Question

Find the equation of the plane passing through the point $(-1,3,2)$ and perpendicular to each of the planes $x+2y+3z=5$ and $3x+3y+z=0$.

Answer 1

The equation of the plane passing through $(-1,2,3)$ will be given by $a(x+1)+b(y-2)+c(z-3)=0$, where $a,b$ and $c$ are the direction ratios of the normal to the plane.

Also, this plane is perpendicular to the planes $x+2y+3z=5$ and $3x+3y+z=0$. According to the perpendicularity condition of planes, the dot product of the direction ratios of the normals to the two planes should be $0$.

$\Rightarrow a+2b+3c=0$ and $3a+3b+c=0$

$\Rightarrow c=-\frac{a+2b}{3}$

Substituting this in the second equation,

$3a+3b-\frac{a+2b}{3}=0$

$\Rightarrow 9a+9b-a-2b=0$

$\Rightarrow 8a+7b=0$

$\Rightarrow b=-\frac{8a}{7}$

$\Rightarrow c$ $=-\frac{a+2b}{3}$

$=-\frac{a-\frac{16a}{7}}{3}$

$=\frac{9a}{7\times 3}$

$=\frac{3a}{7}$

Substituting these values of $b$ and $c$ in the equation of the plane,

$a(x+1)-\frac{8a}{7}(y-2)+\frac{3a}{7}(z-3)=0$

$\Rightarrow 7(x+1)-8(y-2)+3(z-3)=0$

$\Rightarrow 7x-8y+3z+7+16-9=0$

$\Rightarrow 7x-8y+3z+14=0$

Hence, the equation of the required plane is $7x-8y+3z+14=0$