Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 5A (Vectors and Matrices)

$x\cdot y={\sum }_i{x}_i{y}_i$ and $\Vert x\Vert =(x\cdot x{)}^{1/2}$.

Cauchy-Schwarz inequality states $|x\cdot y|\le \Vert x\Vert \Vert y\Vert$. For $y\ne 0$, positivity of $\Vert x-ty{\Vert }^2$ at $t=(x\cdot y)/\Vert y{\Vert }^2$ gives $\Vert x{\Vert }^2-(x\cdot y{)}^2/\Vert y{\Vert }^2\ge 0$; $y=0$ is immediate.

$(b-a)\times (c-a)=(-1,0,0)\times (-1,-1,1)=(0,1,1)$. Thus one choice is $\hat{n}=(0,1,1)/\sqrt{2}$ and $\beta =\sqrt{2}$.

The distance is $|d\cdot \hat{n}-\beta |=|3/\sqrt{2}-\sqrt{2}|=1/\sqrt{2}$.

The centre-to-plane distance is $|g-2|/\sqrt{2}$. Intersection occurs iff this is at most $2\sqrt{2}$, i.e. $-2\le g\le 6$. Equality gives one tangent point; strict inequality gives a circle of radius $\sqrt{8-(g-2{)}^2/2}$.