Let the plane P:8x+ _1y+ _2z+12=0 be parallel to the line L:(x+2)/(2)=(y-3)/(3)=(z+4)/(5).

Question

Let the plane $P:8x+{\alpha }_{1}y+{\alpha }_{2}z+12=0$ be parallel to

the line $L:\frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the

intercept of $P$

on the $y$-axis is 1 , then the distance between $P$ and $L$ is :

Accepted Answer

P: $8x+{\alpha }_{1}y+{\alpha }_{2}z+12=0$

L: $\frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$

$\because P$ is parallel to $L$

$\Rightarrow 8(2)+{\alpha }_{1}(3)+5({\alpha }_{2})=0$

$\Rightarrow 3{\alpha }_{1}+5({\alpha }_{2})=-16$

Also $y$-intercept of plane $P$ is 1

$\Rightarrow {\alpha }_{1}=-12$

And ${\alpha }_{2}=4$

$\Rightarrow$ Equation of plane $P$ is $2x-3y+z+3=0$

$\Rightarrow$ Distance of line L from Plane $P$ is

$=|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}|$

$=\sqrt{14}$

Solutions · 1