 Solution

For nonparallel direction vectors, let $\mathbf{n}=(\mathbf{b}\times \mathbf{d})/|\mathbf{b}\times \mathbf{d}|$. Every joining vector has the same normal component, $\mathbf{R}\cdot \mathbf{n}=(\mathbf{a}-\mathbf{c})\cdot \mathbf{n}$. Its norm is therefore at least the absolute value of that component. Minimizing $|\mathbf{R}{|}^2$ makes it perpendicular to both directions, and hence parallel to $\mathbf{n}$, attaining the bound:The expression printed without absolute values is an oriented separation, not a nonnegative distance. Reversing either line's parametrization can flip its sign. For parallel nonzero directions, this normal is undefined and the distance is instead $|(\mathbf{a}-\mathbf{c})\times \mathbf{b}|/|\mathbf{b}|$.