Question

The equation of the plane equally inclined to the lines $L_1:\frac{x-1}{2}=\frac{y}{-2}=\frac{z+2}{-1}$ and $L_2:\frac{x+3}{8}=\frac{y-4}{1}=\frac{z}{-4}$ and passing through the origin is/are

• A. $14x-5y-7z=0$
• B. $2x+7y-z=0$
• C. $2x+7y+z=0$
• D. $14x+5y-7z=0$

Answer 1

Answer: (A), (B)

The correct options are
A $14x-5y-7z=0$
B $2x+7y-z=0$

Let the equation of the plane is $P:ax+by+cz=0$
Direction ratios of the angle bisector of the lines $L_1$ and $L_2$ is
$(\frac{2}{3},\frac{-2}{3},\frac{-1}{3})\pm (\frac{8}{9},\frac{1}{9},\frac{-4}{9})=(\frac{6\pm 8}{9},\frac{-6\pm 1}{9},\frac{-3\mp 4}{9})$
$\because$ It is parallel to the normal of the plane $P$,
$\therefore 14x-5y-7z=0\cdots \left(1\right)\text{or}2x+7y-z=0\cdots \left(2\right)$