Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 9

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$ be the scalar triple product. The reciprocal basis satisfies $\mathbf{a}\cdot {\mathbf{a}}^{\prime }=1$ and all mismatched dot products vanish, with cyclic analogues. Apply the given vector triple product identity to obtainDotting with ${\mathbf{a}}^{\prime }$ givesThe sign is retained: a negatively oriented original basis has a negatively oriented reciprocal basis. Equivalently, if the original vectors are columns of $B$, their reciprocals are columns of $B^{-T}$, whose determinant is $1/\det B$.

The denominator for taking reciprocals a second time is $D^{\prime }=1/D$. The preceding cross-product result therefore givesCyclic permutation gives $\mathbf{b}$ and $\mathbf{c}$. Thus taking the reciprocal basis twice returns the original basis. The matrix statement is the same reciprocal basis involution: $(B^{-T}{)}^{-T}=B$. Noncoplanarity is essential, since otherwise the defining denominator and the matrix inverse do not exist.

Dot the reciprocal-basis expansion with each member of the original basis. All mismatched dot products vanish and the matching one equals one. ConsequentlyThese are coefficients in the reciprocal basis. They would instead involve dot products with the reciprocal vectors if $\mathbf{r}$ were being expanded in the original basis. The two coefficient conventions should not be interchanged for a nonorthonormal basis.

Use the preceding coefficient identities with the three prescribed plane equations. They immediately giveFor example, its dot product with $\mathbf{a}$ is $s$, because ${\mathbf{a}}^{\prime }\cdot \mathbf{a}=1$ while the other two terms give zero; the other equations follow cyclically. This verifies existence. If two points satisfied all three equations, their difference would be perpendicular to all three members of a basis and therefore zero. Hence the intersection is unique, not merely one possible intersection point.