Question

The equations 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. $17x+8y-11z+13=0$
• C. $x+y+z-2=0$
• D. $x-5y-3z=0$

Answer 1

The correct option is B $17x+8y-11z+13=0$

Solution - Position vectors through which plane is

passing

$\begin{array}{c}\overrightarrow{a}=\hat{ı}-\hat{ȷ}+2\hat{k};\overrightarrow{b}=-3\hat{ı}+2\hat{ȷ}-2\hat{k}\\ \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=-4\hat{ı}+3\hat{j}-4\hat{k}\end{array}$

Equation of given plane is $x+2y+3z+7=0$

$\begin{array}{c}\Rightarrow \overrightarrow{n_1}=\hat{ı}+2\hat{ȷ}+3\hat{k}\\ \overrightarrow{n}={\overrightarrow{n}}_1\times \overrightarrow{AB}=\left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& \hat{k}\\ 1& 2& 3\\ -4& 3& -4\end{array}\right|=(-17)\hat{ı}-8\hat{j}+11\hat{k}\end{array}$

Equation of required plane

$(\overrightarrow{r}-\overrightarrow{a})\cdot \overrightarrow{n}=0$ $\begin{array}{c}\Rightarrow (x\hat{ı}+y\hat{ȷ}+z\hat{k}-\hat{ı}+\hat{ȷ}-2\hat{k})\cdot (-17\hat{ı}-18\hat{j}+11\hat{k})=0\\ \Rightarrow \left(\right(x-1)\hat{ı}+(y+1)\hat{ȷ}+(z-2\left)\hat{k}\right)\cdot (-17\hat{ı}-18\hat{ȷ}+11\hat{k})=0\end{array}$ $\Rightarrow -17x-08y+11z+17-D8+(-22)=0$ $\Rightarrow 17x+By+11z+13=0$

Hence B is the correct option