Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 1

Three vectors in three-dimensional Euclidean space are non-coplanar precisely when their scalar triple product is nonzero. Put $\Delta =a\cdot (b\times c)\ne 0$. In particular, $a\ne 0$. The identity for cross products with a common vector givesFor completeness, this follows from the vector triple product identity: writing $p=a\times b$ gives $p\times (a\times c)=a(p\cdot c)-c(p\cdot a)$, with $p\cdot c=\Delta$ and $p\cdot a=0$. Therefore the new scalar triple product isThus $a,a\times b,a\times c$ are non-coplanar, or equivalently linearly independent.

Take the dot product of the equation with $a$. Since $a\cdot (a\times r)=0$, we obtain $a\cdot r=(a\cdot c)/\lambda$. Next take its cross product with $a$ and apply the vector triple product identity:Substitute $a\times r=c-\lambda r$ and the known dot product, then rearrange:Here the vectors and $\lambda \ne 0$ are real, so the denominator is positive. The formula also covers $a=0$, giving $r=c/\lambda$. To check existence as well as uniqueness, write its numerator as $N$. The vector triple product identity gives $a\times N=\lambda a\times c-a(a\cdot c)+|a{|}^{2}c$, so $a\times N+\lambda N=({\lambda }^2+|a{|}^2)c$. Substitution verifies the original equation. This is an instance of the inverse of a scalar plus a cross-product operator.