Past exam of the mathematical methods I course of the University of Oxford / 2017 resit / 8 / c

For nonzero vectors with angle $\theta$, $|\mathbf{a}\times \mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin \theta$ and $\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos \theta$. Hence ${\sin }^2\theta =1-{\cos }^2\theta$ gives Lagrange's identityIf either vector is zero both sides vanish, so this covers every case. Alternatively, expansion of the three cross product components gives the same identity without introducing an angle.

For the quadratic form of a squared cross product, express the two scalar terms as quadratic forms in $\mathbf{a}$:Then ${\mathbf{a}}^{T}M\mathbf{a}=|\mathbf{b}{|}^2|\mathbf{a}{|}^2-(\mathbf{a}\cdot \mathbf{b}{)}^2$. In particular, for $\mathbf{b}=(b_1,b_2,0)$,The cancellation is structural: $M\mathbf{b}=0$, while its action on every vector perpendicular to a nonzero $\mathbf{b}$ is multiplication by $|\mathbf{b}{|}^2$. Its eigenvalues are therefore $0,|\mathbf{b}{|}^2,|\mathbf{b}{|}^2$; for $\mathbf{b}=0$ the whole matrix is zero.