Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 1 / 5B / i / Solution

The first coordinate of the vector triple product isThe other two coordinates follow by cyclically permuting the indices, proving the identity.

A point on the line has the form $r=b+sm$. Substituting in the plane equation gives $s(m\cdot n)=(a-b)\cdot n$. When $m\cdot n\ne 0$ the parameter, and therefore the intersection point, is unique. The vector triple product gives $n\times (b\times m)=(n\cdot m)b-(n\cdot b)m$, so the unique intersection isFor the parallel case $m\cdot n=0$, the line is entirely in the plane if $(b-a)\cdot n=0$, and otherwise has no intersection. As usual, a genuine line and plane require nonzero direction and normal vectors. This is the line-plane intersection criterion.