Scalar triple product and the Determinate

Question

I noticed something when I was trying to solve a problem today. The scalar triple product is the same as the determinant or a three by three matrix with three vectors as rows:

A = [a, b, c]

det(A) = (a X b) * c

I came across this in Real Timer Rendering, and I can't really figure out why this is, or if its even useful. It seems sort of related to the short cut method of computing the cross product using a determinate where you write the unit vectors along the top of the matrix, but I always thought that was more of a mnemonic and not actually sound math.

Is there a real relationship here, or is this just some kind of happy coincidence?

Answer 1

No coincidence at all; this is a fairly standard result. Note that cross products a X b are often written in determinant form themselves with the top row being the unit vectors i j k, the next row being a1 a2 a3 and the bottom row being b1 b2 b3.

|i  j   k|
|a1 a2 a3|
|b1 b2 b3|

Now taking the dotproduct of that with another vector c, and you get the same thing as if you had just written c in the top row.

|i  j   k|                      |c1 c2 c3|     |c1 c2 c3|    |a1 a2 a3|
|a1 a2 a3| .  (c1,c2,c3)  =     |a1 a2 a3|  = -|a1 a2 a3|  = |b1 b2 b3|
|b1 b2 b3|                      |b1 b2 b3|     |b1 b2 b3|    |c1 c2 c3|

Edit: Also the wikipedia page for scalar triple product says that its equivalent to the determinant of the matrix using the vectors as rows or columns. Q.E.D.

Answer 2

Up to a sign, the determinant of an n-by-n matrix is the volume of the parallelepiped spanned by its n n-dimensional row (or column) vectors (or the volume of a unit cube linearly transformed by that matrix). The (axb).c product does, in three dimensions, exactly the same; axb gives a vector perpendicular to a and b and of length equal to the area of the parallelogram spanned by a and b; (axb).c gives the height of c over that parallelogram, times its area. So, no, it's no coincidence.