13. Vector Product.— The vector product of a and b is a third vector normal to the plane of a and b pointing toward the direction from which rotation from a to b is seen as positive rotation, and of length equal to ab sin 0, the area of the parallelogram formed by a, b. The vector product will be denoted by a x b after Willard Gibbs, and is read a cross b. If a and b are parallel, a x b=0, because 0=0; similarly a x a=0. The vector a x b can be conveniently regarded as the area vector (9) of the parallelogram (a, b), this area being sup posed to face toward the side at which rota tion from a to b appears positive. The com mutative law does not hold in vector multiplica tion for a x b=—b x a.
14. Distributive Law.— Vector multiplica tion is distributive, that is ex(a+b)=cxa+cxb. First suppose that c is not coplanar with a and b; and construct a triangular prism having the vectors a, b, and a + b for sides of base, and c for the parallel edges. The sum of the area vectors of the faces of the prism is zero (10), and the vectors of the two opposite bases cel; but the vectors of the lateral faces pointing outward are c x (a + b),— c x a,— c x b, hence cx (a+b)=cxa+cxb.
If c happens to be coplanar with a and b we can resolve it into two which are not coplanar with a and b, apply the preceding result, and then combine. For the fundamental unit vectors j, k, the vector multiplication table is is i=j xj=k.k=0; ixj=—j xi=k; jxk= k j=1; k x i= k=j.
15. Expansion of a x b.— If a=asi+asj+asks b=lssi bsj + bilc, then by applying the dis tributive law, and the table we get a x b= (a2bs—aab,) i (asb,--a,b.) j + (albs—ado,)k. There are various interpretations of this re sult. From it we can easily obtain the three projections of the area of the parallelogram formed by (a, b) ; also the directions of its normal; also sin B in terms of the direction cosines of a and b; also the area of any tri angle in terms of the co-ordinates of its vertices.
For an application to plane analytic geom etry, suppose a and b both lie in the xy-plane. then as=0, bs=0, and a b=(albs —a,b,)k, hence ab sin 0 =a,h, — a,b2; this gives twice the area of a triangle one of whose vertices is at (0, 0), and whose other vertices are at (ai, a,), (61, b,). Again, if we divide by ab, we can show that sin (a, — ai) = sin cos al cos ai sin al ; and can then find sin (as + al) by changing the sign of xi. Similarly we could find cos (a, — ai) by taking the scalar instead of the vector product.
In mechanics the moment of a force a act ing at 0, taken about a point P whose vector OP=r, is The vector a x r; it follows from the distributive law that the moment of a resultant force about P equals the vector sum of the moments of its components. If there is
a couple formed by the parallel forces a and — a acting at points whose distance apart is rep resented by the vector r, then the moment of the couple is a x r; and the resultant of any number of couples is a single couple whose vector is the sum of the vectors of the given couples.
16. Triple Scalar Product.— It is evident that (b x c) is the volume of the lelopiped whose adjacent edges are a, b, c, hence a- (b x c)=b• (c x a)=c• (a x b), where the letters are in the same cyclic order; if the cyclic order is changed the sign of the triple product is altered. The dot and cross can be interchanged, for a• (b x c)= (a x 17. Triple Vector Product— It can be shown by expansion that the triple vector prod uct can be resolved into scalar multiples of two of the given vectors, in the form a x (b x c)= (a. c) b — (a• b)c.
18. Vector and Scalar Fields.— The the ories of fluid motion, heat, electricity, etc., are largely concerned with vector and scalar fields. With every point (x, y, z) are associ ated such scalars as mass, density, potential, and such vectors as velocity, force, displace ment, flux, all of which are functions of (x.
19. Directional Derivative of a Scalar If a moving point P is displaced from (x, y, z) to a neighboring position in the direction ds, the rate of change of a scalar function V (x, y, z) per unit of displacement is given by the directional derivative dV aVdx OV dy 8Vds av ds ax ds ' ay ds ' az ds ax cos a a v , v a ,av ,av , .
— cos p — cos y =-- -r-j — ay aZ aX ay az (J cos a -1-j cos P + k cos y), in which the second factor is a unit vector in the direction of ds, whose direction angles are a, 0, y.
20. Tangential and Normal Derivatives to a Level Let the family of level sur faces for the function V have the equation V (x, y. z)C, where C is an arbitrary constant; first let the displacement ds be tangential to the level surface that passes through P, then dVids•=0, and the above scalar product is a zero, hence the vector 1 av av + j +)k — v is ax ay as perpendicular to ds, that is, to any tangent to the surface, and is, therefore, normal to the surface. Let N be the length of this normal vector, then 1V3=ax aY az '+ CZ) I and the direction cosines of this vector are toy lay i av o find the rate of change of V ./Ti ax' N ay' N az in this normal direction we have dn a v a ÷i a .0 cos a + j cos 3 + k cos y)= ax ay az av • j ay • (av av av) 1-- 4. j_ ax ay ax ' 'ay Ars , a ax , av , of vector — j — — az • ax ay ' hence this vector is not only normal to the sur face V.= C, but represents in magnitude the rate of change of V in that direction; this vector is called the gradient of V, and is de noted by V V.