Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 11 / c

Let $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ and $\mathbf{w}={\mathbf{v}}_1\times {\mathbf{v}}_2$. The assumed linear independence gives $\mathbf{w}\ne 0$. Every displacement joining one line to the other isIts dot product with $\mathbf{w}$ is always $\mathbf{d}\cdot \mathbf{w}$. The component in that direction is therefore fixed, so orthogonal decomposition gives $\Vert \mathbf{s}\Vert \ge |\mathbf{d}\cdot \mathbf{w}|/\Vert \mathbf{w}\Vert$.

The two linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$ span the plane perpendicular to $\mathbf{w}$. Suitable ${\lambda }_1,{\lambda }_2$ cancel the component of $\mathbf{d}$ in that plane, attaining the lower bound. Thus the distance between skew lines formula, also valid when the lines intersect, gives

The nonzero direction vectors now have linear dependence, so the lines are parallel or coincident. Subtract from $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ its orthogonal projection along ${\mathbf{v}}_1$; the remaining component is the shortest joining displacement. The distance between parallel lines isThe second expression follows from the relation between the cross product and the perpendicular component. It includes distance zero for coincident lines.

For the stipulated linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$, the denominator in part (i) is nonzero. ThusEquivalently, ${\mathbf{a}}_2-{\mathbf{a}}_1$ belongs to the linear span of ${\mathbf{v}}_1,{\mathbf{v}}_2$. Geometrically the lines lie in one plane, and, since their direction vectors are not parallel, they intersect at a unique point. The linear combination expressing ${\mathbf{a}}_2-{\mathbf{a}}_1$ gives the two parameters of that intersection.