# Find the equation of the plane passing through the point

Question:

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

Solution:

The equation of the plane passing through the point (−1, 3, 2) is

(x + 1) + b (y − 3) + c (z − 2) = 0 … (1)

where, abc are the direction ratios of normal to the plane.

It is known that two planes, $a_{1} x+b_{1} y+c_{1} z+d_{1}=0$ and $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$, are perpendicular, if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

Plane (1) is perpendicular to the plane, x + 2y + 3= 5

$\therefore a \cdot 1+b \cdot 2+c \cdot 3=0$

$\Rightarrow a+2 b+3 c=0$          $\cdots(2)$

Also, plane (1) is perpendicular to the plane, 3x + 3y + = 0

$\therefore a \cdot 3+b \cdot 3+c \cdot 1=0$

$\Rightarrow 3 a+3 b+c=0$              ...(3)

From equations (2) and (3), we obtain

$\frac{a}{2 \times 1-3 \times 3}=\frac{b}{3 \times 3-1 \times 1}=\frac{c}{1 \times 3-2 \times 3}$

$\Rightarrow \frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=k($ say $)$

$\Rightarrow a=-7 k, b=8 k, c=-3 k$

Substituting the values of ab, and c in equation (1), we obtain

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

$\Rightarrow(-7 x-7)+(8 y-24)-3 z+6=0$

$\Rightarrow-7 x+8 y-3 z-25=0$

$\Rightarrow 7 x-8 y+3 z+25=0$

This is the required equation of the plane.