Closest approach of two uniformly moving particles ($t_{\ast }=-{\mathbf{d}}_0\cdot \mathbf{v}/|\mathbf{v}{|}^2$)

If the relative position is $\mathbf{d}(t)={\mathbf{d}}_0+t\mathbf{v}$ with $\mathbf{v}\ne 0$, completing the square givesThis proves the minimum distance over all real times. If time is restricted to an interval, project $t_{\ast }$ onto that interval; if $\mathbf{v}=0$, the distance is constant. Both particles are evaluated at the same time, unlike the two independent parameters in the distance between skew lines.