Question

A line having direction ratios $3,4,5$ cuts $2$ planes $2x-3y+6z-12=0$ and $2x-3y+6z+2=0$ at point P & Q, then Find length of PQ

• A. $\frac{35\sqrt{2}}{12}$
• B. $\frac{35\sqrt{2}}{24}$
• C. $\frac{35\sqrt{2}}{6}$
• D. $\frac{35\sqrt{2}}{8}$

Answer 1

The correct option is A $\frac{35\sqrt{2}}{12}$

$3r,4r,5r$ be the point that cuts the two plane.

$\therefore$ for plane $1$, $2x+6z-3y-12=0$

$\Rightarrow 6r+30r-12r-12=0$

$\Rightarrow r=\frac{1}{2}$

for plane $2$,

$2x-3y+6z+2=0$

$\Rightarrow 6r-12r+30r+2=0$

$\Rightarrow r=\frac{-1}{12}$

point of intersection at plane $1$ $\Rightarrow P(\frac{3}{2},2,\frac{5}{2})$

point of intersection at plane $2$ $\Rightarrow Q(\frac{-3}{12},\frac{-1}{3},\frac{-5}{12})$

$\therefore$ distance $\left(PQ\right)$ $==\sqrt{(\frac{3}{2}+\frac{3}{12}{)}^2+(2+\frac{1}{3}{)}^2+(\frac{5}{2}+\frac{5}{12}{)}^2}=\frac{35\sqrt{2}}{12}$

Ans $A$