Question

A plane passes through $\left(1 , - 2 , 1\right)$ and is perpendicular to two planes $2x-2y+z=0$ and $x-y+2z=4.$ The distance of the plane from the point $\left(0 , 2 , 2\right)$ is

• A. $\frac{3}{\sqrt{2}}$ units
• B. $4\sqrt{2}$ units
• C. $3\sqrt{2}$ units
• D. $2\sqrt{2}$ units

Answer 1

Answer: (A)

Normal vector perpendicular to the required plane is
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix}$
$=\hat{i}\left(- 3\right)-\hat{j}\left(3\right)+\hat{k}\left(0\right)$
$=-3\hat{i}-3\hat{j}$
Equation of the plane through $\left(1 , - 2 , 1\right)$ is
$1\left(x - 1\right)+1\left(y + 2\right)+0\left(z - 1\right)=0$
$x+y+1=0$
Distance from $\left(0 , 2 , 2\right)$ is $\left|\frac{2 + 1}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}$