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

Write $\times$ for the paper's $\wedge$. In a right-handed Cartesian coordinate system, the cross product isIt is perpendicular to both vectors, has norm $|\mathbf{a}||\mathbf{b}|\sin \varphi$, where $\varphi$ is their angle, and has direction given by the right-hand rule when nonzero. Its norm is the area of the parallelogram spanned by the two vectors.

If $\mathbf{b}\times \mathbf{c}\ne 0$, use that parallelogram as the base of the parallelepiped. Its perpendicular height is $|\mathbf{a}\cdot \mathbf{n}|$, where $\mathbf{n}=(\mathbf{b}\times \mathbf{c})/|\mathbf{b}\times \mathbf{c}|$. Thus its volume isIf $\mathbf{b}\times \mathbf{c}=0$, both the volume and this scalar triple product vanish. If the cross product is nonzero, the scalar triple product vanishes exactly when $\mathbf{a}$ lies in the plane spanned by $\mathbf{b},\mathbf{c}$. If it is zero, $\mathbf{b},\mathbf{c}$ are linearly dependent, and adjoining $\mathbf{a}$ still spans at most a plane. Consequently the three vectors are coplanar if and only if $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$, including cases involving a zero vector.

Expand the first component using the definition of the cross product:The added $a_1{b}_1{c}_1$ terms cancel in the last step. Cyclically permuting the component indices gives the other two components, proving the vector triple product identity

Put $V=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$. The vector triple product identity givesTaking its dot product with $\mathbf{a}\times \mathbf{b}$ and using cyclic invariance of the scalar triple product yieldsThere was no division by $V$, so this also proves the identity for coplanar vectors.

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 $-\mathbf{a}$.