The equations of the lines of shortest distance between the lines $\frac{x}{2}=\frac{y}{-3}=\frac{z}

Question

The equations of the lines of shortest distance between the lines $\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$ and $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ are:

Professor PC Thomas Classes · Answer

Let $P\left(2 r_1,-3 r_1, r_1\right)$ and $Q\left(3 r_2+2,-5 r_2+1,2 r_2-2\right)$ be the points on the given lines so that PQ is the line of shortest distance
d.r.s of $P Q 2 r_1-3 r_2-2,-3 r_1+5 r_2-1, r_1-2 r_2+2$

Since it is perpendicular to given lines $2\left(2 r_1-3 r_2-2\right)-3\left(3 r_1+5 r_2-1\right)+\left(r_1-2 r_2+2\right)=0$ and $\left(2 r_1-3 r_2-2\right)-5\left(-3 r_1+5 r_2-1\right)+2\left(r_1-2 r_2+2\right)=0$
$\Rightarrow r_1=31 / 3, r_2=19 / 3 \Rightarrow P$ is $\left(\frac{62}{3},-31, \frac{31}{3}\right)$ and $Q$ is $\left(21, \frac{-92}{3}, \frac{32}{3}\right)$
d. $\mathrm{r} . s$ of PQ is $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$