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
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 1& -1& 2\end{array}\right|$
$=\hat{i}\left(-3\right)-\hat{j}\left(3\right)+\hat{k}\left(0\right)$
$=-3\hat{i}-3\hat{j}$
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 $\left|\frac{2+1}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}$