Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 9 / Solution

The three vectors are coplanar precisely when their scalar triple product vanishes:This includes linearly dependent or zero vectors. It expresses that the determinant with those vectors as columns is zero.

For the two moving particles, distinguish closest approach at a common time from the geometric distance between their supporting lines. For fixed reference times, putTheir relative position is ${\mathbf{d}}_0+t\mathbf{v}$. If $\mathbf{v}\ne 0$, the closest approach of two uniformly moving particles occurs atIf $\mathbf{v}=0$, the distance is constant. If only $t\ge 0$ is permitted, replace $t_{\ast }$ by $\max (0,t_{\ast })$ before evaluating the distance.

Now suppose the velocities are nonparallel, and allow the relative starting-time offset to be chosen so that the particles reach the closest points of the two lines simultaneously. Write $\Delta ={\mathbf{a}}_2-{\mathbf{a}}_1$, $A=|{\mathbf{b}}_1{|}^2$, $B={\mathbf{b}}_1\cdot {\mathbf{b}}_2$, $C=|{\mathbf{b}}_2{|}^2$ and $D=AC-B^2>0$. The independent line parameters $s_1,s_2$ at the closest points satisfyThese equations make the connecting vector perpendicular to both velocities. Solving the two-by-two linear system givesThe corresponding distance between skew lines is $|\Delta \cdot ({\mathbf{b}}_1\times {\mathbf{b}}_2)|/|{\mathbf{b}}_1\times {\mathbf{b}}_2|$. Simultaneous arrival requires $t_1+s_1=t_2+s_2$, so $t_1-t_2=s_2-s_1$. Since the vector triple product givesthis condition is preciselyThus the printed formula has a valid interpretation as a synchronization condition for attaining the line-distance lower bound. It is not an identity that must hold for arbitrary fixed reference times at the time of closest approach.

For a counterexample to that stronger reading, take ${\mathbf{a}}_1=0$, ${\mathbf{a}}_2=(1,0,1)$, ${\mathbf{b}}_1=(1,0,0)$, ${\mathbf{b}}_2=(0,1,0)$ and $t_1=t_2=0$. The actual squared distance is $2(t-1/2{)}^2+3/2$, minimized at $t=1/2$. The printed right-hand side is $-1$, whereas the fixed difference $t_1-t_2$ is zero. The line-distance bound is one, but it is not attained by that synchronized motion. Parallel velocities also require separate treatment because the displayed denominator then vanishes.