Question

Find the equation of the plane containing the point $(2,4.-1)$ and perpendicular to each of the plane $3x-4y+2z-7=0$ and $x+2y-3z+2=0.$

Answer 1

Given planes are :

$3x-4y+2z-7=0$ and

$x+2y-3z+2=0$

normal of the planes are

${\overline{n}}_1=3\hat{i}-4\hat{j}+2\hat{k}$ and

${\overrightarrow{n}}_2=\hat{i}+2\hat{j}-3\hat{k}$

The plane is perpendicular to both these planes.

$\Rightarrow$ its normal will also be perpendicular to both the normal.

i.e. $\overrightarrow{n}={\overrightarrow{n}}_1\times {\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -4& 2\\ 1& 2& -3\end{array}\right|$

$=(12-4)\hat{i}+(-11)(-\hat{j})+10\hat{k}$

$=8\hat{i}+11\hat{j}+10\hat{k}$

$\Rightarrow$ Equation of the plane is given by $8x+11y+10z=k$

it pass through the point $(2,4,-1)$

$\Rightarrow 8\left(2\right)+11(-4)+10(-1)=k$

$\Rightarrow k=16+44-10=60-10=50$

$\Rightarrow$ plane is $8x+11y+10z=50$