calculus
posted by matt .
show that if a, b & c are in v3, then (a cross b) dot [(b cross c)×(c cross a)] = [a dot (b cross c)]^2

If you look for some list of identities for vector products, such as wolfram: cross product you will find the following:
A.(BxC) = det(ABC)
(AxB)x(CxD) = det(ABD)C  det(ABC)D
If we adapt this formula to our problem, we need to set
A = b
B = c
C = c
D = a
Now, we plug and chug:
bxc x cxa = det(bca)c  det(bcc)a
Now, a determinant is zero if two rows are the same. So, we can toss out he last term on the right, and we are left with just
bxc x cxa = det(bca)c
AxB.C = C.AxB so,
axb . det(bca)c = det(bca)c . axb
= det(bca) * det(cab)
When we swap columns, the detrminant changes sign. Swapping columns twice, we get
det(bca) * det(cab) = det(abc) * det(abc) = (a.bxc)^2