \beta^\frac{\pi}{2}}
+          {\left( \sum p\ \right)^2 + \left( \sum q \right)^2}\sum R.
+\end{align*}
+
+\smallskip Example.\index{Composition!of coaxial quaternions}---In
+the case of a compound circuit composed of a number of simple
+circuits in parallel
+\begin{equation*}
+I_1 = \frac{r_1 - 2\pi nl_1 \cdot
+  \beta^\frac{\pi}{2}}{r_1^2 + (2\pi n)^2 l_1^2}E, \quad
+I_2 = \frac{r_2 - 2\pi nl_2 \cdot \beta^\frac{\pi}{2}}{r_2^2 +
+  (2\pi n)^2 l_2^2}E, \quad \text{etc.},
+\end{equation*}
+therefore,
+\begin{align*}
+\sum I & = \sum\left\{\frac{r - 2\pi nl \cdot
+   \beta^\frac{\pi}{2}} {r^2 + (2\pi n)^2 l^2}\right\}E \\
+& = \left\{\sum\left(\frac{r}{r^2 + (2\pi n)^2l^2}\right) -
+   2\pi n\sum\frac{l}{r^2 + (2\pi n)^2l^2} \cdot
+   \beta^\frac{\pi}{2}\right\}E,
+\end{align*}
+and reciprocally
+\begin{equation*}
+E = \frac{ \sum\left(\frac{r}{r^2 + (2\pi n)^2 l^2}\right) +
+    2\pi n\sum\left(\frac{l}{r^2 + (2\pi n)^2 l^2}\right) \cdot
+    \beta^\frac{\pi}{2}}
+{\left(\sum\frac{r}{r^2 + (2\pi n)^2 l^2}\right)^2 +
+    (2\pi n)^2\left(\sum\frac{l}{r^2 + (2\pi n)^2 l^2}\right)^2}
+    \sum I.\footnotemark
+\end{equation*}
+\footnotetext{This theorem was discovered by Lord
+Rayleigh\index{Rayleigh}; Philosophical Magazine, May, 1886. See
+also Bedell \& Crehore's Alternating Currents, p.\ 238.}
+
+\smallskip Product of Coaxial Quaternions.%
+\index{Coaxial Quaternions!Product of}%
+\index{Product!of coaxial quaternions}---If the quaternions which
+change $A$ to $R$, and $R$ to $R'$, are given, the quaternion which
+changes $A$ to $R'$ is obtained by taking the product of the given
+quaternions.
+
+Given
+\begin{align*}
+R  & = r\beta^\theta A = \left(p + q \cdot
+                                       \beta^\frac{\pi}{2}\right)A \\
+\intertext{and}
+R' & = r'\beta^{\theta'}A = \left(p' +
+                             q' \cdot \beta^\frac{\pi}{2}\right)R, \\
+\intertext{then}
+R' & = rr'\beta^{\theta+\theta'}A =
+  \left\{(pp'-qq') + (pq'+p'q) \cdot \beta^\frac{\pi}{2}\right\}A.
+\end{align*}
+
+Note that the product is formed by taking the product of the
+magnitudes, and likewise the product of the turning factors. The
+angles are summed because they are indices of the common base
+$\beta$.\footnote{Many writers, such as Hayward in ``Vector Algebra
+and Trigonometry,'' and Stringham in ``Uniplanar Algebra,'' treat
+this product of coaxial quaternions as if it were the product of
+vectors.\index{Hayward}\index{Stringham} This is the fundamental
+error in the Argand method.\index{Argand method}}
+
+\smallskip Quotient of two Coaxial Quaternions.%
+\index{Coaxial Quaternions!Quotient of}%
+\index{Quotient of two coaxial quaternions}---If the given
+quaternions are those which change $A$ to $R$, and $A$ to $R'$, then
+that which changes $R$ to $R'$ is obtained by taking the quotient of
+the latter by the former.
+
+Given
+\begin{align*}
+R & = r\beta^\theta A = (p + q \cdot \beta^\frac{\pi}{2})A \\
+\intertext{and}
+R' & = r'\beta'^{\theta'} A = (p' + q' \cdot \beta^\frac{\pi}{2})A,\\
+\intertext{then}
+R' & = \frac{r'}{r}\beta^{\theta' - \theta}R, \\
+   & = (p' + q' \cdot \beta^\frac{\pi}{2})\frac{1}{p + q \cdot
+       \beta^\frac{\pi}{2}}R, \\
+   & = (p' + q' \cdot \beta^\frac{\pi}{2})\frac{p - q \cdot
+       \beta^\frac{\pi}{2}}{p^2 + q^2}R, \\
+   & = \frac{(pp' + qq') + (pq' - p'q) \cdot
+       \beta^\frac{\pi}{2}}{p^2 + q^2}R.
+\end{align*}
+
+\small \begin{enumerate}
+\item[Prob.~19.] The impressed alternating electromotive force is
+$200$ volts, the resistance of the circuit is $10$ ohms, the
+self-induction is $\frac{1}{100}$ henry, and there are $60$
+alternations per second; required the current. \hfill (Ans. $18.7$
+amperes $\underline{/-20^\circ\,42'}$.)
+
+\item[Prob.~20.] If in the above circuit the current is $10$
+amperes, find the impressed voltage.
+
+\item[Prob.~21.] If the electromotive force is $110$ volts
+$\underline{/\theta}$ and the current is $10$ amperes
+$\underline{/\theta - \frac{1}{4}\pi}$, find the resistance and the
+self-induction, there being $120$ alternations per second.
+
+\item[Prob.~22.] A number of coils having resistances $r_1$, $r_2$,
+etc., and self-inductions $l_1$, $l_2$, etc., are placed in series;
+find the impressed electromotive force in terms of the current, and
+reciprocally.
+\end{enumerate} \normalsize
+
+\chapter{Addition of Vectors in Space.}
+
+A vector in space can be expressed in terms of three independent
+components, and when these form a rectangular set the directions of
+resolution are expressed by $i$, $j$, $k$.%
+\index{Composition!of simultaneous vectors in space}%
+\index{Vector!in space} Any variable vector $R$ may be expressed as
+$R = r\rho = xi + yj + zk$, and any constant vector $B$ may be
+expressed as
+\begin{equation*}
+B = b\beta = b_1i + b_2j + b_3k.
+\end{equation*}
+
+In space the symbol $\rho$ for the direction involves two elements.
+It may be specified as
+\begin{equation*}
+\rho = \frac{xi + yj + zk}{x^2 + y^2 + z^2},
+\end{equation*}
+where the three squares are subject to the condition that their sum
+is unity. Or it may be specified by this notation,
+$\overline{\phi/}\!\underline{/\theta}$, a generalization of the
+notation for a plane.\index{Meaning!of $\overline{\ /}$} The
+additional angle $\overline{\phi/}$ is introduced to specify the
+plane in which the angle from the initial line lies.
+
+If we are given $R$ in the form $r\overline{\phi/}\!
+\underline{/\theta}$, then we deduce the other form thus:
+\begin{equation*}
+R = r \cos\theta \cdot i + r \sin\theta \cos \phi \cdot j
+    + r \sin\theta \sin\phi \cdot k.
+\end{equation*}
+
+If $R$ is given in the form $xi + yj + zk$, we deduce
+\begin{gather*}
+R = \sqrt{x^2 + y^2 + z^2}\
+\overline{\left.\tan^{-1}\frac{z}{y}\right/}\!\!\!\!
+\underline{\left/\tan^{-1}\frac{\sqrt{y^2 + z^2}}{x}\right.}. \\
+\intertext{For example,}
+\begin{aligned}
+B &= 10\ \overline{30^\circ/}\! \underline{/45^\circ} \\
+  &= 10 \cos 45^\circ \cdot i
+     + 10 \sin 45^\circ \cos 30^\circ \cdot j
+     + 10 \sin 45^\circ \sin 30^\circ \cdot k.
+\end{aligned}
+\end{gather*}
+
+Again, from $C = 3i + 4j + 5k$ we deduce
+\begin{align*}
+C & = \sqrt{9 + 16 + 25}\
+      \overline{\left.\tan^{-1}\frac{5}{4}\right/}\!\!\!\!
+      \underline{\left/\tan^{-1}\frac{\sqrt{41}}{3}\right.} \\
+  & = 7.07\ \overline{51^\circ.4/}\! \underline{/64^\circ.9}.
+\end{align*}
+
+To find the resultant of any number of component vectors applied at
+a common point, let $R_1$, $R_2$, $\ldots$ $R_n$ represent the $n$
+vectors or,
+\begin{align*}
+R_1 &= x_1i + y_1j + z_1k, \\
+R_2 &= x_2i + y_2j + z_2k, \\
+\cdots & \cdots\cdots\cdots\cdots\cdots\cdots \\
+R_n &= x_ni + y_nj + z_nk; \\
+\intertext{then}
+\sum R &= \left(\sum x \right)i + \left(\sum y \right)j +
+  \left(\sum z \right)k \\
+\intertext{and}
+r &= \sqrt{\left(\sum x \right)^2 + \left(\sum y \right)^2 +
+  \left(\sum z \right)^2}, \\
+\tan\phi &= \frac{\sum z}{\sum y} \text{ and }
+  \tan\theta = \frac{\sqrt{\left(\sum y \right)^2
+         + \left(\sum z \right)^2}}{\sum x}.
+\end{align*}
+
+\medskip Successive Addition.---When the successive vectors do not
+lie in one plane, the several elements of the area enclosed will lie
+in different planes, but these add by vector addition into a
+resultant directed area.
+
+\small \begin{enumerate}
+\item[Prob.~23.] Express $A = 4i - 5j + 6k$ and $B = 5i + 6j - 7k$ in
+the form $r\overline{\phi/}\!\underline{/\theta}$ \\ (Ans.~$8.8\
+\overline{130^\circ/}\!\underline{/63^\circ}$ and $10.5\
+\overline{311^\circ/}\!\underline{/61^\circ .5}$.)
+
+\item[Prob.~24.] Express $C = 123\
+\overline{57^\circ/}\!\underline{/142^\circ}$ and $D = 456\
+\overline{65^\circ/}\!\underline{/200^\circ}$ in the form $xi + yj +
+zk$.
+
+\item[Prob.~25.] Express $E =
+100\ \overline{\left.\dfrac{\pi}{4}\right/}\!\!\!
+\underline{\left/\dfrac{\pi}{3}\right.}$ and $F = 1000\
+\overline{\left.\dfrac{\pi}{6}\right/}\!\!\!\underline{\left/
+\dfrac{3\pi}{4}\right.}$ in the form $xi + yj + zk$.
+
+\item[Prob.~26.] Find the resultant of $10\ \overline{20^\circ /}\!
+\underline{/30^\circ}$, $20\ \overline{30^\circ /}\!
+\underline{/40^\circ}$, and $30\ \overline{40^\circ /}\!
+\underline{/50^\circ}$.
+
+\item[Prob.~27.] Express in the form $r\ \overline{\phi/}\!
+\underline{/\theta}$ the resultant vector of $1i + 2j - 3k$, $4i -
+5j + 6k$ and $-7i + 8j + 9k$.
+\end{enumerate} \normalsize
+
+\chapter{Product of Two Vectors.}
+
+Rules of Signs for Vectors in Space.\index{Rules!for vectors}---By
+the rules $i^2 =+$, $j^2 = +$, $ij = k$, and $ji =-k$ we obtained
+(p.\ 432) a product of two vectors containing two partial products,
+each of which has the highest importance in mathematical and
+physical analysis. Accordingly, from the symmetry of space we assume
+that the following rules are true for the product of two vectors in
+space:
+\begin{align*}
+i^2 &=  +, & j^2 &=  +, & k^2 &=  + \, , \\
+ij  &=  k, & jk  &=  i, & ki  &=  j \, , \\
+ji  &= -k, & kj  &= -i, & ik  &= -j \, .
+\end{align*}
+
+\begin{center}
+\includegraphics[width=30mm]{fig13.png}
+\end{center}
+
+The square combinations give results which are independent of
+direction, and consequently are summed by simple addition. The area
+vector determined by $i$ and $j$ can be represented in direction by
+$k$, because $k$ is in tri-dimensional space the axis which is
+complementary to $i$ and $j$. We also observe that the three rules
+$ij = k$, $jk = i$, $ki = j$ are derived from one another by
+cyclical permutation; likewise the three rules $ji = -k$, $kj = -i$,
+$ik = -j$. The figure shows that these rules are made to represent
+the relation of the advance to the rotation in the right-handed
+screw. The physical meaning of these rules is made clearer by an
+application to the dynamo and the electric motor. In the dynamo
+three principal vectors have to be considered: the velocity of the
+conductor at any instant, the intensity of magnetic flux, and the
+vector of electromotive force. Frequently all that is demanded is,
+given two of these directions to determine the third. Suppose that
+the direction of the velocity is $i$, and that of the flux $j$, then
+the direction of the electromotive force is $k$. The formula $ij =
+k$ becomes
+\begin{gather*}
+\text{velocity flux} = \text{electromotive-force},\\
+\intertext{from which we deduce}
+\text{flux electromotive-force} = \text{velocity}, \\
+\intertext{and}
+\text{electromotive-force velocity} = \text{flux}.
+\end{gather*}
+
+The corresponding formula for the electric motor is %
+\index{Dynamo rule}%
+\index{Electric motor rule}%
+\index{Rules!for dynamo}
+\begin{equation*}
+\text{current flux} = \text{mechanical-force},
+\end{equation*}
+from which we derive by cyclical permutation
+\begin{equation*}
+   \text{flux force} = \text{current},
+\quad\text{and}\quad
+   \text{force current} = \text{flux}.
+\end{equation*}
+
+The formula $\text{velocity flux} = \text{electromotive-force}$ is much
+handier than any thumb-and-finger rule; for it compares the
+three directions directly with the right-handed screw.
+
+\medskip Example.---Suppose that the conductor is normal to the plane
+of the paper, that its velocity is towards the bottom, and that the
+magnetic flux is towards the left; corresponding to the rotation
+from the velocity to the flux in the right-handed screw we have
+advance into the paper: that then is the direction of the
+electromotive force.%
+\index{Relation of right-handed screw}%
+\index{Screw, relation of right-handed}
+
+Again, suppose that in a motor the direction of the current along
+the conductor is up from the paper, and that the magnetic flux is to
+the left; corresponding to current flux we have advance towards the
+bottom of the page, which therefore must be the direction of the
+mechanical force which is applied to the conductor.
+
+\medskip Complete Product of two Vectors%
+\index{Complete product!of two vectors}%
+\index{Product!complete}.---Let $A = a_1i + a_2j + a_3k$ and
+$B = b_1i + b_2j + b_3k$ be any two vectors, not necessarily of the
+same kind physically, Their product, according to the rules
+(p.~444), is
+\begin{align*}
+AB &= (a_1i + a_2j + a_3k)(b_1i + b_2j + b_3k), \\
+   &= a_1 b_1 ii + a_2 b_2 jj + a_3 b_3 kk \\
+   & \qquad + a_2 b_3 jk + a_3 b_2 kj + a_3 b_1 ki + a_1 b_3 ik
+     + a_1 b_2 ij + a_2 b_1 ji \\
+   &= a_1 b_1 + a_2 b_2 + a_3 b_3 \\
+   & \qquad + (a_2 b_3)i + (a_3 b_1 - a_1 b_3)j
+     + (a_1 b_2 - a_2 b_1)k \\
+   &= a_1 b_1 + a_2 b_2 +a_3 b_3 +
+            \begin{vmatrix}
+               a_1 & a_2 & a_3 \\
+               b_1 & b_2 & b_3 \\
+               i   & j   & k
+            \end{vmatrix}
+\end{align*}
+
+Thus the product breaks up into two partial products%
+\index{Partial products}%
+\index{Product!partial}, namely, $a_1 b_1 + a_2 b_2 + a_3 b_3$,
+which is independent of direction, and
+$\begin{vmatrix} a_1 & a_2 & a_3 \\
+                 b_1 & b_2 & b_3 \\
+                 i   & j   & k
+\end{vmatrix}$, which has the direction normal to the plane of
+$A$ and $B$. The former is called the scalar product, and the latter
+the vector product.%
+\index{Determinant!for vector product of two vectors}
+
+\smallskip In a sum of vectors, the vectors are necessarily
+homogeneous, but in a product the vectors may be heterogeneous. By
+making $a_3 = b_3 = 0$, we deduce the results already obtained for a
+plane.
+
+\begin{center}
+\includegraphics[width=30mm]{fig14.png}
+\end{center}
+
+\smallskip Scalar Product of two Vectors.\index{Product!scalar}---The
+scalar product is denoted as before by $\textrm{S}AB$. Its
+geometrical meaning is the product of $A$ and the orthogonal
+projection of $B$ upon $A$. Let OP represent $A$, and $OQ$ represent
+$B$, and let $OL$, $LM$, and $MN$ be the orthogonal projections upon
+$OP$ of the coordinates $b_1i$, $b_2j$, $b_3k$ respectively. Then
+$ON$ is the orthogonal projection of $OQ$, and
+\begin{align*}
+OP \times ON &= OP \times(OL + LM + MN), \\
+             &= a\left(b_1\frac{a_1}{a}
+                + b_2\frac{a_2}{a}
+                + b_3\frac{a_3}{a}\right), \\
+             &= a_1b_1 + a_2b_2 + a_3b_3 =\mathrm{S}AB.
+\end{align*}
+
+\smallskip Example.---Let the intensity of a magnetic flux be
+$B = b_1i + b_2j + b_3k$, and let the area be $S = s_1i + s_2j +
+s_3k$; then the flux through the area is $\mathrm{S}SB = b_ls_l +
+b_2s_2 + b_3s_3$.
+
+\medskip Corollary 1.---Hence $\mathrm{S}BA = \mathrm{S}AB$. For
+\begin{equation*}
+b_1a_1 + b_2a_2 + b_3a_3 = a_1b_1 + a_2b_2 + a_3b_3.
+\end{equation*}
+
+The product of $B$ and the orthogonal projection on it of $A$ is
+equal to the product of $A$ and the orthogonal projection on it of
+$B$. The product is positive when the vector and the projection have
+the same direction, and negative when they have opposite directions.
+
+\medskip Corollary 2.---Hence $A^2 = {a_1}^2 + {a_2}^2 + {a_3}^2 =
+a^2$. The square of $A$ must be positive; for the two factors have
+the same direction.
+
+\medskip Vector Product of two Vectors.%
+\index{Product!vector}%
+\index{Product!of two vectors in space}---The vector product as
+before is denoted by $\mathrm{V}AB$. It means the product of $A$ and
+the component of $B$ which is perpendicular to $A$, and is
+represented by the area of the parallelogram formed by $A$ and $B$.
+The orthogonal projections of this area upon the planes of $jk$,
+$ki$, and $ij$ represent the respective components of the product.
+For, let $OP$ and $OQ$ (see second figure of Art.\ 3) be the
+orthogonal projections of $A$ and $B$ on the plane of $i$ and $j$;
+then the triangle $OPQ$ is the projection of half of the
+parallelogram formed by $A$ and $B$. But it is there shown that the
+area of the triangle $OPQ$ is ${\frac{1}{2}}(a_1b_2 - a_2b_1)$. Thus
+$(a_1b_2 - a_2b_1)k$ denotes the magnitude and direction of the
+parallelogram formed by the projections of $A$ and $B$ on the plane
+of $i$ and $j$. Similarly $(a_2b_3 - a_3b_2)i$ denotes in magnitude
+and direction the projection on the plane of $j$ and $k$, and
+$(a_3b_1 - a_1b_3)j$ that on the plane of $k$ and $i$.
+
+\medskip Corollary 1.---Hence $\mathrm{V}BA = -\mathrm{V}AB$.
+
+\medskip Example.---Given two lines $A = 7i - 10j + 3k$ and $B = -9i
++ 4j - 6k$; to find the rectangular projections of the parallelogram
+which they define:
+\begin{align*}
+\mathrm{V}AB &= (60 - 12)i + (-27 + 42)j + (28 - 90)k \\
+             &= 48i + 15j - 62k.
+\end{align*}
+
+\medskip Corollary 2.---If $A$ is expressed as $a\alpha$ and $B$ as
+$b\beta$, then $\mathrm{S}AB = ab \cos \alpha\beta$ and
+$\mathrm{V}AB = ab \sin \alpha\beta \cdot \overline{\alpha\beta}$,
+where $\overline{\alpha\beta}$ denotes the direction which is normal
+to both $\alpha$ and $\beta$, and drawn in the sense given by the
+right-handed screw.
+
+\medskip Example.---Given $A = r\,\overline{\phi/}\!
+\underline{/\theta}$ and $B = r'\,\overline{\phi'/}\!
+\underline{/\theta'}$. Then
+\begin{align*}
+\mathrm{S}AB &= rr' \cos \overline{\phi/}\!\underline{/\theta}\:
+  \overline{\phi'/}\!\underline{/\theta'} \\
+             &= rr'\{\cos \theta \cos \theta' +
+                \sin \theta \sin \theta' cos (\phi'-\phi)\}.
+\end{align*}
+
+\medskip Product of two Sums of non-successive Vectors.%
+\index{Product!of two sums of simultaneous vectors}%
+\index{Simultaneous components!product of two sums of}---Let $A$ and
+$B$ be two component vectors, giving the resultant $A + B$, and let
+$C$ denote any other vector having the same point of application.
+
+\begin{center}
+\includegraphics[width=40mm]{fig15.png}
+\end{center}
+
+Let
+\begin{align*}
+A &= a_1j + a_2j + a_3k, \\
+B &= b_1i + b_2j + b_3k, \\
+C &= c_1i + c_2j + c_3k.
+\end{align*}
+
+Since $A$ and $B$ are independent of order,\index{Distributive rule}
+\begin{gather*}
+A + B = (a_1 + b_1)i + (a_2 + b_2)j + (a_3 + b_3)k, \\
+\intertext{consequently by the principle already established}
+\begin{split}
+\mathrm{S}(A + B)C &= (a_1 + b_1)c_1 + (a_2 + b_2)c_2
+                                                  + (a_3 + b_3)c_3 \\
+                   &= a_1c_1 + a_2c_2 + a_3c_3
+                                        + b_1c_1 + b_2c_2 + b_3c_3 \\
+                   &= \mathrm{S}AC + \mathrm{S}BC.
+\end{split}
+\end{gather*}
+
+Similarly
+\begin{align*}
+\mathrm{V}(A + B)C &= \{(a_2 + b_2)c_3 - (a_3 + b_3)c_2\}i+
+                                                       \text{etc.} \\
+                   &= (a_2c_3 - a_3c_2)i + (b_2c_3 - b_3c_2)i
+                                                          + \cdots \\
+                   &= \mathrm{V}AC + \mathrm{V}BC.
+\end{align*}
+
+Hence $(A + B)C = AC + BC$.
+
+In the same way it may be shown that if the second factor consists
+of two components, $C$ and $D$, which are non-successive in their
+nature, then
+\begin{equation*}
+(A+B)(C+D) = AC + AD + BC + BD.
+\end{equation*}
+
+When $A + B$ is a sum of component vectors
+\begin{align*}
+(A+B)^2 & = A^2 + B^2 + AB + BA \\
+        & = A^2 + B^2 + 2\mathrm{S}AB.
+\end{align*}
+
+\small \begin{enumerate}
+\item[Prob.~28.] The relative velocity of a conductor is S.W., and the
+magnetic flux is N.W.; what is the direction of the electromotive
+force in the conductor?
+
+\item[Prob.~29.] The direction of the current is vertically downward,
+that of the magnetic flux is West; find the direction of the
+mechanical force on the conductor.
+
+\item[Prob.~30.] A body to which a force of $2i + 3j + 4k$~pounds is
+applied moves with a velocity of $5i + 6j + 7k$~feet per second;
+find the rate at which work is done.
+
+\item[Prob.~31.] A conductor $8i + 9j + 10k$~inches long is subject to
+an electromotive force of $11 i +12j + 13k$~volts per inch; find the
+difference of potential at the ends. \hfill (Ans.\ 326~volts.)
+
+\item[Prob.~32.] Find the rectangular projections of the area of the
+parallelogram defined by the vectors $A = 12i - 23j - 34k$ and $B =
+-45i - 56j + 67k$.
+
+\item[Prob.~33.] Show that the moment of the velocity of a body with
+respect to a point is equal to the sum of the moments of its
+component velocities with respect to the same point.
+
+\item[Prob.~34.] The arm is $9i + 11j + 13k$~feet, and the force
+applied at either end is $17i + 19j + 23k$~pounds weight; find the
+torque.
+
+\item[Prob.~35.] A body of 1000~pounds mass has linear velocities of
+50~feet per second $\overline{30^\circ/}\!\underline{/45^\circ}$ and
+60~feet per second $\overline{60^\circ/}\!\underline{/22^\circ.5}$;
+find its kinetic energy.
+
+\item[Prob.~36.] Show that if a system of area-vectors can be
+represented by the faces of a polyhedron, their resultant vanishes.
+
+\item[Prob.~37.] Show that work done by the resultant velocity is equal
+to the sum of the works done by its components.
+\end{enumerate} \normalsize
+
+\chapter{Product of Three Vectors.}
+
+Complete Product.\index{Complete product!of three vectors}---Let us
+take $A = a_1i + a_2j + a_3k$, $B = b_1i + b_2j + b_3k$, and $C =
+c_1i + c_2j + c_3k$. By the product of $A$, $B$, and $C$ is meant
+the product of the product of $A$ and $B$ with $C$, according to the
+rules p.~444).%
+\index{Determinant!for second partial product of three vectors}
+Hence
+\begin{align*}
+ABC &= (a_1b_1 + a_2b_2 + a_3b_3)(c_1i + c_2j + c_3k) \notag \\
+&\quad+\Bigl\{(a_2b_3 - a_3b_2)i + (a_3b_1-a_1b_3)j +
+(a_1b_2-a_2b_1)k\Bigr\}
+       (c_1i+c_2j+c_3k) \notag \\
+    &= (a_1b_1+a_2b_2+a_3b_3)(c_1i+c_2j+c_3k) \tag{1} \\
+    &\quad+ \begin{vmatrix}
+              \begin{vmatrix}
+                 a_2 & a_3 \\
+                 b_2 & b_3
+              \end{vmatrix} &
+              \begin{vmatrix}
+                 a_3 & a_1 \\
+                 b_3 & b_1
+              \end{vmatrix} &
+              \begin{vmatrix}
+                 a_1 & a_2 \\
+                 b_1 & b_2
+              \end{vmatrix} \\
+              c_1 & c_2 & c_3 \\
+              i   & j   & k
+            \end{vmatrix} \tag{2} \\
+    &\quad+ \begin{vmatrix}
+               a_1 & a_2 & a_3 \\
+               b_1 & b_2 & b_3 \\
+               c_1 & c_2 & c_3
+             \end{vmatrix} \tag{3}
+\end{align*}
+
+\medskip Example.---Let $A = 1i + 2j + 3k$, $B = 4i + 5j + 6k$, and
+$C = 7i + 8j + 9k$. Then
+\begin{align*}
+(1) &= (4 + 10 + 18)(7i + 8j + 9k) = 32(7i + 8j + 9k).\\
+(2) &= \begin{vmatrix}
+          -3 & 6 & -3 \\
+           7 & 8 &  9 \\
+           i & j & k
+       \end{vmatrix} = 78i + 6j - 66k.\\
+(3) &= \begin{vmatrix}
+           1 & 2 & 3 \\
+           4 & 5 & 6 \\
+           7 & 8 & 9
+       \end{vmatrix} = 0.
+\end{align*}
+
+\smallskip If we write $A = a\alpha$, $B = b\beta$, $C = c\gamma$,
+then
+\begin{align}
+ABC &= abc \cos \alpha\beta \cdot \gamma \tag{1} \\
+    &\quad+ abc \sin \alpha\beta \sin \overline{\alpha\beta\gamma}
+           \cdot \overline{\overline{\alpha\beta}\gamma} \tag{2} \\
+    &\quad+ abc \sin\alpha\beta \cos\overline{\alpha\beta}\gamma,
+                                                              \tag{3}
+\end{align}
+where $\cos\overline{\alpha\beta}\gamma$ denotes the cosine of the
+angle between the directions $\overline{\alpha\beta}$ and $\gamma$,
+and $\overline{\overline{\alpha\beta}\gamma}$ denotes the direction
+which is normal to both $\overline{\alpha\beta}$ and $\gamma$.
+
+We may also write
+\begin{align*}
+ABC &= \mathrm{S}AB \cdot C + \mathrm{V}(\mathrm{V}AB)C
+        + \mathrm{S}(\mathrm{V}AB)C \\
+  &\quad \qquad (1) \qquad \qquad (2) \qquad \qquad (3)
+\end{align*}
+
+\medskip First Partial Product.---It is merely the third vector
+multiplied by the scalar product of the other two, or weighted by
+that product as an ordinary algebraic quantity. If the directions
+are kept constant, each of the three partial products is
+proportional to each of the three magnitudes.%
+\index{Partial products!of three vectors}
+
+\medskip Second Partial Product.---The second partial product may be
+expressed as the difference of two products similar to the
+first.%
+\index{Partial products!resolution of second partial product}%
+\index{Resolution!of second partial product of three vectors} For
+\begin{align*}
+\mathrm{V}(\mathrm{V}AB)C
+  &= \{-(b_2c_2 + b_3c_3)a_1 + (c_2a_2 + c_3a_3)b_1\}i \\
+  &\quad+ \{-(b_3c_3 + b_1c_1)a_2 + (c_3a_3 + c_1a_1)b_2\}j \\
+  &\quad+ \{-(b_1c_1 + b_2c_2)a_3 + (c_1a_1 + c_2a_2)b_3\}k.
+\end{align*}
+
+By adding to the first of these components the null term $(b_1c_1a_1
+- c_1a_1b_1)i$ we get $-\mathrm{S}BC \cdot a_1i + \mathrm{S}CA \cdot
+b_1i$, and by treating the other two components similarly and adding
+the results we obtain
+\begin{equation*}
+\mathrm{V}(\mathrm{V}AB)C = -\mathrm{S}BC \cdot A
+ + \mathrm{S}CA \cdot B.
+\end{equation*}
+
+The principle here proved is of great use in solving equations (see
+p.~455).
+
+\medskip Example.---Take the same three vectors as in the preceding
+example. Then
+\begin{align*}
+\mathrm{V}(\mathrm{V}AB)C & = -(28 + 40 + 54)(1i + 2j + 3k)\\
+  &\quad +(7 + 16 + 27)(4i + 5j + 6k) \\
+  & = 78i + 6j - 66k.
+\end{align*}
+
+\newpage
+The determinant%
+\index{Determinant!for second partial product of three vectors}
+expression for this partial product may also be written in the form
+\begin{equation*}
+\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}
+\begin{vmatrix} c_1 & c_2 \\ i & j \end{vmatrix} +
+\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}
+\begin{vmatrix} c_2 & c_3 \\ j & k \end{vmatrix} +
+\begin{vmatrix} a_3 & a_1 \\ b_3 & b_1 \end{vmatrix}
+\begin{vmatrix} c_3 & c_1 \\ k & i \end{vmatrix}
+\end{equation*}
+It follows that the frequently occurring determinant expression
+\begin{equation*}
+\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}
+\begin{vmatrix} c_1 & c_2 \\ d_1 & d_2 \end{vmatrix} +
+\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}
+\begin{vmatrix} c_2 & c_3 \\ d_2 & d_3 \end{vmatrix} +
+\begin{vmatrix} a_3 & a_1 \\ b_3 & b_1 \end{vmatrix}
+\begin{vmatrix} c_3 & c_1 \\ d_3 & d_1 \end{vmatrix}
+\end{equation*}
+means $\mathrm{S}(\mathrm{V}AB)(\mathrm{V}CD)$.
+
+\medskip Third Partial Product.---From the determinant expression for
+the third product, we know that
+\begin{align*}
+\mathrm{S}(\mathrm{V}AB)C &= \mathrm{S}(\mathrm{V}BC)A =
+  \mathrm{S}(\mathrm{V}CA)B \\
+&= -\mathrm{S}(\mathrm{V}BA)C = -\mathrm{S}(\mathrm{V}CB)A
+  = -\mathrm{S}(\mathrm{V}AC)B.
+\end{align*}
+Hence any of the three former may be expressed by $\mathrm{S}ABC$,
+and any of the three latter by $-\mathrm{S}ABC$.
+
+\begin{center}
+\includegraphics[width=40mm]{fig16.png}
+\end{center}
+
+The third product $\mathrm{S}(\mathrm{V}AB)C$ is represented by the
+volume of the parallelepiped formed by the vectors $A, B, C$
+taken in that order.%
+\index{Determinant!for scalar product of three vectors} The line
+$\mathrm{V}AB$ represents in magnitude and direction the area formed
+by $A$ and $B$, and the product of $\mathrm{V}AB$ with the
+projection of $C$ upon it is the measure of the volume in magnitude
+and sign. Hence the volume formed by the three vectors has no
+direction in space, but it is positive or negative according to the
+cyclical order of the vectors.
+
+In the expression $abc\, \sin \alpha\beta\, \cos \alpha\beta\gamma$
+it is evident that $\sin \alpha\beta$ corresponds to $\sin \theta$,
+and $\cos \alpha\beta\gamma$ to $\cos \phi$, in the usual formula
+for the volume of a parallelepiped.
+
+\medskip Example.---Let the velocity of a straight wire parallel to
+itself be $V = 1000\, \underline{/30^\circ}$ centimeters per second,
+let the intensity of the magnetic flux be $B = 6000\,
+\underline{/90^\circ}$ lines per square centimeter, and let the
+straight wire $L = 15$ centimeters $\overline{60^\circ/}\!
+\underline{/45^\circ}$. Then $\mathrm{V}VB = 6000000 \sin 60^\circ\,
+\overline{90^\circ/}\!\underline{/90^\circ}$ lines per centimeter
+per second. Hence $\mathrm{S}(\mathrm{V}VB)L = 15 \times 6000000
+\sin 60^\circ \cos \phi$ lines per second where $\cos \phi = \sin
+45^\circ\, \sin 60^\circ$.
+
+\medskip Sum of the Partial Vector Products.%
+\index{Total vector product of three vectors}%
+\index{Vector product!of three vectors}---By adding the first and
+second partial products we obtain the total vector product of $ABC$,
+which is denoted by $\mathrm{V}(ABC)$. By decomposing the second
+product we obtain
+\begin{equation*}
+\mathrm{V}(ABC) = \mathrm{S}AB \cdot C - \mathrm{S}BC \cdot A +
+\mathrm{S}CA \cdot B.
+\end{equation*}
+By removing the common multiplier $abc$, we get
+\begin{align*}
+\mathrm{V}(\alpha\beta\gamma) &= \cos \alpha\beta \cdot \gamma -
+\cos \beta\gamma \cdot \alpha + \cos \gamma\alpha \cdot \beta. \\
+\intertext{Similarly}
+\mathrm{V}(\beta\gamma\alpha) &= \cos \beta\gamma \cdot \alpha -
+\cos \gamma \alpha \cdot \beta + \cos \alpha \beta \cdot\gamma \\
+\intertext{and}
+\mathrm{V}(\gamma\alpha\beta) &= \cos \gamma\alpha \cdot \beta -
+\cos \alpha\beta \cdot \gamma + \cos \beta\gamma \cdot \alpha.
+\end{align*}
+
+These three vectors have the same magnitude, for the square of each
+is
+\begin{equation*}
+\cos^2\alpha\beta + \cos^2\beta\gamma + \cos^2\gamma\alpha -
+2\cos\alpha\beta \cos\beta\gamma\cos\gamma\alpha,
+\end{equation*}
+that is, $1-\{\mathrm{S}(\alpha\beta\gamma)\}^2.$
+
+\begin{center}
+\includegraphics[width=30mm]{fig17.png}
+\end{center}
+
+They have the directions respectively of $\alpha'$, $\beta'$,
+$\gamma'$, which are the corners of the triangle whose sides are
+bisected by the corners $\alpha$, $\beta$, $\gamma$ of the given
+triangle.
+
+\small \begin{enumerate}
+\item[Prob.~38.] Find the second partial product of $9\,
+\overline{20^\circ/}\!\underline{/30^\circ}$, $10\,
+\overline{30^\circ/}\!\underline{/40^\circ}$, $11\,
+\overline{45^\circ/}\!\underline{/45^\circ}$. Also the third partial
+product.
+
+\item[Prob.~39.] Find the cosine of the angle between the plane of
+$l_1 i + m_1 j + n_1 k$ and $l_2 i + m_2 j + n_2 k$ and the plane of
+$l_3 i + m_3 j + n_3 k$ and $l_4 i + m_4 j + n_4 k$.
+
+\item[Prob.~40.] Find the volume of the parallelepiped determined by
+the vectors $100i + 50j + 25k$, $50i + 10j + 80k$, and $-75i + 40j -
+80k$.
+
+\item[Prob.~41.] Find the volume of the tetrahedron determined by the
+extremities of the following vectors: $3i - 2j + 1k$, $-4i + 5j -
+7k$, $3i - 7j - 2k$, $8i + 4j - 3k$.
+
+\item[Prob.~42.] Find the voltage at the terminals of a conductor when
+its velocity is 1500 centimeters per second, the intensity of the
+magnetic flux is 7000 lines per square centimeter, and the length of
+the conductor is 20 centimeters, the angle between the first and
+second being $30^\circ$, and that between the plane of the first two
+and the direction of the third $60^\circ$. \hfill (Ans.
+$.91$~volts.)
+
+\item[Prob.~43.] Let $\alpha = \overline{20^\circ/}\!
+\underline{/10^\circ}$, $\beta = \overline{30^\circ/}\!
+\underline{/25^\circ}$, $\gamma=\overline{40^\circ/}\!
+\underline{/35^\circ}$. Find $\mathrm{V}\alpha\beta\gamma$, and
+deduce $\mathrm{V}\beta\gamma\alpha$ and
+$\mathrm{V}\gamma\alpha\beta$.
+\end{enumerate} \normalsize
+
+\chapter{Composition of Quantities.}
+
+A number of homogeneous quantities are simultaneously located at
+different points; it is required to find how to add or compound
+them.
+
+\begin{center}
+\includegraphics[width=40mm]{fig18.png}
+\end{center}
+
+\smallskip Addition of a Located Scalar Quantity.---Let $m_A$ denote a
+mass $m$ situated at the extremity of the radius-vector $A$. A mass
+$m-m$ may be introduced at the extremity of any radius-vector R, so
+that
+\begin{align*}
+m_A &= (m - m)_R + m_A \\
+    &= m_R + m_A - m_R \\
+    &= m_R + m(A - R).
+\end{align*}
+Here $A-R$ is a simultaneous sum, and denotes the radius-vector from
+the extremity of $R$ to the extremity of $A$. The product $m(A - R)$
+is what Clerk Maxwell called a mass-vector, and means the directed
+moment of $m$ with respect to the extremity of $R$.\index{Maxwell}
+The equation states that the mass $m$ at the extremity of the vector
+$A$ is equivalent to the equal mass at the extremity of $R$,
+together with the said mass-vector applied at the extremity of $R$.
+The equation expresses a physical of mechanical
+principle.\index{Composition!of
+mass-vectors}\index{Mass-vector}\index{Mass-vector!composition of}
+
+Hence for any number of masses, $m_1$ at the extremity of $A_1$,
+$m_2$ at the extremity of $A_2$, etc.,
+\begin{equation*}
+\sum m_A = \sum m_R + \sum\Bigl\{m(A - R)\Bigr\},
+\end{equation*}
+where the latter term denotes the sum of the mass-vectors treated as
+simultaneous vectors applied at a common point. Since
+\begin{align*}
+\sum \bigl\{m(A-R)\bigr\} &= \sum m A - \sum mR \\
+                &= \sum m A - R\sum m,
+\end{align*}
+the resultant moment will vanish\index{Couple of forces!condition
+for couple vanishing} if
+\begin{equation*}
+R = \frac{\sum mA}{\sum m},\quad\text{or}\quad R \sum m = \sum mA
+\end{equation*}
+
+\smallskip Corollary.\index{Couple of forces}---Let
+\begin{align*}
+R &= xi + yj + zk, \\
+\intertext{and}
+A &= a_1j + b_1j+c_1k; \\
+\intertext{then the above condition may be written as}
+xi + yj + zk &= \frac{\sum\bigl\{m(ai + bj + ck)\bigr\}}{\sum m} \\
+             &= \frac{\sum (ma)\cdot i}{\sum m} +
+                \frac{\sum (mb)\cdot j}{\sum m} +
+                \frac{\sum (mc)\cdot k}{\sum m}; \\
+\intertext{therefore}
+           x &= \frac{\sum (ma)}{\sum m},\
+           y  = \frac{\sum (mb)}{\sum m},\
+           z  = \frac{\sum (mc)}{\sum m}. \\
+\end{align*}
+
+Example.---Given $5$ pounds at $10$ feet $\overline{45^\circ/}\!
+\underline{/30^\circ}$ and $8$ pounds at $7$ feet
+$\overline{60^\circ/}\!\underline{/45^\circ}$; find the moment when
+both masses are transferred to $12$ feet $\overline{75^\circ/}\!
+\underline{/60^\circ}$.
+\begin{align*}
+m_1A_1 &= 50(\cos 30^\circ i + \sin 30^\circ \cos 45^\circ j
+          + \sin 30^\circ \sin 45^\circ k), \\
+m_1A_1 &= 56(\cos 45^\circ i + \sin 45^\circ \cos 60^\circ j
+          + \sin 45^\circ \sin 60^\circ k), \\
+(m_1 + m_2)R &= 156(\cos 60^\circ i + \sin 60^\circ \cos 75^\circ j
+          + \sin 60^\circ \sin 75^\circ k),\\
+\text{moment} &=  m_1 A_1 + m_2A_2 -(m_1 + m_2)R.
+\end{align*}
+
+\newpage
+\begin{center}
+\includegraphics[width=40mm]{fig19.png}
+\end{center}
+
+\smallskip Composition of a Located Vector
+Quantity.\index{Composition!of located vectors}\index{Located
+vectors}---Let $F_A$ denote a force applied at the extremity of the
+radius-vector $A$. As a force $F-F$ may introduced at the extremity
+of any radius-vector $R$, we have
+\begin{align*}
+F_A &= (F - F) + F_A \\
+    &= F_R + \mathrm{V}(A-R)F.
+\end{align*}
+
+This equation asserts that a force $F$ applied at the extremity of
+$A$ is equivalent to an equal force applied at the extremity of $R$
+together with a couple whose magnitude and direction are given by
+the vector product of the radius-vector from the extremity of $R$ to
+the extremity of $A$ and the force.
+
+Hence for a system of forces applied at different points, such as
+$F_1$ at $A_1$, $F_2$ at $A_2$, etc., we obtain
+\begin{align*}
+\sum \left(F_A\right) &= \sum \left(F_R\right)
+                         + \sum \mathrm{V}\left(A - R\right)F  \\
+           &= \left(\sum F\right)_R
+                         + \sum \mathrm{V}\left(A - R\right)F. \\
+\intertext{Since}
+\sum \mathrm{V}\left(A - R\right)F
+                        &= \sum \mathrm{V}AF - \sum \mathrm{V}RF \\
+                        &= \sum \mathrm{V}AF - \mathrm{V}R \sum F \\
+\intertext{the condition for no resultant couple is} \mathrm{V} R
+\sum F &= \sum \mathrm{V} A F,
+\end{align*}
+which requires $\sum F$ to be normal to $\sum \mathrm{V} A F$.
+
+\medskip Example.---Given a force $1i + 2j + 3k$ pounds weight at $4i
++ 5j + 6k$ feet, and a force of $7i + 9j + 11k$ pounds weight at
+$10i + 12j + 14k$ feet; find the torque which must be supplied when
+both are transferred to $2i + 5j + 3k$, so that the effect may be
+the same as before.\index{Torque}
+\begin{align*}
+\mathrm{V} A_1 F_1  &= 3i - 6j + 3k, \\
+\mathrm{V} A_2 F_2  &= 6i - 12j + 6k, \\
+\sum \mathrm{V} A F &= 9i - 18j + 9k, \\
+\sum F              &= 8i + 11j + 14k, \\
+\mathrm{V} R \sum F &= 37i - 4j - 18k, \\
+\text{Torque}       &= -28i - 14j + 27k.
+\end{align*}
+
+By taking the vector product of the above equal vectors with the
+reciprocal of $\sum F$ we obtain
+\begin{equation*}
+\mathrm{V}\left\{ \left(\mathrm{V} R \sum F\right)
+                \frac{1}{\sum F} \right\}
+= \mathrm{V}\left\{ \left(\sum \mathrm{V} A F \right)
+                \frac{1}{\sum F} \right\}.
+\end{equation*}
+
+By the principle previously established the left member resolves
+into $-R + \mathrm{S}R\dfrac{1}{\sum F} \cdot \sum F$; and the right
+member is equivalent to the complete product on account of the two
+factors being normal to one another; hence
+\begin{align}
+-R &+ \mathrm{S} R \frac{1}{\sum F} \cdot \sum F
+   = \sum \left(\mathrm{V} A F \right) \frac{1}{\sum F}; \notag \\
+\intertext{that is,}
+R &= \frac{1}{\sum F}\sum \left(\mathrm{V}AF \right) \tag{1} \\
+  &\quad+ \mathrm{S}R\frac{1}{\sum F} \cdot \sum F \tag{2}.
+\end{align}
+
+\begin{center}
+\includegraphics[width=25mm]{fig20.png}
+\end{center}
+
+The extremity of $R$ lies on a straight line whose perpendicular is
+the vector (1) and whose direction is that of the resultant force.
+The term (2) means the projection of $R$ upon that line.
+
+The condition for the central axis\index{Central axis} is that the
+resultant force and the resultant couple should have the same
+direction; hence it is given by
+\begin{align*}
+\mathrm{V}\left\{\sum \mathrm{V}AF - \mathrm{V}R\sum F\right\}
+  \sum F = 0; \\
+\intertext{that is}
+\mathrm{V}\left(\mathrm{V}R\sum F \right)\sum F =
+  \mathrm{V}\left(\sum AF \right)\sum F.
+\end{align*}
+
+By expanding the left member according to the same principle as
+above, we obtain
+\begin{equation*}
+-\left(\sum F\right)^2R + \mathrm{S}R\sum F \cdot \sum F
+  = V\left(\sum AF \right)\sum F;
+\end{equation*}
+therefore
+\begin{align*}
+R &= \frac{1}{\left(\sum F \right)^2}\mathrm{V}\sum F
+     \left(\mathrm{V}\sum AF\right) +
+     \frac{\mathrm{S}R\sum F}{\left(\sum F\right)^2} \cdot \sum F \\
+  &= \mathrm{V}\left(\frac{1}{\sum F}\right)(\mathrm{V}\sum AF) +
+       \mathrm{S}R\frac{1}{\sum F} \cdot \sum F.
+\end{align*}
+
+This is the same straight line as before, only no relation is now
+imposed on the directions of $\sum F$ and $\sum \mathrm{V}AF$; hence
+there always is a central axis.
+
+\medskip Example.---Find the central axis for the system of forces in
+the previous example. Since $\sum F = 8i + 11j + 14k$, the direction
+of the line is
+\begin{equation*}
+\frac{8i + 11j + 14k}{\sqrt{64 + 121 + 196}}.
+\end{equation*}
+
+Since $\dfrac{1}{\sum F} = \dfrac{8i + 11j + 14k}{381}$ and $\sum
+\mathrm{V}AF = 9i - 18j + 9k$, the perpendicular to the line is
+\begin{equation*}
+\mathrm{V}\,\frac{8i + 11j + 14k}{381}\, 9i - 18j + 9k =
+  \frac{1}{381}\,\{351i + 54j -243k\}.
+\end{equation*}
+
+\small \begin{enumerate}
+\item[Prob.~44.] Find the moment at $\overline{90^\circ/}\!
+\underline{/270^\circ}$ of 10~pounds at 4~feet
+$\overline{10^\circ/}\!\underline{/20^\circ}$ and 20~pounds at
+5~feet $\overline{30^\circ/}\!\underline{/120^\circ}$.
+
+\item[Prob.~45.] Find the torque for $4i + 3j + 2k$ pounds weight at
+$2i - 3j + 1k$ feet, and $2i - 1k - 1k$ pounds weight at $-3i + 4j +
+5k$~feet when transferred to $-3i -2j -4k$ feet.
+
+\item[Prob.~46.] Find the central axis in the above case.
+
+\item[Prob.~47.] Prove that the mass-vector drawn from any origin to a
+mass equal to that of the whole system placed at the center of mass
+of the system is equal to the sum of the mass-vectors drawn from the
+same origin to all the particles of the system.
+\end{enumerate} \normalsize
+
+\chapter{Spherical Trigonometry.}\index{Spherical trigonometry}
+
+\begin{center}
+\includegraphics[width=40mm]{fig21.png}
+\end{center}
+
+Let $i$, $j$, $k$ denote three mutually perpendicular axes. In order
+to distinguish clearly between an axis and a quadrantal version
+round it, let $i^\frac{\pi}{2}$, $j^\frac{\pi}{2}$,
+$k^\frac{\pi}{2}$ denote quadrantal versions in the positive sense
+about the axes $i$, $j$, $k$ respectively.\index{Meaning!of
+$\frac{1}{2}\pi$ as index} The directions of positive version are
+indicated by the arrows.
+
+By $i^\frac{\pi}{2}i^\frac{\pi}{2}$ is meant the product of two
+quadrantal versions round $i$; it is equivalent to a semicircular
+version round $i$; hence $i^\frac{\pi}{2}i^\frac{\pi}{2} = i^\pi =
+-$. Similarly $j^\frac{\pi}{2}j^\frac{\pi}{2}$ means the product of
+two quadrantal versions round $j$, and
+$j^\frac{\pi}{2}j^\frac{\pi}{2}=j^\pi=-$. Similarly
+$k^\frac{\pi}{2}k^\frac{\pi}{2}=k^\pi=-$.
+
+By $i^\frac{\pi}{2}j^\frac{\pi}{2}$ is meant a quadrant round $i$
+followed by a quadrant round $j$; it is equivalent to the quadrant
+from $j$ to $i$, that is, to $-k^\frac{\pi}{2}$. But
+$j^\frac{\pi}{2}i^\frac{\pi}{2}$ is equivalent to the quadrant from
+$-i$ to $-j$, that is, to $k^\frac{\pi}{2}$. Similarly for the other
+two pairs of products. Hence we obtain the following
+
+\begin{center}
+Rules for Versors.\index{Rules!for versors}\index{Versor!rules for}
+\end{center}
+\begin{gather*}
+i^\frac{\pi}{2}i^\frac{\pi}{2} = -, \quad
+j^\frac{\pi}{2}j^\frac{\pi}{2} = -, \quad
+k^\frac{\pi}{2}k^\frac{\pi}{2} = -, \\
+i^\frac{\pi}{2}j^\frac{\pi}{2} = -k^\frac{\pi}{2}, \quad
+j^\frac{\pi}{2}i^\frac{\pi}{2} =  k^\frac{\pi}{2}, \\
+j^\frac{\pi}{2}k^\frac{\pi}{2} = -i^\frac{\pi}{2}, \quad
+k^\frac{\pi}{2}j^\frac{\pi}{2} =  i^\frac{\pi}{2}  \\
+k^\frac{\pi}{2}i^\frac{\pi}{2} = -j^\frac{\pi}{2}, \quad
+i^\frac{\pi}{2}k^\frac{\pi}{2} =  j^\frac{\pi}{2}.
+\end{gather*}
+
+The meaning of these rules will be seen from the following
+application. Let $li + mj + nk$ denote any axis, then $(li + mj +
+nk)^\frac{\pi}{2}$ denotes a quadrant of angle round that axis. This
+quadrantal version can be decomposed into the three rectangular
+components $li^\frac{\pi}{2}$, $mj^\frac{\pi}{2}$,
+$nk^\frac{\pi}{2}$; and these components are not successive
+versions, but the parts of one version.\index{Versor!components of}
+Similarly any other quadrantal version $(l'i + m'j +
+n'j)^\frac{\pi}{2}$ can be resolved into
+$l'i^\frac{\pi}{2}$, $m'j^\frac{\pi}{2}$, $n'k^\frac{\pi}{2}$.%
+\index{Product!of two quadrantal versors}%
+\index{Rules!for expansion of product of two quadrantal versors} By
+applying the above rules, we obtain
+\begin{align*}
+(li &+ mj + nk)^\frac{\pi}{2}(l'i + m'j + n'k)^\frac{\pi}{2} \\
+    &= (li^\frac{\pi}{2} + mj^\frac{\pi}{2} + nk^\frac{\pi}{2})
+         (l'i^\frac{\pi}{2} + m'j^\frac{\pi}{2} + n'k^\frac{\pi}{2}) \\
+    &= -(ll' + mm' + nn') -(mn' - m'n)i^\frac{\pi}{2}
+            - (nl' - n'l)j^\frac{\pi}{2} -(lm' - l'm)k^\frac{\pi}{2} \\
+    &= -(ll' + mm' + nn')-\bigl\{(mn' - m'n)i + (nl' - n'l)j
+             +(lm' - l'm)k\bigr\}^\frac{\pi}{2}.
+\end{align*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig22.png}
+\end{center}
+
+\smallskip Product of Two Spherical Versors.%
+\index{Product!of two spherical versors}%
+\index{Spherical versor}%
+\index{Spherical versor!product of two}%
+\index{Versor!product of two quadrantal}%
+\index{Versor!product of two general spherical}---Let $\beta$ denote
+the axis and $b$ the ratio of the spherical versor $PA$, then the
+versor itself is expressed by $\beta^b$. Similarly let $\gamma$
+denote the axis and $c$ the ratio of the spherical versor $AQ$, then
+the versor itself is expressed by $\gamma^c$.
+
+Now
+\begin{align*}
+\beta^b &= \cos b + \sin b \cdot \beta^\frac{\pi}{2}, \\
+\intertext{and}
+\gamma^c &= \cos c + \sin c \cdot \gamma^\frac{\pi}{2}; \\
+\intertext{therefore}
+\beta^b\gamma^c &= (\cos b + \sin b \cdot
+\beta^\frac{\pi}{2})(\cos c + \sin c \cdot \gamma^\frac{\pi}{2}) \\
+ &= \cos b \cos c + \cos b \sin c \cdot \gamma^\frac{\pi}{2}
+     + \cos c \sin b \cdot \beta^\frac{\pi}{2}
+     + \sin b \sin c \cdot \beta^\frac{\pi}{2}\gamma^\frac{\pi}{2}.
+\end{align*}
+
+\smallskip But from the preceding paragraph
+\begin{align}
+\beta^\frac{\pi}{2}\gamma^\frac{\pi}{2} &= -\cos\beta\gamma -
+  \sin\beta\gamma \cdot
+                      \overline{\beta\gamma}^\frac{\pi}{2}; \notag \\
+\intertext{therefore}
+\beta^b\gamma^c &=
+            \cos b \cos c - \sin b \sin c \cos \beta\gamma \tag{1} \\
+&\quad+ \{\cos b \sin c \cdot \gamma + \cos c \sin b \cdot \beta -
+\sin b \sin c \sin \beta\gamma \cdot
+\overline{\beta\gamma}\}^\frac{\pi}{2}. \tag{2}
+\end{align}
+
+\smallskip The first term gives the cosine of the product versor; it
+is equivalent to the fundamental theorem of spherical
+trigonometry,\index{Spherical trigonometry!fundamental theorem of}
+namely,
+\begin{equation*}
+\cos a = \cos b \cos c + \sin b \sin c \cos A,
+\end{equation*}
+where $A$ denotes the external angle instead of the angle included
+by the sides.
+
+The second term is the directed sine of the angle; for the square of
+(2) is equal to 1 minus the square of (1), and its direction is
+normal to the plane of the product angle.\footnote{Principles of
+Elliptic and Hyperbolic Analysis, p.~2.}
+
+\medskip Example.---Let $\beta = \overline{30^\circ/}\!
+\underline{/45^\circ}$ and $\gamma = \overline{60^\circ/}\!
+\underline{/30^\circ}$. Then
+\begin{align*}
+\cos \beta\gamma &= \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin
+30^\circ \cos 30^\circ, \\
+\intertext{and}
+\sin \beta\gamma &\cdot \overline{\beta\gamma} =
+  \mathrm{V}\beta\gamma; \\
+\intertext{but}
+\beta &= \cos 45^\circ i + \sin 45^\circ \cos
+30^\circ j + \sin 45^\circ \sin 30^\circ k, \\
+\intertext{and}
+\gamma &= \cos 30^\circ i + \sin 30^\circ \cos
+60^\circ j + \sin 30^\circ \sin 60^\circ k; \\
+\intertext{therefore}
+\mathrm{V}\beta\gamma & = \{\sin 45^\circ \cos 30^\circ
+   \sin 30^\circ \sin 60^\circ -\sin 45^\circ \sin 30^\circ
+   \sin 30^\circ \cos 60^\circ \}i \\
+&\quad+ \{ \sin 45^\circ \sin 30^\circ \cos 30^\circ -
+   \cos 45^\circ \sin 30^\circ \sin 60^\circ \} j \\
+&\quad+ \{ \cos 45^\circ \sin 30^\circ \cos 60^\circ -
+   \sin 45^\circ \cos 30^\circ \cos 30^\circ \} k.
+\end{align*}
+
+\medskip Quotient of Two Spherical Versors.%
+\index{Spherical versor!quotient of two}---The reciprocal of a given
+versor is derived by changing the sign of the index; $\gamma^{-c}$
+is the reciprocal of $\gamma^c$. As $\beta^b = \cos b + \sin b \cdot
+\beta^\frac{\pi}{2}$, and $y^{-c} = \cos c - \sin c \cdot
+\gamma^\frac{\pi}{2}$,
+\begin{align*}
+\beta^b\gamma^{-c} &= \cos b \cos c +
+                                    \sin b \sin c \cos \beta\gamma \\
+  &+\{\cos c \sin b \cdot \beta - \cos b \sin c \cdot \gamma
+      + \sin b \sin c \sin \beta\gamma \cdot
+      \overline{\beta\gamma}\}^\frac{\pi}{2}.
+\end{align*}
+
+\begin{center}
+\includegraphics[width=30mm]{fig23.png}
+\end{center}
+
+\smallskip Product of Three Spherical Versors.%
+\index{Product!of three spherical versors}%
+\index{Spherical versor!product of three}%
+\index{Versor!product of three general spherical}---Let $\alpha^a$
+denote the versor $PQ$, $\beta^b$ the versor $QR$, and $\gamma^c$
+the versor $RS$; then $\alpha^a\beta^b\gamma^c$ denotes $PS$. Now
+$\alpha^a\beta^b\gamma^c$
+\begin{align}
+=&(\cos a + \sin a \cdot\alpha^\frac{\pi}{2})
+  (\cos b + \sin b \cdot\beta^\frac{\pi}{2})
+  (\cos c + \sin c \cdot\gamma^\frac{\pi}{2}) \notag \\
+=& \cos a\cos b\cos c \tag{1} \\
+ & + \cos a \cos b \sin c \cdot \gamma^\frac{\pi}{2} +
+     \cos a \cos c \sin b \cdot \beta^\frac{\pi}{2} +
+     \cos b \cos c \sin a \cdot \alpha^\frac{\pi}{2} \tag{2} \\
+ & + \cos a \sin b \sin c \cdot
+                            \beta^\frac{\pi}{2}\gamma^\frac{\pi}{2} +
+     \cos b \sin a \sin c \cdot
+                   \alpha^\frac{\pi}{2}\gamma^\frac{\pi}{2} \notag \\
+ & \qquad + \cos c \sin a \sin b \cdot
+                              \alpha^\frac{\pi}{2}\beta^\frac{\pi}{2}
+ \tag{3} \\
+ & + \sin a \sin b \sin c \cdot
+          \alpha^\frac{\pi}{2}\beta^\frac{\pi}{2}\gamma^\frac{\pi}{2}
+\tag{4}
+\end{align}
+
+The versors in (3) are expanded by the rule already obtained,
+namely,
+\begin{equation*}
+\beta^\frac{\pi}{2}\gamma^\frac{\pi}{2} = -\cos \beta\gamma -\sin
+\beta\gamma \cdot \overline{\beta\gamma}^\frac{\pi}{2}.
+\end{equation*}
+
+The versor of the fourth term is
+\begin{align*}
+\alpha^\frac{\pi}{2}\beta^\frac{\pi}{2}\gamma^\frac{\pi}{2} &=
+  -(\cos\alpha\beta + \sin\alpha\beta \cdot
+      \overline{\alpha\beta}^\frac{\pi}{2}) \gamma^\frac{\pi}{2} \\
+&= -\cos\alpha\beta \cdot \gamma^\frac{\pi}{2} +
+  \sin\alpha\beta\cos\overline{\alpha\beta}\gamma +
+  \sin\alpha\beta\sin\overline{\alpha\beta}\gamma \cdot
+  \overline{\overline{\alpha\beta}\gamma}^\frac{\pi}{2}.
+\end{align*}
+
+Now $\sin\alpha\beta \sin\overline{\alpha\beta}\gamma \cdot
+\overline{\overline{\alpha\beta}\gamma} = \cos\alpha\gamma \cdot
+\beta - \cos\beta\gamma \cdot \alpha$ (p.~451), hence the last term
+of the product, when expanded, is
+\begin{equation*}
+\sin a\sin b\sin c\left\{-\cos \alpha\beta \cdot
+  \gamma^\frac{\pi}{2}
++ \cos\alpha\gamma \cdot \beta^\frac{\pi}{2}
+- \cos\beta\gamma \cdot \alpha^\frac{\pi}{2}
++ \cos\overline{\alpha\beta}\gamma\right\}.
+\end{equation*}
+
+\newpage
+Hence
+\begin{align*}
+\cos\alpha^a\beta^b\gamma^c &=
+  \cos a\cos b\cos c - \cos a\sin b\sin c\cos \beta\gamma \\
+&- \cos b\sin a\sin c\cos \alpha\gamma -
+                                \cos c\sin a\sin b\cos \alpha\beta \\
+&+ \sin a\sin b\sin c\sin \alpha\beta\cos\alpha\beta\gamma, \\
+\intertext{and, letting Sin denote the directed sine,}
+\Sin \alpha^a\beta^b\gamma^c &=
+  \cos a \cos b \sin c \cdot \gamma +
+                                  \cos a \cos c \sin b \cdot \beta \\
+&+ \cos b \cos c \sin a \cdot \alpha -
+  \cos a \sin b \sin c \sin \beta\gamma \cdot
+                                            \overline{\beta\gamma} \\
+&- \cos b \sin a \sin c \sin \alpha\gamma \cdot
+                                           \overline{\alpha\gamma} \\
+&- \cos c \sin a \sin b \sin \alpha\beta \cdot
+                                            \overline{\alpha\beta} \\
+&- \sin a \sin b \sin c\left\{\cos\alpha\beta \cdot \gamma -
+  \cos \alpha\gamma \cdot  \beta + \cos \beta\gamma \cdot
+  \alpha\right\}.\footnotemark
+\end{align*}
+\footnotetext{In the above case the three axes of the successive
+angles are not perfectly independent, for the third angle must begin
+where the second leaves off. But the theorem remains true when the
+axes are independent; the factors are then quaternions in the most
+general sense.}
+
+Extension of the Exponential Theorem to Spherical
+Trigonometry.\index{Binomial theorem in spherical analysis}%
+\index{Exponential theorem in spherical trigonometry}---It has been
+shown (p.~458) that
+\begin{align*}
+\cos\beta^b\gamma^c &= \cos b\cos c - \sin b\sin c\cos \beta\gamma
+\intertext{and}
+\left(\sin \beta^b\gamma^c\right)^\frac{\pi}{2} &=
+  \cos c \sin b \cdot \beta^\frac{\pi}{2} + \cos b \sin c \cdot
+  \gamma^\frac{\pi}{2} - \sin b \sin c \sin \beta\gamma \cdot
+  \overline{\beta\gamma}^\frac{\pi}{2}.
+\end{align*}
+
+Now
+\begin{align*}
+\cos b &= 1 - \frac{b^2}{2!} + \frac{b^4}{4!} - \frac{b^6}{6!} +
+   \text{ etc.} \\
+\intertext{and} \sin b &= b - \frac{b^3}{3!} + \frac{b^5}{5!} -
+   \text{ etc.}
+\end{align*}\index{Spherical trigonometry!binomial theorem}
+
+\smallskip Substitute these series for $\cos b$, $\sin b$, $\cos c$,
+and $\sin c$ in the above equations, multiply out, and group the
+homogeneous terms together. It will be found that
+\begin{align*}
+\cos\beta^b\gamma^c = 1
+  &- \frac{1}{2!}\{b^2 + 2bc\cos\beta\gamma + c^2\} \\
+  &+ \frac{1}{4!}\{b^4 + 4b^3c\cos\beta\gamma + 6b^2c^2 +
+       4bc^3\cos\beta\gamma + c^4\} \\
+  &- \frac{1}{6!}\{b^6 + 6b^5c\cos\beta\gamma + 15b^4c^2 +
+       20b^3c^3\cos\beta\gamma \\
+  & \qquad \qquad + 15b^2c^4 + 6bc^5\cos\beta\gamma + c^6\} + \ldots,
+\end{align*}
+where the coefficients are those of the binomial theorem, the only
+difference being that $\cos \beta\gamma$ occurs in all the odd terms
+as a factor. Similarly, by expanding the terms of the sine, we
+obtain
+\begin{align*}
+(\Sin \beta^b\gamma^c)^\frac{\pi}{2} &=
+    b \cdot \beta^\frac{\pi}{2} + c \cdot \gamma^\frac{\pi}{2} -
+    bc \sin \beta\gamma \cdot \overline{\beta\gamma}^\frac{\pi}{2} \\
+&\quad- \frac{1}{3!}\{b^3 \cdot \beta^\frac{\pi}{2} + 3b^2c \cdot
+      \gamma^\frac{\pi}{2} +
+    3bc^2 \cdot \beta^\frac{\pi}{2} + c^3 \cdot
+                                            \gamma^\frac{\pi}{2}\} \\
+&\quad+ \frac{1}{3!}\{bc^3 + b^3c\}
+      \sin\beta\gamma \cdot \overline{\beta\gamma}^\frac{\pi}{2} \\
+&\quad+ \frac{1}{5!}\{b^5 \cdot \beta^\frac{\pi}{2} +
+      5b^4c \cdot \gamma^\frac{\pi}{2} + 10b^3c^2 \cdot
+      \beta^\frac{\pi}{2} \\
+&\quad\qquad + 10b^2c^3\cdot\gamma^\frac{\pi}{2} + 5bc^4 \cdot
+      \beta^\frac{\pi}{2}
+      + c^5 \cdot \gamma^\frac{\pi}{2} \\
+&\quad- \frac{1}{5!}\left\{b^5c + \frac{5\cdot 4}{2\cdot 3}b^2c^3 +
+      bc^5\right\} \sin\beta\gamma \cdot
+      \overline{\beta\gamma}^\frac{\pi}{2} - \ldots
+\end{align*}
+
+By adding these two expansions together we get the expansion for
+$\beta^b\gamma^c$, namely,
+\begin{align*}
+\beta^b\gamma^c = 1 &+ b \cdot\beta^\frac{\pi}{2} +
+                                        c\cdot\gamma^\frac{\pi}{2} \\
+&- \frac{1}{2!}\{b^2 + 2bc(\cos\beta\gamma + \sin\beta\gamma \cdot
+  \overline{\beta\gamma}^\frac{\pi}{2}) + c^2\} \\
+&- \frac{1}{3!}\{b^3 \cdot \beta^\frac{\pi}{2} + 3b^2c
+\cdot\gamma^\frac{\pi}{2}
+  + 3bc^2 \cdot \beta^\frac{\pi}{2} + c^3 \cdot
+                                            \gamma^\frac{\pi}{2}\} \\
+&+ \frac{1}{4!}\{b^4 + 4b^3c(\cos\beta\gamma + \sin\beta\gamma \cdot
+    \overline{\beta\gamma}^\frac{\pi}{2}) + 6b^2c^2 \\
+&\qquad + 4bc^3(\cos\beta\gamma+\sin\beta\gamma \cdot
+  \overline{\beta\gamma}^\frac{\pi}{2}) + c^4\} + \ldots
+\end{align*}
+
+By restoring the minus, we find that the terms on the second line
+can be thrown into the form
+\begin{gather*}
+\frac{1}{2!} \left\{ b^2 \cdot \beta^{\pi} + 2bc \cdot
+\beta^{\frac{\pi}{2}}\gamma^{\frac{\pi}{2}} + c^{2} \cdot
+\gamma^{\pi} \right\}, \\
+\intertext{and this is equal to}
+\frac{1}{2!} \left\{ b \cdot \beta^{\frac{\pi}{2}} +
+  c \cdot \gamma^{\frac{\pi}{2}} \right\}^2, \\
+\intertext{where we have the square of a sum of successive terms. In
+a similar manner the terms on the third line can be restored to}
+b^3 \cdot \beta^\frac{3\pi}{2} +
+  3b^2c \cdot \beta^\pi \gamma^\frac{\pi}{2} +
+  3bc^2 \cdot \beta^\frac{\pi}{2}\gamma^\pi +
+  c^3 \cdot \gamma^{3(\frac{\pi}{2})}, \\
+\intertext{that is,}
+\frac{1}{3!} \left\{ b \cdot \beta^\frac{\pi}{2} +
+  c \cdot \gamma^\frac{\pi}{2} \right\} ^3.
+\end{gather*}
+
+Hence
+\begin{align*}
+\beta^b\gamma^c &= 1 + b \cdot \beta^\frac{\pi}{2}
+      + c \cdot \gamma^\frac{\pi}{2}
+    + \frac{1}{2!} \left\{ b \cdot \beta^\frac{\pi}{2}
+      + c \cdot \gamma^\frac{\pi}{2} \right\} ^2 \\
+   &\qquad + \frac{1}{3!} \left\{ b \cdot \beta^\frac{\pi}{2}
+      + c \cdot \gamma^\frac{\pi}{2} \right\} ^3
+      + \frac{1}{4!} \left\{ b \cdot \beta^\frac{\pi}{2}
+      + c \cdot \gamma^\frac{\pi}{2} \right\} ^4 + \ldots \\
+&= e ^{b \cdot \beta^\frac{\pi}{2} + c \cdot
+\gamma^\frac{\pi}{2}}.\footnotemark
+\end{align*}
+\footnotetext{At page 386 of his Elements of Quaternions, Hamilton
+says: ``In the present theory of diplanar quaternions we cannot
+expect to find that the sum of the logarithms of any two proposed
+factors shall be generally equal to the logarithm of the product;
+but for the simpler and earlier case of coplanar quaternions, that
+algebraic property may be considered to exist, with due modification
+for multiplicity of value.'' He was led to this view by not
+distinguishing between vectors and quadrantal quaternions and
+between simultaneous and successive addition. The above
+demonstration was first given in my paper on ``The Fundamental
+Theorems of Analysis generalized for Space.'' It forms the key to
+the higher development of space analysis.}%
+\index{Exponential theorem in spherical trigonometry!Hamilton's
+view}%
+\index{Hamilton's!view of exponential theorem in spherical analysis}
+
+Extension of the Binomial Theorem.---We have proved above that
+$e^{b\beta^\frac{\pi}{2}} e^{c\gamma^\frac{\pi}{2}} =
+e^{b\beta^\frac{\pi}{2} + c\gamma^\frac{\pi}{2}}$ provided that the
+powers of the binomial are expanded as due to a successive sum, that
+is, the order of the terms in the binomial must be preserved. Hence
+the expansion for a power of a successive binomial is given by
+\begin{multline*}
+\left\{ b \cdot \beta^\frac{\pi}{2} +
+        c \cdot \gamma^\frac{\pi}{2} \right\}^n =
+b^n \cdot \beta^{n^\frac{\pi}{2}} + nb^{n-1}c  \cdot
+    \beta^{(n-1)(\frac{\pi}{2})} \gamma^\frac{\pi}{2} \\
++ \frac{n(n-1)}{1 \cdot 2} b^{n-2} c^2 \cdot
+    \beta^{(n-2)(\frac{\pi}{2})} \gamma^\pi + \text{etc.}
+\end{multline*}
+
+\smallskip Example.---Let $b = \frac{1}{10}$ and $c = \frac{1}{5}$,
+$\beta = \overline{30^\circ/}\!\underline{/45^\circ}$, $\gamma =
+\overline{60^\circ/}\!\underline{/30^\circ}$.
+\begin{align*}
+(b \cdot \beta^\frac{\pi}{2} + c \cdot \gamma^\frac{\pi}{2})^2
+&= -\{b^2 + c^2 + 2bc\cos\beta\gamma
+      + 2bc(\sin\beta\gamma)^\frac{\pi}{2} \} \\
+&= -\left(\tfrac{1}{100} + \tfrac{1}{25}
+      + \tfrac{2}{50}\cos\beta\gamma\right)
+      - \tfrac{2}{50}\left(\sin\beta\gamma\right)^\frac{\pi}{2}.
+\end{align*}
+Substitute the calculated values of $\cos \beta\gamma$ and
+$\sin \beta\gamma$ (page 459).
+
+\small \begin{enumerate}
+\item[Prob.~48.] Find the equivalent of a quadrantal version round
+$\dfrac{\sqrt{3}}{2}i + \dfrac{1}{2\sqrt{2}}j +
+\dfrac{1}{2\sqrt{2}}k$ followed by a quadrantal version round
+$\dfrac{1}{2}i + \dfrac{\sqrt{3}}{4}j + \dfrac{3}{4}k$.
+
+\item[Prob.~49.] In the example on p.~459 let $b=25^\circ$ and $c =
+50^\circ$; calculate out the cosine and the directed sine of the
+product angle.
+
+\item[Prob.~50.] In the above example calculate the cosine and the
+directed sine up to and inclusive of the fourth power of the
+binomial. \hfill (Ans.~$\cos =.9735$.)
+
+\item[Prob.~51.] Calculate the first four terms of the series when
+$b = \frac{1}{50}$, $c = \frac{1}{100}$, $\beta =
+\overline{0^\circ/}\! \underline{/0^\circ}$, $\gamma =
+\overline{90^\circ/}\! \underline{/90^\circ}$.
+
+\item[Prob.~52.] From the fundamental theorem of spherical
+trigonometry deduce the polar theorem with respect to both the
+cosine and the directed sine.
+
+\item[Prob.~53.] Prove that if $\alpha^a, \beta^b, \gamma^c$ denote
+the three versors of a spherical triangle, then
+\begin{equation*}
+\frac{\sin\beta\gamma}{\sin a} = \frac{\sin\gamma\alpha}{\sin b} =
+  \frac{\sin\alpha\beta}{\sin c}.
+\end{equation*}
+\end{enumerate} \normalsize
+
+\chapter{Composition of Rotations.}
+
+\begin{center}
+\includegraphics[width=25mm]{fig24.png}
+\end{center}
+
+A version refers to the change of direction of a line, but a
+rotation refers to a rigid body. The composition of rotations is a
+different matter from the composition of versions.%
+\index{Composition!of finite rotations}%
+\index{Rotations, finite}
+
+\medskip Effect of a Finite Rotation on a Line.---Suppose that a
+rigid body rotates $\theta$~radians round the axis $\beta$ passing
+through the point $O$, and that $R$ is the radius vector from $O$ to
+some particle. In the diagram $OB$ represents the axis $\beta$, and
+$OP$ the vector $R$. Draw $OK$ and $OL$, the rectangular components
+of $R$.
+\begin{align*}
+\beta^\theta R &= (\cos\theta + \sin\theta \cdot
+     \beta^\frac{\pi}{2})r\rho \\
+ &= r(\cos\theta \sin \theta \cdot \beta^\frac{\pi}{2})
+  (\cos \beta\rho \cdot \beta +
+  \sin \beta\rho \cdot \overline{\overline{\beta\rho}\beta}) \\
+ &= r\{\cos\beta\rho \cdot \beta +
+  \cos\theta\sin\beta\rho \cdot \overline{\overline{\beta\rho}\beta} +
+  \sin\theta\sin\beta\rho \cdot \overline{\beta\rho}\}.
+\end{align*}
+When $\cos \beta\rho = 0$, this reduces to
+\begin{equation*}
+\beta^\theta R = \cos \theta R + \sin \theta \mathrm{V}(\beta R).
+\end{equation*}
+The general result may be written
+\begin{equation*}
+\beta^\theta R = \mathrm{S}\beta R \cdot \beta +
+  \cos \theta(\mathrm{V}\beta R)\beta + \sin \theta \mathrm{V}\beta R.
+\end{equation*}
+
+Note that $(\mathrm{V}\beta R)\beta$ is equal to
+$\mathrm{V}(\mathrm{V}\beta R)\beta$ because $\mathrm{S}\beta
+R\beta$ is 0, for it involves two coincident directions.
+
+\smallskip Example.---Let $\beta = li + mj + nk$, where
+$l^2 + m^2 + n^2 = 1$ and $R = xi + yj + zk$; then $\mathrm{S}\beta
+R = lx + my + nz$
+\begin{gather*}
+\mathrm{V}(\beta R)\beta = \begin{vmatrix}
+    mz - ny & nx - lz & ly - mx \\
+    l       & m       & n       \\
+    i       & j       & k
+    \end{vmatrix}
+\intertext{and} \mathrm{V}\beta R = \begin{vmatrix}
+  l & m & n \\
+  x & y & z \\
+  i & j & k
+  \end{vmatrix}.
+\intertext{Hence}
+\begin{split}
+\beta^\theta &= (lx + my + nz)(li + mj + nk) \\
+&+ \cos\theta \begin{vmatrix}
+   mz - ny & nx - lz & ly - mx \\
+   l       & m       & n       \\
+   i       & j       & k
+   \end{vmatrix} \\
+&+ \sin\theta\begin{vmatrix}
+   l & m & n  \\
+   x & y & z  \\
+   i & j & k \end{vmatrix}.
+\end{split}
+\end{gather*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig25.png}
+\end{center}
+
+To prove that $\beta^b \rho$ coincides with the axis of
+$\beta^\frac{-b}{2} \rho^\frac{\pi}{2} \beta^\frac{b}{2}$. Take the
+more general versor $\rho^\theta$. Let $OP$ represent the axis
+$\beta$, $AB$ the versor $\beta^\frac{-b}{2}$, $BC$ the versor
+$\rho^\theta$. Then $(AB)(BC) = AC = DA$, therefore $(AB)(BC)(AE) =
+(DA)(AE) = DE$. Now $DE$ has the same angle as $BC$, but its axis
+has been rotated round $P$ by the angle $b$. Hence if $\theta =
+\frac{\pi}{2}$, the axis of $\beta^\frac{-b}{2}
+\rho^\frac{\pi}{2}\beta^\frac{b}{2}$ will coincide with
+$\beta^b\rho$.\footnote{This theorem was discovered by
+Cayley.\index{Cayley} It indicates that quaternion multiplication in
+the most general sense has its physical meaning in the composition
+of rotations.}
+
+The exponential expression for $\beta^\frac{-b}{2}
+\rho^\frac{\pi}{2} \beta^\frac{b}{2}$ is
+$e^{-\frac{1}{2}b\beta^\frac{\pi}{2} +
+\frac{1}{2}\pi\rho^\frac{\pi}{2} + \frac{1}{2}b\beta^\frac{\pi}{2}}$
+which may be expanded according to the exponential theorem, the
+successive powers of the trinomial being formed according to the
+multinomial theorem, the order of the factors being preserved.
+
+\medskip Composition of Finite Rotations round Axes which
+Intersect.---Let $\beta$ and $\gamma$ denote the two axes in space
+round which the successive rotations take place, and let $\beta^b$
+denote the first and $\gamma^c$ the second. Let $\beta^b \times
+\gamma^c$ denote the single rotation which is equivalent to the two
+given rotations applied in succession; the sign $\times$ is
+introduced to distinguish from the product of versors. It has been
+shown in the preceding paragraph that
+\begin{gather*}
+\beta^b\rho = \beta^\frac{-b}{2}\rho^\frac{\pi}{2}\beta^\frac{b}{2}; \\
+\intertext{and as the result is a line, the same principle applies
+to the subsequent rotation. Hence}
+\begin{split}
+\gamma^c(\beta^b\rho) &=
+  \gamma^\frac{-c}{2}(\beta^\frac{-b}{2}\rho^\frac{\pi}{2}
+     \beta^\frac{\pi}{2})\gamma^\frac{c}{2} \\
+&= (\gamma^\frac{-c}{2}\beta^\frac{-b}{2})
+     \rho^\frac{\pi}{2} (\beta^\frac{b}{2}\gamma^\frac{c}{2}),
+\end{split} \\
+\intertext{because the factors in a product of versors can be
+associated in any manner. Hence, reasoning backwards,}
+\beta^b \times \gamma^c = (\beta^\frac{b}{2}\gamma^\frac{c}{2})^2. \\
+\intertext{Let $m$ denote the cosine of
+$\beta^\frac{b}{2}\gamma^\frac{c}{2}$, namely,}
+\cos\frac{b}{2}\,\cos\frac{c}{2}-\sin\frac{b}{2}\,\sin\frac{c}{2}, \\
+\intertext{ and $n \cdot \nu$ their directed sine, namely,}
+\cos \frac{b}{2}\, \sin \frac{c}{2} \cdot \gamma + \cos\frac{c}{2}\,
+\sin \frac{b}{2} \cdot \beta - \sin \frac{b}{2}\,
+  \sin \frac{c}{2}\, \sin \beta\gamma \cdot \overline{\beta\gamma}; \\
+\intertext{then}
+\beta^b \times \gamma^c = m^2 - n^2 + 2mn \cdot \nu.
+\end{gather*}
+
+\newpage
+\begin{center}
+\includegraphics[width=40mm]{fig26.png}
+\end{center}
+
+\smallskip Observation.---The expression
+$(\beta^\frac{b}{2}\gamma^\frac{c}{2})^2$ is not, as might be
+supposed, identical with $\beta^b\gamma^c$. The former reduces to
+the latter only when $\beta$ and $\gamma$ are the same or opposite.
+In the figure $\beta^b$ is represented by $PQ$, $\gamma^c$ by $QR$,
+$\beta^b\gamma^c$ by $PR$, $\beta^\frac{b}{2}\gamma^\frac{c}{2}$ by
+$ST$, and $(\beta^\frac{b}{2}\gamma^\frac{c}{2})^2$ by $SU$, which
+is twice $ST$. The cosine of $SU$ differs from the cosine of $PR$ by
+the term $-(\sin \frac{b}{2}\, \sin \frac{c}{2}\, \sin
+\beta\gamma)^2$ It is evident from the figure that their axes are
+also different.
+
+\medskip Corollary.---When $b$ and $c$ are infinitesimals,
+$\cos\beta^b \times \gamma^c = 1$, and $\Sin \beta^b \times \gamma^c
+= b \cdot \beta + c \cdot \gamma$, which is the parallelogram rule
+for the composition of infinitesimal rotations.
+
+\small \begin{enumerate}
+
+\item[Prob.~54.] Let $\beta = \overline{30^\circ/}\!
+\underline{/45^\circ}$, $\theta = \frac{\pi}{3}$, and $R = 2i - 3j +
+4k$; calculate $\beta^\theta R$.
+
+\item[Prob.~55.] Let $\beta = \overline{90^\circ/}\!
+\underline{/90^\circ}$, $\theta = \frac{\pi}{4}$, $R = -i + 2j -
+3k$; calculate $\beta^\theta R$.
+
+\item[Prob.~56.] Prove by multiplying out that $\beta^\frac{-b}{2}
+\rho^\frac{\pi}{2} \beta^\frac{b}{2} =
+\{\beta^b\rho\}^\frac{\pi}{2}$.
+
+\item[Prob.~57.] Prove by means of the exponential theorem that
+$\gamma^{-c}\beta^b\gamma^c$ has an angle $b$, and that its axis is
+$\gamma^{2c}\beta$.
+
+\item[Prob.~58.] Prove that the cosine of
+$(\beta^\frac{b}{2}\gamma^\frac{c}{2})^2$
+differs from the cosine of $\beta^b\gamma^c$ by \\
+$-(\sin\frac{b}{2} \sin\frac{c}{2} \sin\beta \gamma)^2$.
+
+\item[Prob.~59.] Compare the axes of
+$(\beta^\frac{b}{2}\gamma^\frac{c}{2})^2$ and $\beta^b\gamma^c$.
+
+\item[Prob.~60.] Find the value of $\beta^b\times\gamma^c$ when
+$\beta = \overline{0^\circ/}\!\underline{/90^\circ}$ and
+$\gamma=\overline{90^\circ/}\!\underline{/90^\circ}$.
+
+\item[Prob.~61.] Find the single rotation equivalent to
+$i^\frac{\pi}{2} \times j^\frac{\pi}{2} \times k^\frac{\pi}{2}$.
+
+\item[Prob.~62.] Prove that successive rotations about radii to two
+corners of a spherical triangle and through angles double of those
+of the triangle are equivalent to a single rotation about the radius
+to the third corner, and through an angle double of the external
+angle of the triangle.
+\end{enumerate} \normalsize
+
+\addcontentsline{toc}{chapter}{Index}
+\printindex
+
+\markright{ADVERTISEMENT}
+\begin{center}
+\textbf{
+\Large SHORT-TITLE CATALOGUE \\
+\small OF THE           \\
+\Large PUBLICATIONS          \\
+\small OF               \\
+\Large JOHN WILEY \& SONS, \textsc{Inc.} \\
+\normalsize NEW YORK  \\
+London: CHAPMAN \& HALL, Limited \\
+Montreal, Can.: RENOUF PUB. CO.} \\
+
+\bigskip
+\begin{center}
+\rule{25mm}{.2pt}
+\end{center}
+\medskip
+\normalsize \textbf{ARRANGED UNDER SUBJECTS} \\
+\footnotesize Descriptive circulars sent on application. Books
+marked with an asterisk (*) are sold at \emph{net} prices only. All
+books are bound in cloth unless otherwise stated.
+
+\medskip
+\begin{center}
+\rule{25mm}{.2pt}
+\end{center}
+\bigskip
+
+\normalsize
+\textbf{AGRICULTURE---HORTICULTURE---FORESTRY.}
+\bigskip
+
+{\footnotesize
+\begin{tabular}{lr}
+\textsc{Armsby}---Principles of Animal Nutrition \dotfill 8vo,
+                                                          & \$4 00 \\
+\textsc{Bowman}---Forest Physiography \dotfill 8vo, & *5 00 \\
+\textsc{Bryant}---Hand Book of Logging\dotfill  (Ready, Fall 1913)
+                                                                 & \\
+\textsc{Budd} and \textsc{Hansen}---American Horticultural Manual:
+                                                        \dotfill & \\
+\quad Part I. Propagation, Culture, and Improvement\dotfill 12mo,
+                                                            & 1 50 \\
+\quad Part II. Systematic Pomology \dotfill 12mo, & 1 50 \\
+\textsc{Elliott}---Engineering for Land Drainage \dotfill 12mo,
+                                                            & 2 00 \\
+\quad  Practical Farm Drainage.  (Second Edition, Rewritten)
+                                             \dotfill 12mo, & 1 50 \\
+\textsc{Fuller}---Domestic Water Supplies for the Farm \dotfill 8vo,
+                                                           & *1 50 \\
+\textsc{Graham}---Text-book on Poultry \dotfill
+                                        (\emph{In Preparation.}) & \\
+\quad  Manual on Poultry (Loose Leaf Lab. Manual)
+  \dotfill (\emph{In Preparation.}) & \\
+\textsc{Graves}---Forest Mensuration \dotfill 8vo, & 4 00 \\
+\quad  Principles of Handling Woodlands \dotfill Small 8vo,
+                                                           & *1 50 \\
+\textsc{Green}---Principles of American Forestry \dotfill 12mo,
+                                                            & 1 50 \\
+\textsc{Grotenfelt}---Principles of Modern Dairy Practice.
+(\textsc{Woll}.) \dotfill 12mo, & 2 00 \\
+\textsc{Hawley} and \textsc{Hawes}---Forestry in New England
+\dotfill  8vo, & *3 50 \\
+\textsc{Herrick}---Denatured or Industrial Alcohol \dotfill 8vo,
+                                                           & *4 00 \\
+\textsc{Howe}---Agricultural Drafting \dotfill oblong quarto,
+                                                           & *1 25 \\
+\quad Reference and Problem Sheets to accompany
+                      Agricultural Drafting, \dotfill each & *0 20 \\
+\textsc{Keitt}---Agricultural Chemistry Text-book
+   \dotfill (\emph{In Preparation.}) & \\
+\quad Laboratory and Field Exercises in Agricultural Chemistry
+  \dotfill (\emph{In Preparation.}) & \\
+\textsc{Kemp} and \textsc{Waugh}---Landscape Gardening. (New
+Edition, Rewritten)
+  \dotfill  12mo, & *1 50 \\
+\textsc{Larsen}---Exercises in Dairying (Loose Leaf Field Manual)
+  \dotfill 4to, paper, & *1 00 \\
+\quad  Single Exercises each \dotfill &  *0 02 \\
+\quad  and \textsc{White}---Dairy Technology \dotfill Small 8vo,
+                                                           & *1 50 \\
+\textsc{Levison}---Studies of Trees, Loose Leaf Field Manual, 4to,
+  pamphlet form,  & \\
+\quad  Price from 5--10 cents net, each, according to number
+                                                       of pages. & \\
+\textsc{McCall}---Crops and Soils (Loose Leaf Field Manual)
+  \dotfill (\emph{In Preparation.}) & \\
+\quad  Soils (Text-book) \dotfill (\emph{In Preparation.}) & \\
+\textsc{McKay} and \textsc{Larsen}---Principles and Practice of
+                               Butter-making \dotfill 8vo, & *1 50 \\
+\textsc{Maynard}---Landscape Gardening as Applied to Home Decoration
+  \dotfill 12mo, & 1 50 \\
+\textsc{Moon} and \textsc{Brown}---Elements of Forestry
+  \dotfill (\emph{In Preparation.}) & \\
+\textsc{Record}---Identification of the Economic Woods of the United
+States \dotfill 8vo, & *1 25 \\
+\textsc{Recknagel}---Theory and Practice of Working Plans (Forest
+  Organization) \dotfill 8vo, & *2 00
+\end{tabular}
+} % \end{\footnotesize}
+\end{center}
+\newpage
+\begin{verbatim}
+*** END OF THIS PROJECT GUTENBERG EBOOK VECTOR ANALYSIS AND QUATERNIONS ***
+
+***** This file should be named 13609-t.tex or 13609-t.zip *****
+This and all associated files of various formats will be found in:
+        https://www.gutenberg.org/1/3/6/0/13609/
+
+Updated editions will replace the previous one--the old editions
+will be renamed.
+
+Creating the works from public domain print editions means that no
+one owns a United States copyright in these works, so the Foundation
+(and you!) can copy and distribute it in the United States without
+permission and without paying copyright royalties.  Special rules,
+set forth in the General Terms of Use part of this license, apply to
+copying and distributing Project Gutenberg-tm electronic works to
+protect the PROJECT GUTENBERG-tm concept and trademark.  Project
+Gutenberg is a registered trademark, and may not be used if you
+charge for the eBooks, unless you receive specific permission.  If
+you do not charge anything for copies of this eBook, complying with
+the rules is very easy.  You may use this eBook for nearly any
+purpose such as creation of derivative works, reports, performances
+and research.  They may be modified and printed and given away--you
+may do practically ANYTHING with public domain eBooks.
+Redistribution is subject to the trademark license, especially
+commercial redistribution.
+
+
+
+*** START: FULL LICENSE ***
+
+THE FULL PROJECT GUTENBERG LICENSE PLEASE READ THIS BEFORE YOU
+DISTRIBUTE OR USE THIS WORK
+
+To protect the Project Gutenberg-tm mission of promoting the free
+distribution of electronic works, by using or distributing this work
+(or any other work associated in any way with the phrase "Project
+Gutenberg"), you agree to comply with all the terms of the Full
+Project Gutenberg-tm License (available with this file or online at
+https://gutenberg.org/license).
+
+
+Section 1.  General Terms of Use and Redistributing Project
+Gutenberg-tm electronic works
+
+1.A.  By reading or using any part of this Project Gutenberg-tm
+electronic work, you indicate that you have read, understand, agree
+to and accept all the terms of this license and intellectual
+property (trademark/copyright) agreement.  If you do not agree to
+abide by all the terms of this agreement, you must cease using and
+return or destroy all copies of Project Gutenberg-tm electronic
+works in your possession. If you paid a fee for obtaining a copy of
+or access to a Project Gutenberg-tm electronic work and you do not
+agree to be bound by the terms of this agreement, you may obtain a
+refund from the person or entity to whom you paid the fee as set
+forth in paragraph 1.E.8.
+
+1.B.  "Project Gutenberg" is a registered trademark.  It may only be
+used on or associated in any way with an electronic work by people
+who agree to be bound by the terms of this agreement.  There are a
+few things that you can do with most Project Gutenberg-tm electronic
+works even without complying with the full terms of this agreement.
+See paragraph 1.C below.  There are a lot of things you can do with
+Project Gutenberg-tm electronic works if you follow the terms of
+this agreement and help preserve free future access to Project
+Gutenberg-tm electronic works.  See paragraph 1.E below.
+
+1.C.  The Project Gutenberg Literary Archive Foundation ("the
+Foundation" or PGLAF), owns a compilation copyright in the
+collection of Project Gutenberg-tm electronic works.  Nearly all the
+individual works in the collection are in the public domain in the
+United States.  If an individual work is in the public domain in the
+United States and you are located in the United States, we do not
+claim a right to prevent you from copying, distributing, performing,
+displaying or creating derivative works based on the work as long as
+all references to Project Gutenberg are removed.  Of course, we hope
+that you will support the Project Gutenberg-tm mission of promoting
+free access to electronic works by freely sharing Project
+Gutenberg-tm works in compliance with the terms of this agreement
+for keeping the Project Gutenberg-tm name associated with the work.
+You can easily comply with the terms of this agreement by keeping
+this work in the same format with its attached full Project
+Gutenberg-tm License when you share it without charge with others.
+
+1.D.  The copyright laws of the place where you are located also
+govern what you can do with this work.  Copyright laws in most
+countries are in a constant state of change.  If you are outside the
+United States, check the laws of your country in addition to the
+terms of this agreement before downloading, copying, displaying,
+performing, distributing or creating derivative works based on this
+work or any other Project Gutenberg-tm work.  The Foundation makes
+no representations concerning the copyright status of any work in
+any country outside the United States.
+
+1.E.  Unless you have removed all references to Project Gutenberg:
+
+1.E.1.  The following sentence, with active links to, or other
+immediate access to, the full Project Gutenberg-tm License must
+appear prominently whenever any copy of a Project Gutenberg-tm work
+(any work on which the phrase "Project Gutenberg" appears, or with
+which the phrase "Project Gutenberg" is associated) is accessed,
+displayed, performed, viewed, copied or distributed:
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+1.E.2.  If an individual Project Gutenberg-tm electronic work is
+derived from the public domain (does not contain a notice indicating
+that it is posted with permission of the copyright holder), the work
+can be copied and distributed to anyone in the United States without
+paying any fees or charges.  If you are redistributing or providing
+access to a work with the phrase "Project Gutenberg" associated with
+or appearing on the work, you must comply either with the
+requirements of paragraphs 1.E.1 through 1.E.7 or obtain permission
+for the use of the work and the Project Gutenberg-tm trademark as
+set forth in paragraphs 1.E.8 or 1.E.9.
+
+1.E.3.  If an individual Project Gutenberg-tm electronic work is
+posted with the permission of the copyright holder, your use and
+distribution must comply with both paragraphs 1.E.1 through 1.E.7
+and any additional terms imposed by the copyright holder.
+Additional terms will be linked to the Project Gutenberg-tm License
+for all works posted with the permission of the copyright holder
+found at the beginning of this work.
+
+1.E.4.  Do not unlink or detach or remove the full Project
+Gutenberg-tm License terms from this work, or any files containing a
+part of this work or any other work associated with Project
+Gutenberg-tm.
+
+1.E.5.  Do not copy, display, perform, distribute or redistribute
+this electronic work, or any part of this electronic work, without
+prominently displaying the sentence set forth in paragraph 1.E.1
+with active links or immediate access to the full terms of the
+Project Gutenberg-tm License.
+
+1.E.6.  You may convert to and distribute this work in any binary,
+compressed, marked up, nonproprietary or proprietary form, including
+any word processing or hypertext form.  However, if you provide
+access to or distribute copies of a Project Gutenberg-tm work in a
+format other than "Plain Vanilla ASCII" or other format used in the
+official version posted on the official Project Gutenberg-tm web
+site (www.gutenberg.org), you must, at no additional cost, fee or
+expense to the user, provide a copy, a means of exporting a copy, or
+a means of obtaining a copy upon request, of the work in its
+original "Plain Vanilla ASCII" or other form.  Any alternate format
+must include the full Project Gutenberg-tm License as specified in
+paragraph 1.E.1.
+
+1.E.7.  Do not charge a fee for access to, viewing, displaying,
+performing, copying or distributing any Project Gutenberg-tm works
+unless you comply with paragraph 1.E.8 or 1.E.9.
+
+1.E.8.  You may charge a reasonable fee for copies of or providing
+access to or distributing Project Gutenberg-tm electronic works
+provided that
+
+- You pay a royalty fee of 20% of the gross profits you derive from
+     the use of Project Gutenberg-tm works calculated using the method
+     you already use to calculate your applicable taxes.  The fee is
+     owed to the owner of the Project Gutenberg-tm trademark, but he
+     has agreed to donate royalties under this paragraph to the
+     Project Gutenberg Literary Archive Foundation.  Royalty payments
+     must be paid within 60 days following each date on which you
+     prepare (or are legally required to prepare) your periodic tax
+     returns.  Royalty payments should be clearly marked as such and
+     sent to the Project Gutenberg Literary Archive Foundation at the
+     address specified in Section 4, "Information about donations to
+     the Project Gutenberg Literary Archive Foundation."
+
+- You provide a full refund of any money paid by a user who notifies
+     you in writing (or by e-mail) within 30 days of receipt that s/he
+     does not agree to the terms of the full Project Gutenberg-tm
+     License.  You must require such a user to return or
+     destroy all copies of the works possessed in a physical medium
+     and discontinue all use of and all access to other copies of
+     Project Gutenberg-tm works.
+
+- You provide, in accordance with paragraph 1.F.3, a full refund of
+     any money paid for a work or a replacement copy, if a defect in
+     the electronic work is discovered and reported to you within 90
+     days of receipt of the work.
+
+- You comply with all other terms of this agreement for free
+     distribution of Project Gutenberg-tm works.
+
+1.E.9.  If you wish to charge a fee or distribute a Project
+Gutenberg-tm electronic work or group of works on different terms
+than are set forth in this agreement, you must obtain permission in
+writing from both the Project Gutenberg Literary Archive Foundation
+and Michael Hart, the owner of the Project Gutenberg-tm trademark.
+Contact the Foundation as set forth in Section 3 below.
+
+1.F.
+
+1.F.1.  Project Gutenberg volunteers and employees expend
+considerable effort to identify, do copyright research on,
+transcribe and proofread public domain works in creating the Project
+Gutenberg-tm collection.  Despite these efforts, Project
+Gutenberg-tm electronic works, and the medium on which they may be
+stored, may contain "Defects," such as, but not limited to,
+incomplete, inaccurate or corrupt data, transcription errors, a
+copyright or other intellectual property infringement, a defective
+or damaged disk or other medium, a computer virus, or computer codes
+that damage or cannot be read by your equipment.
+
+1.F.2.  LIMITED WARRANTY, DISCLAIMER OF DAMAGES - Except for the
+"Right of Replacement or Refund" described in paragraph 1.F.3, the
+Project Gutenberg Literary Archive Foundation, the owner of the
+Project Gutenberg-tm trademark, and any other party distributing a
+Project Gutenberg-tm electronic work under this agreement, disclaim
+all liability to you for damages, costs and expenses, including
+legal fees.  YOU AGREE THAT YOU HAVE NO REMEDIES FOR NEGLIGENCE,
+STRICT LIABILITY, BREACH OF WARRANTY OR BREACH OF CONTRACT EXCEPT
+THOSE PROVIDED IN PARAGRAPH F3.  YOU AGREE THAT THE FOUNDATION, THE
+TRADEMARK OWNER, AND ANY DISTRIBUTOR UNDER THIS AGREEMENT WILL NOT
+BE LIABLE TO YOU FOR ACTUAL, DIRECT, INDIRECT, CONSEQUENTIAL,
+PUNITIVE OR INCIDENTAL DAMAGES EVEN IF YOU GIVE NOTICE OF THE
+POSSIBILITY OF SUCH DAMAGE.
+
+1.F.3.  LIMITED RIGHT OF REPLACEMENT OR REFUND - If you discover a
+defect in this electronic work within 90 days of receiving it, you
+can receive a refund of the money (if any) you paid for it by
+sending a written explanation to the person you received the work
+from.  If you received the work on a physical medium, you must
+return the medium with your written explanation.  The person or
+entity that provided you with the defective work may elect to
+provide a replacement copy in lieu of a refund.  If you received the
+work electronically, the person or entity providing it to you may
+choose to give you a second opportunity to receive the work
+electronically in lieu of a refund.  If the second copy is also
+defective, you may demand a refund in writing without further
+opportunities to fix the problem.
+
+1.F.4.  Except for the limited right of replacement or refund set
+forth in paragraph 1.F.3, this work is provided to you 'AS-IS', WITH
+NO OTHER WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT
+NOT LIMITED TO WARRANTIES OF MERCHANTIBILITY OR FITNESS FOR ANY
+PURPOSE.
+
+1.F.5.  Some states do not allow disclaimers of certain implied
+warranties or the exclusion or limitation of certain types of
+damages. If any disclaimer or limitation set forth in this agreement
+violates the law of the state applicable to this agreement, the
+agreement shall be interpreted to make the maximum disclaimer or
+limitation permitted by the applicable state law.  The invalidity or
+unenforceability of any provision of this agreement shall not void
+the remaining provisions.
+
+1.F.6.  INDEMNITY - You agree to indemnify and hold the Foundation,
+the trademark owner, any agent or employee of the Foundation, anyone
+providing copies of Project Gutenberg-tm electronic works in
+accordance with this agreement, and any volunteers associated with
+the production, promotion and distribution of Project Gutenberg-tm
+electronic works, harmless from all liability, costs and expenses,
+including legal fees, that arise directly or indirectly from any of
+the following which you do or cause to occur: (a) distribution of
+this or any Project Gutenberg-tm work, (b) alteration, modification,
+or additions or deletions to any Project Gutenberg-tm work, and (c)
+any Defect you cause.
+
+
+Section  2.  Information about the Mission of Project Gutenberg-tm
+
+Project Gutenberg-tm is synonymous with the free distribution of
+electronic works in formats readable by the widest variety of
+computers including obsolete, old, middle-aged and new computers.
+It exists because of the efforts of hundreds of volunteers and
+donations from people in all walks of life.
+
+Volunteers and financial support to provide volunteers with the
+assistance they need, is critical to reaching Project Gutenberg-tm's
+goals and ensuring that the Project Gutenberg-tm collection will
+remain freely available for generations to come.  In 2001, the
+Project Gutenberg Literary Archive Foundation was created to provide
+a secure and permanent future for Project Gutenberg-tm and future
+generations. To learn more about the Project Gutenberg Literary
+Archive Foundation and how your efforts and donations can help, see
+Sections 3 and 4 and the Foundation web page at
+https://www.pglaf.org.
+
+
+Section 3.  Information about the Project Gutenberg Literary Archive
+Foundation
+
+The Project Gutenberg Literary Archive Foundation is a non profit
+501(c)(3) educational corporation organized under the laws of the
+state of Mississippi and granted tax exempt status by the Internal
+Revenue Service.  The Foundation's EIN or federal tax identification
+number is 64-6221541.  Its 501(c)(3) letter is posted at
+https://pglaf.org/fundraising.  Contributions to the Project
+Gutenberg Literary Archive Foundation are tax deductible to the full
+extent permitted by U.S. federal laws and your state's laws.
+
+The Foundation's principal office is located at 4557 Melan Dr. S.
+Fairbanks, AK, 99712., but its volunteers and employees are
+scattered throughout numerous locations.  Its business office is
+located at 809 North 1500 West, Salt Lake City, UT 84116, (801)
+596-1887, email business@pglaf.org.  Email contact links and up to
+date contact information can be found at the Foundation's web site
+and official page at https://pglaf.org
+
+For additional contact information:
+     Dr. Gregory B. Newby
+     Chief Executive and Director
+     gbnewby@pglaf.org
+
+Section 4.  Information about Donations to the Project Gutenberg
+Literary Archive Foundation
+
+Project Gutenberg-tm depends upon and cannot survive without wide
+spread public support and donations to carry out its mission of
+increasing the number of public domain and licensed works that can
+be freely distributed in machine readable form accessible by the
+widest array of equipment including outdated equipment.  Many small
+donations ($1 to $5,000) are particularly important to maintaining
+tax exempt status with the IRS.
+
+The Foundation is committed to complying with the laws regulating
+charities and charitable donations in all 50 states of the United
+States.  Compliance requirements are not uniform and it takes a
+considerable effort, much paperwork and many fees to meet and keep
+up with these requirements.  We do not solicit donations in
+locations where we have not received written confirmation of
+compliance.  To SEND DONATIONS or determine the status of compliance
+for any particular state visit https://pglaf.org
+
+While we cannot and do not solicit contributions from states where
+we have not met the solicitation requirements, we know of no
+prohibition against accepting unsolicited donations from donors in
+such states who approach us with offers to donate.
+
+International donations are gratefully accepted, but we cannot make
+any statements concerning tax treatment of donations received from
+outside the United States.  U.S. laws alone swamp our small staff.
+
+Please check the Project Gutenberg Web pages for current donation
+methods and addresses.  Donations are accepted in a number of other
+ways including including checks, online payments and credit card
+donations.  To donate, please visit: https://pglaf.org/donate
+
+
+Section 5.  General Information About Project Gutenberg-tm
+electronic works.
+
+Professor Michael S. Hart was the originator of the Project
+Gutenberg-tm concept of a library of electronic works that could be
+freely shared with anyone.  For thirty years, he produced and
+distributed Project Gutenberg-tm eBooks with only a loose network of
+volunteer support.
+
+Project Gutenberg-tm eBooks are often created from several printed
+editions, all of which are confirmed as Public Domain in the U.S.
+unless a copyright notice is included.  Thus, we do not necessarily
+keep eBooks in compliance with any particular paper edition.
+
+Most people start at our Web site which has the main PG search
+facility:
+
+     https://www.gutenberg.org
+
+This Web site includes information about Project Gutenberg-tm,
+including how to make donations to the Project Gutenberg Literary
+Archive Foundation, how to help produce our new eBooks, and how to
+subscribe to our email newsletter to hear about new eBooks.
+\end{verbatim}
+
+\end{document}
diff --git a/13609-t/fig01.png b/13609-t/fig01.png
new file mode 100644
index 0000000..d17f682
--- /dev/null
+++ b/13609-t/fig01.png
diff --git a/13609-t/fig02.png b/13609-t/fig02.png
new file mode 100644
index 0000000..105e875
--- /dev/null
+++ b/13609-t/fig02.png
diff --git a/13609-t/fig03.png b/13609-t/fig03.png
new file mode 100644
index 0000000..9de4bf3
--- /dev/null
+++ b/13609-t/fig03.png
diff --git a/13609-t/fig04.png b/13609-t/fig04.png
new file mode 100644
index 0000000..0810607
--- /dev/null
+++ b/13609-t/fig04.png
diff --git a/13609-t/fig05.png b/13609-t/fig05.png
new file mode 100644
index 0000000..2e9bc7d
--- /dev/null
+++ b/13609-t/fig05.png
diff --git a/13609-t/fig06.png b/13609-t/fig06.png
new file mode 100644
index 0000000..9461495
--- /dev/null
+++ b/13609-t/fig06.png
diff --git a/13609-t/fig07.png b/13609-t/fig07.png
new file mode 100644
index 0000000..b0a2979
--- /dev/null
+++ b/13609-t/fig07.png
diff --git a/13609-t/fig08.png b/13609-t/fig08.png
new file mode 100644
index 0000000..edd52d3
--- /dev/null
+++ b/13609-t/fig08.png
diff --git a/13609-t/fig09.png b/13609-t/fig09.png
new file mode 100644
index 0000000..a681266
--- /dev/null
+++ b/13609-t/fig09.png
diff --git a/13609-t/fig10.png b/13609-t/fig10.png
new file mode 100644
index 0000000..62ab3b7
--- /dev/null
+++ b/13609-t/fig10.png
diff --git a/13609-t/fig11.png b/13609-t/fig11.png
new file mode 100644
index 0000000..06af8ef
--- /dev/null
+++ b/13609-t/fig11.png
diff --git a/13609-t/fig12.png b/13609-t/fig12.png
new file mode 100644
index 0000000..7361f14
--- /dev/null
+++ b/13609-t/fig12.png
diff --git a/13609-t/fig13.png b/13609-t/fig13.png
new file mode 100644
index 0000000..86c40aa
--- /dev/null
+++ b/13609-t/fig13.png
diff --git a/13609-t/fig14.png b/13609-t/fig14.png
new file mode 100644
index 0000000..d3a87d5
--- /dev/null
+++ b/13609-t/fig14.png
diff --git a/13609-t/fig15.png b/13609-t/fig15.png
new file mode 100644
index 0000000..ebb0d96
--- /dev/null
+++ b/13609-t/fig15.png
diff --git a/13609-t/fig16.png b/13609-t/fig16.png
new file mode 100644
index 0000000..c5829fe
--- /dev/null
+++ b/13609-t/fig16.png
diff --git a/13609-t/fig17.png b/13609-t/fig17.png
new file mode 100644
index 0000000..40d7628
--- /dev/null
+++ b/13609-t/fig17.png
diff --git a/13609-t/fig18.png b/13609-t/fig18.png
new file mode 100644
index 0000000..bbc188b
--- /dev/null
+++ b/13609-t/fig18.png
diff --git a/13609-t/fig19.png b/13609-t/fig19.png
new file mode 100644
index 0000000..0191f71
--- /dev/null
+++ b/13609-t/fig19.png
diff --git a/13609-t/fig20.png b/13609-t/fig20.png
new file mode 100644
index 0000000..830b6eb
--- /dev/null
+++ b/13609-t/fig20.png
diff --git a/13609-t/fig21.png b/13609-t/fig21.png
new file mode 100644
index 0000000..b59f5d2
--- /dev/null
+++ b/13609-t/fig21.png
diff --git a/13609-t/fig22.png b/13609-t/fig22.png
new file mode 100644
index 0000000..48f55fc
--- /dev/null
+++ b/13609-t/fig22.png
diff --git a/13609-t/fig23.png b/13609-t/fig23.png
new file mode 100644
index 0000000..e0ef296
--- /dev/null
+++ b/13609-t/fig23.png
diff --git a/13609-t/fig24.png b/13609-t/fig24.png
new file mode 100644
index 0000000..f443d27
--- /dev/null
+++ b/13609-t/fig24.png
diff --git a/13609-t/fig25.png b/13609-t/fig25.png
new file mode 100644
index 0000000..b17b11b
--- /dev/null
+++ b/13609-t/fig25.png
diff --git a/13609-t/fig26.png b/13609-t/fig26.png
new file mode 100644
index 0000000..309bfb1
--- /dev/null
+++ b/13609-t/fig26.png
diff --git a/LICENSE.txt b/LICENSE.txt
new file mode 100644
index 0000000..6312041
--- /dev/null
+++ b/LICENSE.txt
@@ -0,0 +1,11 @@
+This eBook, including all associated images, markup, improvements,
+metadata, and any other content or labor, has been confirmed to be
+in the PUBLIC DOMAIN IN THE UNITED STATES.
+
+Procedures for determining public domain status are described in
+the "Copyright How-To" at https://www.gutenberg.org.
+
+No investigation has been made concerning possible copyrights in
+jurisdictions other than the United States. Anyone seeking to utilize
+this eBook outside of the United States should confirm copyright
+status under the laws that apply to them.
diff --git a/README.md b/README.md
new file mode 100644
index 0000000..17fa885
--- /dev/null
+++ b/README.md
@@ -0,0 +1,2 @@
+Project Gutenberg (https://www.gutenberg.org) public repository for
+eBook #13609 (https://www.gutenberg.org/ebooks/13609)
diff --git a/old/13609-pdf.pdf b/old/13609-pdf.pdf
new file mode 100644
index 0000000..216296c
--- /dev/null
+++ b/old/13609-pdf.pdf
diff --git a/old/13609-pdf.zip b/old/13609-pdf.zip
new file mode 100644
index 0000000..5061b9d
--- /dev/null
+++ b/old/13609-pdf.zip
diff --git a/old/13609-t.tex b/old/13609-t.tex
new file mode 100644
index 0000000..138476a
--- /dev/null
+++ b/old/13609-t.tex
@@ -0,0 +1,3233 @@
+\documentclass[oneside]{book}
+\usepackage[latin1]{inputenc}
+\usepackage[reqno]{amsmath}
+\usepackage{makeidx,graphicx}
+\makeindex
+\renewcommand{\chaptername}{Article}
+\DeclareMathOperator{\Sin}{Sin}
+\begin{document}
+\thispagestyle{empty}
+\small
+\begin{verbatim}
+Project Gutenberg's Vector Analysis and Quaternions, by Alexander Macfarlane
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+
+Title: Vector Analysis and Quaternions
+
+Author: Alexander Macfarlane
+
+Release Date: October 5, 2004 [EBook #13609]
+
+Language: English
+
+Character set encoding: TeX
+
+*** START OF THIS PROJECT GUTENBERG EBOOK VECTOR ANALYSIS AND QUATERNIONS ***
+
+
+
+
+Produced by David Starner, Joshua Hutchinson, John Hagerson, and the 
+Project Gutenberg On-line Distributed Proofreaders.
+
+
+
+
+
+\end{verbatim}
+\normalsize
+\newpage
+
+\frontmatter
+
+\begin{center}
+\noindent \Large MATHEMATICAL MONOGRAPHS.
+
+\bigskip\footnotesize{EDITED BY} \\
+\normalsize \textsc{MANSFIELD MERRIMAN and ROBERT S. WOODWARD.}
+
+\bigskip\bigskip\huge
+No. 8.
+
+\bigskip\bigskip\huge VECTOR ANALYSIS \\
+\bigskip\footnotesize \textsc{and} \\
+\bigskip\huge QUATERNIONS.
+
+\bigskip\bigskip\footnotesize \textsc{by} \\
+\bigskip \large ALEXANDER MACFARLANE, \\
+\footnotesize\textsc{Secretary of International Association for
+Promoting the Study of Quaternions.} \\
+
+\bigskip\bigskip\normalsize NEW YORK: \\
+\medskip JOHN WILEY \& SONS. \\
+\medskip \textsc{London: CHAPMAN \& HALL, Limited.} \\
+\medskip 1906.
+\end{center}
+
+\bigskip\bigskip
+\scriptsize \noindent \textsc{Transcriber's Notes:} \emph{This
+material was originally published in a book by Merriman and Woodward
+titled \emph{Higher Mathematics}. I believe that some of the page
+number cross-references have been retained from that presentation of
+this material.}
+
+\emph{I did my best to recreate the index.} \normalsize
+
+\newpage
+
+\fbox{\parbox{11cm}{
+\begin{center}
+\textbf{MATHEMATICAL MONOGRAPHS.} \\
+\footnotesize\textsc{edited by}\normalsize \\
+\textbf{Mansfield Merriman and Robert S. Woodward.} \\
+\smallskip \footnotesize \textbf{Octavo. Cloth. \$1.00 each.}
+
+\bigskip
+\textbf{No. 1. History of Modern Mathematics.} \\
+By \textsc{David Eugene Smith.}
+
+\smallskip
+\textbf{No. 2. Synthetic Projective Geometry.} \\
+By \textsc{George Bruce Halsted.}
+
+\smallskip
+\textbf{No. 3. Determinants.} \\
+By \textsc{Laenas Gifford Weld.}
+
+\smallskip
+\textbf{No. 4. Hyperbolic Functions.} \\
+By \textsc{James McMahon.}
+
+\smallskip
+\textbf{No. 5. Harmonic Functions.} \\
+By \textsc{William E. Byerly.}
+
+\smallskip
+\textbf{No. 6. Grassmann's Space Analysis.} \\
+By \textsc{Edward W. Hyde.}
+
+\smallskip
+\textbf{No. 7. Probability and Theory of Errors.} \\
+By \textsc{Robert S. Woodward.}
+
+\smallskip
+\textbf{No. 8. Vector Analysis and Quaternions.} \\
+By \textsc{Alexander Macfarlane.}
+
+\smallskip
+\textbf{No. 9. Differential Equations.} \\
+By \textsc{William Woolsey Johnson.}
+
+\smallskip
+\textbf{No. 10. The Solution of Equations.} \\
+By \textsc{Mansfield Merriman.}
+
+\smallskip
+\textbf{No. 11. Functions of a Complex Variable.} \\
+By \textsc{Thomas S. Fiske.}
+
+\normalsize \bigskip PUBLISHED BY \\
+\smallskip \textbf{JOHN WILEY \& SONS, Inc., NEW YORK. \\
+CHAPMAN \& HALL, Limited, LONDON.}
+\end{center}}}
+
+\chapter{Editors' Preface}
+
+The volume called Higher Mathematics, the first edition of which was
+published in 1896, contained eleven chapters by eleven authors, each
+chapter being independent of the others, but all supposing the
+reader to have at least a mathematical training equivalent to that
+given in classical and engineering colleges. The publication of that
+volume is now discontinued and the chapters are issued in separate
+form. In these reissues it will generally be found that the
+monographs are enlarged by additional articles or appendices which
+either amplify the former presentation or record recent advances.
+This plan of publication has been arranged in order to meet the
+demand of teachers and the convenience of classes, but it is also
+thought that it may prove advantageous to readers in special lines
+of mathematical literature.
+
+It is the intention of the publishers and editors to add other
+monographs to the series from time to time, if the call for the same
+seems to warrant it. Among the topics which are under consideration
+are those of elliptic functions, the theory of numbers, the group
+theory, the calculus of variations, and non-Euclidean geometry;
+possibly also monographs on branches of astronomy, mechanics, and
+mathematical physics may be included. It is the hope of the editors
+that this form of publication may tend to promote mathematical study
+and research over a wider field than that which the former volume
+has occupied.
+
+\medskip \footnotesize December, 1905. \normalsize
+
+\chapter{Author's Preface}
+
+Since this Introduction to Vector Analysis and Quaternions was first
+published in 1896, the study of the subject has become much more
+general; and whereas some reviewers then regarded the analysis as a
+luxury, it is now recognized as a necessity for the exact student of
+physics or engineering. In America, Professor Hathaway has published
+a Primer of Quaternions (New York, 1896), and Dr. Wilson has
+amplified and extended Professor Gibbs' lectures on vector analysis
+into a text-book for the use of students of mathematics and physics
+(New York, 1901). In Great Britain, Professor Henrici and Mr. Turner
+have published a manual for students entitled Vectors and Rotors
+(London, 1903); Dr. Knott has prepared a new edition of Kelland and
+Tait's Introduction to Quaternions (London, 1904); and Professor
+Joly has realized Hamilton's idea of a Manual of Quaternions
+(London, 1905). In Germany Dr. Bucherer has published Elemente der
+Vektoranalysis (Leipzig, 1903) which has now reached a second
+edition.\index{Bibliography}
+
+Also the writings of the great masters have been rendered more
+accessible. A new edition of Hamilton's classic, the Elements of
+Quaternions, has been prepared by Professor Joly (London, 1899,
+1901); Tait's Scientific Papers have been reprinted in collected
+form (Cambridge, 1898, 1900); and a complete edition of Grassmann's
+mathematical and physical works has been edited by Friedrich Engel
+with the assistance of several of the eminent mathematicians of
+Germany (Leipzig, 1894--). In the same interval many papers,
+pamphlets, and discussions have appeared. For those who desire
+information on the literature of the subject a Bibliography has been
+published by the Association for the promotion of the study of
+Quaternions and Allied Mathematics (Dublin, 1904).
+
+There is still much variety in the matter of notation, and the
+relation of Vector Analysis to Quaternions is still the subject of
+discussion (see Journal of the Deutsche Mathematiker-Vereinigung for
+1904 and 1905).
+
+\medskip \footnotesize \textsc{Chatham, Ontario, Canada,}
+December, 1905. \normalsize
+
+\tableofcontents
+
+%%  1. INTRODUCTION
+%%  2. ADDITION OF COPLANAR VECTORS
+%%  3. PRODUCTS OF COPLANAR VECTORS
+%%  4. COAXIAL QUATERNIONS
+%%  5. ADDITION OF VECTORS IN SPACE
+%%  6. PRODUCT OF TWO VECTORS
+%%  7. PRODUCT OF THREE VECTORS
+%%  8. COMPOSITION OF LOCATED QUANTITIES
+%%  9. SPHERICAL TRIGONOMETRY
+%% 10. COMPOSITION OF ROTATIONS
+%% INDEX
+
+\mainmatter
+
+\chapter{Introduction.}
+
+By ``Vector Analysis'' is meant a space analysis in which the vector
+is the fundamental idea; by ``Quaternions'' is meant a
+space-analysis in which the quaternion is the fundamental idea.%
+\index{Quaternions!definion of}%
+\index{Quaternions!relation to vector analysis}%
+\index{Space-analysis}%
+\index{Vector analysis!definition of}%
+\index{Vector analysis!relation to Quaternions} They are in truth
+complementary parts of one whole; and in this chapter they will be
+treated as such, and developed so as to harmonize with one another
+and with the Cartesian Analysis\footnote{For a discussion of the
+relation of
+Vector Analysis to Quaternions, see Nature, 1891--1893.}.%
+\index{Cartesian analysis} The subject to be treated is the analysis
+of quantities in space, whether they are vector in nature, or
+quaternion in nature, or of a still different nature, or are of such
+a kind that they can be adequately represented by space quantities.
+
+Every proposition about quantities in space ought to remain true
+when restricted to a plane; just as propositions about quantities in
+a plane remain true when restricted to a straight line. Hence in the
+following articles the ascent to the algebra of space%
+\index{Algebra!of space} is made through the intermediate algebra of
+the plane\index{Algebra!of the plane}. Arts.\ 2--4 treat of the more
+restricted analysis, while Arts.\ 5--10 treat of the general
+analysis.
+
+This space analysis is a universal Cartesian analysis, in the same
+manner as algebra is a universal arithmetic. By providing an
+explicit notation for directed quantities, it enables their general
+properties to be investigated independently of any particular system
+of coordinates, whether rectangular, cylindrical, or polar. It also
+has this advantage that it can express the directed quantity by a
+linear function of the coordinates, instead of in a roundabout way
+by means of a quadratic function.%
+\index{Space-analysis!advantage over Cartesian analysis}
+
+The different views of this extension of analysis which have been
+held by independent writers are briefly indicated by the titles of
+their works:\index{Bibliography}
+
+\small\begin{itemize}
+\item Argand, Essai sur une maniére de représenter les
+quantités imaginaires dans les constructions géométriques, 1806.
+
+\item Warren, Treatise on the geometrical representation of the
+square roots of negative quantities, 1828.
+
+\item Moebius, Der barycentrische Calcul, 1827.
+
+\item Bellavitis, Calcolo delle Equipollenze, 1835.
+
+\item Grassmann, Die lineale Ausdehnungslehre, 1844.
+
+\item De~Morgan, Trigonometry and Double Algebra, 1849.
+
+\item O'Brien, Symbolic Forms derived from the conception of the
+translation of a directed magnitude. Philosophical Transactions,
+1851.
+
+\item Hamilton, Lectures on Quaternions, 1853, and Elements of
+Quaternions, 1866.
+
+\item Tait, Elementary Treatise on Quaternions, 1867.
+
+\item Hankel, Vorlesungen über die complexen Zahlen und ihre
+Functionen, 1867.
+
+\item Schlegel, System der Raumlehre, 1872.
+
+\item Hoüel, Théorie des quantités complexes, 1874.
+
+\item Gibbs, Elements of Vector Analysis, 1881--4.
+
+\item Peano, Calcolo geometrico, 1888.
+
+\item Hyde, The Directional Calculus, 1890.
+
+\item Heaviside, Vector Analysis, in ``Reprint of Electrical
+Papers,'' 1885--92.
+
+\item Macfarlane, Principles of the Algebra of Physics, 1891. Papers
+on Space Analysis, 1891--3.
+\end{itemize}
+
+An excellent synopsis is given by Hagen in the second volume of his
+``Synopsis der höheren Mathematik.'' \normalsize
+
+\chapter{Addition of Coplanar Vectors.}
+
+By a ``vector'' is meant a quantity which has magnitude and
+direction.\index{Vector!definition of} It is graphically represented
+by a line whose length represents the magnitude on some convenient
+scale, and whose direction coincides with or represents the
+direction of the vector. Though a vector is represented by a line,
+its physical dimensions may be different from that of a line.
+Examples are a linear velocity which is of one dimension in length,
+a directed area which is of two dimensions in length, an axis which
+is of no dimensions in length.
+
+A vector will be denoted by a capital italic letter, as
+$B$,\footnote{This notation is found convenient by electrical
+writers in order to harmonize with the Hospitalier system of symbols
+and abbreviations.\index{Hospitalier system}} its magnitude by a
+small italic letter, as $b$, and its direction by a small Greek
+letter, as $\beta$.\index{Notation for vector}%
+\index{Vector!dimensions of}%
+\index{Vector!notation for} For example, $B = b\beta$, $R = r\rho$.
+Sometimes it is necessary to introduce a dot or a mark $\angle$ to
+separate the specification of the direction from the expression for
+the magnitude;\footnote{The dot was used for this purpose in the
+author's Note on Plane Algebra, 1883; Kennelly has since used
+$\angle$ for the same purpose in his electrical
+papers.\index{Kennelly's notation}}%
+\index{Meaning!of dot}%
+\index{Meaning!of $\angle$} but in such simple expressions as the
+above, the difference is sufficiently indicated by the difference of
+type. A system of three mutually rectangular axes will be indicated,
+as usual, by the letters $i$, $j$, $k$.\index{Unit-vector}
+
+The analysis of a vector here supposed is that into magnitude and
+direction. According to Hamilton and Tait and other writers on
+Quaternions, the vector is analyzed into tensor and unit-vector,
+which means that the tensor is a mere ratio destitute of dimensions,
+while the unit-vector is the physical magnitude.%
+\index{Hamilton's!analysis of vector}%
+\index{Tait's analysis of vector} But it will be found that the
+analysis into magnitude and direction is much more in accord with
+physical ideas, and explains readily many things which are difficult
+to explain by the other analysis.
+
+A vector quantity may be such that its components have a common
+point of application and are applied simultaneously;%
+\index{Simultaneous components}%
+\index{Vector!simultaneous} or it may be such that its components
+are applied in succession, each component
+starting from the end of its predecessor.%
+\index{Successive components}%
+\index{Vector!successive} An example of the former is found in two
+forces applied simultaneously at the same point, and an example of
+the latter in two rectilinear displacements made in succession to
+one another.
+
+\begin{center}
+\includegraphics[width=40mm]{fig01.png}
+\end{center}
+
+\smallskip Composition of Components having a common Point of
+Application.%
+\index{Composition!of two simultaneous components}%
+\index{Parallelogram of simultaneous components}%
+\index{Simultaneous components!composition of}%
+\index{Simultaneous components!parallelogram of}---Let $OA$ and $OB$
+represent two vectors of the same kind simultaneously applied at the
+point $O$. Draw $BC$ parallel to $OA$, and $AC$ parallel to $OB$,
+and join $OC$. The diagonal $OC$ represents in magnitude and
+direction and point of application the resultant of $OA$ and $OB$.
+This principle was discovered with reference to force, but it
+applies to any vector quantity coming under the above conditions.
+
+Take the direction of $OA$ for the initial direction; the direction
+of any other vector will be sufficiently denoted by the angle round
+which the initial direction has to be turned in order to coincide
+with it. Thus $OA$ may be denoted by $f_1\underline{/0}$, $OB$ by
+$f_2\underline{/\theta_2}$, $OC$ by $f\underline{/\theta}$. From the
+geometry of the figure it follows that
+\begin{gather*}
+f^2 = f_1^2 + f_2^2 + 2f_1f_2 \cos \theta_2 \\
+\intertext{and}
+\tan \theta =\frac{f_2\sin\theta_2}{f_1 + f_2\cos\theta_2}; \\
+\intertext{hence}
+OC = \sqrt{f_1^2 + f_2^2 + 2f_1f_2\cos\theta^2}
+  \underline{\left/\tan^{-1}
+    \frac{f_2\sin \theta_2}{f_1 + f_2\cos\theta^2}\right.}.
+\end{gather*}
+
+\smallskip Example.---Let the forces applied at a point be
+$2\underline{/0^{\circ}}$ and $3\underline{/60^{\circ}}$. Then the
+resultant is $\sqrt{4 + 9 + 12 \times \frac{1}{2}}\,
+\underline{\left/\tan^{-1}\frac{3\sqrt{3}}{7}\right.} =
+4.36\underline{/36^{\circ}\,30'}$.
+
+\smallskip If the first component is given as
+$f_1\underline{/\theta_1}$, then we have the more symmetrical
+formula
+\begin{equation*}
+OC = \sqrt{f_1^2 + f_2^2 + 2f_1f_2\cos(\theta_2 - \theta_1)}\,
+     \underline{\left/\tan^{-1} \frac{f_1\sin\theta_1 +
+       f_2\sin\theta_2}{f_1\cos\theta_1 + f_2\cos\theta_2}\right.}.
+\end{equation*}
+
+When the components are equal, the direction of the resultant
+bisects the angle formed by the vectors; and the magnitude of the
+resultant is twice the projection of either component on the
+bisecting line. The above formula reduces to
+\begin{equation*}
+OC = 2f_1\cos\frac{\theta_2}{2}
+  \underline{\left/\frac{\theta_2}{2}\right.}.
+\end{equation*}
+
+\smallskip Example.---The resultant of two equal alternating
+electromotive forces which differ $120^\circ$ in phase is equal in
+magnitude to either and has a phase of $60^\circ$.
+
+\begin{center}
+\includegraphics[width=40mm]{fig02.png}
+\end{center}
+
+\smallskip Given a vector and one component, to find the other
+component.---Let $OC$ represent the resultant, and $OA$ the
+component. Join $AC$ and draw $OB$ equal and parallel to $AC$. The
+line $OB$ represents the component required, for it is the only line
+which combined with $OA$ gives $OC$ as resultant. The line $OB$ is
+identical with the diagonal of the parallelogram formed by $OC$ and
+$OA$ reversed; hence the rule is, ``Reverse the direction of the
+component, then compound it with the given resultant to find the
+required component.'' Let $f\underline{/\theta}$ be the vector and
+$f_1\underline{/0}$ one component; then the other component is
+\begin{equation*}
+f_2\underline{/\theta_2} = \sqrt{f^2 + f_1^2 - 2ff_1\cos\theta}
+  \underline{\left/\tan^{-1}
+             \frac{f\sin\theta}{-f_1 + f\cos\theta}\right.}
+\end{equation*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig03.png}
+\end{center}
+
+\smallskip Given the resultant and the directions of the two
+components, to find the magnitude of the components.%
+\index{Resolution!of a vector}%
+\index{Simultaneous components!resolution of}---The resultant is
+represented by $OC$, and the directions by $OX$ and $OY$. From C
+draw $CA$ parallel to $OY$, and $CB$ parallel to $OX$; the lines
+$OA$ and $OB$ cut off represent the required components. It is
+evident that $OA$ and $OB$ when compounded produce the given
+resultant $OC$, and there is only one set of two components which
+produces a given resultant; hence they are the only pair of
+components having the given directions.
+
+\smallskip Let $f\underline{/\theta}$ be the vector and
+$\underline{/\theta_1}$ and $\underline{/\theta_2}$ the given
+directions. Then
+\begin{align*}
+f_1 + f_2\cos(\theta_2 - \theta_1) &= f\cos(\theta - \theta_1), \\
+f_1\cos(\theta_2 - \theta_1) + f_2 &= f\cos(\theta_2 - \theta),
+\end{align*}
+from which it follows that
+\begin{equation*}
+f_1 = f\frac{ \{\cos(\theta - \theta_1)
+        - \cos(\theta_2 - \theta)\cos(\theta_2 - \theta_1)\} }
+      {1 - \cos^2(\theta_2 - \theta_1)}.
+\end{equation*}
+
+For example, let $100\underline{/60^\circ}$,
+$\underline{/30^\circ}$, and $\underline{/90^\circ}$ be given; then
+\begin{equation*}
+f_1 = 100 \frac{\cos 30^\circ}{1 + \cos 60^\circ} .
+\end{equation*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig04.png}
+\end{center}
+
+\smallskip Composition of any Number of Vectors applied at a common
+Point.%
+\index{Composition!of any number of simultaneous components}%
+\index{Polygon of simultaneous components}%
+\index{Simultaneous components!polygon of}---The resultant may be
+found by the following graphic construction: Take the vectors in any
+order, as $A$, $B$, $C$. From the end of $A$ draw $B'$ equal and
+parallel to $B$, and from the end of $B'$ draw $C'$ equal and
+parallel to $C$; the vector from the beginning of $A$ to the end of
+$C'$ is the resultant of the given vectors. This follows by
+continued application of the parallelogram construction. The
+resultant obtained is the same, whatever the order; and as the order
+is arbitrary, the area enclosed has no physical meaning.
+
+The result may be obtained analytically as follows:
+
+Given
+\begin{gather*}
+f_1\underline{/\theta_1} + f_2\underline{/\theta_2} +
+  f_3\underline{/\theta_3} + \cdots + f_n\underline{/\theta_n}. \\
+\intertext{Now}
+f_1\underline{/\theta_1} = f_1\cos\theta_1\underline{/0} +
+  f_1\sin\theta_1\underline{\left/ \frac{\pi}{2}\right.}. \\
+\intertext{Similarly}
+f_2\underline{/\theta_2} = f_2\cos\theta_2\underline{/0}
+  + f_2\sin\theta_2\underline{\left/ \frac{\pi}{2}\right.}. \\
+\intertext{and}
+f_n\underline{/\theta_n} = f_n\cos\theta_n\underline{/0} +
+  f_n\sin\theta_n\underline{\left/ \frac{\pi}{2}\right.}. \\
+\intertext{Hence}
+\begin{align*}
+\sum\Bigl\lbrace f\underline{/\theta} \Bigr\rbrace & =
+  \Bigl\lbrace \sum f\cos\theta \Bigr\rbrace \underline{/0} +
+  \Bigl\lbrace \sum f\sum\theta \Bigr\rbrace
+    \underline{\left/ \frac{\pi}{2}\right.} \\
+& =\sqrt{\left( \sum f\cos\theta \right)^2 +
+     \left( \sum f\sin\theta \right )^2} \cdot
+  \tan^{-1}\frac{\sum f\sin\theta}{\sum f\cos\theta}.
+\end{align*}
+\end{gather*}
+
+In the case of a sum of simultaneous vectors applied at a common
+point, the ordinary rule about the transposition of a term in an
+equation holds good. For example, if $A + B + C = 0$, then $A + B =
+-C$, and $A + C = -B$, and $B + C = -A$, etc. This is permissible
+because there is no real order of succession among the given
+components.\footnote{ This does not hold true of a sum of vectors
+having a real order of succession. It is a mistake to attempt to
+found space-analysis upon arbitrary formal laws; the fundamental
+rules must be made to express universal properties of the thing
+denoted. In this chapter no attempt is made to apply formal laws to
+directed quantities. What is attempted is an analysis of these
+quantities.}\index{Space-analysis!foundation of}
+
+\begin{center}
+\includegraphics[width=50mm]{fig05.png} \qquad
+\includegraphics[width=20mm]{fig06.png}
+\end{center}
+
+\smallskip Composition of Successive Vectors.%
+  \index{Composition!of successive components}%
+  \index{Successive components!composition of}---The
+composition of successive vectors partakes more of the nature of
+multiplication than of addition. Let $A$ be a vector starting from
+the point $O$, and $B$ a vector starting from the end of $A$. Draw
+the third side $OP$, and from $O$ draw a vector equal to $B$, and
+from its extremity a vector equal to $A$. The line $OP$ is not the
+complete equivalent of $A + B$; if it were so, it would also be the
+complete equivalent of $B + A$. But $A + B$ and $B + A$ determine
+different paths; and as they go oppositely around, the areas they
+determine with $OP$ have different signs. The diagonal $OP$
+represents $A + B$ only so far as it is considered independent of
+path. For any number of successive vectors, the sum so far as it is
+independent of path is the vector from the initial point of the
+first to the final point of the last. This is also true when the
+successive vectors become so small as to form a continuous curve.
+The area between the curve $OPQ$ and the vector $OQ$ depends on the
+path, and has a physical meaning. \bigskip
+
+\small \begin{enumerate}
+
+\item[Prob.~1.] The resultant vector is $123\underline{/45^\circ}$,
+and one component is $100\underline{/0^\circ}$; find the other
+component.
+
+\item[Prob.~2.] The velocity of a body in a given plane is
+$200\underline{/75^\circ}$, and one component is
+$100\underline{/25^\circ}$; find the other component.
+
+\item[Prob.~3.] Three alternating magnetomotive forces are of equal
+virtual value, but each pair differs in phase by $120^\circ$; find
+the resultant. \hfill (Ans.~Zero.)
+
+\item[Prob.~4.] Find the components of the vector
+$100\underline{/70^\circ}$ in the directions $20^\circ$ and
+$100^\circ$.
+
+\item[Prob.~5.] Calculate the resultant vector of
+$1\underline{/10^\circ}$, $2\underline{/20^\circ}$,
+$3\underline{/30^\circ}$, $4\underline{/40^\circ}$.
+
+\item[Prob.~6.] Compound the following magnetic fluxes: $h \sin nt + h
+\sin (nt - 120^\circ)\underline{/120^\circ} + h \sin (nt -
+240^\circ)\underline{/240^\circ}$. \hfill
+(Ans.~$\frac{3}{2}h\underline{/nt}$.)
+
+\item[Prob.~7.] Compound two alternating magnetic fluxes at a point $a
+\cos nt \underline{/0}$ and $a \sin nt \underline{/\frac{\pi}{2}}$.
+(Ans.~$a \underline{/nt}$.)
+
+\item[Prob.~8.] Find the resultant of two simple alternating
+electromotive forces $100\underline{/20^\circ}$ and
+$50\underline{/75^\circ}$.
+
+\item[Prob.~9.] Prove that a uniform circular motion is obtained by
+compounding two equal simple harmonic motions which have the
+space-phase of their angular positions equal to the supplement of
+the time-phase of their motions.
+\end{enumerate} \normalsize
+
+\chapter{Products of Coplanar Vectors.}%
+\index{Coplanar vectors}%
+\index{Product!of two coplanar vectors}%
+\index{Rules!for vectors}%
+\index{Scalar product!of two coplanar vectors}%
+\index{Vector!co-planar}
+
+When all the vectors considered are confined to a common plane, each
+may be expressed as the sum of two rectangular components. Let $i$
+and $j$ denote two directions in the plane at right angles to one
+another; then $A = a_1i + a_2j$, $B = b_1i + b_2j$, $R = xi + yj$.
+Here $i$ and $j$ are not unit-vectors, but rather signs of
+direction.
+
+\smallskip Product of two Vectors.---Let $A = a_1i + a_2j$ and $B =
+b_1i + b_2j$ be any two vectors, not necessarily of the same kind
+physically. We assume that their product is obtained by applying the
+distributive law, but we do not assume that the order of the factors
+is indifferent. Hence
+\begin{equation*}
+AB = (a_1i + a_2j)(b_1i+b_2j) = a_1b_1ii + a_2b_2jj +
+        a_1b_2ij + a_2b_2ji.
+\end{equation*}
+
+If we assume, as suggested by ordinary algebra, that the
+square of a sign of direction is $+$, and further that the product
+of two directions at right angles to one another is the direction
+normal to both, then the above reduces to
+\begin{equation*}
+AB = a_1b_1 + a_2b_2 + (a_1b_2 - a_2b_1)k.
+\end{equation*}
+
+Thus the complete product%
+\index{Complete product!of two vectors}%
+\index{Product!complete} breaks up into two partial products%
+\index{Partial products}%
+\index{Product!partial}, namely, $a_1b_1 + a_2b_2$ which is
+independent of direction, and $(a_1b_2 - a_2b_1)k$ which has the
+axis of the plane for direction.\footnote{A common explanation which
+is given of $ij = k$ is that $i$ is an operator, $j$ an operand, and
+$k$ the result. The kind of operator which $i$ is supposed to denote
+is a quadrant of turning round the axis $i$; it is supposed not to
+be an axis, but a quadrant of rotation round an axis. This explains
+the result $ij = k$, but unfortunately it does not explain $ii = +$;
+for it would give $ii = i$.}
+
+\begin{center}
+\includegraphics[width=40mm]{fig07.png}
+\end{center}
+
+\smallskip Scalar Product of two Vectors.%
+\index{Product!scalar}%
+\index{Scalar product}---By a scalar quantity is meant a quantity
+which has magnitude and may be positive or negative but is destitute
+of direction. The former partial product is so called because it is
+of such a nature. It is denoted by $\mathrm{S}AB$ where the symbol
+S, being in Roman type, denotes, not a vector, but a function of the
+vectors $A$ and $B$.\index{Meaning!of S} The geometrical meaning of
+$\mathrm{S}AB$ is the product of $A$ and the orthogonal projection
+of $B$ upon $A$.\index{Scalar product!geometrical meaning} Let $OP$
+and $OQ$ represent the vectors $A$ and $B$; draw $QM$ and $NL$
+perpendicular to $OP$. Then
+\begin{align*}
+(OP)(OM) &= (OP)(OL) + (OP)(LM),  \\
+         &= a\left\{ b_1\frac{a_1}{a} + b_2\frac{a_2}{a} \right\},  \\
+         &= a_1b_1 + a_2b_2.
+\end{align*}
+
+\smallskip Corollary 1.---$\mathrm{S}BA = \mathrm{S}AB$. For instance,
+let $A$ denote a force and $B$ the velocity of its point of
+application; then $\mathrm{S}AB$ denotes the rate of working of the
+force. The result is the same whether the force is projected on the
+velocity or the velocity on the force.
+
+\medskip Example 1.---A force of $2$ pounds East + $3$ pounds North
+is moved with a velocity of $4$ feet East per second + $5$ feet
+North per second; find the rate at which work is done.
+\begin{equation*}
+  2\times 4 + 3\times 5 = 23 \text{ foot-pounds per second.}
+\end{equation*}
+
+\smallskip Corollary 2.---$A^2 = a_1^2 + a_2^2 = a^2$. The square of
+any vector is independent of direction;\index{Square!of a vector} it
+is an essentially positive or signless quantity; for whatever the
+direction of $A$, the direction of the other $A$ must be the same;
+hence the scalar product cannot be negative.
+
+\medskip Example 2.---A stone of $10$ pounds mass is moving with a
+velocity $64$ feet down per second + $100$ feet horizontal per
+second. Its kinetic energy then is
+\begin{equation*}
+  \frac{10}{2} (64^2 + 100^2) \text{ foot-poundals,}
+\end{equation*}
+a quantity which has no direction. The kinetic energy due to the
+downward velocity is $10\times\dfrac{64^2}{2}$ and that due to the
+horizontal velocity is $\dfrac{10}{2} \times 100^2$; the whole
+kinetic energy is obtained, not by vector, but by simple addition,
+when the components are rectangular.
+
+\begin{center}
+\includegraphics[width=40mm]{fig08.png}
+\end{center}
+
+\smallskip Vector Product of two Vectors.%
+\index{Product!vector}%
+\index{Vector product}%
+\index{Vector product!of two vectors}---The other partial product
+from its nature is called the vector product, and is denoted by
+$\mathrm{V}AB$.\index{Meaning!of V} Its geometrical meaning is the
+product of $A$ and the projection of $B$ which is perpendicular to
+$A$, that is, the area of the parallelogram formed upon $A$ and $B$.
+Let $OP$ and $OQ$ represent the vectors $A$ and $B$, and draw the
+lines indicated by the figure. It is then evident that the area of
+the triangle $OPQ = a_1 b_2 - \frac{1}{2} a_2 a_2 - \frac{1}{2} b_1
+b_2 - \frac{1}{2} (a_1 - b_1)(b_2 - a_2) = \frac{1}{2}(a_1 b_2 - a_2
+b_1)$.
+
+Thus $(a_1 b_2 - a_2 b_1)k$ denotes the magnitude of the
+parallelogram formed by $A$ and $B$ and also the axis of the plane
+in which it lies.
+
+It follows that $\mathrm{V}BA = -\mathrm{V}AB$. It is to be observed
+that the coordinates of $A$ and $B$ are mere component vectors,
+whereas $A$ and $B$ themselves are taken in a real order.
+
+\medskip Example.---Let $A = (10i + 11j)$~inches and $B = (5i +
+12j)$~inches, then $\mathrm{V}AB = (120-55)k$~square inches; that
+is, 65~square inches in the plane which has the direction $k$ for
+axis.
+
+\medskip If $A$ is expressed as $a\alpha$ and $B$ as $b\beta$, then
+$\mathrm{S}AB = ab \cos \alpha\beta$, where $\alpha\beta$ denotes
+the angle between the directions $\alpha$ and $\beta$.
+
+\medskip Example.---The effective electromotive force of $100$~volts
+per inch $\underline{/90^\circ}$ along a conductor $8$~inch
+$\underline{/45^\circ}$ is $\mathrm{S}AB = 8 \times 100\,
+\cos\underline{/45^\circ}\underline{/90^\circ}$~volts, that is, $800
+\cos 45^\circ$ volts. Here $\underline{/45^\circ}$ indicates the
+direction $\alpha$ and $\underline{/90^\circ}$ the direction
+$\beta$, and $\underline{/45^\circ}\underline{/90^\circ}$ means the
+angle between the direction of $45^\circ$ and the direction of
+$90^\circ$.
+
+\smallskip Also $\mathrm{V}AB = ab \sin \alpha\beta \cdot
+\overline{\alpha\beta}$, where $\overline{\alpha\beta}$ denotes the
+direction which is normal to both $\alpha$ and $\beta$, that is,
+their pole.\index{Meaning!of vinculum over two axes}
+
+\smallskip Example.---At a distance of $10$ feet
+$\underline{/30^\circ}$ there is a force of $100$ pounds
+$\underline{/60^\circ}$ The moment is $\mathrm{V}AB$
+\begin{align*}
+&= 10 \times 100 \sin \underline{/30^\circ} \underline{/60^\circ}
+  \text{ pound-feet } \overline{90^\circ/}\underline{/90^\circ} \\
+&= 1000 \sin 30^\circ \text{ pound feet }
+  \overline{90^\circ/}\underline{/90^\circ}
+\end{align*}
+
+Here $\overline{90^\circ/}$ specifies the plane of the angle and
+$\underline{/90^\circ}$ the angle. The two together written as above
+specify the normal $k$.
+
+\medskip Reciprocal of a Vector.%
+\index{Reciprocal!of a vector}%
+\index{Vector!reciprocal of}---By the reciprocal of a vector is
+meant the vector which combined with the original vector produces
+the product $+1$. The reciprocal of $A$ is denoted by $A^{-1}$.
+Since $AB = ab (\cos \alpha\beta + \sin \alpha\beta \cdot
+\overline{\alpha\beta})$, $b$ must equal $a^{-1}$ and $\beta$ must
+be identical with $\alpha$ in order that the product may be $1$. It
+follows that
+\begin{equation*}
+A^{-1} = \frac{1}{a}\alpha = \frac{a\alpha}{a^2} = \frac{a_1i +
+a_2j}{a_1^2 + a_2^2}.
+\end{equation*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig09.png}
+\end{center}
+
+The reciprocal and opposite vector is $-A^{-1}$.%
+\index{Opposite vector}%
+\index{Vector!opposite of} In the figure let $OP = 2\beta$ be the
+given vector; then $OQ = \frac{1}{2}\beta$ is its reciprocal, and
+$OR =\frac{1}{2}(-\beta)$ is its reciprocal and
+opposite.\footnote{Writers who identify a vector with a quadrantal
+versor\index{Quadrantal versor} are logically led to define the
+reciprocal of a vector as being opposite in direction as well as
+reciprocal in magnitude.}
+
+\smallskip Example.---If $A = 10 \text{ feet East} + 5 \text{ feet
+North}$, $A^{-1}= \dfrac{10}{125} \text{ feet East} \ +$ \\
+$\dfrac{5}{125} \text{ feet North}$ and $-A^{-1}=-\dfrac{10}{125}
+\text{ feet East} - \dfrac{5}{125}\text{ feet North}$.
+
+\smallskip Product of the reciprocal of a vector and another
+vector.---
+\begin{align*}
+A^{-1}B &= \frac{1}{a^2}AB, \\
+  &= \frac{1}{a^2}\left\{a_1b_1 + a_2b_2 + (a_1b_2 - a_2b_1)
+       \overline{\alpha\beta}\right\}, \\
+  &= \frac{b}{a}(\cos \alpha\beta + \sin\alpha\beta \cdot
+       \overline{\alpha\beta}).
+\end{align*}
+
+Hence $\mathrm{S}A^{-1}B = \dfrac{b}{a}\cos \alpha\beta$ and
+$\mathrm{V}A^{-1}B = \dfrac{b}{a} \sin \alpha\beta \cdot
+\overline{\alpha\beta}$.
+
+\newpage
+\medskip Product of three Coplanar Vectors.%
+\index{Association of three vectors}%
+\index{Product!of three coplanar vectors}---Let $A = a_1i + a_2j$,
+$B = b_1i + b_2j$, $C = c_1i + c_2j$ denote any three vectors in a
+common plane. Then
+\begin{align*}
+(AB)C &= \bigl\{(a_1b_1 + a_2b_2) + (a_1b_2 - a_2b_1)k \bigr\}
+            (c_1i + c_2j) \\
+      &= (a_1b_1 + a_2b_2)(c_1i + c_2j) +
+            (a_1b_2 - a_2b_1)(-c_2i + c_1j).
+\end{align*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig10.png}
+\end{center}
+
+The former partial product means the vector $C$ multiplied by the
+scalar product of $A$ and $B$; while the latter partial product
+means the complementary vector of $C$ multiplied by the magnitude of
+the vector product of $A$ and $B$. If these partial products
+(represented by $OP$ and $OQ$) unite to form a total product, the
+total product will be represented by $OR$, the resultant of $OP$ and
+$OQ$.
+
+The former product is also expressed by $\mathrm{S}AB \cdot C$,
+where the point separates the vectors to which the $\mathrm{S}$
+refers; and more analytically by $abc \cos \alpha\beta \cdot
+\gamma$.
+
+The latter product is also expressed by $(\mathrm{V}AB)C$, which is
+equivalent to $\mathrm{V}(\mathrm{V}AB)C$, because $\mathrm{V}AB$ is
+at right angles to $C$. It is also expressed by $abc \sin
+\alpha\beta \cdot \overline{\overline{\alpha\beta}\gamma}$, where
+$\overline{\overline{\alpha\beta}\gamma}$ denotes the direction
+which is perpendicular to the perpendicular to $\alpha$ and $\beta$
+and $\gamma$.
+
+If the product is formed after the other mode of association
+we have
+\begin{align*}
+A(BC) &= (a_1i + a_2j)(b_1c_1 + b_2c_2) +
+           (a_1i + a_2j)(b_1c_2 - b_2c_1)k \\
+      &= (b_1c_1 + b_2c_2)(a_1i + a_2j) +
+           (b_1c_2 - b_2c_1)(a_2i - a_1j) \\
+      &= \mathrm{S}BC \cdot A + \mathrm{V}A(\mathrm{V}BC).
+\end{align*}
+
+The vector $a_2i - a_1j$ is the opposite of the complementary
+vector of $a_1i + a_2j$. Hence the latter partial product differs
+with the mode of association.
+
+\smallskip Example.---Let $A = 1\underline{/0^\circ} +
+2\underline{/90^\circ}$, $B = 3\underline{/0^\circ} +
+4\underline{/90^\circ}$, $C = 5\underline{/0^\circ} +
+6\underline{/90^\circ}$. The fourth proportional to $A, B, C$ is
+\begin{align*}
+(A^{-1}B)C &=
+   \frac{1 \times 3 + 2 \times 4}{1^2 + 2^2}
+   \left\{ 5 \underline{/0^\circ}
+         + 6 \underline{/90^\circ}\right\} \\
+&\quad + \frac{1 \times 4 - 2 \times 3}{1^2 + 2^2}
+     \left\{ -6 \underline{/0^\circ} +
+                                  5 \underline{/90^\circ}\right \} \\
+&= 13.4 \underline{/0^\circ} + 11.2 \underline{/90^\circ}.
+\end{align*}
+
+\medskip Square of a Binomial of Vectors.%
+\index{Square!of two simultaneous components}---If $A + B$
+denotes a sum of non-successive vectors, it is entirely equivalent
+to the resultant vector $C$. But the square of any vector is a
+positive scalar, hence the square of $A + B$ must be a positive
+scalar. Since $A$ and $B$ are in reality components of one vector,
+the square must be formed after the rules for the products of
+rectangular components (p.\ 432). Hence
+\begin{align*}
+(A + B)^2 &= (A + B)(A + B), \\
+          &= A^2 + AB + BA + B^2, \\
+          &= A^2 + B^2 + \mathrm{S}AB + \mathrm{S}BA +
+               \mathrm{V}AB + \mathrm{V}BA, \\
+          &= A^2 + B^2 + 2\mathrm{S}AB.
+\end{align*}
+This may also be written in the form
+\begin{equation*}
+a^2 + b^2 + 2ab\cos\alpha\beta.
+\end{equation*}
+
+But when $A + B$ denotes a sum of successive vectors, there is no
+third vector $C$ which is the complete equivalent; and consequently
+we need not expect the square to be a scalar quantity.%
+\index{Square!of two successive components} We observe that there is
+a real order, not of the factors, but of the terms in the binomial;
+this causes both product terms to be $AB$, giving
+\begin{align*}
+(A + B)^2 &=  A^2 + 2AB + B^2 \\
+          &=  A^2+B^2 + 2\mathrm{S}AB + 2\mathrm{V}AB.
+\end{align*}
+
+The scalar part gives the square of the length of the third
+side, while the vector part gives four times the area included
+between the path and the third side.
+
+\smallskip Square of a Trinomial of Coplanar Vectors.%
+\index{Square!of three successive components}---Let $A + B + C$
+denote a sum of successive vectors. The product terms must be formed
+so as to preserve the order of the vectors in the trinomial; that
+is, $A$ is prior to $B$ and $C$, and $B$ is prior to $C$. Hence
+\begin{align}
+(A + B + C)^2 &= A^2 + B^2 + C^2 + 2AB + 2AC + 2BC, \notag \\
+              &= A^2 + B^2 + C^2 + 2(\mathrm{S}AB +
+                  \mathrm{S}AC + \mathrm{S}BC), \tag{1} \\
+              &\qquad + 2(\mathrm{V}AB + \mathrm{V}AC +
+                  \mathrm{V}BC). \tag{2}
+\end{align}
+Hence
+\begin{gather*}
+\mathrm{S}(A+B+C)^2 = (1) \\
+= a^2 + b^2 + c^2 + 2ab\cos \alpha\beta
+  + 2ac\cos \alpha\gamma + 2bc\cos\beta\gamma \\
+\intertext{and}
+\mathrm{V}(A+B+C)^2 = (2) \\
+= \{ 2ab\sin\alpha\beta + 2ac\sin\alpha\gamma + 2bc\sin\beta\gamma\}
+  \cdot \overline{\alpha\beta}
+\end{gather*}
+
+\begin{center}
+\includegraphics[width=35mm]{fig11.png}
+\end{center}
+
+The scalar part gives the square of the vector from the beginning of
+$A$ to the end of $C$ and is all that exists when the vectors are
+non-successive. The vector part is four times the area included
+between the successive sides and the resultant side of the polygon.
+
+Note that it is here assumed that $\mathrm{V}(A + B)C = \mathrm{V}AC
++ \mathrm{V}BC$, which is the theorem of moments. Also that the
+product terms are not formed in cyclical order, but in accordance
+with the order of the vectors in the trinomial.%
+\index{Cyclical and natural order}%
+\index{Natural order}
+
+\smallskip Example.---Let $A = 3\underline{/0^\circ}$,
+$B = 5\underline{/30^\circ}$, $C = 7\underline{/45^\circ}$; find the
+area of the polygon.
+\begin{align*}
+\frac{1}{2}\mathrm{V}(AB + AC + BC) &=
+\frac{1}{2}\{15\sin\underline{/0^\circ}\underline{/30^\circ} +
+        21\sin\underline{/0^\circ}\underline{/45^\circ} +
+        35\underline{/30^\circ}\underline{/45^\circ}\}, \\
+&= 3.75 + 7.42 + 4.53 = 15.7.
+\end{align*}
+
+\small \begin{enumerate}
+\item[Prob.~10.] At a distance of $25$ centimeters
+$\underline{/20^\circ}$ there is a force of 1000~dynes
+$\underline{/80^\circ}$; find the moment.
+
+\item[Prob.~11.] A conductor in an armature has a velocity of
+240~inches per second $\underline{/300^\circ}$ and the magnetic flux
+is 50,000~lines per square inch $\underline{/0^\circ}$; find the
+vector product. (Ans.~$1.04 \times 10^7$~lines per inch per second.)
+
+\item[Prob.~12.] Find the sine and cosine of the angle between the
+directions 0.8141~E.\ + 0.5807~N., and 0.5060~E.\ + 0.8625~N.
+
+\item[Prob.~13.] When a force of 200~pounds $\underline{/270^\circ}$ is
+displaced by 10~feet $\underline{/30^\circ}$, what is the work done
+(scalar product)? What is the meaning of the negative sign in the
+scalar product?
+
+\item[Prob.~14.] A mass of $100$ pounds is moving with a velocity of
+30 feet E.\ per second + 50 feet SE.\ per second; find its kinetic
+energy.
+
+\item[Prob.~15.] A force of $10$ pounds $\underline{/45^\circ}$ is
+acting at the end of $8$ feet $\underline{/200^\circ}$; find the
+torque, or vector product.
+
+\item[Prob.~16.] The radius of curvature of a curve is
+$2\underline{/0^\circ} + 5\underline{/90^\circ}$; find the
+curvature. \\ (Ans.~$.03\underline{/0^\circ} +
+.17\underline{/90^\circ}$.)
+
+\item[Prob.~17.] Find the fourth proportional to
+$10\underline{/0^\circ} + 2\underline{/90^\circ}$,
+$8\underline{/0^\circ} - 3\underline{/90^\circ}$, and
+$6\underline{/0^\circ} + 5\underline{/90^\circ}$.
+
+\item[Prob.~18.] Find the area of the polygon whose successive sides
+are $10\underline{/30^\circ}$, $9\underline{/100^\circ}$,
+$8\underline{/180^\circ}$, $7\underline{/225^\circ}$.
+\end{enumerate} \normalsize
+
+\chapter{Coaxial Quaternions.}%
+\index{Coaxial Quaternions}%
+\index{Quaternions!Coaxial}
+
+By a ``quaternion'' is meant the operator which changes one vector
+into another. It is composed of a magnitude and a turning factor.%
+\index{Components!of versor}%
+\index{Quaternion!definition of}%
+\index{Versor!components of} The magnitude may or may not be a mere
+ratio, that is, a quantity destitute of physical dimensions; for the
+two vectors may or may not be of the same physical kind. The turning
+is in a plane, that is to say, it is not conical. For the present
+all the vectors considered lie in a common plane; hence all the
+quaternions considered have a common axis.\footnote{The idea of the
+``quaternion'' is due to Hamilton.\index{Hamilton's!idea of
+quaternion} Its importance may be judged from the fact that it has
+made solid trigonometrical analysis possible. It is the most
+important key to the extension of analysis to space. Etymologically
+``quaternion'' means defined by four elements; which is true in
+space; in plane analysis it is defined by two.%
+\index{Quaternion!etymology of}}
+
+\begin{center}
+\includegraphics[width=30mm]{fig12.png}
+\end{center}
+
+Let $A$ and $R$ be two coinitial vectors; the direction normal to
+the plane may be denoted by $\beta$. The operator which changes $A$
+into $R$ consists of a scalar multiplier and a turning round the
+axis $\beta$. Let the former be denoted by $r$ and the latter by
+$\beta^\theta$, where $\theta$ denotes the angle in radians. Thus $R
+= r\beta^\theta A$ and reciprocally $A =
+\dfrac{1}{r}\beta^{-\theta}R$. Also $\dfrac{1}{A}R = r\beta^\theta$
+and $\dfrac{1}{R}A = \dfrac{1}{r}\beta^{-\theta}$.
+
+The turning factor $\beta^\theta$ may be expressed as the sum of two
+component operators, one of which has a zero angle and the other an
+angle of a quadrant. Thus
+\begin{equation*}
+\beta^\theta = \cos\theta \cdot \beta^\theta + \sin\theta \cdot
+\beta^\frac{\pi}{2}.
+\end{equation*}
+
+When the angle is naught, the turning-factor may be omitted; but the
+above form shows that the equation is homogeneous, and expresses
+nothing but the equivalence of a given quaternion to two component
+quaternions.\footnote{In the method of complex numbers
+$\beta^\frac{\pi}{2}$ is expressed by $i$, which stands for
+$\sqrt{-1}$.%
+\index{Algebraic imaginary}%
+\index{Imaginary algebraic} The advantages of using the above
+notation are that it is capable of being applied to space, and that
+it also serves to specify the general turning factor $\beta^\theta$
+as well as the quadrantal turning factor
+$\beta^\frac{\pi}{2}$.}\index{Components!of quaternion}
+
+Hence
+\begin{align*}
+r\beta^\theta & = r\cos\theta +
+                             r\sin\theta \cdot \beta^\frac{\pi}{2} \\
+& = p + q \cdot \beta^\frac{\pi}{2} \\
+\intertext{and}
+r\beta^\theta A & = pA + q \beta^\frac{\pi}{2} A \\
+& = pa \cdot \alpha + qa \cdot \beta^\frac{\pi}{2} \alpha.
+\end{align*}
+
+The relations between $r$ and $\theta$, and $p$ and $q$, are given by
+\begin{equation*}
+r = \sqrt{p^2 + q^2}, \quad \theta = \tan^{-1} \frac{p}{q}.
+\end{equation*}
+
+\medskip Example.---Let $E$ denote a sine alternating electromotive
+force in magnitude and phase, and $I$ the alternating current in
+magnitude and phase, then
+\begin{equation*}
+E = \left(r + 2\pi n l \cdot \beta^\frac{\pi}{2} \right) I,
+\end{equation*}
+where $r$ is the resistance, $l$ the self-induction, $n$ the
+alternations per unit of time, and $\beta$ denotes the axis of the
+plane of representation. It follows that $E = rI + 2\pi n l \cdot
+\beta^\frac{\pi}{2} I$; also that
+\begin{equation*}
+I^{-1} E = r + 2\pi n l \cdot \beta^\frac{\pi}{2},
+\end{equation*}
+that is, the operator which changes the current into the
+electromotive force is a quaternion. The resistance is the scalar
+part of the quaternion, and the inductance is the vector part.
+
+\medskip Components of the Reciprocal of a Quaternion.%
+\index{Components!of reciprocal of quaternion}%
+\index{Quaternion!reciprocal of}%
+\index{Reciprocal!of a quaternion}---Given
+\begin{equation*}
+R = \left(p + q \cdot \beta^\frac{\pi}{2} \right) A,
+\end{equation*}
+then
+\begin{align*}
+A & = \frac{1}{p + q \cdot \beta^\frac{\pi}{2}} R \\
+  & = \frac{p - q \cdot \beta^\frac{\pi}{2}}
+                       {\left(p + q \cdot \beta^\frac{\pi}{2} \right)
+      \left(p - q \cdot \beta^\frac{\pi}{2} \right)} R \\
+  & = \frac{p - q \cdot \beta^\frac{\pi}{2}}{p^2 + q^2} R \\
+  & = \left\{ \frac{p}{p^2 + q^2} - \frac{q}{p^2 + q^2} \cdot
+    \beta^\frac{\pi}{2} \right\} R.
+\end{align*}
+
+\smallskip Example.---Take the same application as above. It is
+important to obtain $I$ in terms of $E$. By the above we deduce that
+from $E = (r + 2\pi nl \cdot \beta^\frac{\pi}{2})I$
+\begin{equation*}
+I = \left\{\frac{r}{r^2+(2\pi nl)^2} -
+  \frac{2\pi nl}{r^2+(2\pi nl)^2}\cdot \beta^\frac{\pi}{2}\right\}E.
+\end{equation*}
+
+\medskip Addition of Coaxial Quaternions.%
+\index{Coaxial Quaternions!Addition of}---If the ratio of each of
+several vectors to a constant vector $A$ is given, the ratio of
+their resultant to the same constant vector is obtained by taking
+the sum of the ratios. Thus, if
+\begin{align*}
+R_1 &= (p_1 + q_1 \cdot \beta^\frac{\pi}{2}) A, \\
+R_2 &= (p_2 + q_2 \cdot \beta^\frac{\pi}{2}) A, \\
+\qquad \qquad \vdots & \qquad \vdots \qquad \vdots \qquad \vdots \\
+R_n &= (p_n + q_n \cdot \beta^\frac{\pi}{2}) A, \\
+\intertext{then}
+\sum R &= \left\{\sum p + \left(\sum q\right) \cdot
+                                   \beta^\frac{\pi}{2}\right\}A, \\
+\intertext{and reciprocally}
+A &= \frac{\sum p - \left(\sum q \right) \cdot \beta^\frac{\pi}{2}}
+          {\left( \sum p\ \right)^2 + \left( \sum q \right)^2}\sum R.
+\end{align*}
+
+\smallskip Example.\index{Composition!of coaxial quaternions}---In
+the case of a compound circuit composed of a number of simple
+circuits in parallel
+\begin{equation*}
+I_1 = \frac{r_1 - 2\pi nl_1 \cdot
+  \beta^\frac{\pi}{2}}{r_1^2 + (2\pi n)^2 l_1^2}E, \quad
+I_2 = \frac{r_2 - 2\pi nl_2 \cdot \beta^\frac{\pi}{2}}{r_2^2 +
+  (2\pi n)^2 l_2^2}E, \quad \text{etc.},
+\end{equation*}
+therefore,
+\begin{align*}
+\sum I & = \sum\left\{\frac{r - 2\pi nl \cdot
+   \beta^\frac{\pi}{2}} {r^2 + (2\pi n)^2 l^2}\right\}E \\
+& = \left\{\sum\left(\frac{r}{r^2 + (2\pi n)^2l^2}\right) -
+   2\pi n\sum\frac{l}{r^2 + (2\pi n)^2l^2} \cdot
+   \beta^\frac{\pi}{2}\right\}E,
+\end{align*}
+and reciprocally
+\begin{equation*}
+E = \frac{ \sum\left(\frac{r}{r^2 + (2\pi n)^2 l^2}\right) +
+    2\pi n\sum\left(\frac{l}{r^2 + (2\pi n)^2 l^2}\right) \cdot
+    \beta^\frac{\pi}{2}}
+{\left(\sum\frac{r}{r^2 + (2\pi n)^2 l^2}\right)^2 +
+    (2\pi n)^2\left(\sum\frac{l}{r^2 + (2\pi n)^2 l^2}\right)^2}
+    \sum I.\footnotemark
+\end{equation*}
+\footnotetext{This theorem was discovered by Lord
+Rayleigh\index{Rayleigh}; Philosophical Magazine, May, 1886. See
+also Bedell \& Crehore's Alternating Currents, p.\ 238.}
+
+\smallskip Product of Coaxial Quaternions.%
+\index{Coaxial Quaternions!Product of}%
+\index{Product!of coaxial quaternions}---If the quaternions which
+change $A$ to $R$, and $R$ to $R'$, are given, the quaternion which
+changes $A$ to $R'$ is obtained by taking the product of the given
+quaternions.
+
+Given
+\begin{align*}
+R  & = r\beta^\theta A = \left(p + q \cdot
+                                       \beta^\frac{\pi}{2}\right)A \\
+\intertext{and}
+R' & = r'\beta^{\theta'}A = \left(p' +
+                             q' \cdot \beta^\frac{\pi}{2}\right)R, \\
+\intertext{then}
+R' & = rr'\beta^{\theta+\theta'}A =
+  \left\{(pp'-qq') + (pq'+p'q) \cdot \beta^\frac{\pi}{2}\right\}A.
+\end{align*}
+
+Note that the product is formed by taking the product of the
+magnitudes, and likewise the product of the turning factors. The
+angles are summed because they are indices of the common base
+$\beta$.\footnote{Many writers, such as Hayward in ``Vector Algebra
+and Trigonometry,'' and Stringham in ``Uniplanar Algebra,'' treat
+this product of coaxial quaternions as if it were the product of
+vectors.\index{Hayward}\index{Stringham} This is the fundamental
+error in the Argand method.\index{Argand method}}
+
+\smallskip Quotient of two Coaxial Quaternions.%
+\index{Coaxial Quaternions!Quotient of}%
+\index{Quotient of two coaxial quaternions}---If the given
+quaternions are those which change $A$ to $R$, and $A$ to $R'$, then
+that which changes $R$ to $R'$ is obtained by taking the quotient of
+the latter by the former.
+
+Given
+\begin{align*}
+R & = r\beta^\theta A = (p + q \cdot \beta^\frac{\pi}{2})A \\
+\intertext{and}
+R' & = r'\beta'^{\theta'} A = (p' + q' \cdot \beta^\frac{\pi}{2})A,\\
+\intertext{then}
+R' & = \frac{r'}{r}\beta^{\theta' - \theta}R, \\
+   & = (p' + q' \cdot \beta^\frac{\pi}{2})\frac{1}{p + q \cdot
+       \beta^\frac{\pi}{2}}R, \\
+   & = (p' + q' \cdot \beta^\frac{\pi}{2})\frac{p - q \cdot
+       \beta^\frac{\pi}{2}}{p^2 + q^2}R, \\
+   & = \frac{(pp' + qq') + (pq' - p'q) \cdot
+       \beta^\frac{\pi}{2}}{p^2 + q^2}R.
+\end{align*}
+
+\small \begin{enumerate}
+\item[Prob.~19.] The impressed alternating electromotive force is
+$200$ volts, the resistance of the circuit is $10$ ohms, the
+self-induction is $\frac{1}{100}$ henry, and there are $60$
+alternations per second; required the current. \hfill (Ans. $18.7$
+amperes $\underline{/-20^\circ\,42'}$.)
+
+\item[Prob.~20.] If in the above circuit the current is $10$
+amperes, find the impressed voltage.
+
+\item[Prob.~21.] If the electromotive force is $110$ volts
+$\underline{/\theta}$ and the current is $10$ amperes
+$\underline{/\theta - \frac{1}{4}\pi}$, find the resistance and the
+self-induction, there being $120$ alternations per second.
+
+\item[Prob.~22.] A number of coils having resistances $r_1$, $r_2$,
+etc., and self-inductions $l_1$, $l_2$, etc., are placed in series;
+find the impressed electromotive force in terms of the current, and
+reciprocally.
+\end{enumerate} \normalsize
+
+\chapter{Addition of Vectors in Space.}
+
+A vector in space can be expressed in terms of three independent
+components, and when these form a rectangular set the directions of
+resolution are expressed by $i$, $j$, $k$.%
+\index{Composition!of simultaneous vectors in space}%
+\index{Vector!in space} Any variable vector $R$ may be expressed as
+$R = r\rho = xi + yj + zk$, and any constant vector $B$ may be
+expressed as
+\begin{equation*}
+B = b\beta = b_1i + b_2j + b_3k.
+\end{equation*}
+
+In space the symbol $\rho$ for the direction involves two elements.
+It may be specified as
+\begin{equation*}
+\rho = \frac{xi + yj + zk}{x^2 + y^2 + z^2},
+\end{equation*}
+where the three squares are subject to the condition that their sum
+is unity. Or it may be specified by this notation,
+$\overline{\phi/}\!\underline{/\theta}$, a generalization of the
+notation for a plane.\index{Meaning!of $\overline{\ /}$} The
+additional angle $\overline{\phi/}$ is introduced to specify the
+plane in which the angle from the initial line lies.
+
+If we are given $R$ in the form $r\overline{\phi/}\!
+\underline{/\theta}$, then we deduce the other form thus:
+\begin{equation*}
+R = r \cos\theta \cdot i + r \sin\theta \cos \phi \cdot j
+    + r \sin\theta \sin\phi \cdot k.
+\end{equation*}
+
+If $R$ is given in the form $xi + yj + zk$, we deduce
+\begin{gather*}
+R = \sqrt{x^2 + y^2 + z^2}\
+\overline{\left.\tan^{-1}\frac{z}{y}\right/}\!\!\!\!
+\underline{\left/\tan^{-1}\frac{\sqrt{y^2 + z^2}}{x}\right.}. \\
+\intertext{For example,}
+\begin{aligned}
+B &= 10\ \overline{30^\circ/}\! \underline{/45^\circ} \\
+  &= 10 \cos 45^\circ \cdot i
+     + 10 \sin 45^\circ \cos 30^\circ \cdot j
+     + 10 \sin 45^\circ \sin 30^\circ \cdot k.
+\end{aligned}
+\end{gather*}
+
+Again, from $C = 3i + 4j + 5k$ we deduce
+\begin{align*}
+C & = \sqrt{9 + 16 + 25}\
+      \overline{\left.\tan^{-1}\frac{5}{4}\right/}\!\!\!\!
+      \underline{\left/\tan^{-1}\frac{\sqrt{41}}{3}\right.} \\
+  & = 7.07\ \overline{51^\circ.4/}\! \underline{/64^\circ.9}.
+\end{align*}
+
+To find the resultant of any number of component vectors applied at
+a common point, let $R_1$, $R_2$, $\ldots$ $R_n$ represent the $n$
+vectors or,
+\begin{align*}
+R_1 &= x_1i + y_1j + z_1k, \\
+R_2 &= x_2i + y_2j + z_2k, \\
+\cdots & \cdots\cdots\cdots\cdots\cdots\cdots \\
+R_n &= x_ni + y_nj + z_nk; \\
+\intertext{then}
+\sum R &= \left(\sum x \right)i + \left(\sum y \right)j +
+  \left(\sum z \right)k \\
+\intertext{and}
+r &= \sqrt{\left(\sum x \right)^2 + \left(\sum y \right)^2 +
+  \left(\sum z \right)^2}, \\
+\tan\phi &= \frac{\sum z}{\sum y} \text{ and }
+  \tan\theta = \frac{\sqrt{\left(\sum y \right)^2
+         + \left(\sum z \right)^2}}{\sum x}.
+\end{align*}
+
+\medskip Successive Addition.---When the successive vectors do not
+lie in one plane, the several elements of the area enclosed will lie
+in different planes, but these add by vector addition into a
+resultant directed area.
+
+\small \begin{enumerate}
+\item[Prob.~23.] Express $A = 4i - 5j + 6k$ and $B = 5i + 6j - 7k$ in
+the form $r\overline{\phi/}\!\underline{/\theta}$ \\ (Ans.~$8.8\
+\overline{130^\circ/}\!\underline{/63^\circ}$ and $10.5\
+\overline{311^\circ/}\!\underline{/61^\circ .5}$.)
+
+\item[Prob.~24.] Express $C = 123\
+\overline{57^\circ/}\!\underline{/142^\circ}$ and $D = 456\
+\overline{65^\circ/}\!\underline{/200^\circ}$ in the form $xi + yj +
+zk$.
+
+\item[Prob.~25.] Express $E =
+100\ \overline{\left.\dfrac{\pi}{4}\right/}\!\!\!
+\underline{\left/\dfrac{\pi}{3}\right.}$ and $F = 1000\
+\overline{\left.\dfrac{\pi}{6}\right/}\!\!\!\underline{\left/
+\dfrac{3\pi}{4}\right.}$ in the form $xi + yj + zk$.
+
+\item[Prob.~26.] Find the resultant of $10\ \overline{20^\circ /}\!
+\underline{/30^\circ}$, $20\ \overline{30^\circ /}\!
+\underline{/40^\circ}$, and $30\ \overline{40^\circ /}\!
+\underline{/50^\circ}$.
+
+\item[Prob.~27.] Express in the form $r\ \overline{\phi/}\!
+\underline{/\theta}$ the resultant vector of $1i + 2j - 3k$, $4i -
+5j + 6k$ and $-7i + 8j + 9k$.
+\end{enumerate} \normalsize
+
+\chapter{Product of Two Vectors.}
+
+Rules of Signs for Vectors in Space.\index{Rules!for vectors}---By
+the rules $i^2 =+$, $j^2 = +$, $ij = k$, and $ji =-k$ we obtained
+(p.\ 432) a product of two vectors containing two partial products,
+each of which has the highest importance in mathematical and
+physical analysis. Accordingly, from the symmetry of space we assume
+that the following rules are true for the product of two vectors in
+space:
+\begin{align*}
+i^2 &=  +, & j^2 &=  +, & k^2 &=  + \, , \\
+ij  &=  k, & jk  &=  i, & ki  &=  j \, , \\
+ji  &= -k, & kj  &= -i, & ik  &= -j \, .
+\end{align*}
+
+\begin{center}
+\includegraphics[width=30mm]{fig13.png}
+\end{center}
+
+The square combinations give results which are independent of
+direction, and consequently are summed by simple addition. The area
+vector determined by $i$ and $j$ can be represented in direction by
+$k$, because $k$ is in tri-dimensional space the axis which is
+complementary to $i$ and $j$. We also observe that the three rules
+$ij = k$, $jk = i$, $ki = j$ are derived from one another by
+cyclical permutation; likewise the three rules $ji = -k$, $kj = -i$,
+$ik = -j$. The figure shows that these rules are made to represent
+the relation of the advance to the rotation in the right-handed
+screw. The physical meaning of these rules is made clearer by an
+application to the dynamo and the electric motor. In the dynamo
+three principal vectors have to be considered: the velocity of the
+conductor at any instant, the intensity of magnetic flux, and the
+vector of electromotive force. Frequently all that is demanded is,
+given two of these directions to determine the third. Suppose that
+the direction of the velocity is $i$, and that of the flux $j$, then
+the direction of the electromotive force is $k$. The formula $ij =
+k$ becomes
+\begin{gather*}
+\text{velocity flux} = \text{electromotive-force},\\
+\intertext{from which we deduce}
+\text{flux electromotive-force} = \text{velocity}, \\
+\intertext{and}
+\text{electromotive-force velocity} = \text{flux}.
+\end{gather*}
+
+The corresponding formula for the electric motor is %
+\index{Dynamo rule}%
+\index{Electric motor rule}%
+\index{Rules!for dynamo}
+\begin{equation*}
+\text{current flux} = \text{mechanical-force},
+\end{equation*}
+from which we derive by cyclical permutation
+\begin{equation*}
+   \text{flux force} = \text{current},
+\quad\text{and}\quad
+   \text{force current} = \text{flux}.
+\end{equation*}
+
+The formula $\text{velocity flux} = \text{electromotive-force}$ is much
+handier than any thumb-and-finger rule; for it compares the
+three directions directly with the right-handed screw.
+
+\medskip Example.---Suppose that the conductor is normal to the plane
+of the paper, that its velocity is towards the bottom, and that the
+magnetic flux is towards the left; corresponding to the rotation
+from the velocity to the flux in the right-handed screw we have
+advance into the paper: that then is the direction of the
+electromotive force.%
+\index{Relation of right-handed screw}%
+\index{Screw, relation of right-handed}
+
+Again, suppose that in a motor the direction of the current along
+the conductor is up from the paper, and that the magnetic flux is to
+the left; corresponding to current flux we have advance towards the
+bottom of the page, which therefore must be the direction of the
+mechanical force which is applied to the conductor.
+
+\medskip Complete Product of two Vectors%
+\index{Complete product!of two vectors}%
+\index{Product!complete}.---Let $A = a_1i + a_2j + a_3k$ and
+$B = b_1i + b_2j + b_3k$ be any two vectors, not necessarily of the
+same kind physically, Their product, according to the rules
+(p.~444), is
+\begin{align*}
+AB &= (a_1i + a_2j + a_3k)(b_1i + b_2j + b_3k), \\
+   &= a_1 b_1 ii + a_2 b_2 jj + a_3 b_3 kk \\
+   & \qquad + a_2 b_3 jk + a_3 b_2 kj + a_3 b_1 ki + a_1 b_3 ik
+     + a_1 b_2 ij + a_2 b_1 ji \\
+   &= a_1 b_1 + a_2 b_2 + a_3 b_3 \\
+   & \qquad + (a_2 b_3)i + (a_3 b_1 - a_1 b_3)j
+     + (a_1 b_2 - a_2 b_1)k \\
+   &= a_1 b_1 + a_2 b_2 +a_3 b_3 +
+            \begin{vmatrix}
+               a_1 & a_2 & a_3 \\
+               b_1 & b_2 & b_3 \\
+               i   & j   & k
+            \end{vmatrix}
+\end{align*}
+
+Thus the product breaks up into two partial products%
+\index{Partial products}%
+\index{Product!partial}, namely, $a_1 b_1 + a_2 b_2 + a_3 b_3$,
+which is independent of direction, and
+$\begin{vmatrix} a_1 & a_2 & a_3 \\
+                 b_1 & b_2 & b_3 \\
+                 i   & j   & k
+\end{vmatrix}$, which has the direction normal to the plane of
+$A$ and $B$. The former is called the scalar product, and the latter
+the vector product.%
+\index{Determinant!for vector product of two vectors}
+
+\smallskip In a sum of vectors, the vectors are necessarily
+homogeneous, but in a product the vectors may be heterogeneous. By
+making $a_3 = b_3 = 0$, we deduce the results already obtained for a
+plane.
+
+\begin{center}
+\includegraphics[width=30mm]{fig14.png}
+\end{center}
+
+\smallskip Scalar Product of two Vectors.\index{Product!scalar}---The
+scalar product is denoted as before by $\textrm{S}AB$. Its
+geometrical meaning is the product of $A$ and the orthogonal
+projection of $B$ upon $A$. Let OP represent $A$, and $OQ$ represent
+$B$, and let $OL$, $LM$, and $MN$ be the orthogonal projections upon
+$OP$ of the coordinates $b_1i$, $b_2j$, $b_3k$ respectively. Then
+$ON$ is the orthogonal projection of $OQ$, and
+\begin{align*}
+OP \times ON &= OP \times(OL + LM + MN), \\
+             &= a\left(b_1\frac{a_1}{a}
+                + b_2\frac{a_2}{a}
+                + b_3\frac{a_3}{a}\right), \\
+             &= a_1b_1 + a_2b_2 + a_3b_3 =\mathrm{S}AB.
+\end{align*}
+
+\smallskip Example.---Let the intensity of a magnetic flux be
+$B = b_1i + b_2j + b_3k$, and let the area be $S = s_1i + s_2j +
+s_3k$; then the flux through the area is $\mathrm{S}SB = b_ls_l +
+b_2s_2 + b_3s_3$.
+
+\medskip Corollary 1.---Hence $\mathrm{S}BA = \mathrm{S}AB$. For
+\begin{equation*}
+b_1a_1 + b_2a_2 + b_3a_3 = a_1b_1 + a_2b_2 + a_3b_3.
+\end{equation*}
+
+The product of $B$ and the orthogonal projection on it of $A$ is
+equal to the product of $A$ and the orthogonal projection on it of
+$B$. The product is positive when the vector and the projection have
+the same direction, and negative when they have opposite directions.
+
+\medskip Corollary 2.---Hence $A^2 = {a_1}^2 + {a_2}^2 + {a_3}^2 =
+a^2$. The square of $A$ must be positive; for the two factors have
+the same direction.
+
+\medskip Vector Product of two Vectors.%
+\index{Product!vector}%
+\index{Product!of two vectors in space}---The vector product as
+before is denoted by $\mathrm{V}AB$. It means the product of $A$ and
+the component of $B$ which is perpendicular to $A$, and is
+represented by the area of the parallelogram formed by $A$ and $B$.
+The orthogonal projections of this area upon the planes of $jk$,
+$ki$, and $ij$ represent the respective components of the product.
+For, let $OP$ and $OQ$ (see second figure of Art.\ 3) be the
+orthogonal projections of $A$ and $B$ on the plane of $i$ and $j$;
+then the triangle $OPQ$ is the projection of half of the
+parallelogram formed by $A$ and $B$. But it is there shown that the
+area of the triangle $OPQ$ is ${\frac{1}{2}}(a_1b_2 - a_2b_1)$. Thus
+$(a_1b_2 - a_2b_1)k$ denotes the magnitude and direction of the
+parallelogram formed by the projections of $A$ and $B$ on the plane
+of $i$ and $j$. Similarly $(a_2b_3 - a_3b_2)i$ denotes in magnitude
+and direction the projection on the plane of $j$ and $k$, and
+$(a_3b_1 - a_1b_3)j$ that on the plane of $k$ and $i$.
+
+\medskip Corollary 1.---Hence $\mathrm{V}BA = -\mathrm{V}AB$.
+
+\medskip Example.---Given two lines $A = 7i - 10j + 3k$ and $B = -9i
++ 4j - 6k$; to find the rectangular projections of the parallelogram
+which they define:
+\begin{align*}
+\mathrm{V}AB &= (60 - 12)i + (-27 + 42)j + (28 - 90)k \\
+             &= 48i + 15j - 62k.
+\end{align*}
+
+\medskip Corollary 2.---If $A$ is expressed as $a\alpha$ and $B$ as
+$b\beta$, then $\mathrm{S}AB = ab \cos \alpha\beta$ and
+$\mathrm{V}AB = ab \sin \alpha\beta \cdot \overline{\alpha\beta}$,
+where $\overline{\alpha\beta}$ denotes the direction which is normal
+to both $\alpha$ and $\beta$, and drawn in the sense given by the
+right-handed screw.
+
+\medskip Example.---Given $A = r\,\overline{\phi/}\!
+\underline{/\theta}$ and $B = r'\,\overline{\phi'/}\!
+\underline{/\theta'}$. Then
+\begin{align*}
+\mathrm{S}AB &= rr' \cos \overline{\phi/}\!\underline{/\theta}\:
+  \overline{\phi'/}\!\underline{/\theta'} \\
+             &= rr'\{\cos \theta \cos \theta' +
+                \sin \theta \sin \theta' cos (\phi'-\phi)\}.
+\end{align*}
+
+\medskip Product of two Sums of non-successive Vectors.%
+\index{Product!of two sums of simultaneous vectors}%
+\index{Simultaneous components!product of two sums of}---Let $A$ and
+$B$ be two component vectors, giving the resultant $A + B$, and let
+$C$ denote any other vector having the same point of application.
+
+\begin{center}
+\includegraphics[width=40mm]{fig15.png}
+\end{center}
+
+Let
+\begin{align*}
+A &= a_1j + a_2j + a_3k, \\
+B &= b_1i + b_2j + b_3k, \\
+C &= c_1i + c_2j + c_3k.
+\end{align*}
+
+Since $A$ and $B$ are independent of order,\index{Distributive rule}
+\begin{gather*}
+A + B = (a_1 + b_1)i + (a_2 + b_2)j + (a_3 + b_3)k, \\
+\intertext{consequently by the principle already established}
+\begin{split}
+\mathrm{S}(A + B)C &= (a_1 + b_1)c_1 + (a_2 + b_2)c_2
+                                                  + (a_3 + b_3)c_3 \\
+                   &= a_1c_1 + a_2c_2 + a_3c_3
+                                        + b_1c_1 + b_2c_2 + b_3c_3 \\
+                   &= \mathrm{S}AC + \mathrm{S}BC.
+\end{split}
+\end{gather*}
+
+Similarly
+\begin{align*}
+\mathrm{V}(A + B)C &= \{(a_2 + b_2)c_3 - (a_3 + b_3)c_2\}i+
+                                                       \text{etc.} \\
+                   &= (a_2c_3 - a_3c_2)i + (b_2c_3 - b_3c_2)i
+                                                          + \cdots \\
+                   &= \mathrm{V}AC + \mathrm{V}BC.
+\end{align*}
+
+Hence $(A + B)C = AC + BC$.
+
+In the same way it may be shown that if the second factor consists
+of two components, $C$ and $D$, which are non-successive in their
+nature, then
+\begin{equation*}
+(A+B)(C+D) = AC + AD + BC + BD.
+\end{equation*}
+
+When $A + B$ is a sum of component vectors
+\begin{align*}
+(A+B)^2 & = A^2 + B^2 + AB + BA \\
+        & = A^2 + B^2 + 2\mathrm{S}AB.
+\end{align*}
+
+\small \begin{enumerate}
+\item[Prob.~28.] The relative velocity of a conductor is S.W., and the
+magnetic flux is N.W.; what is the direction of the electromotive
+force in the conductor?
+
+\item[Prob.~29.] The direction of the current is vertically downward,
+that of the magnetic flux is West; find the direction of the
+mechanical force on the conductor.
+
+\item[Prob.~30.] A body to which a force of $2i + 3j + 4k$~pounds is
+applied moves with a velocity of $5i + 6j + 7k$~feet per second;
+find the rate at which work is done.
+
+\item[Prob.~31.] A conductor $8i + 9j + 10k$~inches long is subject to
+an electromotive force of $11 i +12j + 13k$~volts per inch; find the
+difference of potential at the ends. \hfill (Ans.\ 326~volts.)
+
+\item[Prob.~32.] Find the rectangular projections of the area of the
+parallelogram defined by the vectors $A = 12i - 23j - 34k$ and $B =
+-45i - 56j + 67k$.
+
+\item[Prob.~33.] Show that the moment of the velocity of a body with
+respect to a point is equal to the sum of the moments of its
+component velocities with respect to the same point.
+
+\item[Prob.~34.] The arm is $9i + 11j + 13k$~feet, and the force
+applied at either end is $17i + 19j + 23k$~pounds weight; find the
+torque.
+
+\item[Prob.~35.] A body of 1000~pounds mass has linear velocities of
+50~feet per second $\overline{30^\circ/}\!\underline{/45^\circ}$ and
+60~feet per second $\overline{60^\circ/}\!\underline{/22^\circ.5}$;
+find its kinetic energy.
+
+\item[Prob.~36.] Show that if a system of area-vectors can be
+represented by the faces of a polyhedron, their resultant vanishes.
+
+\item[Prob.~37.] Show that work done by the resultant velocity is equal
+to the sum of the works done by its components.
+\end{enumerate} \normalsize
+
+\chapter{Product of Three Vectors.}
+
+Complete Product.\index{Complete product!of three vectors}---Let us
+take $A = a_1i + a_2j + a_3k$, $B = b_1i + b_2j + b_3k$, and $C =
+c_1i + c_2j + c_3k$. By the product of $A$, $B$, and $C$ is meant
+the product of the product of $A$ and $B$ with $C$, according to the
+rules p.~444).%
+\index{Determinant!for second partial product of three vectors}
+Hence
+\begin{align*}
+ABC &= (a_1b_1 + a_2b_2 + a_3b_3)(c_1i + c_2j + c_3k) \notag \\
+&\quad+\Bigl\{(a_2b_3 - a_3b_2)i + (a_3b_1-a_1b_3)j +
+(a_1b_2-a_2b_1)k\Bigr\}
+       (c_1i+c_2j+c_3k) \notag \\
+    &= (a_1b_1+a_2b_2+a_3b_3)(c_1i+c_2j+c_3k) \tag{1} \\
+    &\quad+ \begin{vmatrix}
+              \begin{vmatrix}
+                 a_2 & a_3 \\
+                 b_2 & b_3
+              \end{vmatrix} &
+              \begin{vmatrix}
+                 a_3 & a_1 \\
+                 b_3 & b_1
+              \end{vmatrix} &
+              \begin{vmatrix}
+                 a_1 & a_2 \\
+                 b_1 & b_2
+              \end{vmatrix} \\
+              c_1 & c_2 & c_3 \\
+              i   & j   & k
+            \end{vmatrix} \tag{2} \\
+    &\quad+ \begin{vmatrix}
+               a_1 & a_2 & a_3 \\
+               b_1 & b_2 & b_3 \\
+               c_1 & c_2 & c_3
+             \end{vmatrix} \tag{3}
+\end{align*}
+
+\medskip Example.---Let $A = 1i + 2j + 3k$, $B = 4i + 5j + 6k$, and
+$C = 7i + 8j + 9k$. Then
+\begin{align*}
+(1) &= (4 + 10 + 18)(7i + 8j + 9k) = 32(7i + 8j + 9k).\\
+(2) &= \begin{vmatrix}
+          -3 & 6 & -3 \\
+           7 & 8 &  9 \\
+           i & j & k
+       \end{vmatrix} = 78i + 6j - 66k.\\
+(3) &= \begin{vmatrix}
+           1 & 2 & 3 \\
+           4 & 5 & 6 \\
+           7 & 8 & 9
+       \end{vmatrix} = 0.
+\end{align*}
+
+\smallskip If we write $A = a\alpha$, $B = b\beta$, $C = c\gamma$,
+then
+\begin{align}
+ABC &= abc \cos \alpha\beta \cdot \gamma \tag{1} \\
+    &\quad+ abc \sin \alpha\beta \sin \overline{\alpha\beta\gamma}
+           \cdot \overline{\overline{\alpha\beta}\gamma} \tag{2} \\
+    &\quad+ abc \sin\alpha\beta \cos\overline{\alpha\beta}\gamma,
+                                                              \tag{3}
+\end{align}
+where $\cos\overline{\alpha\beta}\gamma$ denotes the cosine of the
+angle between the directions $\overline{\alpha\beta}$ and $\gamma$,
+and $\overline{\overline{\alpha\beta}\gamma}$ denotes the direction
+which is normal to both $\overline{\alpha\beta}$ and $\gamma$.
+
+We may also write
+\begin{align*}
+ABC &= \mathrm{S}AB \cdot C + \mathrm{V}(\mathrm{V}AB)C
+        + \mathrm{S}(\mathrm{V}AB)C \\
+  &\quad \qquad (1) \qquad \qquad (2) \qquad \qquad (3)
+\end{align*}
+
+\medskip First Partial Product.---It is merely the third vector
+multiplied by the scalar product of the other two, or weighted by
+that product as an ordinary algebraic quantity. If the directions
+are kept constant, each of the three partial products is
+proportional to each of the three magnitudes.%
+\index{Partial products!of three vectors}
+
+\medskip Second Partial Product.---The second partial product may be
+expressed as the difference of two products similar to the
+first.%
+\index{Partial products!resolution of second partial product}%
+\index{Resolution!of second partial product of three vectors} For
+\begin{align*}
+\mathrm{V}(\mathrm{V}AB)C
+  &= \{-(b_2c_2 + b_3c_3)a_1 + (c_2a_2 + c_3a_3)b_1\}i \\
+  &\quad+ \{-(b_3c_3 + b_1c_1)a_2 + (c_3a_3 + c_1a_1)b_2\}j \\
+  &\quad+ \{-(b_1c_1 + b_2c_2)a_3 + (c_1a_1 + c_2a_2)b_3\}k.
+\end{align*}
+
+By adding to the first of these components the null term $(b_1c_1a_1
+- c_1a_1b_1)i$ we get $-\mathrm{S}BC \cdot a_1i + \mathrm{S}CA \cdot
+b_1i$, and by treating the other two components similarly and adding
+the results we obtain
+\begin{equation*}
+\mathrm{V}(\mathrm{V}AB)C = -\mathrm{S}BC \cdot A
+ + \mathrm{S}CA \cdot B.
+\end{equation*}
+
+The principle here proved is of great use in solving equations (see
+p.~455).
+
+\medskip Example.---Take the same three vectors as in the preceding
+example. Then
+\begin{align*}
+\mathrm{V}(\mathrm{V}AB)C & = -(28 + 40 + 54)(1i + 2j + 3k)\\
+  &\quad +(7 + 16 + 27)(4i + 5j + 6k) \\
+  & = 78i + 6j - 66k.
+\end{align*}
+
+\newpage
+The determinant%
+\index{Determinant!for second partial product of three vectors}
+expression for this partial product may also be written in the form
+\begin{equation*}
+\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}
+\begin{vmatrix} c_1 & c_2 \\ i & j \end{vmatrix} +
+\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}
+\begin{vmatrix} c_2 & c_3 \\ j & k \end{vmatrix} +
+\begin{vmatrix} a_3 & a_1 \\ b_3 & b_1 \end{vmatrix}
+\begin{vmatrix} c_3 & c_1 \\ k & i \end{vmatrix}
+\end{equation*}
+It follows that the frequently occurring determinant expression
+\begin{equation*}
+\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}
+\begin{vmatrix} c_1 & c_2 \\ d_1 & d_2 \end{vmatrix} +
+\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}
+\begin{vmatrix} c_2 & c_3 \\ d_2 & d_3 \end{vmatrix} +
+\begin{vmatrix} a_3 & a_1 \\ b_3 & b_1 \end{vmatrix}
+\begin{vmatrix} c_3 & c_1 \\ d_3 & d_1 \end{vmatrix}
+\end{equation*}
+means $\mathrm{S}(\mathrm{V}AB)(\mathrm{V}CD)$.
+
+\medskip Third Partial Product.---From the determinant expression for
+the third product, we know that
+\begin{align*}
+\mathrm{S}(\mathrm{V}AB)C &= \mathrm{S}(\mathrm{V}BC)A =
+  \mathrm{S}(\mathrm{V}CA)B \\
+&= -\mathrm{S}(\mathrm{V}BA)C = -\mathrm{S}(\mathrm{V}CB)A
+  = -\mathrm{S}(\mathrm{V}AC)B.
+\end{align*}
+Hence any of the three former may be expressed by $\mathrm{S}ABC$,
+and any of the three latter by $-\mathrm{S}ABC$.
+
+\begin{center}
+\includegraphics[width=40mm]{fig16.png}
+\end{center}
+
+The third product $\mathrm{S}(\mathrm{V}AB)C$ is represented by the
+volume of the parallelepiped formed by the vectors $A, B, C$
+taken in that order.%
+\index{Determinant!for scalar product of three vectors} The line
+$\mathrm{V}AB$ represents in magnitude and direction the area formed
+by $A$ and $B$, and the product of $\mathrm{V}AB$ with the
+projection of $C$ upon it is the measure of the volume in magnitude
+and sign. Hence the volume formed by the three vectors has no
+direction in space, but it is positive or negative according to the
+cyclical order of the vectors.
+
+In the expression $abc\, \sin \alpha\beta\, \cos \alpha\beta\gamma$
+it is evident that $\sin \alpha\beta$ corresponds to $\sin \theta$,
+and $\cos \alpha\beta\gamma$ to $\cos \phi$, in the usual formula
+for the volume of a parallelepiped.
+
+\medskip Example.---Let the velocity of a straight wire parallel to
+itself be $V = 1000\, \underline{/30^\circ}$ centimeters per second,
+let the intensity of the magnetic flux be $B = 6000\,
+\underline{/90^\circ}$ lines per square centimeter, and let the
+straight wire $L = 15$ centimeters $\overline{60^\circ/}\!
+\underline{/45^\circ}$. Then $\mathrm{V}VB = 6000000 \sin 60^\circ\,
+\overline{90^\circ/}\!\underline{/90^\circ}$ lines per centimeter
+per second. Hence $\mathrm{S}(\mathrm{V}VB)L = 15 \times 6000000
+\sin 60^\circ \cos \phi$ lines per second where $\cos \phi = \sin
+45^\circ\, \sin 60^\circ$.
+
+\medskip Sum of the Partial Vector Products.%
+\index{Total vector product of three vectors}%
+\index{Vector product!of three vectors}---By adding the first and
+second partial products we obtain the total vector product of $ABC$,
+which is denoted by $\mathrm{V}(ABC)$. By decomposing the second
+product we obtain
+\begin{equation*}
+\mathrm{V}(ABC) = \mathrm{S}AB \cdot C - \mathrm{S}BC \cdot A +
+\mathrm{S}CA \cdot B.
+\end{equation*}
+By removing the common multiplier $abc$, we get
+\begin{align*}
+\mathrm{V}(\alpha\beta\gamma) &= \cos \alpha\beta \cdot \gamma -
+\cos \beta\gamma \cdot \alpha + \cos \gamma\alpha \cdot \beta. \\
+\intertext{Similarly}
+\mathrm{V}(\beta\gamma\alpha) &= \cos \beta\gamma \cdot \alpha -
+\cos \gamma \alpha \cdot \beta + \cos \alpha \beta \cdot\gamma \\
+\intertext{and}
+\mathrm{V}(\gamma\alpha\beta) &= \cos \gamma\alpha \cdot \beta -
+\cos \alpha\beta \cdot \gamma + \cos \beta\gamma \cdot \alpha.
+\end{align*}
+
+These three vectors have the same magnitude, for the square of each
+is
+\begin{equation*}
+\cos^2\alpha\beta + \cos^2\beta\gamma + \cos^2\gamma\alpha -
+2\cos\alpha\beta \cos\beta\gamma\cos\gamma\alpha,
+\end{equation*}
+that is, $1-\{\mathrm{S}(\alpha\beta\gamma)\}^2.$
+
+\begin{center}
+\includegraphics[width=30mm]{fig17.png}
+\end{center}
+
+They have the directions respectively of $\alpha'$, $\beta'$,
+$\gamma'$, which are the corners of the triangle whose sides are
+bisected by the corners $\alpha$, $\beta$, $\gamma$ of the given
+triangle.
+
+\small \begin{enumerate}
+\item[Prob.~38.] Find the second partial product of $9\,
+\overline{20^\circ/}\!\underline{/30^\circ}$, $10\,
+\overline{30^\circ/}\!\underline{/40^\circ}$, $11\,
+\overline{45^\circ/}\!\underline{/45^\circ}$. Also the third partial
+product.
+
+\item[Prob.~39.] Find the cosine of the angle between the plane of
+$l_1 i + m_1 j + n_1 k$ and $l_2 i + m_2 j + n_2 k$ and the plane of
+$l_3 i + m_3 j + n_3 k$ and $l_4 i + m_4 j + n_4 k$.
+
+\item[Prob.~40.] Find the volume of the parallelepiped determined by
+the vectors $100i + 50j + 25k$, $50i + 10j + 80k$, and $-75i + 40j -
+80k$.
+
+\item[Prob.~41.] Find the volume of the tetrahedron determined by the
+extremities of the following vectors: $3i - 2j + 1k$, $-4i + 5j -
+7k$, $3i - 7j - 2k$, $8i + 4j - 3k$.
+
+\item[Prob.~42.] Find the voltage at the terminals of a conductor when
+its velocity is 1500 centimeters per second, the intensity of the
+magnetic flux is 7000 lines per square centimeter, and the length of
+the conductor is 20 centimeters, the angle between the first and
+second being $30^\circ$, and that between the plane of the first two
+and the direction of the third $60^\circ$. \hfill (Ans.
+$.91$~volts.)
+
+\item[Prob.~43.] Let $\alpha = \overline{20^\circ/}\!
+\underline{/10^\circ}$, $\beta = \overline{30^\circ/}\!
+\underline{/25^\circ}$, $\gamma=\overline{40^\circ/}\!
+\underline{/35^\circ}$. Find $\mathrm{V}\alpha\beta\gamma$, and
+deduce $\mathrm{V}\beta\gamma\alpha$ and
+$\mathrm{V}\gamma\alpha\beta$.
+\end{enumerate} \normalsize
+
+\chapter{Composition of Quantities.}
+
+A number of homogeneous quantities are simultaneously located at
+different points; it is required to find how to add or compound
+them.
+
+\begin{center}
+\includegraphics[width=40mm]{fig18.png}
+\end{center}
+
+\smallskip Addition of a Located Scalar Quantity.---Let $m_A$ denote a
+mass $m$ situated at the extremity of the radius-vector $A$. A mass
+$m-m$ may be introduced at the extremity of any radius-vector R, so
+that
+\begin{align*}
+m_A &= (m - m)_R + m_A \\
+    &= m_R + m_A - m_R \\
+    &= m_R + m(A - R).
+\end{align*}
+Here $A-R$ is a simultaneous sum, and denotes the radius-vector from
+the extremity of $R$ to the extremity of $A$. The product $m(A - R)$
+is what Clerk Maxwell called a mass-vector, and means the directed
+moment of $m$ with respect to the extremity of $R$.\index{Maxwell}
+The equation states that the mass $m$ at the extremity of the vector
+$A$ is equivalent to the equal mass at the extremity of $R$,
+together with the said mass-vector applied at the extremity of $R$.
+The equation expresses a physical of mechanical
+principle.\index{Composition!of
+mass-vectors}\index{Mass-vector}\index{Mass-vector!composition of}
+
+Hence for any number of masses, $m_1$ at the extremity of $A_1$,
+$m_2$ at the extremity of $A_2$, etc.,
+\begin{equation*}
+\sum m_A = \sum m_R + \sum\Bigl\{m(A - R)\Bigr\},
+\end{equation*}
+where the latter term denotes the sum of the mass-vectors treated as
+simultaneous vectors applied at a common point. Since
+\begin{align*}
+\sum \bigl\{m(A-R)\bigr\} &= \sum m A - \sum mR \\
+                &= \sum m A - R\sum m,
+\end{align*}
+the resultant moment will vanish\index{Couple of forces!condition
+for couple vanishing} if
+\begin{equation*}
+R = \frac{\sum mA}{\sum m},\quad\text{or}\quad R \sum m = \sum mA
+\end{equation*}
+
+\smallskip Corollary.\index{Couple of forces}---Let
+\begin{align*}
+R &= xi + yj + zk, \\
+\intertext{and}
+A &= a_1j + b_1j+c_1k; \\
+\intertext{then the above condition may be written as}
+xi + yj + zk &= \frac{\sum\bigl\{m(ai + bj + ck)\bigr\}}{\sum m} \\
+             &= \frac{\sum (ma)\cdot i}{\sum m} +
+                \frac{\sum (mb)\cdot j}{\sum m} +
+                \frac{\sum (mc)\cdot k}{\sum m}; \\
+\intertext{therefore}
+           x &= \frac{\sum (ma)}{\sum m},\
+           y  = \frac{\sum (mb)}{\sum m},\
+           z  = \frac{\sum (mc)}{\sum m}. \\
+\end{align*}
+
+Example.---Given $5$ pounds at $10$ feet $\overline{45^\circ/}\!
+\underline{/30^\circ}$ and $8$ pounds at $7$ feet
+$\overline{60^\circ/}\!\underline{/45^\circ}$; find the moment when
+both masses are transferred to $12$ feet $\overline{75^\circ/}\!
+\underline{/60^\circ}$.
+\begin{align*}
+m_1A_1 &= 50(\cos 30^\circ i + \sin 30^\circ \cos 45^\circ j
+          + \sin 30^\circ \sin 45^\circ k), \\
+m_1A_1 &= 56(\cos 45^\circ i + \sin 45^\circ \cos 60^\circ j
+          + \sin 45^\circ \sin 60^\circ k), \\
+(m_1 + m_2)R &= 156(\cos 60^\circ i + \sin 60^\circ \cos 75^\circ j
+          + \sin 60^\circ \sin 75^\circ k),\\
+\text{moment} &=  m_1 A_1 + m_2A_2 -(m_1 + m_2)R.
+\end{align*}
+
+\newpage
+\begin{center}
+\includegraphics[width=40mm]{fig19.png}
+\end{center}
+
+\smallskip Composition of a Located Vector
+Quantity.\index{Composition!of located vectors}\index{Located
+vectors}---Let $F_A$ denote a force applied at the extremity of the
+radius-vector $A$. As a force $F-F$ may introduced at the extremity
+of any radius-vector $R$, we have
+\begin{align*}
+F_A &= (F - F) + F_A \\
+    &= F_R + \mathrm{V}(A-R)F.
+\end{align*}
+
+This equation asserts that a force $F$ applied at the extremity of
+$A$ is equivalent to an equal force applied at the extremity of $R$
+together with a couple whose magnitude and direction are given by
+the vector product of the radius-vector from the extremity of $R$ to
+the extremity of $A$ and the force.
+
+Hence for a system of forces applied at different points, such as
+$F_1$ at $A_1$, $F_2$ at $A_2$, etc., we obtain
+\begin{align*}
+\sum \left(F_A\right) &= \sum \left(F_R\right)
+                         + \sum \mathrm{V}\left(A - R\right)F  \\
+           &= \left(\sum F\right)_R
+                         + \sum \mathrm{V}\left(A - R\right)F. \\
+\intertext{Since}
+\sum \mathrm{V}\left(A - R\right)F
+                        &= \sum \mathrm{V}AF - \sum \mathrm{V}RF \\
+                        &= \sum \mathrm{V}AF - \mathrm{V}R \sum F \\
+\intertext{the condition for no resultant couple is} \mathrm{V} R
+\sum F &= \sum \mathrm{V} A F,
+\end{align*}
+which requires $\sum F$ to be normal to $\sum \mathrm{V} A F$.
+
+\medskip Example.---Given a force $1i + 2j + 3k$ pounds weight at $4i
++ 5j + 6k$ feet, and a force of $7i + 9j + 11k$ pounds weight at
+$10i + 12j + 14k$ feet; find the torque which must be supplied when
+both are transferred to $2i + 5j + 3k$, so that the effect may be
+the same as before.\index{Torque}
+\begin{align*}
+\mathrm{V} A_1 F_1  &= 3i - 6j + 3k, \\
+\mathrm{V} A_2 F_2  &= 6i - 12j + 6k, \\
+\sum \mathrm{V} A F &= 9i - 18j + 9k, \\
+\sum F              &= 8i + 11j + 14k, \\
+\mathrm{V} R \sum F &= 37i - 4j - 18k, \\
+\text{Torque}       &= -28i - 14j + 27k.
+\end{align*}
+
+By taking the vector product of the above equal vectors with the
+reciprocal of $\sum F$ we obtain
+\begin{equation*}
+\mathrm{V}\left\{ \left(\mathrm{V} R \sum F\right)
+                \frac{1}{\sum F} \right\}
+= \mathrm{V}\left\{ \left(\sum \mathrm{V} A F \right)
+                \frac{1}{\sum F} \right\}.
+\end{equation*}
+
+By the principle previously established the left member resolves
+into $-R + \mathrm{S}R\dfrac{1}{\sum F} \cdot \sum F$; and the right
+member is equivalent to the complete product on account of the two
+factors being normal to one another; hence
+\begin{align}
+-R &+ \mathrm{S} R \frac{1}{\sum F} \cdot \sum F
+   = \sum \left(\mathrm{V} A F \right) \frac{1}{\sum F}; \notag \\
+\intertext{that is,}
+R &= \frac{1}{\sum F}\sum \left(\mathrm{V}AF \right) \tag{1} \\
+  &\quad+ \mathrm{S}R\frac{1}{\sum F} \cdot \sum F \tag{2}.
+\end{align}
+
+\begin{center}
+\includegraphics[width=25mm]{fig20.png}
+\end{center}
+
+The extremity of $R$ lies on a straight line whose perpendicular is
+the vector (1) and whose direction is that of the resultant force.
+The term (2) means the projection of $R$ upon that line.
+
+The condition for the central axis\index{Central axis} is that the
+resultant force and the resultant couple should have the same
+direction; hence it is given by
+\begin{align*}
+\mathrm{V}\left\{\sum \mathrm{V}AF - \mathrm{V}R\sum F\right\}
+  \sum F = 0; \\
+\intertext{that is}
+\mathrm{V}\left(\mathrm{V}R\sum F \right)\sum F =
+  \mathrm{V}\left(\sum AF \right)\sum F.
+\end{align*}
+
+By expanding the left member according to the same principle as
+above, we obtain
+\begin{equation*}
+-\left(\sum F\right)^2R + \mathrm{S}R\sum F \cdot \sum F
+  = V\left(\sum AF \right)\sum F;
+\end{equation*}
+therefore
+\begin{align*}
+R &= \frac{1}{\left(\sum F \right)^2}\mathrm{V}\sum F
+     \left(\mathrm{V}\sum AF\right) +
+     \frac{\mathrm{S}R\sum F}{\left(\sum F\right)^2} \cdot \sum F \\
+  &= \mathrm{V}\left(\frac{1}{\sum F}\right)(\mathrm{V}\sum AF) +
+       \mathrm{S}R\frac{1}{\sum F} \cdot \sum F.
+\end{align*}
+
+This is the same straight line as before, only no relation is now
+imposed on the directions of $\sum F$ and $\sum \mathrm{V}AF$; hence
+there always is a central axis.
+
+\medskip Example.---Find the central axis for the system of forces in
+the previous example. Since $\sum F = 8i + 11j + 14k$, the direction
+of the line is
+\begin{equation*}
+\frac{8i + 11j + 14k}{\sqrt{64 + 121 + 196}}.
+\end{equation*}
+
+Since $\dfrac{1}{\sum F} = \dfrac{8i + 11j + 14k}{381}$ and $\sum
+\mathrm{V}AF = 9i - 18j + 9k$, the perpendicular to the line is
+\begin{equation*}
+\mathrm{V}\,\frac{8i + 11j + 14k}{381}\, 9i - 18j + 9k =
+  \frac{1}{381}\,\{351i + 54j -243k\}.
+\end{equation*}
+
+\small \begin{enumerate}
+\item[Prob.~44.] Find the moment at $\overline{90^\circ/}\!
+\underline{/270^\circ}$ of 10~pounds at 4~feet
+$\overline{10^\circ/}\!\underline{/20^\circ}$ and 20~pounds at
+5~feet $\overline{30^\circ/}\!\underline{/120^\circ}$.
+
+\item[Prob.~45.] Find the torque for $4i + 3j + 2k$ pounds weight at
+$2i - 3j + 1k$ feet, and $2i - 1k - 1k$ pounds weight at $-3i + 4j +
+5k$~feet when transferred to $-3i -2j -4k$ feet.
+
+\item[Prob.~46.] Find the central axis in the above case.
+
+\item[Prob.~47.] Prove that the mass-vector drawn from any origin to a
+mass equal to that of the whole system placed at the center of mass
+of the system is equal to the sum of the mass-vectors drawn from the
+same origin to all the particles of the system.
+\end{enumerate} \normalsize
+
+\chapter{Spherical Trigonometry.}\index{Spherical trigonometry}
+
+\begin{center}
+\includegraphics[width=40mm]{fig21.png}
+\end{center}
+
+Let $i$, $j$, $k$ denote three mutually perpendicular axes. In order
+to distinguish clearly between an axis and a quadrantal version
+round it, let $i^\frac{\pi}{2}$, $j^\frac{\pi}{2}$,
+$k^\frac{\pi}{2}$ denote quadrantal versions in the positive sense
+about the axes $i$, $j$, $k$ respectively.\index{Meaning!of
+$\frac{1}{2}\pi$ as index} The directions of positive version are
+indicated by the arrows.
+
+By $i^\frac{\pi}{2}i^\frac{\pi}{2}$ is meant the product of two
+quadrantal versions round $i$; it is equivalent to a semicircular
+version round $i$; hence $i^\frac{\pi}{2}i^\frac{\pi}{2} = i^\pi =
+-$. Similarly $j^\frac{\pi}{2}j^\frac{\pi}{2}$ means the product of
+two quadrantal versions round $j$, and
+$j^\frac{\pi}{2}j^\frac{\pi}{2}=j^\pi=-$. Similarly
+$k^\frac{\pi}{2}k^\frac{\pi}{2}=k^\pi=-$.
+
+By $i^\frac{\pi}{2}j^\frac{\pi}{2}$ is meant a quadrant round $i$
+followed by a quadrant round $j$; it is equivalent to the quadrant
+from $j$ to $i$, that is, to $-k^\frac{\pi}{2}$. But
+$j^\frac{\pi}{2}i^\frac{\pi}{2}$ is equivalent to the quadrant from
+$-i$ to $-j$, that is, to $k^\frac{\pi}{2}$. Similarly for the other
+two pairs of products. Hence we obtain the following
+
+\begin{center}
+Rules for Versors.\index{Rules!for versors}\index{Versor!rules for}
+\end{center}
+\begin{gather*}
+i^\frac{\pi}{2}i^\frac{\pi}{2} = -, \quad
+j^\frac{\pi}{2}j^\frac{\pi}{2} = -, \quad
+k^\frac{\pi}{2}k^\frac{\pi}{2} = -, \\
+i^\frac{\pi}{2}j^\frac{\pi}{2} = -k^\frac{\pi}{2}, \quad
+j^\frac{\pi}{2}i^\frac{\pi}{2} =  k^\frac{\pi}{2}, \\
+j^\frac{\pi}{2}k^\frac{\pi}{2} = -i^\frac{\pi}{2}, \quad
+k^\frac{\pi}{2}j^\frac{\pi}{2} =  i^\frac{\pi}{2}  \\
+k^\frac{\pi}{2}i^\frac{\pi}{2} = -j^\frac{\pi}{2}, \quad
+i^\frac{\pi}{2}k^\frac{\pi}{2} =  j^\frac{\pi}{2}.
+\end{gather*}
+
+The meaning of these rules will be seen from the following
+application. Let $li + mj + nk$ denote any axis, then $(li + mj +
+nk)^\frac{\pi}{2}$ denotes a quadrant of angle round that axis. This
+quadrantal version can be decomposed into the three rectangular
+components $li^\frac{\pi}{2}$, $mj^\frac{\pi}{2}$,
+$nk^\frac{\pi}{2}$; and these components are not successive
+versions, but the parts of one version.\index{Versor!components of}
+Similarly any other quadrantal version $(l'i + m'j +
+n'j)^\frac{\pi}{2}$ can be resolved into
+$l'i^\frac{\pi}{2}$, $m'j^\frac{\pi}{2}$, $n'k^\frac{\pi}{2}$.%
+\index{Product!of two quadrantal versors}%
+\index{Rules!for expansion of product of two quadrantal versors} By
+applying the above rules, we obtain
+\begin{align*}
+(li &+ mj + nk)^\frac{\pi}{2}(l'i + m'j + n'k)^\frac{\pi}{2} \\
+    &= (li^\frac{\pi}{2} + mj^\frac{\pi}{2} + nk^\frac{\pi}{2})
+         (l'i^\frac{\pi}{2} + m'j^\frac{\pi}{2} + n'k^\frac{\pi}{2}) \\
+    &= -(ll' + mm' + nn') -(mn' - m'n)i^\frac{\pi}{2}
+            - (nl' - n'l)j^\frac{\pi}{2} -(lm' - l'm)k^\frac{\pi}{2} \\
+    &= -(ll' + mm' + nn')-\bigl\{(mn' - m'n)i + (nl' - n'l)j
+             +(lm' - l'm)k\bigr\}^\frac{\pi}{2}.
+\end{align*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig22.png}
+\end{center}
+
+\smallskip Product of Two Spherical Versors.%
+\index{Product!of two spherical versors}%
+\index{Spherical versor}%
+\index{Spherical versor!product of two}%
+\index{Versor!product of two quadrantal}%
+\index{Versor!product of two general spherical}---Let $\beta$ denote
+the axis and $b$ the ratio of the spherical versor $PA$, then the
+versor itself is expressed by $\beta^b$. Similarly let $\gamma$
+denote the axis and $c$ the ratio of the spherical versor $AQ$, then
+the versor itself is expressed by $\gamma^c$.
+
+Now
+\begin{align*}
+\beta^b &= \cos b + \sin b \cdot \beta^\frac{\pi}{2}, \\
+\intertext{and}
+\gamma^c &= \cos c + \sin c \cdot \gamma^\frac{\pi}{2}; \\
+\intertext{therefore}
+\beta^b\gamma^c &= (\cos b + \sin b \cdot
+\beta^\frac{\pi}{2})(\cos c + \sin c \cdot \gamma^\frac{\pi}{2}) \\
+ &= \cos b \cos c + \cos b \sin c \cdot \gamma^\frac{\pi}{2}
+     + \cos c \sin b \cdot \beta^\frac{\pi}{2}
+     + \sin b \sin c \cdot \beta^\frac{\pi}{2}\gamma^\frac{\pi}{2}.
+\end{align*}
+
+\smallskip But from the preceding paragraph
+\begin{align}
+\beta^\frac{\pi}{2}\gamma^\frac{\pi}{2} &= -\cos\beta\gamma -
+  \sin\beta\gamma \cdot
+                      \overline{\beta\gamma}^\frac{\pi}{2}; \notag \\
+\intertext{therefore}
+\beta^b\gamma^c &=
+            \cos b \cos c - \sin b \sin c \cos \beta\gamma \tag{1} \\
+&\quad+ \{\cos b \sin c \cdot \gamma + \cos c \sin b \cdot \beta -
+\sin b \sin c \sin \beta\gamma \cdot
+\overline{\beta\gamma}\}^\frac{\pi}{2}. \tag{2}
+\end{align}
+
+\smallskip The first term gives the cosine of the product versor; it
+is equivalent to the fundamental theorem of spherical
+trigonometry,\index{Spherical trigonometry!fundamental theorem of}
+namely,
+\begin{equation*}
+\cos a = \cos b \cos c + \sin b \sin c \cos A,
+\end{equation*}
+where $A$ denotes the external angle instead of the angle included
+by the sides.
+
+The second term is the directed sine of the angle; for the square of
+(2) is equal to 1 minus the square of (1), and its direction is
+normal to the plane of the product angle.\footnote{Principles of
+Elliptic and Hyperbolic Analysis, p.~2.}
+
+\medskip Example.---Let $\beta = \overline{30^\circ/}\!
+\underline{/45^\circ}$ and $\gamma = \overline{60^\circ/}\!
+\underline{/30^\circ}$. Then
+\begin{align*}
+\cos \beta\gamma &= \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin
+30^\circ \cos 30^\circ, \\
+\intertext{and}
+\sin \beta\gamma &\cdot \overline{\beta\gamma} =
+  \mathrm{V}\beta\gamma; \\
+\intertext{but}
+\beta &= \cos 45^\circ i + \sin 45^\circ \cos
+30^\circ j + \sin 45^\circ \sin 30^\circ k, \\
+\intertext{and}
+\gamma &= \cos 30^\circ i + \sin 30^\circ \cos
+60^\circ j + \sin 30^\circ \sin 60^\circ k; \\
+\intertext{therefore}
+\mathrm{V}\beta\gamma & = \{\sin 45^\circ \cos 30^\circ
+   \sin 30^\circ \sin 60^\circ -\sin 45^\circ \sin 30^\circ
+   \sin 30^\circ \cos 60^\circ \}i \\
+&\quad+ \{ \sin 45^\circ \sin 30^\circ \cos 30^\circ -
+   \cos 45^\circ \sin 30^\circ \sin 60^\circ \} j \\
+&\quad+ \{ \cos 45^\circ \sin 30^\circ \cos 60^\circ -
+   \sin 45^\circ \cos 30^\circ \cos 30^\circ \} k.
+\end{align*}
+
+\medskip Quotient of Two Spherical Versors.%
+\index{Spherical versor!quotient of two}---The reciprocal of a given
+versor is derived by changing the sign of the index; $\gamma^{-c}$
+is the reciprocal of $\gamma^c$. As $\beta^b = \cos b + \sin b \cdot
+\beta^\frac{\pi}{2}$, and $y^{-c} = \cos c - \sin c \cdot
+\gamma^\frac{\pi}{2}$,
+\begin{align*}
+\beta^b\gamma^{-c} &= \cos b \cos c +
+                                    \sin b \sin c \cos \beta\gamma \\
+  &+\{\cos c \sin b \cdot \beta - \cos b \sin c \cdot \gamma
+      + \sin b \sin c \sin \beta\gamma \cdot
+      \overline{\beta\gamma}\}^\frac{\pi}{2}.
+\end{align*}
+
+\begin{center}
+\includegraphics[width=30mm]{fig23.png}
+\end{center}
+
+\smallskip Product of Three Spherical Versors.%
+\index{Product!of three spherical versors}%
+\index{Spherical versor!product of three}%
+\index{Versor!product of three general spherical}---Let $\alpha^a$
+denote the versor $PQ$, $\beta^b$ the versor $QR$, and $\gamma^c$
+the versor $RS$; then $\alpha^a\beta^b\gamma^c$ denotes $PS$. Now
+$\alpha^a\beta^b\gamma^c$
+\begin{align}
+=&(\cos a + \sin a \cdot\alpha^\frac{\pi}{2})
+  (\cos b + \sin b \cdot\beta^\frac{\pi}{2})
+  (\cos c + \sin c \cdot\gamma^\frac{\pi}{2}) \notag \\
+=& \cos a\cos b\cos c \tag{1} \\
+ & + \cos a \cos b \sin c \cdot \gamma^\frac{\pi}{2} +
+     \cos a \cos c \sin b \cdot \beta^\frac{\pi}{2} +
+     \cos b \cos c \sin a \cdot \alpha^\frac{\pi}{2} \tag{2} \\
+ & + \cos a \sin b \sin c \cdot
+                            \beta^\frac{\pi}{2}\gamma^\frac{\pi}{2} +
+     \cos b \sin a \sin c \cdot
+                   \alpha^\frac{\pi}{2}\gamma^\frac{\pi}{2} \notag \\
+ & \qquad + \cos c \sin a \sin b \cdot
+                              \alpha^\frac{\pi}{2}\beta^\frac{\pi}{2}
+ \tag{3} \\
+ & + \sin a \sin b \sin c \cdot
+          \alpha^\frac{\pi}{2}\beta^\frac{\pi}{2}\gamma^\frac{\pi}{2}
+\tag{4}
+\end{align}
+
+The versors in (3) are expanded by the rule already obtained,
+namely,
+\begin{equation*}
+\beta^\frac{\pi}{2}\gamma^\frac{\pi}{2} = -\cos \beta\gamma -\sin
+\beta\gamma \cdot \overline{\beta\gamma}^\frac{\pi}{2}.
+\end{equation*}
+
+The versor of the fourth term is
+\begin{align*}
+\alpha^\frac{\pi}{2}\beta^\frac{\pi}{2}\gamma^\frac{\pi}{2} &=
+  -(\cos\alpha\beta + \sin\alpha\beta \cdot
+      \overline{\alpha\beta}^\frac{\pi}{2}) \gamma^\frac{\pi}{2} \\
+&= -\cos\alpha\beta \cdot \gamma^\frac{\pi}{2} +
+  \sin\alpha\beta\cos\overline{\alpha\beta}\gamma +
+  \sin\alpha\beta\sin\overline{\alpha\beta}\gamma \cdot
+  \overline{\overline{\alpha\beta}\gamma}^\frac{\pi}{2}.
+\end{align*}
+
+Now $\sin\alpha\beta \sin\overline{\alpha\beta}\gamma \cdot
+\overline{\overline{\alpha\beta}\gamma} = \cos\alpha\gamma \cdot
+\beta - \cos\beta\gamma \cdot \alpha$ (p.~451), hence the last term
+of the product, when expanded, is
+\begin{equation*}
+\sin a\sin b\sin c\left\{-\cos \alpha\beta \cdot
+  \gamma^\frac{\pi}{2}
++ \cos\alpha\gamma \cdot \beta^\frac{\pi}{2}
+- \cos\beta\gamma \cdot \alpha^\frac{\pi}{2}
++ \cos\overline{\alpha\beta}\gamma\right\}.
+\end{equation*}
+
+\newpage
+Hence
+\begin{align*}
+\cos\alpha^a\beta^b\gamma^c &=
+  \cos a\cos b\cos c - \cos a\sin b\sin c\cos \beta\gamma \\
+&- \cos b\sin a\sin c\cos \alpha\gamma -
+                                \cos c\sin a\sin b\cos \alpha\beta \\
+&+ \sin a\sin b\sin c\sin \alpha\beta\cos\alpha\beta\gamma, \\
+\intertext{and, letting Sin denote the directed sine,}
+\Sin \alpha^a\beta^b\gamma^c &=
+  \cos a \cos b \sin c \cdot \gamma +
+                                  \cos a \cos c \sin b \cdot \beta \\
+&+ \cos b \cos c \sin a \cdot \alpha -
+  \cos a \sin b \sin c \sin \beta\gamma \cdot
+                                            \overline{\beta\gamma} \\
+&- \cos b \sin a \sin c \sin \alpha\gamma \cdot
+                                           \overline{\alpha\gamma} \\
+&- \cos c \sin a \sin b \sin \alpha\beta \cdot
+                                            \overline{\alpha\beta} \\
+&- \sin a \sin b \sin c\left\{\cos\alpha\beta \cdot \gamma -
+  \cos \alpha\gamma \cdot  \beta + \cos \beta\gamma \cdot
+  \alpha\right\}.\footnotemark
+\end{align*}
+\footnotetext{In the above case the three axes of the successive
+angles are not perfectly independent, for the third angle must begin
+where the second leaves off. But the theorem remains true when the
+axes are independent; the factors are then quaternions in the most
+general sense.}
+
+Extension of the Exponential Theorem to Spherical
+Trigonometry.\index{Binomial theorem in spherical analysis}%
+\index{Exponential theorem in spherical trigonometry}---It has been
+shown (p.~458) that
+\begin{align*}
+\cos\beta^b\gamma^c &= \cos b\cos c - \sin b\sin c\cos \beta\gamma
+\intertext{and}
+\left(\sin \beta^b\gamma^c\right)^\frac{\pi}{2} &=
+  \cos c \sin b \cdot \beta^\frac{\pi}{2} + \cos b \sin c \cdot
+  \gamma^\frac{\pi}{2} - \sin b \sin c \sin \beta\gamma \cdot
+  \overline{\beta\gamma}^\frac{\pi}{2}.
+\end{align*}
+
+Now
+\begin{align*}
+\cos b &= 1 - \frac{b^2}{2!} + \frac{b^4}{4!} - \frac{b^6}{6!} +
+   \text{ etc.} \\
+\intertext{and} \sin b &= b - \frac{b^3}{3!} + \frac{b^5}{5!} -
+   \text{ etc.}
+\end{align*}\index{Spherical trigonometry!binomial theorem}
+
+\smallskip Substitute these series for $\cos b$, $\sin b$, $\cos c$,
+and $\sin c$ in the above equations, multiply out, and group the
+homogeneous terms together. It will be found that
+\begin{align*}
+\cos\beta^b\gamma^c = 1
+  &- \frac{1}{2!}\{b^2 + 2bc\cos\beta\gamma + c^2\} \\
+  &+ \frac{1}{4!}\{b^4 + 4b^3c\cos\beta\gamma + 6b^2c^2 +
+       4bc^3\cos\beta\gamma + c^4\} \\
+  &- \frac{1}{6!}\{b^6 + 6b^5c\cos\beta\gamma + 15b^4c^2 +
+       20b^3c^3\cos\beta\gamma \\
+  & \qquad \qquad + 15b^2c^4 + 6bc^5\cos\beta\gamma + c^6\} + \ldots,
+\end{align*}
+where the coefficients are those of the binomial theorem, the only
+difference being that $\cos \beta\gamma$ occurs in all the odd terms
+as a factor. Similarly, by expanding the terms of the sine, we
+obtain
+\begin{align*}
+(\Sin \beta^b\gamma^c)^\frac{\pi}{2} &=
+    b \cdot \beta^\frac{\pi}{2} + c \cdot \gamma^\frac{\pi}{2} -
+    bc \sin \beta\gamma \cdot \overline{\beta\gamma}^\frac{\pi}{2} \\
+&\quad- \frac{1}{3!}\{b^3 \cdot \beta^\frac{\pi}{2} + 3b^2c \cdot
+      \gamma^\frac{\pi}{2} +
+    3bc^2 \cdot \beta^\frac{\pi}{2} + c^3 \cdot
+                                            \gamma^\frac{\pi}{2}\} \\
+&\quad+ \frac{1}{3!}\{bc^3 + b^3c\}
+      \sin\beta\gamma \cdot \overline{\beta\gamma}^\frac{\pi}{2} \\
+&\quad+ \frac{1}{5!}\{b^5 \cdot \beta^\frac{\pi}{2} +
+      5b^4c \cdot \gamma^\frac{\pi}{2} + 10b^3c^2 \cdot
+      \beta^\frac{\pi}{2} \\
+&\quad\qquad + 10b^2c^3\cdot\gamma^\frac{\pi}{2} + 5bc^4 \cdot
+      \beta^\frac{\pi}{2}
+      + c^5 \cdot \gamma^\frac{\pi}{2} \\
+&\quad- \frac{1}{5!}\left\{b^5c + \frac{5\cdot 4}{2\cdot 3}b^2c^3 +
+      bc^5\right\} \sin\beta\gamma \cdot
+      \overline{\beta\gamma}^\frac{\pi}{2} - \ldots
+\end{align*}
+
+By adding these two expansions together we get the expansion for
+$\beta^b\gamma^c$, namely,
+\begin{align*}
+\beta^b\gamma^c = 1 &+ b \cdot\beta^\frac{\pi}{2} +
+                                        c\cdot\gamma^\frac{\pi}{2} \\
+&- \frac{1}{2!}\{b^2 + 2bc(\cos\beta\gamma + \sin\beta\gamma \cdot
+  \overline{\beta\gamma}^\frac{\pi}{2}) + c^2\} \\
+&- \frac{1}{3!}\{b^3 \cdot \beta^\frac{\pi}{2} + 3b^2c
+\cdot\gamma^\frac{\pi}{2}
+  + 3bc^2 \cdot \beta^\frac{\pi}{2} + c^3 \cdot
+                                            \gamma^\frac{\pi}{2}\} \\
+&+ \frac{1}{4!}\{b^4 + 4b^3c(\cos\beta\gamma + \sin\beta\gamma \cdot
+    \overline{\beta\gamma}^\frac{\pi}{2}) + 6b^2c^2 \\
+&\qquad + 4bc^3(\cos\beta\gamma+\sin\beta\gamma \cdot
+  \overline{\beta\gamma}^\frac{\pi}{2}) + c^4\} + \ldots
+\end{align*}
+
+By restoring the minus, we find that the terms on the second line
+can be thrown into the form
+\begin{gather*}
+\frac{1}{2!} \left\{ b^2 \cdot \beta^{\pi} + 2bc \cdot
+\beta^{\frac{\pi}{2}}\gamma^{\frac{\pi}{2}} + c^{2} \cdot
+\gamma^{\pi} \right\}, \\
+\intertext{and this is equal to}
+\frac{1}{2!} \left\{ b \cdot \beta^{\frac{\pi}{2}} +
+  c \cdot \gamma^{\frac{\pi}{2}} \right\}^2, \\
+\intertext{where we have the square of a sum of successive terms. In
+a similar manner the terms on the third line can be restored to}
+b^3 \cdot \beta^\frac{3\pi}{2} +
+  3b^2c \cdot \beta^\pi \gamma^\frac{\pi}{2} +
+  3bc^2 \cdot \beta^\frac{\pi}{2}\gamma^\pi +
+  c^3 \cdot \gamma^{3(\frac{\pi}{2})}, \\
+\intertext{that is,}
+\frac{1}{3!} \left\{ b \cdot \beta^\frac{\pi}{2} +
+  c \cdot \gamma^\frac{\pi}{2} \right\} ^3.
+\end{gather*}
+
+Hence
+\begin{align*}
+\beta^b\gamma^c &= 1 + b \cdot \beta^\frac{\pi}{2}
+      + c \cdot \gamma^\frac{\pi}{2}
+    + \frac{1}{2!} \left\{ b \cdot \beta^\frac{\pi}{2}
+      + c \cdot \gamma^\frac{\pi}{2} \right\} ^2 \\
+   &\qquad + \frac{1}{3!} \left\{ b \cdot \beta^\frac{\pi}{2}
+      + c \cdot \gamma^\frac{\pi}{2} \right\} ^3
+      + \frac{1}{4!} \left\{ b \cdot \beta^\frac{\pi}{2}
+      + c \cdot \gamma^\frac{\pi}{2} \right\} ^4 + \ldots \\
+&= e ^{b \cdot \beta^\frac{\pi}{2} + c \cdot
+\gamma^\frac{\pi}{2}}.\footnotemark
+\end{align*}
+\footnotetext{At page 386 of his Elements of Quaternions, Hamilton
+says: ``In the present theory of diplanar quaternions we cannot
+expect to find that the sum of the logarithms of any two proposed
+factors shall be generally equal to the logarithm of the product;
+but for the simpler and earlier case of coplanar quaternions, that
+algebraic property may be considered to exist, with due modification
+for multiplicity of value.'' He was led to this view by not
+distinguishing between vectors and quadrantal quaternions and
+between simultaneous and successive addition. The above
+demonstration was first given in my paper on ``The Fundamental
+Theorems of Analysis generalized for Space.'' It forms the key to
+the higher development of space analysis.}%
+\index{Exponential theorem in spherical trigonometry!Hamilton's
+view}%
+\index{Hamilton's!view of exponential theorem in spherical analysis}
+
+Extension of the Binomial Theorem.---We have proved above that
+$e^{b\beta^\frac{\pi}{2}} e^{c\gamma^\frac{\pi}{2}} =
+e^{b\beta^\frac{\pi}{2} + c\gamma^\frac{\pi}{2}}$ provided that the
+powers of the binomial are expanded as due to a successive sum, that
+is, the order of the terms in the binomial must be preserved. Hence
+the expansion for a power of a successive binomial is given by
+\begin{multline*}
+\left\{ b \cdot \beta^\frac{\pi}{2} +
+        c \cdot \gamma^\frac{\pi}{2} \right\}^n =
+b^n \cdot \beta^{n^\frac{\pi}{2}} + nb^{n-1}c  \cdot
+    \beta^{(n-1)(\frac{\pi}{2})} \gamma^\frac{\pi}{2} \\
++ \frac{n(n-1)}{1 \cdot 2} b^{n-2} c^2 \cdot
+    \beta^{(n-2)(\frac{\pi}{2})} \gamma^\pi + \text{etc.}
+\end{multline*}
+
+\smallskip Example.---Let $b = \frac{1}{10}$ and $c = \frac{1}{5}$,
+$\beta = \overline{30^\circ/}\!\underline{/45^\circ}$, $\gamma =
+\overline{60^\circ/}\!\underline{/30^\circ}$.
+\begin{align*}
+(b \cdot \beta^\frac{\pi}{2} + c \cdot \gamma^\frac{\pi}{2})^2
+&= -\{b^2 + c^2 + 2bc\cos\beta\gamma
+      + 2bc(\sin\beta\gamma)^\frac{\pi}{2} \} \\
+&= -\left(\tfrac{1}{100} + \tfrac{1}{25}
+      + \tfrac{2}{50}\cos\beta\gamma\right)
+      - \tfrac{2}{50}\left(\sin\beta\gamma\right)^\frac{\pi}{2}.
+\end{align*}
+Substitute the calculated values of $\cos \beta\gamma$ and
+$\sin \beta\gamma$ (page 459).
+
+\small \begin{enumerate}
+\item[Prob.~48.] Find the equivalent of a quadrantal version round
+$\dfrac{\sqrt{3}}{2}i + \dfrac{1}{2\sqrt{2}}j +
+\dfrac{1}{2\sqrt{2}}k$ followed by a quadrantal version round
+$\dfrac{1}{2}i + \dfrac{\sqrt{3}}{4}j + \dfrac{3}{4}k$.
+
+\item[Prob.~49.] In the example on p.~459 let $b=25^\circ$ and $c =
+50^\circ$; calculate out the cosine and the directed sine of the
+product angle.
+
+\item[Prob.~50.] In the above example calculate the cosine and the
+directed sine up to and inclusive of the fourth power of the
+binomial. \hfill (Ans.~$\cos =.9735$.)
+
+\item[Prob.~51.] Calculate the first four terms of the series when
+$b = \frac{1}{50}$, $c = \frac{1}{100}$, $\beta =
+\overline{0^\circ/}\! \underline{/0^\circ}$, $\gamma =
+\overline{90^\circ/}\! \underline{/90^\circ}$.
+
+\item[Prob.~52.] From the fundamental theorem of spherical
+trigonometry deduce the polar theorem with respect to both the
+cosine and the directed sine.
+
+\item[Prob.~53.] Prove that if $\alpha^a, \beta^b, \gamma^c$ denote
+the three versors of a spherical triangle, then
+\begin{equation*}
+\frac{\sin\beta\gamma}{\sin a} = \frac{\sin\gamma\alpha}{\sin b} =
+  \frac{\sin\alpha\beta}{\sin c}.
+\end{equation*}
+\end{enumerate} \normalsize
+
+\chapter{Composition of Rotations.}
+
+\begin{center}
+\includegraphics[width=25mm]{fig24.png}
+\end{center}
+
+A version refers to the change of direction of a line, but a
+rotation refers to a rigid body. The composition of rotations is a
+different matter from the composition of versions.%
+\index{Composition!of finite rotations}%
+\index{Rotations, finite}
+
+\medskip Effect of a Finite Rotation on a Line.---Suppose that a
+rigid body rotates $\theta$~radians round the axis $\beta$ passing
+through the point $O$, and that $R$ is the radius vector from $O$ to
+some particle. In the diagram $OB$ represents the axis $\beta$, and
+$OP$ the vector $R$. Draw $OK$ and $OL$, the rectangular components
+of $R$.
+\begin{align*}
+\beta^\theta R &= (\cos\theta + \sin\theta \cdot
+     \beta^\frac{\pi}{2})r\rho \\
+ &= r(\cos\theta \sin \theta \cdot \beta^\frac{\pi}{2})
+  (\cos \beta\rho \cdot \beta +
+  \sin \beta\rho \cdot \overline{\overline{\beta\rho}\beta}) \\
+ &= r\{\cos\beta\rho \cdot \beta +
+  \cos\theta\sin\beta\rho \cdot \overline{\overline{\beta\rho}\beta} +
+  \sin\theta\sin\beta\rho \cdot \overline{\beta\rho}\}.
+\end{align*}
+When $\cos \beta\rho = 0$, this reduces to
+\begin{equation*}
+\beta^\theta R = \cos \theta R + \sin \theta \mathrm{V}(\beta R).
+\end{equation*}
+The general result may be written
+\begin{equation*}
+\beta^\theta R = \mathrm{S}\beta R \cdot \beta +
+  \cos \theta(\mathrm{V}\beta R)\beta + \sin \theta \mathrm{V}\beta R.
+\end{equation*}
+
+Note that $(\mathrm{V}\beta R)\beta$ is equal to
+$\mathrm{V}(\mathrm{V}\beta R)\beta$ because $\mathrm{S}\beta
+R\beta$ is 0, for it involves two coincident directions.
+
+\smallskip Example.---Let $\beta = li + mj + nk$, where
+$l^2 + m^2 + n^2 = 1$ and $R = xi + yj + zk$; then $\mathrm{S}\beta
+R = lx + my + nz$
+\begin{gather*}
+\mathrm{V}(\beta R)\beta = \begin{vmatrix}
+    mz - ny & nx - lz & ly - mx \\
+    l       & m       & n       \\
+    i       & j       & k
+    \end{vmatrix}
+\intertext{and} \mathrm{V}\beta R = \begin{vmatrix}
+  l & m & n \\
+  x & y & z \\
+  i & j & k
+  \end{vmatrix}.
+\intertext{Hence}
+\begin{split}
+\beta^\theta &= (lx + my + nz)(li + mj + nk) \\
+&+ \cos\theta \begin{vmatrix}
+   mz - ny & nx - lz & ly - mx \\
+   l       & m       & n       \\
+   i       & j       & k
+   \end{vmatrix} \\
+&+ \sin\theta\begin{vmatrix}
+   l & m & n  \\
+   x & y & z  \\
+   i & j & k \end{vmatrix}.
+\end{split}
+\end{gather*}
+
+\begin{center}
+\includegraphics[width=40mm]{fig25.png}
+\end{center}
+
+To prove that $\beta^b \rho$ coincides with the axis of
+$\beta^\frac{-b}{2} \rho^\frac{\pi}{2} \beta^\frac{b}{2}$. Take the
+more general versor $\rho^\theta$. Let $OP$ represent the axis
+$\beta$, $AB$ the versor $\beta^\frac{-b}{2}$, $BC$ the versor
+$\rho^\theta$. Then $(AB)(BC) = AC = DA$, therefore $(AB)(BC)(AE) =
+(DA)(AE) = DE$. Now $DE$ has the same angle as $BC$, but its axis
+has been rotated round $P$ by the angle $b$. Hence if $\theta =
+\frac{\pi}{2}$, the axis of $\beta^\frac{-b}{2}
+\rho^\frac{\pi}{2}\beta^\frac{b}{2}$ will coincide with
+$\beta^b\rho$.\footnote{This theorem was discovered by
+Cayley.\index{Cayley} It indicates that quaternion multiplication in
+the most general sense has its physical meaning in the composition
+of rotations.}
+
+The exponential expression for $\beta^\frac{-b}{2}
+\rho^\frac{\pi}{2} \beta^\frac{b}{2}$ is
+$e^{-\frac{1}{2}b\beta^\frac{\pi}{2} +
+\frac{1}{2}\pi\rho^\frac{\pi}{2} + \frac{1}{2}b\beta^\frac{\pi}{2}}$
+which may be expanded according to the exponential theorem, the
+successive powers of the trinomial being formed according to the
+multinomial theorem, the order of the factors being preserved.
+
+\medskip Composition of Finite Rotations round Axes which
+Intersect.---Let $\beta$ and $\gamma$ denote the two axes in space
+round which the successive rotations take place, and let $\beta^b$
+denote the first and $\gamma^c$ the second. Let $\beta^b \times
+\gamma^c$ denote the single rotation which is equivalent to the two
+given rotations applied in succession; the sign $\times$ is
+introduced to distinguish from the product of versors. It has been
+shown in the preceding paragraph that
+\begin{gather*}
+\beta^b\rho = \beta^\frac{-b}{2}\rho^\frac{\pi}{2}\beta^\frac{b}{2}; \\
+\intertext{and as the result is a line, the same principle applies
+to the subsequent rotation. Hence}
+\begin{split}
+\gamma^c(\beta^b\rho) &=
+  \gamma^\frac{-c}{2}(\beta^\frac{-b}{2}\rho^\frac{\pi}{2}
+     \beta^\frac{\pi}{2})\gamma^\frac{c}{2} \\
+&= (\gamma^\frac{-c}{2}\beta^\frac{-b}{2})
+     \rho^\frac{\pi}{2} (\beta^\frac{b}{2}\gamma^\frac{c}{2}),
+\end{split} \\
+\intertext{because the factors in a product of versors can be
+associated in any manner. Hence, reasoning backwards,}
+\beta^b \times \gamma^c = (\beta^\frac{b}{2}\gamma^\frac{c}{2})^2. \\
+\intertext{Let $m$ denote the cosine of
+$\beta^\frac{b}{2}\gamma^\frac{c}{2}$, namely,}
+\cos\frac{b}{2}\,\cos\frac{c}{2}-\sin\frac{b}{2}\,\sin\frac{c}{2}, \\
+\intertext{ and $n \cdot \nu$ their directed sine, namely,}
+\cos \frac{b}{2}\, \sin \frac{c}{2} \cdot \gamma + \cos\frac{c}{2}\,
+\sin \frac{b}{2} \cdot \beta - \sin \frac{b}{2}\,
+  \sin \frac{c}{2}\, \sin \beta\gamma \cdot \overline{\beta\gamma}; \\
+\intertext{then}
+\beta^b \times \gamma^c = m^2 - n^2 + 2mn \cdot \nu.
+\end{gather*}
+
+\newpage
+\begin{center}
+\includegraphics[width=40mm]{fig26.png}
+\end{center}
+
+\smallskip Observation.---The expression
+$(\beta^\frac{b}{2}\gamma^\frac{c}{2})^2$ is not, as might be
+supposed, identical with $\beta^b\gamma^c$. The former reduces to
+the latter only when $\beta$ and $\gamma$ are the same or opposite.
+In the figure $\beta^b$ is represented by $PQ$, $\gamma^c$ by $QR$,
+$\beta^b\gamma^c$ by $PR$, $\beta^\frac{b}{2}\gamma^\frac{c}{2}$ by
+$ST$, and $(\beta^\frac{b}{2}\gamma^\frac{c}{2})^2$ by $SU$, which
+is twice $ST$. The cosine of $SU$ differs from the cosine of $PR$ by
+the term $-(\sin \frac{b}{2}\, \sin \frac{c}{2}\, \sin
+\beta\gamma)^2$ It is evident from the figure that their axes are
+also different.
+
+\medskip Corollary.---When $b$ and $c$ are infinitesimals,
+$\cos\beta^b \times \gamma^c = 1$, and $\Sin \beta^b \times \gamma^c
+= b \cdot \beta + c \cdot \gamma$, which is the parallelogram rule
+for the composition of infinitesimal rotations.
+
+\small \begin{enumerate}
+
+\item[Prob.~54.] Let $\beta = \overline{30^\circ/}\!
+\underline{/45^\circ}$, $\theta = \frac{\pi}{3}$, and $R = 2i - 3j +
+4k$; calculate $\beta^\theta R$.
+
+\item[Prob.~55.] Let $\beta = \overline{90^\circ/}\!
+\underline{/90^\circ}$, $\theta = \frac{\pi}{4}$, $R = -i + 2j -
+3k$; calculate $\beta^\theta R$.
+
+\item[Prob.~56.] Prove by multiplying out that $\beta^\frac{-b}{2}
+\rho^\frac{\pi}{2} \beta^\frac{b}{2} =
+\{\beta^b\rho\}^\frac{\pi}{2}$.
+
+\item[Prob.~57.] Prove by means of the exponential theorem that
+$\gamma^{-c}\beta^b\gamma^c$ has an angle $b$, and that its axis is
+$\gamma^{2c}\beta$.
+
+\item[Prob.~58.] Prove that the cosine of
+$(\beta^\frac{b}{2}\gamma^\frac{c}{2})^2$
+differs from the cosine of $\beta^b\gamma^c$ by \\
+$-(\sin\frac{b}{2} \sin\frac{c}{2} \sin\beta \gamma)^2$.
+
+\item[Prob.~59.] Compare the axes of
+$(\beta^\frac{b}{2}\gamma^\frac{c}{2})^2$ and $\beta^b\gamma^c$.
+
+\item[Prob.~60.] Find the value of $\beta^b\times\gamma^c$ when
+$\beta = \overline{0^\circ/}\!\underline{/90^\circ}$ and
+$\gamma=\overline{90^\circ/}\!\underline{/90^\circ}$.
+
+\item[Prob.~61.] Find the single rotation equivalent to
+$i^\frac{\pi}{2} \times j^\frac{\pi}{2} \times k^\frac{\pi}{2}$.
+
+\item[Prob.~62.] Prove that successive rotations about radii to two
+corners of a spherical triangle and through angles double of those
+of the triangle are equivalent to a single rotation about the radius
+to the third corner, and through an angle double of the external
+angle of the triangle.
+\end{enumerate} \normalsize
+
+\addcontentsline{toc}{chapter}{Index}
+\printindex
+
+\markright{ADVERTISEMENT}
+\begin{center}
+\textbf{
+\Large SHORT-TITLE CATALOGUE \\
+\small OF THE           \\
+\Large PUBLICATIONS          \\
+\small OF               \\
+\Large JOHN WILEY \& SONS, \textsc{Inc.} \\
+\normalsize NEW YORK  \\
+London: CHAPMAN \& HALL, Limited \\
+Montreal, Can.: RENOUF PUB. CO.} \\
+
+\bigskip
+\begin{center}
+\rule{25mm}{.2pt}
+\end{center}
+\medskip
+\normalsize \textbf{ARRANGED UNDER SUBJECTS} \\
+\footnotesize Descriptive circulars sent on application. Books
+marked with an asterisk (*) are sold at \emph{net} prices only. All
+books are bound in cloth unless otherwise stated.
+
+\medskip
+\begin{center}
+\rule{25mm}{.2pt}
+\end{center}
+\bigskip
+
+\normalsize
+\textbf{AGRICULTURE---HORTICULTURE---FORESTRY.}
+\bigskip
+
+{\footnotesize
+\begin{tabular}{lr}
+\textsc{Armsby}---Principles of Animal Nutrition \dotfill 8vo,
+                                                          & \$4 00 \\
+\textsc{Bowman}---Forest Physiography \dotfill 8vo, & *5 00 \\
+\textsc{Bryant}---Hand Book of Logging\dotfill  (Ready, Fall 1913)
+                                                                 & \\
+\textsc{Budd} and \textsc{Hansen}---American Horticultural Manual:
+                                                        \dotfill & \\
+\quad Part I. Propagation, Culture, and Improvement\dotfill 12mo,
+                                                            & 1 50 \\
+\quad Part II. Systematic Pomology \dotfill 12mo, & 1 50 \\
+\textsc{Elliott}---Engineering for Land Drainage \dotfill 12mo,
+                                                            & 2 00 \\
+\quad  Practical Farm Drainage.  (Second Edition, Rewritten)
+                                             \dotfill 12mo, & 1 50 \\
+\textsc{Fuller}---Domestic Water Supplies for the Farm \dotfill 8vo,
+                                                           & *1 50 \\
+\textsc{Graham}---Text-book on Poultry \dotfill
+                                        (\emph{In Preparation.}) & \\
+\quad  Manual on Poultry (Loose Leaf Lab. Manual)
+  \dotfill (\emph{In Preparation.}) & \\
+\textsc{Graves}---Forest Mensuration \dotfill 8vo, & 4 00 \\
+\quad  Principles of Handling Woodlands \dotfill Small 8vo,
+                                                           & *1 50 \\
+\textsc{Green}---Principles of American Forestry \dotfill 12mo,
+                                                            & 1 50 \\
+\textsc{Grotenfelt}---Principles of Modern Dairy Practice.
+(\textsc{Woll}.) \dotfill 12mo, & 2 00 \\
+\textsc{Hawley} and \textsc{Hawes}---Forestry in New England
+\dotfill  8vo, & *3 50 \\
+\textsc{Herrick}---Denatured or Industrial Alcohol \dotfill 8vo,
+                                                           & *4 00 \\
+\textsc{Howe}---Agricultural Drafting \dotfill oblong quarto,
+                                                           & *1 25 \\
+\quad Reference and Problem Sheets to accompany
+                      Agricultural Drafting, \dotfill each & *0 20 \\
+\textsc{Keitt}---Agricultural Chemistry Text-book
+   \dotfill (\emph{In Preparation.}) & \\
+\quad Laboratory and Field Exercises in Agricultural Chemistry
+  \dotfill (\emph{In Preparation.}) & \\
+\textsc{Kemp} and \textsc{Waugh}---Landscape Gardening. (New
+Edition, Rewritten)
+  \dotfill  12mo, & *1 50 \\
+\textsc{Larsen}---Exercises in Dairying (Loose Leaf Field Manual)
+  \dotfill 4to, paper, & *1 00 \\
+\quad  Single Exercises each \dotfill &  *0 02 \\
+\quad  and \textsc{White}---Dairy Technology \dotfill Small 8vo,
+                                                           & *1 50 \\
+\textsc{Levison}---Studies of Trees, Loose Leaf Field Manual, 4to,
+  pamphlet form,  & \\
+\quad  Price from 5--10 cents net, each, according to number
+                                                       of pages. & \\
+\textsc{McCall}---Crops and Soils (Loose Leaf Field Manual)
+  \dotfill (\emph{In Preparation.}) & \\
+\quad  Soils (Text-book) \dotfill (\emph{In Preparation.}) & \\
+\textsc{McKay} and \textsc{Larsen}---Principles and Practice of
+                               Butter-making \dotfill 8vo, & *1 50 \\
+\textsc{Maynard}---Landscape Gardening as Applied to Home Decoration
+  \dotfill 12mo, & 1 50 \\
+\textsc{Moon} and \textsc{Brown}---Elements of Forestry
+  \dotfill (\emph{In Preparation.}) & \\
+\textsc{Record}---Identification of the Economic Woods of the United
+States \dotfill 8vo, & *1 25 \\
+\textsc{Recknagel}---Theory and Practice of Working Plans (Forest
+  Organization) \dotfill 8vo, & *2 00
+\end{tabular}
+} % \end{\footnotesize}
+\end{center}
+\newpage
+\chapter{PROJECT GUTENBERG ``SMALL PRINT''}
+\small \pagenumbering{gobble}
+\begin{verbatim}
+
+End of the Project Gutenberg EBook of Vector Analysis and Quaternions, 
+by Alexander Macfarlane
+
+*** END OF THIS PROJECT GUTENBERG EBOOK VECTOR ANALYSIS AND QUATERNIONS ***
+
+***** This file should be named 13609-t.tex or 13609-t.zip *****
+This and all associated files of various formats will be found in:
+        https://www.gutenberg.org/1/3/6/0/13609/
+
+Produced by David Starner, Joshua Hutchinson, John Hagerson, and the 
+Project Gutenberg On-line Distributed Proofreaders.
+
+
+
+Updated editions will replace the previous one--the old editions
+will be renamed.
+
+Creating the works from public domain print editions means that no
+one owns a United States copyright in these works, so the Foundation
+(and you!) can copy and distribute it in the United States without
+permission and without paying copyright royalties.  Special rules,
+set forth in the General Terms of Use part of this license, apply to
+copying and distributing Project Gutenberg-tm electronic works to
+protect the PROJECT GUTENBERG-tm concept and trademark.  Project
+Gutenberg is a registered trademark, and may not be used if you
+charge for the eBooks, unless you receive specific permission.  If
+you do not charge anything for copies of this eBook, complying with
+the rules is very easy.  You may use this eBook for nearly any
+purpose such as creation of derivative works, reports, performances
+and research.  They may be modified and printed and given away--you
+may do practically ANYTHING with public domain eBooks.
+Redistribution is subject to the trademark license, especially
+commercial redistribution.
+
+
+
+*** START: FULL LICENSE ***
+
+THE FULL PROJECT GUTENBERG LICENSE PLEASE READ THIS BEFORE YOU
+DISTRIBUTE OR USE THIS WORK
+
+To protect the Project Gutenberg-tm mission of promoting the free
+distribution of electronic works, by using or distributing this work
+(or any other work associated in any way with the phrase "Project
+Gutenberg"), you agree to comply with all the terms of the Full
+Project Gutenberg-tm License (available with this file or online at
+https://gutenberg.org/license).
+
+
+Section 1.  General Terms of Use and Redistributing Project
+Gutenberg-tm electronic works
+
+1.A.  By reading or using any part of this Project Gutenberg-tm
+electronic work, you indicate that you have read, understand, agree
+to and accept all the terms of this license and intellectual
+property (trademark/copyright) agreement.  If you do not agree to
+abide by all the terms of this agreement, you must cease using and
+return or destroy all copies of Project Gutenberg-tm electronic
+works in your possession. If you paid a fee for obtaining a copy of
+or access to a Project Gutenberg-tm electronic work and you do not
+agree to be bound by the terms of this agreement, you may obtain a
+refund from the person or entity to whom you paid the fee as set
+forth in paragraph 1.E.8.
+
+1.B.  "Project Gutenberg" is a registered trademark.  It may only be
+used on or associated in any way with an electronic work by people
+who agree to be bound by the terms of this agreement.  There are a
+few things that you can do with most Project Gutenberg-tm electronic
+works even without complying with the full terms of this agreement.
+See paragraph 1.C below.  There are a lot of things you can do with
+Project Gutenberg-tm electronic works if you follow the terms of
+this agreement and help preserve free future access to Project
+Gutenberg-tm electronic works.  See paragraph 1.E below.
+
+1.C.  The Project Gutenberg Literary Archive Foundation ("the
+Foundation" or PGLAF), owns a compilation copyright in the
+collection of Project Gutenberg-tm electronic works.  Nearly all the
+individual works in the collection are in the public domain in the
+United States.  If an individual work is in the public domain in the
+United States and you are located in the United States, we do not
+claim a right to prevent you from copying, distributing, performing,
+displaying or creating derivative works based on the work as long as
+all references to Project Gutenberg are removed.  Of course, we hope
+that you will support the Project Gutenberg-tm mission of promoting
+free access to electronic works by freely sharing Project
+Gutenberg-tm works in compliance with the terms of this agreement
+for keeping the Project Gutenberg-tm name associated with the work.
+You can easily comply with the terms of this agreement by keeping
+this work in the same format with its attached full Project
+Gutenberg-tm License when you share it without charge with others.
+
+1.D.  The copyright laws of the place where you are located also
+govern what you can do with this work.  Copyright laws in most
+countries are in a constant state of change.  If you are outside the
+United States, check the laws of your country in addition to the
+terms of this agreement before downloading, copying, displaying,
+performing, distributing or creating derivative works based on this
+work or any other Project Gutenberg-tm work.  The Foundation makes
+no representations concerning the copyright status of any work in
+any country outside the United States.
+
+1.E.  Unless you have removed all references to Project Gutenberg:
+
+1.E.1.  The following sentence, with active links to, or other
+immediate access to, the full Project Gutenberg-tm License must
+appear prominently whenever any copy of a Project Gutenberg-tm work
+(any work on which the phrase "Project Gutenberg" appears, or with
+which the phrase "Project Gutenberg" is associated) is accessed,
+displayed, performed, viewed, copied or distributed:
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+1.E.2.  If an individual Project Gutenberg-tm electronic work is
+derived from the public domain (does not contain a notice indicating
+that it is posted with permission of the copyright holder), the work
+can be copied and distributed to anyone in the United States without
+paying any fees or charges.  If you are redistributing or providing
+access to a work with the phrase "Project Gutenberg" associated with
+or appearing on the work, you must comply either with the
+requirements of paragraphs 1.E.1 through 1.E.7 or obtain permission
+for the use of the work and the Project Gutenberg-tm trademark as
+set forth in paragraphs 1.E.8 or 1.E.9.
+
+1.E.3.  If an individual Project Gutenberg-tm electronic work is
+posted with the permission of the copyright holder, your use and
+distribution must comply with both paragraphs 1.E.1 through 1.E.7
+and any additional terms imposed by the copyright holder.
+Additional terms will be linked to the Project Gutenberg-tm License
+for all works posted with the permission of the copyright holder
+found at the beginning of this work.
+
+1.E.4.  Do not unlink or detach or remove the full Project
+Gutenberg-tm License terms from this work, or any files containing a
+part of this work or any other work associated with Project
+Gutenberg-tm.
+
+1.E.5.  Do not copy, display, perform, distribute or redistribute
+this electronic work, or any part of this electronic work, without
+prominently displaying the sentence set forth in paragraph 1.E.1
+with active links or immediate access to the full terms of the
+Project Gutenberg-tm License.
+
+1.E.6.  You may convert to and distribute this work in any binary,
+compressed, marked up, nonproprietary or proprietary form, including
+any word processing or hypertext form.  However, if you provide
+access to or distribute copies of a Project Gutenberg-tm work in a
+format other than "Plain Vanilla ASCII" or other format used in the
+official version posted on the official Project Gutenberg-tm web
+site (www.gutenberg.org), you must, at no additional cost, fee or
+expense to the user, provide a copy, a means of exporting a copy, or
+a means of obtaining a copy upon request, of the work in its
+original "Plain Vanilla ASCII" or other form.  Any alternate format
+must include the full Project Gutenberg-tm License as specified in
+paragraph 1.E.1.
+
+1.E.7.  Do not charge a fee for access to, viewing, displaying,
+performing, copying or distributing any Project Gutenberg-tm works
+unless you comply with paragraph 1.E.8 or 1.E.9.
+
+1.E.8.  You may charge a reasonable fee for copies of or providing
+access to or distributing Project Gutenberg-tm electronic works
+provided that
+
+- You pay a royalty fee of 20% of the gross profits you derive from
+     the use of Project Gutenberg-tm works calculated using the method
+     you already use to calculate your applicable taxes.  The fee is
+     owed to the owner of the Project Gutenberg-tm trademark, but he
+     has agreed to donate royalties under this paragraph to the
+     Project Gutenberg Literary Archive Foundation.  Royalty payments
+     must be paid within 60 days following each date on which you
+     prepare (or are legally required to prepare) your periodic tax
+     returns.  Royalty payments should be clearly marked as such and
+     sent to the Project Gutenberg Literary Archive Foundation at the
+     address specified in Section 4, "Information about donations to
+     the Project Gutenberg Literary Archive Foundation."
+
+- You provide a full refund of any money paid by a user who notifies
+     you in writing (or by e-mail) within 30 days of receipt that s/he
+     does not agree to the terms of the full Project Gutenberg-tm
+     License.  You must require such a user to return or
+     destroy all copies of the works possessed in a physical medium
+     and discontinue all use of and all access to other copies of
+     Project Gutenberg-tm works.
+
+- You provide, in accordance with paragraph 1.F.3, a full refund of
+     any money paid for a work or a replacement copy, if a defect in
+     the electronic work is discovered and reported to you within 90
+     days of receipt of the work.
+
+- You comply with all other terms of this agreement for free
+     distribution of Project Gutenberg-tm works.
+
+1.E.9.  If you wish to charge a fee or distribute a Project
+Gutenberg-tm electronic work or group of works on different terms
+than are set forth in this agreement, you must obtain permission in
+writing from both the Project Gutenberg Literary Archive Foundation
+and Michael Hart, the owner of the Project Gutenberg-tm trademark.
+Contact the Foundation as set forth in Section 3 below.
+
+1.F.
+
+1.F.1.  Project Gutenberg volunteers and employees expend
+considerable effort to identify, do copyright research on,
+transcribe and proofread public domain works in creating the Project
+Gutenberg-tm collection.  Despite these efforts, Project
+Gutenberg-tm electronic works, and the medium on which they may be
+stored, may contain "Defects," such as, but not limited to,
+incomplete, inaccurate or corrupt data, transcription errors, a
+copyright or other intellectual property infringement, a defective
+or damaged disk or other medium, a computer virus, or computer codes
+that damage or cannot be read by your equipment.
+
+1.F.2.  LIMITED WARRANTY, DISCLAIMER OF DAMAGES - Except for the
+"Right of Replacement or Refund" described in paragraph 1.F.3, the
+Project Gutenberg Literary Archive Foundation, the owner of the
+Project Gutenberg-tm trademark, and any other party distributing a
+Project Gutenberg-tm electronic work under this agreement, disclaim
+all liability to you for damages, costs and expenses, including
+legal fees.  YOU AGREE THAT YOU HAVE NO REMEDIES FOR NEGLIGENCE,
+STRICT LIABILITY, BREACH OF WARRANTY OR BREACH OF CONTRACT EXCEPT
+THOSE PROVIDED IN PARAGRAPH F3.  YOU AGREE THAT THE FOUNDATION, THE
+TRADEMARK OWNER, AND ANY DISTRIBUTOR UNDER THIS AGREEMENT WILL NOT
+BE LIABLE TO YOU FOR ACTUAL, DIRECT, INDIRECT, CONSEQUENTIAL,
+PUNITIVE OR INCIDENTAL DAMAGES EVEN IF YOU GIVE NOTICE OF THE
+POSSIBILITY OF SUCH DAMAGE.
+
+1.F.3.  LIMITED RIGHT OF REPLACEMENT OR REFUND - If you discover a
+defect in this electronic work within 90 days of receiving it, you
+can receive a refund of the money (if any) you paid for it by
+sending a written explanation to the person you received the work
+from.  If you received the work on a physical medium, you must
+return the medium with your written explanation.  The person or
+entity that provided you with the defective work may elect to
+provide a replacement copy in lieu of a refund.  If you received the
+work electronically, the person or entity providing it to you may
+choose to give you a second opportunity to receive the work
+electronically in lieu of a refund.  If the second copy is also
+defective, you may demand a refund in writing without further
+opportunities to fix the problem.
+
+1.F.4.  Except for the limited right of replacement or refund set
+forth in paragraph 1.F.3, this work is provided to you 'AS-IS', WITH
+NO OTHER WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT
+NOT LIMITED TO WARRANTIES OF MERCHANTIBILITY OR FITNESS FOR ANY
+PURPOSE.
+
+1.F.5.  Some states do not allow disclaimers of certain implied
+warranties or the exclusion or limitation of certain types of
+damages. If any disclaimer or limitation set forth in this agreement
+violates the law of the state applicable to this agreement, the
+agreement shall be interpreted to make the maximum disclaimer or
+limitation permitted by the applicable state law.  The invalidity or
+unenforceability of any provision of this agreement shall not void
+the remaining provisions.
+
+1.F.6.  INDEMNITY - You agree to indemnify and hold the Foundation,
+the trademark owner, any agent or employee of the Foundation, anyone
+providing copies of Project Gutenberg-tm electronic works in
+accordance with this agreement, and any volunteers associated with
+the production, promotion and distribution of Project Gutenberg-tm
+electronic works, harmless from all liability, costs and expenses,
+including legal fees, that arise directly or indirectly from any of
+the following which you do or cause to occur: (a) distribution of
+this or any Project Gutenberg-tm work, (b) alteration, modification,
+or additions or deletions to any Project Gutenberg-tm work, and (c)
+any Defect you cause.
+
+
+Section  2.  Information about the Mission of Project Gutenberg-tm
+
+Project Gutenberg-tm is synonymous with the free distribution of
+electronic works in formats readable by the widest variety of
+computers including obsolete, old, middle-aged and new computers.
+It exists because of the efforts of hundreds of volunteers and
+donations from people in all walks of life.
+
+Volunteers and financial support to provide volunteers with the
+assistance they need, is critical to reaching Project Gutenberg-tm's
+goals and ensuring that the Project Gutenberg-tm collection will
+remain freely available for generations to come.  In 2001, the
+Project Gutenberg Literary Archive Foundation was created to provide
+a secure and permanent future for Project Gutenberg-tm and future
+generations. To learn more about the Project Gutenberg Literary
+Archive Foundation and how your efforts and donations can help, see
+Sections 3 and 4 and the Foundation web page at
+https://www.pglaf.org.
+
+
+Section 3.  Information about the Project Gutenberg Literary Archive
+Foundation
+
+The Project Gutenberg Literary Archive Foundation is a non profit
+501(c)(3) educational corporation organized under the laws of the
+state of Mississippi and granted tax exempt status by the Internal
+Revenue Service.  The Foundation's EIN or federal tax identification
+number is 64-6221541.  Its 501(c)(3) letter is posted at
+https://pglaf.org/fundraising.  Contributions to the Project
+Gutenberg Literary Archive Foundation are tax deductible to the full
+extent permitted by U.S. federal laws and your state's laws.
+
+The Foundation's principal office is located at 4557 Melan Dr. S.
+Fairbanks, AK, 99712., but its volunteers and employees are
+scattered throughout numerous locations.  Its business office is
+located at 809 North 1500 West, Salt Lake City, UT 84116, (801)
+596-1887, email business@pglaf.org.  Email contact links and up to
+date contact information can be found at the Foundation's web site
+and official page at https://pglaf.org
+
+For additional contact information:
+     Dr. Gregory B. Newby
+     Chief Executive and Director
+     gbnewby@pglaf.org
+
+Section 4.  Information about Donations to the Project Gutenberg
+Literary Archive Foundation
+
+Project Gutenberg-tm depends upon and cannot survive without wide
+spread public support and donations to carry out its mission of
+increasing the number of public domain and licensed works that can
+be freely distributed in machine readable form accessible by the
+widest array of equipment including outdated equipment.  Many small
+donations ($1 to $5,000) are particularly important to maintaining
+tax exempt status with the IRS.
+
+The Foundation is committed to complying with the laws regulating
+charities and charitable donations in all 50 states of the United
+States.  Compliance requirements are not uniform and it takes a
+considerable effort, much paperwork and many fees to meet and keep
+up with these requirements.  We do not solicit donations in
+locations where we have not received written confirmation of
+compliance.  To SEND DONATIONS or determine the status of compliance
+for any particular state visit https://pglaf.org
+
+While we cannot and do not solicit contributions from states where
+we have not met the solicitation requirements, we know of no
+prohibition against accepting unsolicited donations from donors in
+such states who approach us with offers to donate.
+
+International donations are gratefully accepted, but we cannot make
+any statements concerning tax treatment of donations received from
+outside the United States.  U.S. laws alone swamp our small staff.
+
+Please check the Project Gutenberg Web pages for current donation
+methods and addresses.  Donations are accepted in a number of other
+ways including including checks, online payments and credit card
+donations.  To donate, please visit: https://pglaf.org/donate
+
+
+Section 5.  General Information About Project Gutenberg-tm
+electronic works.
+
+Professor Michael S. Hart was the originator of the Project
+Gutenberg-tm concept of a library of electronic works that could be
+freely shared with anyone.  For thirty years, he produced and
+distributed Project Gutenberg-tm eBooks with only a loose network of
+volunteer support.
+
+Project Gutenberg-tm eBooks are often created from several printed
+editions, all of which are confirmed as Public Domain in the U.S.
+unless a copyright notice is included.  Thus, we do not necessarily
+keep eBooks in compliance with any particular paper edition.
+
+Most people start at our Web site which has the main PG search
+facility:
+
+     https://www.gutenberg.org
+
+This Web site includes information about Project Gutenberg-tm,
+including how to make donations to the Project Gutenberg Literary
+Archive Foundation, how to help produce our new eBooks, and how to
+subscribe to our email newsletter to hear about new eBooks.
+\end{verbatim}
+
+\end{document}
diff --git a/old/13609-t.zip b/old/13609-t.zip
new file mode 100644
index 0000000..9af736f
--- /dev/null
+++ b/old/13609-t.zip