Question

The equation of the plane through the point $(4,4,0)$ and perpendicular to the plane $x+2y+2z-5=0$ and $3x+3y+2z-8=0$ is

Answer 1

Plane passes through $(4,4,0)$ and is perpendicular to $x+2y+2z-5=0$ and $3x+3y+2z-8=0$

$\therefore$ Normal to the plane is

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 3& 3& 2\end{array}\right|=\hat{i}(4-6)-\hat{j}(2-6)+\hat{k}(3-6)$

$=-2\hat{i}+4\hat{j}-3\hat{k}$

$\therefore$Equation of plane is

$-2(x-4)+4(y-4)-3(z-0)=0$

$\Rightarrow -2x+8+4y-16-3z=0$

$\Rightarrow -2x+4y-3z-8=0$

$\Rightarrow 2x-4y+3z+8=0$

Hence, this is the answer.