Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV resit / 1

The scalar product, also called the dot product, and the vector product, also called the cross product, areFor nonzero vectors, let $\vartheta \in [0,\pi ]$ be their angle. Their Euclidean norms satisfyThe cross product is orthogonal to both vectors, with its orientation determined by the right-hand rule. If either vector is zero, both products vanish, although the angle is undefined.

A real square matrix $A$ is an orthogonal matrix whenHere $A^T$ is the transpose and $I$ the identity matrix; equivalently, its columns form an orthonormal basis. If this holds, the squared Euclidean norm obeysso the linear transformation preserves length.

Conversely, suppose $|A\mathbf{v}|=|\mathbf{v}|$ for every vector $\mathbf{v}$. Put $B=A^{T}A-I$, a symmetric matrix. Then ${\mathbf{v}}^{T}B\mathbf{v}=0$ for every vector. Taking $\mathbf{v}={\mathbf{e}}_i$, the standard basis vectors, gives $B_{ii}=0$. Taking $\mathbf{v}={\mathbf{e}}_i+{\mathbf{e}}_j$ for $i\ne j$ gives $2B_{ij}=0$. Thus $B=0$, proving that the linear transformation preserves length if and only if $A$ is an orthogonal matrix.

Choose the first orthonormal basis vector to be ${\mathbf{e}}_1=\mathbf{a}$. If $\mathbf{b}$ is not parallel to $\mathbf{a}$, choose ${\mathbf{e}}_2$ along $\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$; otherwise choose any unit vector orthogonal to $\mathbf{a}$. Complete to an orthonormal basis with ${\mathbf{e}}_3$. The matrix with rows ${\mathbf{e}}_i^T$ is an orthogonal matrix, and expressing the three unit vectors in this basis givesThe last two components of ${\mathbf{c}}^{\prime }$ have polar coordinates of the displayed form; we may choose $w\ge 0$. If they vanish, $\theta$ is arbitrary.

An orthogonal transformation preserves dot products. Taking dot products with ${\mathbf{a}}^{\prime }$, and then using the unit Euclidean norms, givesThe remaining dot product givesAn orthogonal matrix has determinant $\pm 1$, so it preserves the absolute scalar triple product. In the new coordinates this scalar triple product is $yw\sin \theta$. ConsequentlyTaking the nonnegative square root provesNo division by $y$ or $w$ was used, so the proof includes parallel vectors. Equivalently, the expression under the square root is the Gram determinant of the three unit vectors.