Question

The equation of the plane which contains the line of intersection of the planes $x+y+z-6=0$ and $2x+3y+z+5=0$ and perpendicular to the $xy$ plane is:

• A. $x-2y+11=0$
• B. $x+2y+11=0$
• C. $x+2y-11=0$
• D. $x-2y-11=0$

Answer 1

Answer: (B)

$x+2y+11=0$

Equation of a plane which passes through the line interaction of two planes is

$(a_{1}x+b_{1}y+c_{1}z+d_1)+\lambda (a_{2}x+b_{2}y+c_{2}z+d_2)$

$\therefore$ Our equation of plane will be

$(x+y+z-6)+\lambda (2x+3y+z+5)$

$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1+\lambda )z-6+5\lambda =0$ …..$\left(1\right)$

Now this plane is perpendicular to XY plane

i.e., $1+\lambda =0$

$\Rightarrow \lambda =-1$

$\therefore$ Required plane

$-x-2y-6-5=0$

$\Rightarrow x+2y+11=0$.