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

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