Let $L_1$ be the line passing through the point $P_1=(4, −2, −3)$ with direction vector $\overrightarrow{d}=[−2, 1, 3]T$, and let $L_2$ be the line passing through the point $P_2=(−2, 3, −2)$ with the same direction vector.
Find the shortest distance d between these two lines, and find a point $Q_1$ on $L_1$ and a point $Q_2$ on $L_2$ so that d(Q1,Q2) = d. Use the square root symbol '√' where needed to give an exact value for your answer.
I tried $p_1p_2$, and projected to direction vector $d$, calculated distance wrong. I don't know how to solve the problem.
Best Answer
Notice, a parametric point lying on the line $L_1$ is $A_1(-2t_1+4, t_1-2, 3t_1-3)$ &
parametric point lying on the line $L_2$ is $A_2(-2t_2-2, t_2+3, 3t_2-2)$
Since, $\vec{A_1A_2}=[-2(t_2-t_1)-6, (t_2-t_1)+5, 3(t_2-t_1)+1 ]$ is perpendicular to the direction vector $\vec d=[-2, 1, 3]$ hence we have $(\vec{A_1A_2})\cdot \vec d=0$ (dot product of perpendicular vectors is zero) $$\implies [-2(t_2-t_1)-6, (t_2-t_1)+5, 3(t_2-t_1)+1]\cdot [-2, 1, 3]=0$$ $$14(t_2-t_1)+20=0$$$$\implies t_2-t_1=-\frac{10}{7}$$ setting the value of $t_2-t_1$, we get $$\vec{A_1A_2}=\left[\frac{-22}{7}, \frac{25}{7}, \frac{-23}{7} \right]$$
hence, the shortest distance between the parallel lines $L_1$ & $L_2$ $$=\left|\vec{A_1A_2}\right|=\sqrt{\left(\frac{-22}{7}\right)^2+\left(\frac{25}{7}\right)^2+\left(\frac{-23}{7}\right)^2}$$ $$=\frac{\sqrt{1638}}{7}$$