(Brand) Vector and Tensor Analysis (1947)

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VECTOR and TENSOR

ANALYSIS

By

LOUIS BRAND, Ch.E., E.E., Ph.D.

PROFESSOR OF MATHEMATICS

UNIVERSITY OF CINCINNATI

New York JOHN WILEY & SONS, Inc.

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VECTOR AND TENSOR ANALYSIS. By Louis Brand. 439 pages. 5Y2 by

8%. Cloth.

VECTORIAL MECHANICS. By Louis

Brand. 544 pages. 5% by 8%.

Cloth.

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VECTOR and TENSOR

ANALYSIS

By

LOUIS BRAND, Ch.E., E.E., Ph.D.

PROFESSOR OF MATHEMATICS UNIVERSITY OF CINCINNATI

New York JOHN WILEY & SONS, Inc.

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COPYRIGHT, 1947 BY

Louis BRAND

the writ+en permission of the putlisher.

THIRD PRINTING, NOVEMBER, 1948

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PREFACE

The vector analysis of Gibbs and Heaviside and the more

general tensor analysis of Ricci are now recognized as standard

tools in mechanics, hydrodynamics, and electrodynamics. These

disciplines have also proved their worth in pure mathematics, especially in differential geometry. Their use not only materially simplifies and condenses. the exposition, but also makes

mathe-matical concepts more tangible and easy to grasp.

Moreover tensor analysis provides a simple automatic method for

construct-ing invariants. Since a tensor equation has precisely the same

form in all coordinate systems, the desirability of stating physical

laws or geometrical properties in tensor form is manifest.. The perfect adaptability of the tensor calculus to the theory of

rela-tivity was responsible for its original renown. It has since won a firm place in mathematical physics and engineering technology Thus the British analyst E. T. Whittaker rates the discovery of

the tensor calculus as one of the three principal mathematical

advances in the last quarter of the 19th century.

The first volume of this work not only comprises the standard vector analysis of Gibbs, including dyadics or tensors of valence two, but also supplies an introduction to the algebra of motors,

which is apparently destined to play an important role in mechanics

as well as in line geometry. The entire theory is illustrated by

many significant applications; and surface geometry and

hydro-dynamics * are treated at some length by vector methods in

separate chapters.

For the sake of concreteness, tensor analysis is first developed

in 3-space, then extended to space of n dimensions. As in the case

of vectors and dyadics, I have distinguished the invariant tensor

from its components. This leads to a straightforward treatment

of the affine connection and of covariant differentiation; and also

to a simple introduction of the curvature tensor. Applications of

tensor analysis to relativity, electrodynamics and rotating

elec-* For a systematic development of mechanics in vector notation see the author's Vectorial Mechanics, John Wiley & Sons, New York, 1930.

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viii PREFACE

tric machines are reserved for the second volume. The present volume concludes with a brief introduction to quaternions, the

source of vector analysis, and their use in dealing

with finite

rotations.

Nearly all of the important results are formulated as theorems,

in which the essential conditions are explicitly stated.

In this

connection the student should observe the distinction between

necessary and sufficient conditions. If the assumption of a certain property P leads deductively to a condition C, the condition is necessary. But if the assumption of the condition C leads

deduc-tively to the property P, the condition C is sufficient. Thus

we-have symbolically

P -* C (necessary), C (sufficient) -> P.

When P C, the condition C is necessary and sufficient.

The problems at the end of each chapter have been chosen not

only to develop the student's technical skill, but also to introduce new and important applications. Some of the problems are mathe-matical projects which the student may carry through step by step

and thus arrive at really important results.

As very full cross references are given in this book, an article as well as a page number is given at the top of each page.

Equa-tions are numbered serially (1), (2), . . .

in each article. A

ref-erence to an equation in another article is made by giving article

and number to the left and right of a point; thus (24.9) means

article 2.4, equation 9. Figures are given the number of the article

in which they appear followed by a serial letter; Fig. 6d, for

ex-ample, is the fourth figure in article 6.

Bold-face type is used in the text to denote vectors or tensors

of higher valence with their complement of base vectors. Scalar components of vectors and tensors are printed in italic type.

The rich and diverse field amenable to vector and tensor methods is one of the most fascinating in applied mathematics. It is hoped

that the reasoning will not only appeal to the mind but also im-pinge on the reader's aesthetic sense. For mathematics, which Gauss esteemed as "the queen of the sciences" is also one of the

great arts. For, in the eloquent words of Bertrand Russell:

"The true spirit of delight, the exaltation, the sense of being

more than man, which is the touchstone of the highest excellence,

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PREFACE is

best in mathematics deserves not merely to be learned as a task, but also to be assimilated as a part of daily thought, and brought again and again before the mind with ever-renewed

encourage-ment. Real life is, to most men, a long second-best, a perpetual compromise between the real and the possible; but the world of pure reason knows no compromise, no practical limitations, no

barrier to the creative activity embodying in splendid edifices the

passionate aspiration after the perfect from which all great work

springs."

The material in this book may be adapted to several short

courses.

Thus Chapters I, III, IV, V, and VI may serve as a

course in vector analysis; and Chapters I (in part), IV, V, and IX as one in tensor analysis. I But the prime purpose of the author was to cover the theory and simpler applications of vector and tensor analysis in ordinary space, and to weave into this fabric

such concepts as dyadics, matrices, motors, and quaternions.

The author wishes, finally, to express his thanks to his

col-leagues, Professor J. W. Surbaugh and Mr. Louis Doty for their

help with the figures. Mr. Doty also suggested the notation used

in the problems dealing with air navigation and read the entire

page proof.

Louis BRAND

University of Cincinnati January 15, 1947

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85. Differentiation Formulas . . . 186 86. Gradient of a Tensor . . . 187 87. Functional Dependence . . . 190 88. Curvilinear Coordinates . . . 191 89. Orthogonal Coordinates . . . 194 90. Total Differential. . . 197 91. Irrotational Vectors . . . 198 92. Solenoidal Vectors . . . 201 93. Surfaces . . . 203

94. First Fundamental Form . . . 204

95. Surface Gradients . . . 206

96. Surface Divergence and Rotation . . . 207

97. Spatial and Surface Invariants . . . 209

98. Summary: Differential Invariants . . . 211

Problems . . . 213

CHAPTER VI INTEGRAL TRANSFORMATIONS 99. Green's Theorem in the Plane . . . 216

100. Reduction of Surface to Line Integrals . . . 218

101. Alternative Form of Transformation . . . 221

102. Line Integrals . . . . 222

103. Line Integrals on a Surface . . . 224

104. Field Lines of a Vector . . . 226

105. Pfaff's Problem. . . 230

106. Reduction of Volume to Surface Integrals . . . 233

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xiv CONTENTS

ARTICLE PAGE

108. Green's Identities . . . 237

109. Harmonic Functions . . . 239

110. Electric Point Charges . . . 241

111. Surface Charges . . . 242

112. Doublets and Double Layers . . . 243

113. Space Charges . . . 245

114. Heat Conduction . . . 247

115. Summary: Integral Transformations . . . 248

Problems . . . 250

CHAPTER VII HYDRODYNAMICS 116. Stress Dyadic . . . 253

117. Equilibrium of a Deformable Body . . . 255

118. Equilibrium of a Fluid . . . 257

119. Floating Body . . . 257

120. Equation of Continuity . . . 258

121. Eulerian Equation for a Fluid in Motion. . . 260

122. Vorticity . . . 262

123. Lagrangian Equation of Motion. . . 263

124. Flow and Circulation . . . 266

125. Irrotational Motion. . . 267 126. Steady Motion . . . 268 127. Plane Motion . . . 270 128. Kutta-Joukowsky Formulas . . . 273 129. Summary: Hydrodynamics. . . 278 Problems . . . 280 CHAPTER VIII GEOMETRY ON A SURFACE 130. Curvature of Surface Curves . . . 283

131. The Dyadic On. . . 285

132. Fundamental Forms . . . 289

133. Field of Curves. . . 293

134. The Field Dyadic . . . 293

135. Geodesics . . . 297

136. Geodesic Field . . . 299

137. Equations of Codazzi and Gauss . . . 300

138. Lines of Curvature . . . 302

139. Total Curvature . . . 306

140. Bonnet's Integral Formula . . . 308

141. Normal Systems . . . 313

142. Developable Surfaces . . . 314

143. Minimal Surfaces . . . 316

144. Summary: Surface Geometry . . . 319

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CONTENTS _xv

CHAPTER IX

TENSOR ANALYSIS

ARTICLE PAGE

145. The Summation Convention . . . 328

146. Determinants . . . 329

147. Contragredient Transformations . . . 332

148. Covariance and Contravariance. . . 334

149. Orthogonal Transformations . . . 336

150. Quadratic Forms . . . 337

151. The Metric . . . 339

152. Relations between Reciprocal Bases. . . 339

153. The Affine Group. . . 340

154. Dyadics . . . 342

155. Absolute Tensors . . . 343

156. Relative Tensors . . . 345

157. General Transformations . . . 346

158. Permutation Tensor . . . 350

159. Operations with Tensors. . . 350

160. Symmetry and Antisymmetry . . . 352

161. Kronecker Deltas. . . 353

162. Vector Algebra in Index Notation. . . 354

163. The Affine Connection . . . 356

164. Kinematics of a Particle. . . 359

165. Derivatives of e` and E . . . 360

166. Relation between Affine Connection and Metric Tensor . . . 361

167. Covariant Derivative . . . 362

168. Rules of Covariant Differentiation. . . 365

169. Riemannian Geometry . . . 366

170. Dual of a Tensor . . . 370

171. Divergence . . . 371

172. Stokes Tensor . . . 372

173. Curl . . . 374

174. Relation between Divergence and Curl . . . 376

175. Parallel Displacement . . . 376

176. Curvature Tensor . . . 380

177. Identities of Ricci and Bianchi . . . 384

178. Euclidean Geometry . . . 385

179. Surface Geometry in Tensor Notation. . . 388

180. Summary: Tensor Analysis. . . 392

Problems . . . 396

CHAPTER X QUATERNIONS 181. Quaternion Algebra . . . 403

182. Conjugate and Norm . . . 406

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xvi CONTENTS

ARTICLE PAGE

184. Product of Vectors . . . 410

185. Roots of a Quaternion . . . 412

186. Great Circle Arcs . . . 414

187. Rotations . . . 417

188. Plane Vector Analysis . . . 421

189. Summary: Quaternion Algebra . . . 426

Problems . . . 427,

INDEX . . . . ... . . . 431

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CHAPTER I

VECTOR ALGEBRA

1. Scalars and Vectors. There are certain physical quantities,

such as length, time, mass, temperature, electric charge, that may

be specified by a single real number. A mass, for example, may

be specified by the positive number equal to the ratio of the given mass to the unit mass. Similarly an electric charge may be speci-fied by a number, positive or negative, according as the charge is

"positive" or "negative." Quantities of this sort are called scalar

quantities; and the numbers that represent them are often called

scalars.

On the other hand, some physical quantities require a direction

as well as magnitude for their specification. Thus a rectilinear

displacement can only be completely specified by its length and direction. A displacement may be represented graphically by a segment of a straight line having a definite length and direction.

We shall call such a directed segment a vector. Any physical

quantity that involves both magnitude and direction, so that it

may be represented by a line segment of definite length and

direc-tion, and which moreover conforms to the parallelogram law of

addition (§ 2), is called a vector quantity. Velocity, acceleration, force, electric and magnetic field intensities are examples of vector quantities. It is customary, however, in applied mathematics, to

speak of vector quantities as vectors.

DEFINITIONS: A vector is a segment of a straight line regarded as having a definite length and direction. Thus we may represent a vector by an arrow. The vector directed from the point A to the

point B is denoted by the symbol AB. With this notation AB

--and BA denote different vectors; they have the same length but

opposite directions.

Besides the proper vectors just defined, we extend the term vector

to include the zero vector, an "arrow" of length zero but devoid

of direction.

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2 VECTOR ALGEBRA §1

If the initial point of a vector may be chosen at pleasure, the

vector is said to be free. If, however, its initial point is restricted

to a certain set of points, the vector is said to be localized in this

set. When the set consists of a single point (initial point fixed),

the vector is said to be a bound vector. If the vector is restricted

to the line of which it forms a part, it is called a line vector. For example, the forces acting upon rigid bodies must be regarded as line vectors; they may only be shifted along their line of action.

Two vectors are said to be equal when they have the same length and direction.

Vectors are said to be collinear when they are parallel to the same

line. In this sense two parallel vectors are collinear.

Vectors are said to be coplanar when they are parallel to the same plane. In this sense any two vectors are coplanar.

A unit vector is a vector of unit length.

In addition to the foregoing notation, in which we denote a

vector by giving its end points, we also shall employ single letters in heavy (bold-face) type to denote vectors. Thus, in Fig. 2b the

vectors forming the opposite sides of the parallelogram are equal

(they have the same length and direction) and may therefore be

represented by the same symbol,

AB=DC=u,

AD=BC=v.

A vector symbol between vertical bars, as I AB I or I u 1,

de-notes the length of the vector. We shall also, on occasion, denote

the lengths of vectors, u, v, F by the corresponding letters u, v, F in italic type. Bars about real numbers denote their positive

mag-nitudes: thus I -3 1 = 3.

The preceding definition of a vector is adequate for the

elemen-tary applications to Euclidean space of three dimensions. For

purposes of generalization, however, it is far better to define a vector as a new type of number-a hypernumber-which is given

by a set of real numbers written in a definite order. Thus, in our

ordinary space it will be -seen that, when a suitable system of

reference has been adopted, a vector can be represented by a set

of three real numbers, [a, b, c], called the components of the vector.

With this definition the zero vector is denoted by [0, 0, 0]. In

order to complete this definition of a vector, a rule must be given to enable us to compute the components when the system of

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§ 2 ADDITION OF VECTORS 3 2. Addition of Vectors. To obtain a rule for adding vectors, let

us regard them, for the moment, as representing rectilinear

dis-placements in space. If a particle is given two rectilinear

displace-ments, one from A to B, and a second from B to C, the result is

the same as if the particle were given a-single displacement from A to C. This equivalence may be represented by the notation,

(1)

AB + BC = AC.

We shall regard this equation as the definition of vector addition. The sum of two vectors, u, v, therefore is defined by the following triangle construction (Fig. 2a) :

Draw v from the end of u; then the vector directed from the begin-ning of u to the end of v is the sum of u and v and is written u + v.

C

FIG. 2a Fia. 2b

Since any side of a triangle is less than the sum of the other

two sides,

lu+vl _ lul +lvl,

the equal sign holding only when u and v have the same direction. Vector addition obeys both the commutative and associative laws:

(2)

u+v = v+u,

(3)

(u + v) + w = u + (v + w).

In the parallelogram formed with u and v as sides (Fig. 2b),

u+v=AB+BC=AC,

v+u=AD+DC=AC.

This proves (2). In view of this construction, the rule for vector

addition is called the parallelogram law.

To find u + v when u and v are line vectors whose lines intersect

at A, shift the vectors along their lines so that both issue from A

(Fig. 2b) and complete the parallelogram ABCD; then

---3

--3

---9

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Since the diagonal of the parallelogram on u, v gives the line of

action of u + v, the term "parallelogram law" is especially appro-priate for the addition of intersecting line vectors. We shall speak of the addition of line vectors as statical addition.

The associative law (3) is evident from Fig. 2c:

(u + v) + w = (AB -i- BC) -f- CD = AC -i- CD = AD,

u + (v -F w) = AB -f- (BC + CD) = AB + BD = AD.

Since the grouping of the vectors is immaterial, the preceding swa

is simply written u + v + w.

Fia. 2c

B FIG. 2d

From the commutative and associative laws we may deduce the following general result: The sum of any number of vectors is inde-pendent of the order in which they are added, and of their grouping to form partial sums.

To construct the sum of any number of vectors, form a broken line whose segments, in length and direction, are these vectors taken in any order whatever; then the vector directed from the beginning to the end of the broken line will be the required sum. The figure formed by the vectors and their sum is called a vector

polygon. If A, B, C, , G, H are the successive vertices of a

vector polygon (Fig. 2d), then

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When the vectors to be added are all parallel, the vector "polygon" becomes a portion of a straight line described twice.

If, in the construction of a vector sum, the end point of the last vector coincides with the origin of the first, we say that the sum

of the vectors is zero. Thus, if in (4) the point H coincides with A, we write

-3

-4

-- 3

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§ 3 SUBTRACTION OF VECTORS 5 This equation may be regarded as a special case of (4) if we agree

--->

that AA = 0.

The zero vector AA (or BB, etc.) is not a vector in the proper sense since it has no definite direction; it is an extension of our

original vector concept. From

--- f ----*

--3

-*

-3 -3

AB+BB=AB,

AA+AB=AB

we have, on writing AB = u,

u+0=u,

0+u=u.

We shall refer to vectors which are not zero as proper vectors. 3. Subtraction of Vectors. The sum of two vectors is zero when, and only when, they have the same length and opposite directions:

AB + BA = 0.

If AB = u, it is natural to write BA = -u in

order that the characteristic equation for negatives,

(1)

u + (-u) = 0,

will hold for vectors as well as for numbers. Hence by definition: The negative of a vector is a vector of the same length but opposite direction.

Note also that - (-u) = u.

The difference u - v of two vectors is defined by the equation,

(2)

(u - v) + v = U.

Adding -v to both sides of (2), we have

(3)

u-v=u+(-v);

that is, subtracting a vector is the same as adding its negative. The

construction of u - v is shown in Fig. 3a.

A

Fia. 3a

0

Fia. 3b

If 0 is chosen as a point of reference, any point P in space may

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be expressed in terms of the position vectors of its end points

(Fig. 3b),

AB = AO + OB = OB + (-OA),

and, from (3), (4)

-4

--

-4

AB = OB - OA.

4. Multiplication of Vectors by Numbers. The vector u + u is

naturally denoted by 2u; similarly, we write -u + (-u) = -2u.

Thus, both 2u and -2u denote vectors twice as long as u; the

former has the same direction as u, the latter the opposite

direc-tion. This definition is generalized as follows:

The product au or ua of a vector u and a real number a is defined as a vector a times as long as u, and having the same direction as u, or the opposite, according as a is positive or negative.

If a = 0,

au=0.

In accordance with this definition,

a(-u) = (-a)u = -au,

(-a)(-u) = au.

These relations have the same form as the rules for multiplying numbers. Moreover, the multiplication of a vector by numbers is

commutative (by definition), associative, and distributive:

(1)

au = ua,

(2) (a/3)u = a(/3u),

(3)

(a + a)u = au + (3u.

The product of the sum of two vectors by a given number is also

distributive:

(4)

a(u + v) = au + av.

The proof of (4) follows immediately from the theorem that the

corresponding sides of similar triangles are proportional. Figure 4

applies to the case when a > 0.

The quotient u/a of a vector by a number

u+V av _{a (not zero) is defined as the product of u}

v

U

au by 1/a.

Fia. 4

The developments thus far show that:

As far as addition, subtraction, and

multi-plication by numbers are concerned, vectors may be treated formally in accordance with the rules of ordinary algebra.

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§5 LINEAR, DEPENDENCE 7

5. Linear Dependence. The n vectors u1, u2i , un are said

to be linearly dependent if there exist n real numbers X1, X2, ,

An, not all zero, such that

(1) X1U1 + X2U2 + ... + XnUn = 0.

If the vectors are not linearly dependent, they are said to be

linearly independent. Consequently, if a relation (1) exists be-tween n linearly independent vectors, all the constants must be

zero.

If m vectors ul, U2, , um are linearly dependent, any greater number n of vectors including these are also linearly dependent. For

if u1i u2, , um satisfy

X1U1 + A2U2 + ... + XrUm = 0,

we can give ?1, X2, , Xm the preceding values (at least one of

these is not zero) and take X,n+1 = X m+2 = _ X n = 0. Then (1) is satisfied, and the n vectors ui are linearly dependent.

If Xu = 0 and X 0, u = 0; hence one vector is linearly depend-ent only when it is the zero vector. Hence the vectors of any set

that includes the zero vector are linearly dependent.

Conse-quently, we need only consider sets of proper vectors in the theo-rems following.

If X1u1 + X2u2 = 0 and X1 0 0, we can write ul = au2 i hence

ul and u2 are collinear. Conversely, if ul and u2 are collinear,

U1 = au2 (a 0 0). Therefore:

A necessary and sufficient condition that two proper vectors be lin-early dependent is that they be collinear.

If X1u1 + A2u2 + X3u3 = 0 and X1

0, we can write ul =

aU2 + ,13u3; the parallelogram construction (Fig. 5a) now shows

that ul is parallel to the plane of u2 and u3.

Conversely, if U1, U2, u3 are coplanar, they are linearly dependent.

For (a) if

two of the vectors are collinear, they are linearly dependent, and

the same is true of all three; and (b) if no two vectors are

col-linear, we can construct a parallelogram on ul as diagonal whose sides are parallel to u2 and u3 (Fig. 5a), so that

ul =AC=AB+BC=au2+$u3.

Therefore:

A necessary and sufficient condition that three proper vectors be linearly dependent is that they be coplanar.

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In space of three dimensions, four vectors ul, u2i u3, u4 are al-ways linearly dependent. For (a) if three of the vectors are co-planar, they are linearly dependent, and the same is true of all

four; and (b) if no three vectors are coplanar", we can construct a

FIG. 5a FIG. 5b

parallelepiped on ut as a diagonal whose edges are parallel to

U2i n3, U4 (Fig. 5b), so that

-3

---3 --a

- 3

ut = AD = AB + BC -}- CD = au2 + 9u3 -I- yu4.

Therefore: Any four vectors are linearly dependent.

6. Collinear Points.

If A, B, P are points of a straight line,

P is said to divide the segment AB in the ratio X when

(1)

-->

- 3

AP=APB.

As P passes from A to B (Fig. 6a), X increases through all positive values from 0 to infinity. If P describes the line to the left of A,

0

FIG. 6a

X varies from 0 to -1; and, as P

de-scribes the line to the right of B, X

varies from -oo to -1.

Thus X = 0,

X = =L00,

A = -1 correspond,

respec-tively, to the points A, B, and the

in-finitely distant "point" of the line. The

ratio A is positive or negative, according

as P lies within or without the segment

AB.

To find the position vector of P, relative to an origin 0, write

(1) in the form, Then (2)

OP - OA = X(OB - OP).

--

OA + A OB

OP =

1 + A

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§6 COLLINEAR POINTS

or, if we write X = (3/a,

(3)

- aOA +aOB

OP =

-9

a+0

In particular, if X = 1, P is the mid-point of AB.

In the following we shall denote the position vectors of the

points A, B, C, ,

P by a, b, c,

, p. Thus if C divides AB

in the ratio /3/a,

(4) c=ash-(3b

a+0

Thus the mid-point of AB has the position vector

2

(a + b).

When the points C, D divide a segment AB internally and ex-ternally in the same numerical ratios ±X, we have

a+Ab

a - Xb

c =1+x,

d=

_1-x

If we solve these equations for a and b, we find that the points A, B also divide the segment CD in the same numerical ratios

f(1 - X)/(l + X).

Pairs of points A, B and C, D having this

property are said to be harmonic conjugates; either pair is the

harmonic conjugate of the other.

A useful test for collinearity is given by the following:

THEOREM. Three distinct points A, B, C lie on a straight line

when, and only when, there exist three numbers a, /3, y, different from zero, such that

(5)

aa+/3b-+-yc=0,

a+$+y=0.

Proof. If A, B, C are collinear, C divides AB in some ratio /3/a;

hence on putting y = - (a + ,B) in (4) we obtain (5). Conversely,

from (5) we can deduce (4) since a +

-y

0; hence C lies

on the line AB.

From (5) we conclude that C, A, B divide AB, BC, CA,

re-spectively, in the ratios S/a, y/l3, a/y whose product is 1.

If an equation of the form (5) subsists between three distinct

non-collinear points, we must conclude that a = (3 = y = 0.

For at

least one coefficient y = 0; and from

as+3b=0,

a+/3=0,

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Another criterion for collinearity may be based on the statical

addition of line vectors (§ 2).

THEOREM. The points A, B, C are collinear when the line vectors AB, BC, CA are statically equal to zero:

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AB +BC +CA =O.

Proof. If we use = to denote statical equivalence, AB + BC

BD, a vector through B; and, since BD + CA = 0, B, C, A

are collinear.

Example 1. In the parallelogram ABCD, E and F are the middle points of the sides AB, BC. Show that the lines DE, DF divide the diagonal AC into thirds and that AC cuts off a third of each line (Fig. 6b).

E

"

B

Fia. 6b Fia. 6c

The hypotheses of our problem are expressed by the equations:

d - a=c - b,

2e=a+b,

2f =b+c.

Let DE cut AC at X. To find x, eliminate b from the first and second equa-tions. Thus + c 2e = 2a

=a;

d-a+2e=a+c and

d + 3 3

for the first member represents a point on DE, the second member a point on AC, and, since the points are the same, the point is at the intersection X of these lines. Comparison with (4) now shows that X divides DE in the ratio 2/1, AC in the ratio 1/2.

Let DF cut AC at Y. To find y, eliminate b from the first and third equa-tions. Thus

d-a+2f=2c and

d+2f-a+2c_

3 3

Hence Y divides DF in the ratio 2/1, AC in the ratio 2/1.

Example 2. In a plane quadrilateral ABCD, the diagonals AC, BD

inter-sect at P, the sides AB CD interinter-sect at Q (Fig. 6c). If P divides AC and BD in the ratios 3/2 and 1/2, respectively, in what ratios does Q divide the segments A B, CD?

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§6 COLLINEAR POINTS 11 By hypothesis hence 2a + 3c 2b + d p 5 3 6a + 9c = 10b + 5d and - 5d = lOb - 6a _

for the first fraction represents a point

on CD, the second a point on AB, and both points are the same, that is,

the point Q. Therefore Q divides CD

in the ratio -5/9, AB in the ratio

-10/6.

Example 3. Prove that the

mid-points of the diagonals of a complete quadrilateral are collinear.

In the complete quadrilateral

ABCDLM (Fig. 6d) let P, Q, R be the mid-points of the diagonals AC, BD, LM. The sum of the line vectors,

4

--,

--->

--,

--- --i --> --->

AB + AD = 2 AQ, CB + CD = 2 CQ, QA + QC = 2 QP; hence we have the statical equivalence,

AB +AD +CB +CD =4PQ,

for the quadrilateral ABCD with diagonals AC, BD. Similarly, for the ad-rilateral BLDM with diagonals BD, LM,

--> --9 - 4 --+

BL + BM + DL + D31 = 4QR;

and, for the quadrilateral LA.11C with diagonals LM, AC,

--4 --4 ----+ --+

-,

LA + LC + MA + MC = 4 RP.

On adding these three equations, we find that the entire left member is stati-cally equal to zero; for

--+ --- --> ---*

-9

---=0, DM+MA = 0,

-+

- 4 -+

CB +BM +MC-0,

CD +DL +LC -O,

are statical equations, since the vectors in each are collinear. Hence

PQ +QR+RP =0,

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7. Coplanar Points. THEOREM. If no three of the points A, B,

C, D are collinear, they will lie in a plane when, and only when,

there exist four numbers a, /3, y, S, different from zero, such that

(1)

as+$b+yc+Sd = 0,

a+0+y+S = 0.

Proof. If A, B, C, D are coplanar, either AB is parallel to CD,

or AB cuts CD in a point P (not A, B, C, or D). In the respective

cases, we have

b - a=K(d - c);

a+Ab

c+A'd

1+A = 1+A,

-P,

where A, A' are neither 0 nor -1. In both cases, a, b, c, d are

con-nected by a linear relation of the form (1). Conversely, let us

assume that equations (1) hold good.

If a + /3 = 0 (and hence

y + S = 0), we have

a(a - b) + y(c - d) = 0

and the lines AB and CD are parallel. If a + /3 5,16 0 (and hence

y+6

0),

as+/3b

yc+Sd

(2)

a+/i

y+6

-P

where P is a point common to the lines AB and CD.

In both

cases, A, B, C, D are coplanar.

Note that (2) states that the point P in which AB, CD

inter-sect divides AB and CD in the ratios #/a, 6/y.

Similarly, if

a+y; 0,

as + yc

/3b + 6d

(3)

a -f-.y

#+ 6

-

qi

thus Q, the point in which AC and BD intersect, divides AC and BD in the ratios y/a, S//i.

What conclusion can be drawn if a + 3 F6 0?

If an equation of the form (1) subsists between four distinct

non-coplanar points, we must conclude that a = a = y = S = 0. For

at least one coefficient 6 = 0; then, from

as+/3b+yc = 0,

a+0+y = 0,

and the fact that A, B, C are not collinear, we deduce (§ 6) that

a=0=y=0.

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§ 7 COPLANAR POINTS 13

Example 1. The Trapezoid. If ABCD is a trapezoid with AB parallel to DC (Fig. 7a), then

AB=XDC or b-a=A(c-d).

Hence we have

b+Xd=a+Xc or b-Xc=a-Xd.

These equations may be written

b+ad_a+Xc_

b - Ac_a - Ad

+a

1+a

-P

-q;

for the former expressions represent the point P where the diagonals BD, AC meet, and the latter the point Q where the sides BC, AD meet. Evidently P divides both BD and AC in the same ratio A; and Q divides both BC and AD in the same ratio -A. In what ratio does the line PQ divide AB?

In particular, if A = 1, the trapezoid becomes a parallelogram. The diag-onals then bisect each other at P, while Q recedes to infinity.

Q

C

Fia. 7a

Example 2. Theorem of Menelaus. If a line s cuts the sides BC, CA, AB of the triangle ABC in the points P, Q, R, respectively, the product of the ratios in which P, Q, R divide these sides equals - 1. Conversely, if P, Q, R divide the sides of the triangle in ratios whose product is - 1, the points are collinear.

Proof. (Fig. 7b.) We lose no generality if we assume that P, Q divide BC, CA in the ratios -y//3, -a/y:

(i)

(/3-y)P=$b-yc,

(ii)

(y-a)q=yc-aa.

In order to locate R, which lies on the lines PQ, AB, we seek a linear relation between p, q, a, b. Add (i) and (ii) to eliminate c and divide by S - a; then

(R - y)P + (y - a)q Sb - as

Thus R divides AB in the ratio -S/a. The product of the division ratios

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14 VECTOR ALGEBRA §7 Conversely, let us assume that P, Q, R divide BC, CA, AB in the ratios

-y/S, -a/y, -a/a whose product is -1. Then we have equations (i), (ii)

and also

(iii) (a - fl)r = as - 13b.

From these we deduce the linear relation,

(iv)

(0-y)p+(y-a)q+(a-$)r=0,

in which the sum of the coefficients is zero. The points P, Q, R are therefore collinear. *

Note. From (i), (ii), (iii) it is easily proved that the three pairs of lines BQ, CR; CR, AP; AP, BQ meet in the points A', B', C' given by

(-a+0+y)a' _ -aa+ftb+yc,

(a-S+y)b' =aa-$b+yc,

(a+9-y)c'=as+9b-yc.

If we add 2aa, 2gb, 2yc, respectively, to these equations, we find that the point S given by

(v) s =

as +Ab+yc

a+#+y

is common to the lines AA', BB', CC'. Thus to every line s given by (iv)

C Hence (i) _ Sb + ye as + SS a

0+y

a+s Rb + as yc as _ (ii) b' =

y+a

+b '

c'

=as+Sb -yc+Ss.

a+R

y+S

*This conclusion is obvious; for the line PQ must meet AB in the point R for which the product of the division ratios is -1.

we have a corresponding point S given by (v) -the pole of s relative to the triangle ABC.

Example 3. Theorem of Ceva.

If S is a

point in the plane of the triangle ABC, and the lines SA, SB, SC cut the sides opposite in the points A', B', C', then the product of the ratios in which A', B', C' divide the sides BC, CA, AB equals 1. Conversely, if A', B', C' divide the sides BC, CA, AB in ratios whose product

B is 1, the lines AA' BB' CC meet in a oint

C, , , p .

Proof. (Fig. 7c.) Since A, B, C, S are Fte.7c _coplanar,

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§7 COPLANAR POINTS 15

These equations state that A', B', C' divide BC, CA, AB in the ratios -y/fl, a/y, fl/a, whose product is 1. Incidentally, A', B', C' divide SA, SB, SC in the ratios a/a, fl/S, y/3 whose sum is -1.

Conversely, let us assume that A', B', C' divide BC, CA, AB in the ratios

y/$, a/-j, fl/a whose product is 1. Then we have equations (i), (ii), (iii).

From these we find that the vectors as + (0 + y)a', fib + (y + a)b', yc + (a + fl)c' are all equal to as + lb + yc; the point,

as+pb+yc

(iv) s _{a + fl + y}

is therefore common to the lines AA', BY, CC'.

Note. From (i), (ii), (iii) it is easily proved that the three pairs of lines BC, B'C'; CA, C'A'; AB, A'B' meet in the points P, Q, R given by

(Q-y)p=Rb-yc,

(v) (y - a)q = yc - aa,

(a-fl)r=as-/3b.

From these equations we deduce the linear relation,

(vi)

(fi-y)p+(y-a)q+(a-a)r=0,

in which the sum of the coefficients is zero. The points P, Q, R therefore lie on a line s. Thus to every point S given by (iv) we have a corresponding line s whose points P, Q, R are given by (v)-the polar of S relative to the triangle ABC.

Example 4. Let ABC and A'B'C' be two triangles, in the same or different planes, so that the vertices A, B, C correspond to A', B', C', and the sides AB, BC, CA correspond to A'B', B'C', C'A'. We then have (Fig. 7d)

P

FIG. 7d

DESARGUES' THEOREM. If the lines joining the corresponding vertices of two triangles are concurrent, the three pairs of corresponding sides intersect in col-linear points, and conversely.

Let the lines AA', BY, CC' intersect at S; then

as + a'a' = fib + fl'b' = yc + y'c' = s,

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16 VECTOR ALGEBRA §7 From these equations we find in the usual manner the points P Q, R in which BC, B'C'; CA, C'A'; AB, A'B' intersect:

(i) = fib - yc = fl'b' - y'c/

(ii) Jr fl_y yC - as =

'6',

y'C' - a'a' 1 q

y_a

y,_«,

(iii)

_«a-3b

«a'-fl'b'

Hence

at - of

(iv) (l; - 'y)p + (-y - a)q + (a - 6)r = 0,

(v) 'y')p + ('y' - a')q + (a' - 6')r = 0;

either equation shows that P, Q, R are collinear.

To prove the converse, we may start with the expressions (i), (ii), (iii) for p, q, r. These ensure that P, Q, R are collinear; but, in order that (iv) and (v) determine the same division ratios for P, Q, R, we must have

,-.y'

'y'-a'

= h,

y y - a a

-or

(vii)

a'-ha =fl'-hfi=y'-hy=k,

h and k representing the values of the equal members of (vi) and (vii). In

the usual way we now find from (i), (ii), (iii) that

aa'-haa=6'b'-hfib=y'c'-hyc=ks,

where S is a point common to AA', BB', CC'.

FIG. 7e

Example 5. The Complete Quadrangle. A complete quadrangle consists of four coplanar points, its vertices, no three collinear, and the six lines, its sides,

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3 7 COPLANAR POINTS 17

arc said to be opposite; and the three points in which they meet are called

rliagon<zl points. In the complete quadrangle ABCD (Fig. 7e) the pairs of opposite sides (BC, AD), (CA, BD), (AB, CD), meet at the diagonal points

L, :11, N, respectively.

The properties of this configuration of points and lines must all be con-sequences of the fact that A, B, C, D are coplanar points, that is,

(i) CA +(3b+yc+Sd =0,

a+$+y+a =0.

Since no three vertices are collinear, none of the scalars a, ,, y, S are zero. From equations (i) we locate at once the diagonal points at the intersections of the opposite sides:

Go +yc-aa+ad

0+y

a+S

_yc+aa

/3b+5d m

y+a

13+S , _aa+13b yc+Sd a+13 _'Y+s

We here assume that no two of the scalars a, fl, y, S have a zero sum; the diagonal points L, M, N are then all in the finite plane. But if for example,

a + 0 = y + S = 0, AB and CD are parallel, and N is the point at infinity

in their common direction.

To find the points X1, X2 where LM cuts AB and CD, we seek linear rela-tions connecting 1, m, a, b and 1, m, c, d, respectively.

Thus, from (ii) and (iii),

(vi) X2

('Y+a)m- (S +y)1 as -/3b

a-/3

a-0 '

(0+y)1-(S+S)m

yc - Sd

y-S

Equations (iv) and (v) show that N and X1 divide AB in the ratios '6/a and -/3/a; hence N, X1 are harmonic conjugates of A, B.

Equations (iv) and (vi) show that N and X2 divide CD in the ratios S/y and -S/y; hence N, X2 are harmonic conjugates of C, D.

Equation (v) shows that X1 divides LM in the ratio -(y + a)/(3 + y);

and (vi) shows that X2 divides LM in the ratio -(S + 5)/(3 + y) _

(y + a)/(3 + y); hence X1, X2 are harmonic conjugates of L, M.

We may now state the following harmonic properties of the complete quad-rangle in the

THEOREM. Two vertices of a complete quadrangle are separated harmonically by the diagonal point on their side and by a point on the line joining the other two diagonal points.

Two diagonal points are separated harmonically by points on the sides passing through the third diagonal point.

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From (ii), (iii), (iv) we next find the points X1, Y1, Z1 in which AB, BC, CA are cut by LM, MN, NL, respectively; thus

as - $b 3b - yc yc - as

$1 = I y1 = , Z1 =

a-/3

_0-y

y - a

Since

(a-/3)xl+(f -y)yl+(y-a)Z1 =0,

a-8+0-y+y-a =0,

the points Xl, Y1, Zl are collinear. This is also a consequence of Desargues'

Theorem applied to the triangles ABC, LMN, which are in perspective

from D.

Each vertex of the quadrangle is the center of perspective of the triangles formed by the other three vertices and by the three diagonal points. Thus, with careful regard to exact correspondence, the triangle LMN is in

perspec-tive with DCB, CDA, BAD, ABC from A, B, C, D, respecperspec-tively. Hence the coliesponding sides of these triangles intersect in four lines a, b, c, d (only d, the line X1Y1Z1, is shown in Fig. 7e). These lines are the polars of A, B, C, D with respect to the triangle LMN (ex. 3).

Corresponding to the complete quadrangle given by four points A, B, C, D, we now have a complete quadrilateral given by four lines a, b, c, d. In these configurations the roles of points and lines are interchanged (Fig. 7f):

The quadrangle has four vertices A, B, C, D and six sides consisting of three opposite pairs (BC, AD), (CA, BD), (AB, CD) which meet in the three diag-onal points L, M, N.

The quadrilateral has four sides a, b, c, d and six vertices consisting of three opposite pairs (bc, ad), (ca, bd), (ab, cd) which lie on the three diagonal lines 1, m, n.

The quadrangle and quadrilateral have the same diagonal triangle LMN or lmn; the sides 1, m, n are opposite the vertices L, M, N.

8. Linear Relations Independent of the Origin. We have seen that the position vectors of collinear or coplanar points satisfy a

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§9 CENTROID 19 linear equation in which the sum of the scalar coefficients is zero. The significance of such relations is given by the

THEOREM. A linear relation of the form,

(1) XiP1 + X2P2 + ... + XnPn = 0,

connecting the position vectors of the points P1, P2, , Pn will be

independent of the position of the origin 0 when, and only when, the sum of the scalar coefficients is zero:

(2) X1 + X2 -{- ... + an = 0.

Proof. Change from 0 to a new origin 0'. Writing OPi = pi,

O'Pi = pi, 00' = d, we have pi = d + pi; hence (1) becomes

(X1 + t2 +. +Xn)d+X1p1 +X2P2

_{XnPn' = 0.}

This equation will have the same form as (1) when and only when (2) is satisfied.

For two, three, and four points relations independent of the

origin have a simple geometric meaning; namely, the points are

coincident, collinear, or coplanar, respectively. The question now arises: What geometric property relates five or more points whose position vectors satisfy a linear relation independent of the origin?

9. Centroid. We shall encounter problems in which each point of a given set is associated with a certain number. The points, for example, may represent particles of matter and the numbers,

their masses or electric charges. In the latter case, the numbers

may be positive or negative. We shall now define a point P*

called the centroid of a set of n points P1, P2, , Pn associated with the numbers ml, m2i , mn, respectively. Denote any one

of these "weighted" points by the symbol m.iPi; then, if the sum

of the numbers mi is not zero, the centroid of the entire set is defined as the point for which the sum of all the vectors m1P*Pi is zero. The defining equation for the centroid is thus

(1) 2;miP*Pi = 0, provided Mmi Fl- 0.

As we wish P* to be uniquely defined, the case Emi = 0 must

be excluded; for then (1) is a relation independent of the position of P* (j 8).

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20 VECTOR ALGEBRA

i9

When 2 mi 0 0 there is always a unique point P* which

satis-fies (1).

For if we choose an origin 0 at pleasure, (1) can be

written

---4

-4

F.mi(OPi - OP*) = 0,

or

(2) (2mi)OP* = 1m1OPi.

This relation, independent of the origin, fixes the position of P*

relative to 0.

The point P* thus determined is the only point

that satisfies (1) ; for if Q is a second point for which

EmiQPi = 0,

we have, on subtraction from (1),

2;mi(P*Pi

- QPi) =

Timi(P*Pi + PQ) = (2;mi)P*Q = 0;

hence P*Q = 0, and Q coincides with P*.

If the position vectors of P*, Pi are written p* and pi, (2)

be-comes

(3) (2;mi)P* = Emipi.

The centroid of the points miP1 is not altered when the numbers

mi are replaced by any set of numbers cmi proportional to them; for, in (2), the constant c may be canceled from numerator and

denominator. In particular, if the numbers mi are all equal, we may replace them all by unity; the centroid of the n points then is called their mean center and is given by the equation:

-3

1 --

1

(4) OP* _ - ZOP1, or

p* _ - Zpi.

n n

In finding the centroid P* of any set of weighted points miP1, we may replace any subset of points for which the sum of the weights is a number m' 0 0 by their centroid P with the weight

m'. For, if V and 1", respectively, denote summations extended

over the points of the subset and over all the remaining points,

we may write (3) in the form,

(2;'mi + Z"mi)P* = 7,'mipi + E"miPi,

or

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§9 CENTROID 21 This equation shows that P* is also the centroid of the point m'P'

and the points miPi not included in the subset.

Finally, let us consider the nature of a set of n points miPi

(?Pi F4- 0) for which Ximi = 0.

If we attempt to find a point P*

which satisfies

(5) ZmiP*Pi = 0

when Xmi = 0,

we find that there are two possibilities.

(a) For any arbitrary choice of the points miPi, subject only to the condition Emi = 0, there will in general be no point P* that

will satisfy (5). Thus in the cases n = 2, 3, 4, equations (5) imply,

respectively, that P1 and P2 coincide; P1, P2, P3 are collinear;

P1, P2, P3, P4 are coplanar. Hence, if these conditions are not

fulfilled, P* does not exist.

(h) If, however, a point P* can be found which satisfies (5), any point whatever will serve (§ 8).

In fact, we see, from (2), that

the existence of P* implies that

(6) ,6dmiOPi = 0 1mi = 0,

for any choice of 0. In particular, we may take any one of the

given points Pi as origin. Thus, if we take 0 at P1, (6) becomes

miP1 Pi = 0,

i=2

V

2; mi = -m1 0.

i=2

Hence, from the defining equation (1), we see that P1 is the

cen-troid of the points m2P2, m3P3, . . . , mnPn. Precisely the same

conclusion may be drawn for P2, P3, , Pn. We now can answer

the question raised at the end of § 8. Any set of weighted points

miPi (mi F6 0) whose position vectors satisfy a linear relation (6) independent of the origin has the intrinsic property that any point of the set is the centroid of all the remaining weighted points.

A set of n points having this property may be readily

con-structed. Take any set of n - 1 weighted points miPi, such that

m1 + m2 + + mn_1 0, and adjoin to the set their centroid

P* = Pn with the weight Mn = -(m1 + m2 +

+ mn-1).

Then, since -mnPn = m1P1 + m2P2 + + mn-1Pn-1, we have n n X mipi = 0, 2;mi = 0. 1 t

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22 VECTOR ALGEBRA

Thus the relation,

(7)

as+Sb+yc=0,

a+$+-y=0,

§9

between three collinear points shows that each point

of the set

aA, ,BB, yC is the centroid of the other two. Any point C on the

line of A and B satisfies a relation of the type (7) and hence is the centroid of these points when suitably weighted.

The relation,

(8)

as+,9b+yc+8d = 0,

a+a+7+S = O,

between four coplanar points shows that each point of the set

aA, ,BB, yC, 8D is the centroid of the other three. Any point D

in the plane of A, B, C satisfies a relation of the type (8) and hence is the centroid of these points when suitably weighted.

Example 1. Centroid of Two Points. The centroid of aA and pB is given by

*

as+Rb

P

a+13

Hence P* divides AB in the inverse ratio f/a of their weights. In particular, if a = S, P* divides AB in half. The mean center of two points lies midway

between them.

Example 2. Mean Center of Three Points A, B, C. Let L, M, N be the mid-points of BC, CA, AB (Fig. 9a). Then the mean center P* of A, B, C is the centroid of A and 2L, B and 2M, C and 2N. Hence (ex. 1) the segments AL, BM, CN are all divided by P* in the ratio of 2/1.

If A, B, C are not collinear, they are the vertices of a triangle. Its medians AL, BM, CN intersect at P* and are there divided in the ratio of 2/1.

C

N

a

M2

Fia. 9a FIG. 9b

Example 3. Mean Center of Four Points A, B, C, D. Let L1, L2; M1, M2; N1, N2 be the mid-points of A B, CD; BC, DA; AC, BD, respectively (Fig. 9b). To find P* we may replace A, B, C, D by 2L1 and 2L2, or by 2M1 and 2M2, or by 2N1 and 2N2. Therefore P* is the mid-point of the segments L1L2, M1M2, N1N2. The bisectors of the three pairs of opposite sides of the

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quad-§ 10 BARYCENTRIC COORDINATES 23

rangle ABCD (which may be plane or skew) intersect at P* and are there divided in half.

If A, B, C, D are not coplanar, they determine a tetrahedron. Let A', B', C', D' be the mean centers of the triads BCD, CDA, DAB, ABC. Then P* is the centroid of A and 3A', B and 3B', C and 3C', D and 3D', and hence

divides each of the segments AA', BB', CC', DD' in the ratio of 3/1. The

preceding result also shows that the bisectors of the three pairs of opposite sides of the tetrahedron meet at P*.

Example 4. The sum of n vectors AiB1 (i = 1, 2, , n) is given by

Z(bi-a;) =nb-na = n(b* - a*)

where A* and B* are the mean centers of the initial points Ai and the terminal points Bi, respectively; hence

In particular, 2;A;Bi = n A*B*.

-4

--A1B1 + A2B2 = 2 A*B*,

where A*, B* are the mid-points of A1A2 and B1B2, respectively.

10. Barycentric Coordinates. If P is any point in the plane of

-3-3 --3

the reference triangle ABC, the vectors AP, BP, CP are linearly

dependent (§ 5) ; hence

«(p - a) + /3(p - b) + y(p - c) = 0,

(1)

p=as+9b+-ic

«+6 + y

The denominator is not zero; for, if

a+a+y=0, then aa+(3b+yc=0,

and A, B, C would be collinear, contrary to hypothesis. Thus P

is the centroid of the weighted points aA, $B, yC.

The three

numbers a, /3, y are called the barycentric coordinates of P; as they

all may be multiplied by the same number without altering p,

their ratios determine P. The vertices of the reference triangle

have the coordinates A(1, 0, 0), B(0, 1, 0), C(0, 0, 1); the unit point (1, 1, 1) is the mean center of A, B, C.

If P is a point in space and ABCD a reference tetrahedron, the vectors AP, BP, CP, DP are linearly dependent (§ 5); hence

(2)

«(p-a)+$(pp-b)+y(p-c)+a(p-d) =0,

as+Ab+yc+Sd

p

_a+O+y+S

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24 VECTOR ALGEBRA

The denominator is not zero; for, if

§ 12

a+8+y+S = 0,

then as+,lib+yc+Sd =0

and the points A, B, C, D would be coplanar, contrary to hypoth-esis. Thus P is the centroid of the weighted points aA, $B, yC, W. The four numbers a, 0, y, S are called barycentric coordinates of P;

as they all may be multiplied by the same number without

alter-ing p, their ratios determine P. The vertices of the reference

tetra-Fla. 11

hedron have the barycentric

co-ordinates A(1, 0, 0, 0), B(0, 1, 0,

0), C(0, 0, 1, 0), D(0, 0, 0, 1);

the unit point (1, 1,

1, 1) is the

mean center of A, B, C, D.

11. Projection of a Vector. To

find the projection of a vector

PQ upon a line x, a director plane 7r must be specified. Pass planes

through P and Q parallel to 7r and let them cut x in the points

P1i Ql (Fig. 11). Then the vector P1Q1 is the 7r-projection of PQ upon x.

. If no director plane 7r is specified, we tacitly assume that IT is

perpendicular to x; the projection is then orthogonal.

Let the vectors PQ, QR and their sum PR have the projections

P1Q1, Q1R1 and P1R1 on the line x; then, since

P1Q1 + Q1R1 = P1R1,

the projection of the sum of two vectors on a line is equal to the sum of their projection on this line.

12. Base Vectors. Let e1, e2, e3 be three linearly independent

vectors; they are then non-coplanar.

If the vectors are drawn

from a common origin 0 (Fig. 12a), we may pass a closed plane

curve through their end points El, E2, E3. If this curve is viewed

from the side of its plane opposite to that on which 0 lies, the

order E1E2E3 defines a sense of circulation. If this sense is counter-clockwise, the set e1, e2, e3 is said to be right-handed or dextral; for it is then possible to extend the thumb, index, and middle fingers

(39)

§ 12 BASE VECTORS 2'r

respectively. If the sense defined by E1E2E3 is clockwise the set is said to be left-handed or sinistral.

Any vector u = PQ may be expressed as the sum of three

vec-tors parallel to el, e2, e3, respectively. For, if we construct a

parallelepiped on PQ as diagonal by passing planes through P and

R

Fia. 12a FIG 12b

Q parallel to e2 and e3, e3 and el, el and e2 (Fig. 12b), its edges

will be parallel to el, e2, e3, and

PQ = PQ1 + Q1R + RQ = PQ1 + PQ2 + PQ3

_ y

Since PQi is a scalar multiple of ei, say u`ei,

(1) u = ulel + u2e2 + u3e3,

where ul, u2, u3 are numbers called the components of u with

re-spect to the basis el, e2, e3. Their indices are not exponents but

mere identification tags; they are written as superscripts for reasons

given in Chapter IX. The vector u is often written [ul, u2, u3],

with brackets to enclose its components.

If u is the position vector OP, the components ui are called the

Cartesian coordinates of the point P with respect to the basis

e1, e2,

e3-A straight line upon which two directions are distinguished is

called an axis. One direction is called positive, the other negative.

In a figure the positive direction is marked with an arrowhead. With a given basis, lines drawn through the origin 0 parallel to

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26 VECTOR ALGEBRA § 13

el, e2, e3 and with their directions positive are called the

coordi-nate axes. The numbers u', u2, u3 often are called the components of u on these axes.

The components of a zero vector are all zero; for, since el, e2, e3

are linearly independent, u = 0 implies u' = u2 = u3 = 0.

If A is any scalar,

(2) Au = Aulel + Au2e2 + Au3e3. If v = vie1 + v2e2 +.v3e3,

(3)

_{u + v = (u1 + v')el + (u}

2

+ v2)e2 + (u3 + v3)e3

With the bracket notation, (2) and (3) become

(4) A[ul, u2, u3] = [Au', Au2, Au3],

(5) [ul, u2, u3] + [v', v2, v3] = [u' + v1, u2 + v2, u3 + v3].

To multiply a vector by a number, multiply its components by that number; to add vectors, add their corresponding components. In

par-ticular,

(6) -[u', u2, u3]

= [-u',

-u2, -u3],

(7) [ul, u2, u3]

- [v',

v2, v3] = [u'

-

v', u2

-

v2, u3

-

v3].

If u = v, then u - v = 0; hence

(8)

U=V implies u' = v',

u2 = v2, u3 = v3.

When referred to the same basis, equal vectors have their corresponding components equal.

13. Rectangular Components. Let i, j, k denote a dextral

sys-tem of mutually perpendicular unit vectors.

From an origin 0

draw the coordinate axes x, y, z with positive directions given by

i, j, k (Fig. 13).

Any vector u now is determined by giving its

components u1, u2i u3 on these rectangular axes:

(1)

u = uli + u2j + u3k.

Draw OP = u from the origin, and let OP1, OP2i OP3 be its orthogonal projections on the axes. Then ul is the length of OPI, taken positive or negative according as OPI has the direction of

i or -i; hence

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§ 13 RECTANGULAR COMPONENTS 27

where j u j denotes the length of u and (i, u) the angle between

i and u.

The rectangular components of u thus are obtained by

multiplying its length by the corresponding direction cosines:

(2) ul = I u I cos (i, u), u2 = I U I cos (j, u), u3 = I u I cos (k, u).

R Fic. 13

The Pythagorean Theorem gives the length of u in terms of its

components; from

0P2 = (OR)2 + (RP)2 = (OPl)2 + (OP2)2 + (OP3)2,

(3) I U 12 = (ul)2 + (u2)2 + (u3)2.

If we substitute from (2) in (3), we obtain the relation, cost (i, u) + cost (j, u) + cost (k, u) = 1,

satisfied by the direction cosines of any vector.

The rectangular coordinates of any point P are defined as the components of its position vector OP on the x-, y-, and z- axes.

Thus, if

(4)

OP=xi+yj+zk,

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28 VECTOR ALGEBRA

- OPI, this gives

(5)

PIP2 = (x2 - xl)i + (Y2 - yl)j + (z2 - zl)k

§13

if (xi, y,, zi) are the coordinates of Pi. The components of PIP2

are found by subtracting the coordinates of PI from the corresponding coordinates of P2.

With an orthogonal basis i, j, k, the components may be written with subscripts (as previously) or superscripts. Separately, they

are called the x-, y-, and z- components and often are written

ul, U2, U3- Just as with a general basis, vectors are specified by

giving their components: u = -[ul,

u2, u3].

Thus the point

(x, y, z) has the position vector [x, y, z]; and (5) may be written

PIP2 = [x2 - xl, Y2 - yl, z2 -

Z11-The unit vectors i, j form a basis for all vectors u in their plane:

(6)

u = uli + u2j.

If e is a unit vector in the plane and 0 is the angle (i, e), reckoned

positive in the sense from i to j, we define cos 0 and sin 0 as the

components of e :

(7)

e=icos0+jsin0.

Any vector in tine plane may be written

(8)

u= Iule= Jul {icos(i,u)+jsin(i,u)};

its rectangular components are

(9) ul = I U I cos (i, u), u2 = I u I sin (i, u), u3 = 0.

Example. Addition Theorems for the Sine and Cosine. Let a and b be two unit vectors such that the angles (i, a) = a, (a, b) = 0; then the angle (i, b)

= a + 14, and, from (7),

a =icosa+jsina,

b = i cos (a + fl) + j sin (a + 0).

If we refer b to the new basis i = a, j (a unit vector 90° ahead of a), we have

b=icos0+isin 0

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15 SCALAR PRODUCT 29

Comparing the components of b in the two preceding expressions, we find

cos (a + 13) = cos a cos 8 + cos (a +

2)

sin [3,

sin (a + /3) = sin a cos l3 + sin

(a

+ 2) sin

With a =7r/2, these equations give

cos(/3

_{+ 2)}

sin $, sin (p + 2) = cos 0;

hence

cos (a + 0) = cos a cos 0 - sin a sin 0, sin (a + )3) = sin a cos g + cos a sin g.

These addition theorems hold for all values of a and fl, positive or negative.

14. Products of Two Vectors. Hitherto we have considered

only the products of vectors by numbers. Next we shall define

two operations between vectors, which are known as "products,"

because they have some properties in common with the products of numbers. These products of vectors, however, will also prove

to have properties in striking disagreement with those of numbers.

Since one of these products is a scalar and the other a vector,

they are called the scalar product and vector product, respectively.

The definitions of these new products may seem rather arbitrary to one unfamiliar with the history of vector algebra. We present

this algebra in the form and notation due to the American mathe-matical physicist, J. Willard Gibbs (1839-1903).t It is an offshoot of the algebra of quaternions, adapted to the uses of geometry and physics. In Chapter X quaternion algebra is developed briefly,

and the origin of the foregoing products is revealed.

15. Scalar Product. The scalar product of two vectors u and v,

written u v, is defined as the product of their lengths and the cosine of their included angle:

(1)

Jul IvI cos(u,v).

The scalar product is therefore a number which for proper

(non-zero) vectors is positive, zero, or negative, according as the angle (u, v) is acute, right, or obtuse. Hence, for proper vectors,

(2)

means u1v.

t Professor of mathematical physics at Yale University. His pamphlet on the Elements of Vector Analysis was privately printed in 1881. A more com-plete treatise on Vector Analysis (New Haven, Yale University Press, 1901) based on Gibbs's lectures, was written by Professor E. B. Wilson.

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30 VECTOR ALGEBRA

When u and v are parallel,

Jul lvl,

or

-Jul lvI,

115

according as the vectors have the same or opposite directions; thus

From (1) we see that

(3) (au) ($v) = ag u . V.

The last result is obvious when a and i3 are positive numbers; the other cases then follow from the equations preceding.

Besides the components of a vector on the coordinate axes

(§§ 12, 13) we shall also use the orthogonal projection and

compo-nent of a vector on an arbitrary directed line 1. If the positive

direction of 1 is given by the unit vector e, the projection of u

upon 1 (proj 1 u) is a scalar multiple of e. This scalar is called the component of u upon 1 (comp1 u) ; its defining equation is therefore

(4) e comp j u = proj 1 u. As in § 13, we compute comp j u as

(5) comps u = l u i cos (e, u);

since I e 1, this may also be written

(6) comps u = e u.

From the projection theorem of § 11,

(7) proj 1 (u + v) = proj 1 u + proj 1 v;

hence, from (3),

(8) comps (u + v) = comp] u + comps v.

The operations expressed by proj1 and comp] are distributive with respect to addition.

The definition (1) of u v now may be written

(9) U V= I u l compu V=IV l comp u.

Scalar or "dot" multiplication is commutative and distributive:

(10)

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§ 15 SCALAR, PRODUCT

The last equation is nothing more than (8) multiplied by

when 1 is taken along w.

If c 0 in the equation, or

31

we can conclude either that a - b = 0 or that a - b and c are

perpendicular.

We cannot "cancel" c to obtain a = b unless

a - b and c are not perpendicular.

It is obvious that, in general, (u v)w

u(v w).

Since i, j, k are mutually perpendicular unit vectors,

(12) i

Hence, if we expand the product,

u ' v = lull + u2j + u3k)

(v1i + v2j + v3k) we obtain

(13) u v = u1v1 + u2v2 + u3v3

The scalar product of two vectors is equal to the sum of the products of their corresponding rectangular components.

Example 1. If u = [2, -1, 3], v = [0, 2, 4], u v = 0 - 2 + 12 = 10. Moreover,

comp,, u =

u v

rv7

= 10 = 2 236, compu v = uuv = 10 = 2.673;

/20

luI

14

U

v 10

cos (u, v) = I =

u v /280 = 0.5979, angle (u, v) = 530 18'. Example 2. Identities involving scalar products may be given a geometric i.iterpretation. Thus (Fig. 15a)

gives cdcosw=a2-b2,

and, if we write c = a + b, d = a - b, w =

an-L,Ie (c, d). If a = b, c d = 0; then PQRS is a rhombus and the angle PRT may be inscribed in a semicircle about Q. We thus have two geo-metric theorems:

1. The diagonals of a rhombus cut at right

angles.

2. An angle inscribed in a semicircle is a right P a Q

angle. _{Fro. 15a}

Moreover, from

(a-b) (a - b) =

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32 VECTOR ALGEBRA §15 Example 3. We also may interpret identities involving scalar products by regarding the vectors a, b, as position vectors OA, OB, from an arbitrary origin O. Then a - b = BA and a + b = 2 011 where M is the

mid-point of AB.

Consider, for example, the identity,

(b-a)2+(c-b)2+(d-c)2+(a-d)2

= (c-a)2+(d-b)2+(a+c-b-d)2,

which can be verified on expansion; u2 means u u. If we regard a, b, c, d as the position vectors of the vertices of a space quadrilateral ABCD, p =

(a + c), q =

2(b + d) locate the mid-points P, Q of the diagonals AC, BD; hence

(AB)2 + (BC)2 + (CD)2 + (DA)2 = (AC)2 + (BD)2 + 4(QP)2. The sum of the squares of the sides of any space quadrilateral equals the sum of

C

Fia. 15b

the squares of its diagonals plus four times the square of the segment joining their middle points.

Example 4. The identity,

(a-b) (h-c)

0,

B

shows that the altitudes of a triangle

ABC meet in a point H, the

ortho-center of the triangle (Fig. 15b); for, if two terms of this equation are zero, the third is likewise.

Similarly the identity,

\

(a-b) k-

a 2 + b b + c 2 C

c+a\

shows that the perpendicular bisectors of the triangle ABC meet in a point K, the circumcenter of the triangle.

For the orthocenter H and circumcenter K the individual terms of the fore-going equations vanish; for example,

(a -

0,

(a -

0.

On adding these, we obtain

=0;

or on writing g = 3 (a + b + c) for the position vector of the mean center G of the triangle ABC.