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

The correct option is B $x+2y+11=0$

Equation of the plane passing through the line of intersection of planes $x+y+z-6=0$ and $2x+3y+z+5=0$ is
$(x+y+z-6)+\lambda (2x+3y+z+5)=0$
$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1+\lambda )z+(-6+5\lambda )=0$
Dr's of normal to this plane $\equiv (1+2\lambda ,1+3\lambda ,1+\lambda )$
Dr's of normal to $xy-$plane $\equiv (0,0,1)$
By condition of perpendicularity,
$(1+2\lambda )\cdot 0+(1+3\lambda )\cdot 0+(1+\lambda )\cdot 1=0\Rightarrow \lambda =-1$

$\therefore$ Equation of required plane is
$\Rightarrow (1-2)x+(1-3)y+(1-1)z+(-6-5)=0$
$\Rightarrow x+2y+11=0.$