Calculus: Analytic Geometry and Calculus, with Vectors

106 Vectors and geometry in three dimensions

When ,r/2 < 0 < 7r, the scalar triple product is the negative of the volume of the
parallelepiped.
12 Prove that if i, j, k and i', j', k' are two right-handed orthonormal sets
of vectors, then the determinant of the coefficients in the system

i.' = alli + a12j + a13k

Y = a211 + a22] + a23k

k' = a31i + a32j + a33k

must be +1. Outline of solution: The hypotheses imply that j' X k' = i' and

hence that X k') = 1. But

x k') = (alli + ai2j + ai3k).

i j k

a21 a22 a23 = A,

I a31 a32 a33

where A is the determinant of the coefficients, and the result follows.

13 It is clear from geometrical considerations that if i, j, k and i', j', k' are

two right-handed orthonormal sets of vectors in E3 for which k' = k, then there

must be an angle 0 such that the "new" vectors i' and j' are related to the "old"

vectors i and j as in Figure 2.692. The new vectors and new z' axis and new y?

Figure 2.692

axis are dashed in the figure, and the vectors k and k' are not shown. It is easy

to see, with the aid of the figure, that

(1) i' = (cos 4,)i + (sin ¢)j

j' = -(sin 4,)i + (cos 4,)j.

Check up on this story by using (1) and vector methods to prove that, whatever

0 may be, it is actually true that 10 = 1, Ij'l = 1, i' X j' = k.

Solve the equations (1) for i and j to obtain the formulas

(2) i = (cos 4,)i' - (sin 4,)j'

j = (sin 4,)i' + (cos ¢)j'

which give the new vectors in terms of the old. Observe that changing the sign

of 0 converts one system of equations into the other. Finally, show that if P

lies in the plane of Figure 2.692 and if

(3) OP=xi+Yj=x'i'+Y]',

then

(4) OP = x'f(cos ¢)i + (sin ¢)j] + Y'[-(sin 4,)i + (cos O)jl

=[x'cos0-y'sin¢li+[x'sin0+y'cos4jj