Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the normal vector of the required plane be $\vec{n}$. Since the plane is perpendicular to the given planes, $\vec{n}$ is the cross product of their normal vectors ${\vec{n}}_1$ and ${\vec{n}}_2$.

${\vec{n}}_1=2\hat{i}-2\hat{j}+\hat{k}$

${\vec{n}}_2=\hat{i}-\hat{j}+2\hat{k}$

$\vec{n}={\vec{n}}_1\times {\vec{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 1& -1& 2\end{array}\right|$

$\vec{n}=\hat{i}(-4+1)-\hat{j}(4-1)+\hat{k}(-2+2)=-3\hat{i}-3\hat{j}$

The direction ratios of the normal to the plane are proportional to $(1,1,0)$.

The equation of the plane passing through $(1,-2,1)$ with normal direction ratios $(1,1,0)$ is:

$1(x-1)+1(y-(-2))+0(z-1)=0$

$x-1+y+2=0$

$x+y+1=0$

The perpendicular distance of the point $(1,2,2)$ from the plane $x+y+1=0$ is:

$d=\frac{|1(1)+1(2)+1|}{\sqrt{1^2+1^2+0^2}}$

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

Answer: $2\sqrt{2}$