Distance between parallel lines ($d=\Vert \mathbf{d}\times \mathbf{v}\Vert /\Vert \mathbf{v}\Vert$)

For two parallel lines ${\mathbf{a}}_1+s\mathbf{v}$ and ${\mathbf{a}}_2+t\mathbf{v}$ in three-dimensional Euclidean space, with $\mathbf{v}\ne 0$ and $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$, their shortest distance isEvery joining displacement is $\mathbf{d}+(t-s)\mathbf{v}$. Its component perpendicular to $\mathbf{v}$ is fixed, and its parallel component can be cancelled by a suitable $t-s$. The Pythagorean theorem proves that this perpendicular component has the shortest Euclidean norm. The cross product expression follows from $\Vert \mathbf{d}\times \mathbf{v}\Vert =\Vert \mathbf{d}\Vert \Vert \mathbf{v}\Vert \sin \theta$. For coincident lines the distance is zero. The perpendicular-component formula also applies in any finite-dimensional Euclidean space, where a three-dimensional cross product need not be available.