Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 5A (Vectors and Matrices)

Both sides are antisymmetric in $j,k$ and in $p,q$. If either pair has equal indices, both sides vanish. It therefore suffices to take $j\ne k$ and $p\ne q$. The sum over $i$ is nonzero exactly when $(p,q)=(j,k)$ or $(k,j)$; it is respectively $1$ or $-1$. This is exactlythe contraction of two Levi-Civita symbols.

Expanding a sum of squares gives the $n$-dimensional Lagrange identity for the cross product:The left-hand side is nonnegative, so

In components, part (i) givesTherefore the vector triple product is

PutSince $x,y,z$ are unit vectors,The feasible inner products with two unit vectors satisfyThe desired pair $p=q=1/2$ is feasible exactly when $c\ge -1/2$. In that case every maximizer has these inner products. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point isUsing the vector triple product,Hence

Rotations preserve dot products and cross-product lengths, while $z^{\prime }=2z$. ThusLetThe target pair is feasible precisely whenorOn this interval,

Outside this interval, let $\lambda >0$ be the unique number for whichand write $r=(p,q{)}^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing inner products solely from $c=y\cdot z$. In the remaining cases,Neither answer depends on ${\theta }_1$ or ${\theta }_2$.