How do you find the range of the scalar triple product of the vectors <cos alpha, cos beta, 0>, <0, cos beta, cos gamma> and <cos alpha, 0, cos gamma>?

1 Answer
Aug 19, 2016

[-2, 2]

Explanation:

The triple scalar product of vectors vec(u), vec(v), vec(w) in RR^3 is given by vec(u)*(vec(v)xxvec(w)). A useful fact is that this can also be calculated as the determinant of a 3xx3 matrix with the given vectors as the rows or columns. That is,

vec(u)*(vec(v)xxvec(w)) = det((u_1, u_2, u_3), (v_1, v_2, v_3), (w_1, w_2, w_3))

We will use that to calculate the given scalar triple product.

Let
vec(u) = < cos(alpha), cos(beta), 0 >
vec(v) = < 0, cos(beta), cos(gamma) >
vec(w) = < cos(alpha), 0, cos(gamma) >

Then we have:

vec(u)*(vec(v)xxvec(w)) = det((u_1, u_2, u_3), (v_1, v_2, v_3), (w_1, w_2, w_3))

=det((cos(alpha),cos(beta),0),(0,cos(beta),cos(gamma)),(cos(alpha),0,cos(gamma)))

=cos(alpha)(cos(beta)cos(gamma)-0cos(gamma))

-cos(beta)(0cos(gamma)-cos(alpha)cos(gamma))

+0(0*0-cos(alpha)cos(beta))

=2cos(alpha)cos(beta)cos(gamma)

Thus, it suffices to find the range of 2cos(alpha)cos(beta)cos(gamma).

With no restrictions on alpha, beta, gamma beyond being in RR, we have cos(alpha), cos(beta), cos(gamma) in [-1, 1].
Thus their product will also have the range cos(alpha)cos(beta)cos(gamma) in [-1, 1].
Multiplying by 2 gives our desired range:
2cos(alpha)cos(beta)cos(gamma) in [-2, 2]

Thus the range of the scalar triple product of < cos(alpha), cos(beta), 0 >, < 0, cos(beta), cos(gamma) >, and < cos(alpha), 0, cos(gamma) >, is [-2, 2]