Question

The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ is

• A. $2\sqrt{3}$
• B. $4\sqrt{3}$
• C. $3\sqrt{6}$
• D. $5\sqrt{6}$

Answer 1

The correct option is B $4\sqrt{3}$

Let DC’s of shortest distance line are l, m, n which is perpendicular to both the given lines
$\therefore$ $2l-7m+5n=0$ ......(i)
and 2l+m-3n=0 .....(ii)

$\therefore \frac{l}{16}=\frac{m}{16}=\frac{n}{16}\Rightarrow \frac{l}{1}=\frac{m}{1}=\frac{n}{1}=\frac{\sqrt{(l^2+m^2+n^2)}}{\sqrt{(1^2+1^2+1^2)}}=\frac{1}{\sqrt{3}}\Rightarrow l=\frac{1}{\sqrt{3}},m=\frac{1}{\sqrt{3}},n=\frac{1}{\sqrt{3}}$

$\therefore$ Required shortest distance = Projection of PQ on RS
$=|(3-(-1\left)\right)l+(-15-1)m+(9-9)n|=|4l-16m|=|\frac{-12}{\sqrt{3}}|=4\sqrt{3}$