Question

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>?`

Answer 1

`[-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]`