Equations of Lines — Question 6

Question 6 – Solution

Strategy The connector between closest points must be perpendicular to both line directions.

Closest points Let $P=(s,s,0)$ and $Q=(t,1-t,1)$. Requiring $(Q-P)\cdot\left\langle 1,1,0\right\rangle=0$ gives $1-2s=0$, so $s=1/2$. Requiring $(Q-P)\cdot\left\langle 1,-1,0\right\rangle=0$ gives $2t-1=0$, so $t=1/2$.

Distance Thus $P=(1/2,1/2,0)$ and $Q=(1/2,1/2,1)$, with $$\boxed{d(L_1,L_2)=|PQ|=1}.$$

Verification $Q-P=\left\langle 0,0,1\right\rangle$ is perpendicular to both direction vectors.