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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 Binary files differnew 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. 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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 Binary files differnew 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 Binary files differnew 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. 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