THEOREM OF MOMENTS.

93

Next, suppose o to be out of the plane. Then the vector representing oab will be a line am perpendicular to the plane oab, which may be resolved into components an, nm, of which an is perpendicular to the plane of the parallelogram and nm parallel to that plane. Now an represents on the same scale the projection pab of oab on the plane abcd, and nm its projection opq on the perpendicular plane. For the

NA

triangles oab, pab, opq, being on the same base ab or pq and having the heights respectively ao, ap, po, which are proportional to am, an, nm, must have their areas proportional to the lengths of these lines.

Suppose, then, the vector representing each of the areas oab, oac, oad to be resolved into components perpendicular and parallel to the plane; the theorem will be proved if it is true separately for the components perpendicular and for those parallel to the plane. Now for the perpendicular components the theorem has been already proved, because they represent the triangles pab, pac, pad, which are projections on the plane abcd of oab, oac, oad.

d

For the components parallel to the plane, observe that mn represents opq; it is at right angles to ab and proportional to the product of ab by op the distance of o from the plane. Hence the components parallel to the plane are lines ab', ac', ad' respectively at right angles to ab, ac, ad, and proportional to their lengths multiplied by the distance of o from the plane. Thus the figure ab'c'd' is merely

α

d'

abcd turned through a right angle and altered in scale; whence it is obvious that ad' = ab' + ac'.

Thus the proposition is proved in general. It is well worth noticing, however, that the proof given for the special case of o in the plane applies word for word and symbol for symbol to the general case, if only we interpret

+- and sum as relating to the composition of vectors. Thus oad = ocd+cad+oac, or one face of a tetrahedron is equal to the vector-sum of the other three faces. It is, of course, the sum of their projections upon it; and the components of their representative vectors which are parallel to its plane are respectively perpendicular and proportional to oa, ad, do, so that their vector sum is zero. Again, ocd+cadoab, because the height of oab is the vectorsum of the heights of ocd and cad.

For proving theorems about areas, the following consideration is of great use. We have seen that the projection of an area on any plane is represented by the projection of its representative vector on a line at right angles to the plane. In fact, the angle oap between the two planes is equal to the angle man between the two lines respectively perpendicular to them; if we call this angle 0, the projection of the area A is A cos 0, and the corresponding projection of the line of length A is also A cos 0. Now it is easy to see that if the projections of two vectors on every line whatever are equal, then the two vectors are equal in magnitude and direction. Hence it follows that if the projections of two areas on any plane whatever are equal, then the areas are equal in magnitude and aspect. For example, the areas oad and oac+oab (figure on p. 92) are such that their projections on any plane are equal; this projection is, in fact, the case of the theorem of moments in which o is in the plane abcd. Hence the general theorem may be deduced in this way from that particular case.

PRODUCT OF TWO VECTORS.

On account of the importance of the theorem of moments, we shall present it under yet another aspect. The area of the parallelogram abde may be supposed to be generated by the motion of ab over the step ac, or by the motion of ac over the step ab. Hence it seems natural to speak of it as the product of the two steps ab, ac. We have been accustomed to identify a rectangle with the product of its two sides, when their lengths only are

VECTOR AND SCALAR PRODUCT.

95

taken into account; we shall now make just such an extension of the meaning of a product as we formerly made of the meaning of a sum, and still regard the parallelogram contained by two steps as their product, when their directions are taken into account. The magnitude of this product is ab. ac sin bac; like any other area, it is to be regarded as a directed quantity.

Suppose, however, that one of the two steps, say ac, represents an area perpendicular to it; then to multiply this by ab, we must naturally make that area take the step of translation ab. In so doing it will generate a volume, which may be regarded as the product of ac and ab. But the magnitude of this volume is ab multiplied by the area into the sine of the angle it makes with ab, that is, into the cosine of the angle that ac makes with ab. This kind of product therefore has the magnitude ab. ac cos bac; being a volume, it can only be greater or less; that is, it is a scalar quantity.

We are thus led to two different kinds of product of two vectors ab, ac; a vector product, which may be written V. ab. ac, and which is the area of the parallelogram of which they are two sides, being both regarded as steps; and a scalar product, which may be written S.ab.ac, and which is the volume traced out by an area represented by one, when made to take the step represented by the other.

Now the moment of ab about o is V. oa. ab; that of ac is V.oa.ac; and that of ad is V.oa. ad, which is Voa. (ab+ ac). Hence the theorem tells us that

V. oa (ab + ac) V.oa.ab + V.oa.ac;

=

or if, for shortness, we write oa = a, ab = ẞ, ac=y, the theorem is that

Va (B+y)= Vaß + Vay.

We may state this in words thus: the vector product is distributive. And in this form the proposition may be seen at once in the figure on p. 93, if we make ab=a, ap = ß, po= y; it asserts that

area abqp + area pqro = area abro,

and this is obviously true of their projections on any plane.

The corresponding theorem for the scalar product, that Sa (B+ y) = Saß+ Say, is obvious if we regard a as an area made to take the steps B, y.

But there is a very important difference between a vector product and a product of two scalar quantities. Namely, the sign of an area depends upon the way it is gone round; an area gone round counter-clockwise is positive, gone round clockwise is negative. Now if V. ab. ac area abdc, we must have by symmetry V.ac.ab: = area acdb, and therefore V. ac. ab = − V. ab. ac, or Vyẞ- Vẞy. Hence the sign of a vector product is changed by inverting the order of the terms. It is agreed upon that Vaß shall be a vector facing to that side from which the rotation from a to B appears to be counterclockwise.

It will be found, however, that Saß = Sẞx, so that the scalar product of two vectors resembles in this respect the product of scalar quantities.

MOMENT OF VELOCITY OF A MOVING POINT.

The flux of the moment of velocity of a moving point p about a fixed point o is equal to the moment of the acceleration about o. For suppose that during a certain interval of time the velocity has changed from p to p1, so that P1p is the change of the velocity; then the sum of the moments of p and p1- p is equal to the moment of p1, that is the moment of the change in the velocity is equal to the change in the moment of velocity. Dividing each of these by the interval of time, we see that the moment of the mean flux of velocity is equal to the mean flux of the moment of velocity, during any interval. Consequently the moment of acceleration is equal to the flux of the moment of velocity.

The same thing may be shewn in symbols, as follows, supposing the motion to take place in one plane. We

MOMENT OF ACCELERATION.

may write p=reie, where r is the length of op, and the angle Xop. Then pre+rẻ.ie, or the velocity consists of two parts, along op and re perpendicular to it. The

moment of the velocity is the sum of the moments of these parts; but the part along op (radial component) has no moment, and the part perpendicular (transverse component) has moment rẻ. Next, taking the flux of p, we find for the acceleration the value

ï=Ÿe1o + Ör. ieio + rÒ.ieio + rÒa ̧ï2eio + rö. ieio

=

=

· (ï − rẻ2) eið + (2řẻ +rë) ieio. Or the acceleration consists of a radial component r — rẻ2, and a transverse component 2r+rë. The moment of the acceleration is r times the transverse component, namely 2rrė +r2Ö. But this is precisely the flux of the moment of velocity r2ė.

Observe that the radial acceleration consists of two parts, due to the change in magnitude of the radial velocity, and re due to the change in direction of the transverse velocity.

We may also make this proposition depend upon the flux of a vector product. The moment of the velocity is Vpp, and the moment of the acceleration is Vpp; we have therefore to prove that Vpp is the rate of change of Vpp. Now upon referring to the investigation of the flux of a product, p. 64, the reader will see that every step of it applies with equal justice to a product of two vectors, whether the product be vector or scalar. In fact, the only property used is that the product is distributive. Hence the rate of change of Vaß is Vaß+ Väß. (Observe that the order of the factors must be carefully kept.) Applying this rule to Vpp, we find that its rate of change is Vpp+ Vpp. Now the vector product of two parallel vectors is necessarily zero, because they cannot include any area; thus Vpp=0. Therefore (Vpp)= Vpp. This demonstration does not require the motion to be in one plane.