Question

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

• A. 0
• B. 1
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is D $2\sqrt{2}$

The plane is $\perp$ to the planes $2x-2y+z+2=0$ and $x-y+2z=4$ normal vector $\perp$ to the plane is given by

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 1& -1& 2\end{array}\right|=\hat{i}(-4+1)-\hat{j}(4-1)+\hat{k}(-2+2)=-3\hat{i}-3\hat{j}$

and the plane passes through $(1,-2,1)$

$\therefore$ Equation of plane is

$-3(x-1)-3(y+2)+0(z-1)=0\Rightarrow -3x+3-3y-6=0\Rightarrow 3x+3y+3=0\Rightarrow x+y+1=0$

Distance of $(1,2,2)$ from the plane $x+y+1=0$ is

$\frac{1+2+1}{\sqrt{1+1}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$units