Past exam of the mathematical methods I course of the University of Oxford / 2021 / 8

The linear span is $\{\lambda \mathbf{a}+\mu \mathbf{b}:\lambda ,\mu \in \mathbb{R}\}$, a vector subspace of ${\mathbb{R}}^3$. Its dimension is zero if both vectors vanish, one if at least one is nonzero and they are linearly dependent, and two if they are linearly independent. The corresponding geometric sets are the origin, a line through the origin, and a plane through the origin.

Contract the Levi-Civita symbol identity ${\epsilon }_{ijk}{\epsilon }_{i\ell m}={\delta }_{j\ell }{\delta }_{km}-{\delta }_{jm}{\delta }_{k\ell }$. Using Einstein summation convention,This Lagrange identity for the cross product includes zero and parallel vectors without a separate division by their norms.

The set is the kernel of the linear map $L(\mathbf{r})=\mathbf{v}\cdot \mathbf{r}$. It contains zero, and linearity gives $L(\alpha \mathbf{r}+\beta \mathbf{s})=0$ whenever $L(\mathbf{r})=L(\mathbf{s})=0$, proving closure under linear combinations. If $\mathbf{v}\ne 0$, it is a plane through the origin; if $\mathbf{v}=0$, it is all of ${\mathbb{R}}^3$. An arbitrary plane need not be a vector subspace: for example $z=1$ omits the zero vector. An affine subspace and a vector subspace are not interchangeable.

The dimension formula for a sum of subspaces givesSince the sum lies in ${\mathbb{R}}^3$ and each plane has dimension two, the possibilities areThe first case means the planes coincide; two distinct planes through the origin intersect in a line.

Taking cross products of the two independent spanning vectors gives plane normal vectors proportional to $(-4,1,1)$ and $(4,-6,-1)$. Their cross product is $(5,0,20)$, so the intersection is the line $s\mathbf{d}$, with $\mathbf{d}=(1,0,4)$. The closest point to $\mathbf{p}=(1,1,1)$ is its orthogonal projection,The squared distance is $|\mathbf{p}{|}^2-(\mathbf{p}\cdot \mathbf{d}{)}^2/|\mathbf{d}{|}^2=3-25/17$, givingBoth plane normals are orthogonal to $\mathbf{d}$, and $(\mathbf{p}-{\mathbf{p}}_{\ast })\cdot \mathbf{d}=0$, checking the geometric constraints.