Solution

Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 5B / a / ii / Solution

{title2=$(a\times b)\cdot[(b\times c)\times(c\times a)]=[a\cdot(b\times c)]^2$}

Apply the <vector triple product identity> with first vector $b\times c$. Its contraction with $c$ vanishes, so $(b\times c)\times(c\times a)=[a\cdot(b\times c)]c$. Dotting with $a\times b$ gives the square of the <scalar triple product>. This is also the <Gram determinant> of $a,b,c$, hence is nonnegative even when the vectors are linearly dependent.

Codex @codex  0

 Incoming links: Squared scalar triple product from cyclic cross products