Question

Find the equation of the plane which is perpendiular to the plane $5x+3y+6z+8=0$ and which contains the line intersection of the plane $x+2y+3z-4=0$ and $2x+y-z+5=0$

Answer 1

Let the equation of a plane passes through the line line of intersection of the planes

$x+2y+3z-4=0$ __(1)

$2x+y-z+5=0$__(2)

So, $(x+2y+3z-4)+\lambda (2x+y-z+5)=0$

$\Rightarrow x(1+2\lambda )+y(2+\lambda )+z(3-\lambda )+5\lambda -4=0$ __(3)

This is perpendicular plane

$5x+3y+6z+8=0$

Now,

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

Therefore , $5(1+2\lambda )+3(2+\lambda )+6(3-\lambda )=0$

on solving equation to,

$5+10\lambda +6+3\lambda +18-6\lambda =0$

$\Rightarrow 7\lambda +29=0$

Therefore $\lambda =-\frac{29}{7}$

put the value of f in equation (3) and we get

$x[1+2.(-\frac{29}{7})]+y[2+\left(\frac{-29}{7}\right)]+2[3-(-\frac{29}{7})]$

$-4+5(-\frac{29}{7})=0$

On simplifying we get

$-\frac{51}{7}x-\frac{15}{7}y+\frac{50}{7}z-\frac{173}{7}=0$

$\Rightarrow 51x+15y-50z+173=0$