Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 1 / d

Let $V=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. The scalar triple product is invariant under a cyclic permutation, so the definitions of the reciprocal basis give

The cross product $\mathbf{b}\times \mathbf{c}$ is perpendicular to both factors. ThusCyclic permutations give all the off-diagonal dot products between the original basis and its reciprocal basis.

Apply the squared scalar triple product from cyclic cross products proved in c:Hence the reciprocal basis has scalar triple product $\boxed{{\mathbf{a}}^{\prime }\cdot ({\mathbf{b}}^{\prime }\times {\mathbf{c}}^{\prime })=1/V}$, with its sign retained.

Since $V\ne 0$, iii gives ${\mathbf{a}}^{\prime }\cdot ({\mathbf{b}}^{\prime }\times {\mathbf{c}}^{\prime })=1/V\ne 0$. The criterion in a therefore proves that ${\mathbf{a}}^{\prime },{\mathbf{b}}^{\prime },{\mathbf{c}}^{\prime }$ are non-coplanar. In particular they form a basis of ${\mathbb{R}}^3$.

The vector triple product identity givesEliminate $V$ with iii to obtain the reciprocal basis involution:This formula actually holds without the unit-length assumptions. Under those assumptions it also says $|{\mathbf{b}}^{\prime }\times {\mathbf{c}}^{\prime }|=1/|V|$; normalizing this cross product alone would lose the sign of $V$ and could return $-\mathbf{a}$.