Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 6 / d / Solution

The orthogonal change of coordinates preserves distance to the origin. On the required quadric surface,Thus the squared distance is at least $1/4$. Equality holds precisely when $u=v=0$, and the constraint then gives $w=\pm 1/2$. Using $e_w=(1,0,-3)/\sqrt{10}$ from part (b), the nearest points in the original coordinates areSubstitution checks the constraint directly: $5x^2-3z^2+6xz=(5-27-18)/40=-1$. The inequality proves these are global nearest points, rather than merely stationary candidates.