Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 15

The vector triple product identity givesIt is the requested linear combination of $b$ and $c$. The notation $\wedge$ in this question denotes the three-dimensional cross product.

Expand the same double cross product in two ways. First expanding with $c\times d$ as the inner cross product givesAlternatively, use $(a\times b)\times w=(a\cdot w)b-(b\cdot w)a$ with $w=c\times d$ to getEquating these expressions and moving every term to one side provesThis four-vector scalar triple product identity is valid without assuming that any three of the four vectors are independent.

At an intersection, the two position vectors agree, so $a-b=tn-sm$. The scalar triple product with $m,n$ is thereforesince a determinant with repeated columns vanishes. Geometrically the displacement of the two anchor points must lie in the same plane as the two line directions.

This is a necessary condition. If $m,n$ are not parallel, it is also sufficient, because their span is exactly the plane perpendicular to $m\times n$. For parallel directions the triple product vanishes automatically, even for distinct lines, so the condition alone cannot guarantee intersection.

Put $D=[a,m,n]\ne 0$ and rewrite the intersection equation as $b=a+sm-tn$. Take scalar triple products with $a$ and $n$:Taking the triple product with $a$ and $m$ instead givesThus the intersection parameters from scalar triple products areThe assumed nonzero denominator makes these expressions valid and guarantees nonparallel directions. The resulting point is $r=a+sm=b+tn$; these formulas are conditional on the intersection assumed in the question.