Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 1 / 6C / Solution

A positively oriented planar rotation matrix acts byFor a unit vector $n$ in three dimensions, the parallel and perpendicular orthogonal projections areOn the plane perpendicular to $n$, the map $v\mapsto n\times v$ preserves lengths and turns vectors through a positive right angle about $n$. It squares to minus the identity there. Thus $\cos \theta x_{\perp }+\sin \theta (n\times x_{\perp })$ rotates that component, while the parallel component remains fixed. This proves the Rodrigues rotation formulaIts component form uses the Levi-Civita symbol:Since $n_i{n}_i=1$ and ${ϵ}_{iik}=0$, its matrix trace is $3\cos \theta +1-\cos \theta =1+2\cos \theta$.

For the given matrix, the matrix trace is $1+2r$ with $r=1/\sqrt{2}$, so $\cos \theta =1/\sqrt{2}$. In the stated angular range, $\theta =\pm \pi /4$. The antisymmetric entries giveTherefore the two axis-angle pairs areReversing the axis and the angle leaves the Rodrigues rotation formula unchanged, so these describe the same rotation.