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 plane passing through $(x_1,y_1,z_1)$ is given by

$A(x-x_1)+B(y-y_1)+C(z-z_1)=0$

where, A,B,C are the direction ratios of normal to the plane.

Now the plane passes through $(-1,3,2)$

So, equation of plane is

$A(x+1)+B(y-3)+C(z-2)=\mathrm{0.......}\left(1\right)$

Also, the plane is perpendicular to the given two planes.

So, their normal to plane would be perpendicular to normals of both planes.

We know that

$\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$

So, required normal is cross product of normals of plane $x+2y+3z=5$ and $3x+3y+z=0$

Required normal $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 3& 3& 1\end{array}\right|$

$=\hat{i}(2\times 1-3\times 3)-\hat{j}(1\times 1-3\times 3)+\hat{k}(1\times 3-3\times 2)$

$=\hat{i}(2-9)-\hat{j}(1-9)+\hat{k}(3-6)$

$=-7\hat{i}+8\hat{j}-3\hat{k}$

Hence, direction ratios $=(-7,8,-3)$

$\therefore A=-7,B=8,C=-3$

Putting above values in (1)

$A(x+1)+B(y-3)+C(z-2)=0$

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

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

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

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

Therefore equation of the required plane is $7x-8y+3z+25=0$