Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the normal vectors of the given planes be ${\vec{n}}_1$ and ${\vec{n}}_2$.

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

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

Since the required plane is perpendicular to both given planes, its normal vector $\vec{n}$ is parallel to the cross product of ${\vec{n}}_1$ and ${\vec{n}}_2$.

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

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

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

The equation of the plane passing through $(1,2,1)$ with normal vector $\hat{i}+\hat{j}$ is:

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

$x+y-3=0$

The perpendicular distance $d$ of the point $(2,3,4)$ from the plane $x+y-3=0$ is given by:

$d=\frac{|2+3-3|}{\sqrt{1^2+1^2+0^2}}$

$d=\frac{2}{\sqrt{2}}=\sqrt{2}$ units

Answer: $\sqrt{2}$ units