Question

Find the equation of the plane that contains the point (1,-1,2) and is perpendicular to both the planes 2x+3y-2z=5 and x+2y-3z=8.

Answer 1

Equation of plane containing the point $(1,-1,2)$ is

$a(x-1)+b(y+1)+c(z-2)=0$ .....(1)

Plane 1 is perpendicular to plane $2x+3y-2z=5$.

Therefore, $2a+3b-2c=0$ .......(2)

Also, plane 1 is perpendicular to plane $x+2y-3z=8$.

Therefore, $a+2b-3c=0$ ...........(3)

From equation 2 and 3, we get,

$\frac{a}{-9+4}=\frac{b}{-2+6}=\frac{c}{4-3}$

$\frac{a}{-5}=\frac{b}{4}=\frac{c}{1}=\lambda$

$a=-5\lambda ,b=4\lambda ,c=\lambda$

On putting these values in equation 1,

we get,

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

$-5(x-1)+4(y+1)+(z-2)=0$

$5x-4y-z-7=0$

This is the required equation of the plane.