Equations of Lines — Question 3

Question 3 – Solution

Strategy At the closest point $Q$, the connector $P-Q$ is perpendicular to the line’s direction.

Perpendicular condition Write $Q=(1+2t,-1+t,2-2t)$. Then $$(P-Q)\cdot\left\langle 2,1,-2\right\rangle=0.$$ Since $P-Q=\left\langle 2-2t,2-t,-4+2t\right\rangle$, the equation is $14-9t=0$, so $t=14/9$.

Closest point and distance Thus $Q=\boxed{(37/9,5/9,-10/9)}$. Also $P-Q=\frac 19\left\langle-10,4,-8\right\rangle$, so $$\boxed{d(P,L)=\frac{\sqrt{180}}9=\frac{2\sqrt 5}{3}}.$$

Verification The dot product of $\left\langle-10,4,-8\right\rangle$ with $\left\langle 2,1,-2\right\rangle$ is zero.