Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics IV resit / 1

Use the vector triple product identitySince $\mathbf{n}$ is a unit vector, putThus $\mathbf{w}$ is the component of $\mathbf{v}$ in the orthogonal complement of $\mathbf{n}$, and $\mathbf{n}\cdot \mathbf{w}=0$. A cross product is orthogonal to both its factors, so $\mathbf{n}\cdot \mathbf{q}=0$ and $\mathbf{v}\cdot \mathbf{q}=0$. ConsequentlyAll three pairs are orthogonal. This remains true when $\mathbf{v}$ is parallel to $\mathbf{n}$: then $\mathbf{q}$ and $\mathbf{w}$ are zero vectors. Pairwise orthogonality does not assert linear independence in this case.

Use the convention $[\mathbf{a},\mathbf{b},\mathbf{c}]=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ for the scalar triple product. With $\mathbf{q},\mathbf{w}$ as above, $\mathbf{q}=\mathbf{n}\times \mathbf{w}$. The vector triple product identity givesThe Lagrange identity for the cross product, $|\mathbf{a}\times \mathbf{b}{|}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2-(\mathbf{a}\cdot \mathbf{b}{)}^2$, also givesTaking the dot product with the unit vector $\mathbf{n}$ proves

Decompose $\mathbf{v}$ by orthogonal projection onto the straight line spanned by $\mathbf{n}$ and its orthogonal complement:The first term is fixed by the rotation. If $\mathbf{w}\ne 0$, the vectors $\mathbf{w}/|\mathbf{w}|$ and $(\mathbf{n}\times \mathbf{w})/|\mathbf{w}|$ form an orthonormal basis of the perpendicular plane, oriented in the positive sense about $\mathbf{n}$. A planar rotation through $\theta$ therefore sendsHere $\mathbf{n}\times \mathbf{w}=\mathbf{n}\times \mathbf{v}$. Adding the fixed parallel component proves the Rodrigues rotation formula in the required form:When $\mathbf{w}=0$, both perpendicular terms vanish, so the same formula fixes $\mathbf{v}$ as required. Positive angles use the right-hand convention about $\mathbf{n}$.

The straight line $2x=y=z$ has direction vector $(1,2,2)$, so take the unit vector $\mathbf{n}=(1,2,2{)}^T/3$. A half-turn has $\cos \pi =-1$ and $\sin \pi =0$. The preceding rotation formula becomesThus the rotation matrix in the standard basis isIt fixes $(1,2,2{)}^T$ and negates every vector in its orthogonal complement, which also checks the half-turn geometrically.