Question

Find the equation of the plane passing through the straight line $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5}$ and perpendicular to the plane $x-y+z+2=0$.

Answer 1

Line$:\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5},plane;x-y+z=0\therefore \overrightarrow{a}=\hat{i}+(-2)\hat{j}+0\hat{k}$

is a point on the given line where direction is

$\overrightarrow{b}=2\hat{i}-3\hat{j}+5\hat{k}\therefore \overrightarrow{c}=\hat{i}-\hat{j}+\hat{k}$ is normal to given plane.

The new plane contains line and is $\perp$ to plane.

$\therefore$ Its normal $\perp$ to the direction line & also to normal of given plane, if $\overrightarrow{n}$ be the normal of required plane,

$\overrightarrow{n}=\overrightarrow{a}\times \overrightarrow{b}=(\hat{i}+(-2)\hat{j}+0\hat{k})\times (2\hat{i}-3\hat{j}+5\hat{k})=-2\hat{i}-3\hat{j}-\hat{k}$

d.r's of normal of new plane$v$

$\therefore$ Equation of plane through $(1,-2,0)$&d.r's of normal $(2,3,1)$ is,

$\Rightarrow 2(x-1)+3(y+2)+(z-0)=0\Rightarrow 2x+3y+z+4=0\to requiredequationofplane.$