Question

The equation of the plane through the points $(1,-1,2)$, $(-3,2,-2)$ and perpendicular to the plane $x+2y-3z+7=0$ is

• A. $x+16y+11z-7=0$
• B. $x+16y-11z+37=0$
• C. $x+y+z-2=0$
• D. $x-5y-3z=0$

Answer 1

The correct option is A $x+16y+11z-7=0$

$A(1,-1,2),B(1,2,-3)\overrightarrow{AB}=-4\hat{i}+3\hat{j}-4\hat{k}$

Plane $:x+2y-3z+7=0-\left(i\right)$

Directions of normal to plane $(1,2,-3)$

The new plane is perpendicular to $\left(i\right)$ and contains $\overrightarrow{AB}$.

$\therefore$ The normal of new plane is $\perp$ to both $\overrightarrow{AB}$ and normal of $\left(i\right)$,

$\therefore \overrightarrow{n}=(\hat{i}+2\hat{j}-3\hat{k})\times (-4\hat{i}+3\hat{j}-4\hat{k})\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -3\\ -4& 3& -4\end{array}\right|=\hat{i}+16\hat{j}+11\hat{k}$

$\therefore$ Equation of plane with normal $\overrightarrow{n}$ and point $(1,-1,2)$

$(x-1)+16(y+1)+11(z-2)=0$

$x+16y+11z=7$