AQ as it rotates about a through the angle PAQ, then if we marked off AQ a distance A P' equal to AP, P'Q would be the scalar growth of AP; that is, its growth

in the direction of its length. But if AP be treated as a vector (see p. 153)

AQAP + PQ,

or the directed step P Q must be added to AP in order to convert it into a Q; P Q may be thus termed the directed growth of A P. If we join P P', we shall have PQ equal to the sum of P P′ and P' Q. Now if the angle PA P' be taken very small P P' will be ultimately perpendicular to AP, and this part of the growth PQ might be represented by -1. PP. Hence we are led to represent a growth perpendicular to a rotating line by a scalar quantity multiplied by the symbol −1.

We can now consider the case of our circle of unit radius. Let op be a radius which has revolved through

Q

M

A

an angle

FIG. 79.

from a fixed radius o A, and let oq be an

very

adjacent position of o P such that the angle QOP is small. Then P Q will be a small arc sensibly coincident

with the straight line P Q, and the line P Q will be to all intents and purposes at right angles to o P. Hence to obtain oq we must take a step P Q at right angles to ОР. This we represent by √-1 Q P. Since the radius of the circle is unity the arc QP, which equals the radius multiplied by the angle QOP (see p. 143), must equal the numerical value of the angle QOP. Or the growth of op is given by √-1x angle QOP.

Now since op remains of constant length as it revolves about o, it is equally multiplied (i. e., by the factor unity) in describing equal angles. It thus satisfies our definition of growth at logarithmic rate (see p. 176). In this case what value shall we give to the rate for unit angle?

It must equal divided by the ratio of the angle

PQ
ОР

QOP to unit angle =

PQ
OP X angleQOP

=-1-since o P

-1

is unity. Thus o P is growing at logarithmic rate as it describes unit angle; that is to say, the result of turning op through unit angle might be symbolically expressed by e. Hence the result of turning OP through an angle ✪ must be e√10. We may then write OP=0 A. e√—10

Drop Pм perpendicular to o A and produce it to meet the circle again in p', then by symmetry м P=M P', and we have

OP = OM + √1M P.

OP' OM√1 MP'.
=

Now since o P and op' are of unit magnitude,.

Also the angle P'oм equals the angle м OP, but, according

to our convention as to the measurement of angles, it is of opposite sense, or equals - 0. Thus we must write

These values for cose and sine in terms of the exponential e were first discovered by Euler. They are meaningless in the form (ii) when cose and sine are interpreted as mere numerical ratios; but they have a perfectly clear and definite meaning when we treat each side of the equation in form (i) as a symbol of operation. Thus cose +-1 sine applied to unit step directs us to turn that step without altering its length through an angle ; on the other hand, e√--10 applied to the same step causes it to grow at logarithmic rate unity perpendicular to itself, while it is turned through the angle 0. The two processes give the same result.

§ 15. On the Multiplication of Vectors.

We have discussed how vector steps are to be added, and proved that the order of addition is indifferent; we have also examined the operation denoted

by the ratio of two vectors. The reader will naturally ask: Can no meaning be given to the product of two vectors ?

If both the vectors be treated as complex numbers, or as denoting operations, we have interpreted their product (see p. 193) as another complex number or as a resultant operation. Or again we have interpreted the product of two vectors when one denotes an operation and the other a step of position; the product in this case is a direction to spin the step through a certain angle and then stretch it in a certain ratio. But neither of these cases explains what we are to understand by the product of two steps of position.

Let A P, AQ be two such steps: What is the meaning of the product AP.AQ? Were AP and AQ merely

A

FIG. 80.

scalar quantities then their product would be purely scalar, and we should have no difficulty in interpreting the result AP.PQ as another scalar quantity. But when we consider the steps A P, P Q to possess not only

R

A

FIG. 81.

magnitude but direction, the meaning of their product is by no means so obvious.

If A Q were at right angles to A P (see fig. 81), we should naturally interpret the product AP.AQ as the

area of the rectangle on a Q and A P, or as the area of the figure QA PR. Now let us see how this area might be generated. Were we to move the step a Q parallel to itself and so that its end a always remained in the step A P, it would describe the rectangle Q A P R while its foot A described the step A P. Hence if A P and A Q are at right angles we might interpret their product as follows:

The product AP.AQ bids us move the step ▲ Q parallel to itself so that its end a traverses the step A P; the area traced out by a Q during this motion is the value of the product A P. AQ.

It will be noted at once that this interpretation, although suggested by the case of the angle QA P being a right angle, is entirely independent of what that angle may be. If QA P be not a right angle the area traced out according to the above rule would be the parallelogram on a P, A Q as sides. Hence the interpretation we have discovered for the product A P.AQ gives us an intelligible meaning, whatever be the angle Q A P.

There is, however, a difficulty which we have not yet solved. An area is a directed quantity (see p. 134), and its direction depends on how we go round its perimeter. Now the area QA PR will be positive if we go round its perimeter counter-clockwise, or from A to P; that is, in

R

FIG. 82.

the direction of the first step of the product or in the direction of motion of the second or moving step. Thus the product A P. A Q will be the area Q A P R taken with the sign suggested by the step A P. The product AQ.AP