Question

A line from the origin meets the lines $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}$ at $P$ and $Q$ respectively. If length $PQ=d$, then $d^2$ is equal to

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is D $6$

Let the line $\frac{X}{a}=\frac{Y}{b}=\frac{Z}{c}$ intersects the given lines

$\Rightarrow$ S.D $=0$
$\therefore a+3b+5c=0$ and $3a+b-5c=0$
$\Rightarrow a:b:c=5r:-5r:2r$
On solving with the given lines, we get points intersection $P\equiv (5,-5,2)$

$Q\equiv (\frac{10}{3},-\frac{10}{3},\frac{8}{3})$

$\Rightarrow P{Q}^2=d^2=6$