Equations of Lines — Question 6

Question 6 – Solution

Strategy For closest points $P\in L_1$ and $Q\in L_2$, the connector $Q-P$ must be perpendicular to both line directions.

Classification The directions $\left\langle 1,2,0\right\rangle$ and $\left\langle 2,-1,1\right\rangle$ are not parallel. Solving the first two intersection equations gives $s=1/5$, $t=-2/5$, but the third coordinates are then $0$ and $13/5$, so the lines do not intersect. They are skew.

Closest points Let $$P=(s,2s,0),\qquad Q=(1+2t,-t,3+t).$$ The conditions $(Q-P)\cdot\left\langle 1,2,0\right\rangle=0$ and $(Q-P)\cdot\left\langle 2,-1,1\right\rangle=0$ yield $$s=\frac 15,\qquad t=-\frac 56.$$ Thus $$\boxed{P=\left(\tfrac 15,\tfrac 25,0\right)},\qquad \boxed{Q=\left(-\tfrac 23,\tfrac 56,\tfrac{13}{6}\right)}.$$

Distance and verification Since $Q-P=\tfrac{13}{30}\left\langle -2,1,5\right\rangle$, $$\boxed{d(L_1,L_2)=\frac{13\sqrt{30}}{30}=\frac{13}{\sqrt{30}}}.$$ The vector $\left\langle -2,1,5\right\rangle$ has zero dot product with both line directions.