Past exam of the mathematical methods I course of the University of Oxford / 2017 / 9 / c / Solution

The vector triple product with $|u|=1$ givesThis is the orthogonal projection onto a plane perpendicular to $u$, so it is a linear map. Its image is the plane $u^{\perp }$ through the origin, while its kernel is the normal line $\mathrm{span}\{u\}$. In components,The equation $Ar=r$ is equivalent to $u(u\cdot r)=0$, hence to $u\cdot r=0$. Its solution set is precisely the two-dimensional plane $u^{\perp }$, with infinitely many nonzero vectors, not merely the normal line. These are the eigenvectors with eigenvalue one; the normal direction has eigenvalue zero.