For any three vectors a b c in 3D space prove that a times b

For any three vectors a, b, c in 3D space, prove that (a times b) middot [(b times c) times (c times a)] = [a middot (b times c)]^2

Solution

Let N = b X c,

then N X (c X a) = c(N . a) - a(N . c) (by vector triple product definition)

Therefore, (b X c) X (c X a) = c(b X c . a) - a(b X c . c)

= c(a . b X c) - a(b . c X c)

= c(a . b X c) (as we know that c X c = 0)

Now, (a X b) . (b X c) X (c X a) = (a X b) . c(a . b X c)

= (a X b . c)(a . b X c)

= (a . b X c)(a . b X c)

= (a . b X c)² (by cyclic property)