Calculus: Analytic Geometry and Calculus, with Vectors

98 Vectors and geometry in three dimensions

involving the angle 0 and the unit normal (thumb vector) n of Figure

2.612. We now work out a convenient formula which gives u X v in

Figure 2.612

terms of the scalar components of u and v. Remembering that the

vector product of two vectors depends upon the order of the factors, we

shall be very careful. We have

u X v = (uii + u2j + u3k) X (vii + v2j + flak)

= uli x (v11 + 021 + 03k)

uj x (vii + vzj + vak)

u3k x(vii + vzj + v3k)

so

u X V= uivli X i + ulv2i X j+ ulvai x k

142011 X i + 142V,j X j+ u2vij x k

-I- uavlk x i + uav2k x j + u3v3k x k.

With the aid of the helpful fact about the vector product of two con-

secutive vectors in the parade ijkijk, given in (2.234), we obtain the

unlovely formula

(2.613) u X v = i(u03 - u3V2) - j(u103 - ua01) + k(u1v2 - 14201)

This looks better when we put it in the form

142 ua

02 va

-i^141 ua

V1 0a

+ k

141 142

01 02

Our next step is to allow ourselves the liberty of putting vectors into the

first row of a determinant so we may put this in the form

(2.62) U X V =

i j k

(^141) U2 ua
V1^02 0a
which is very easily remembered. It is the fashion to remember (2.62)
and to expand the determinant in terms of elements of the first row when-
ever this is desirable. When we must find the vector product of two
vectors u and v defined by
u=i-2j+3k, v=21-j-k,
all persons except typists and printers are happy to solve the problem