Question

Find the equation of the plane though the points $(1,-1,2)$ and $(2,-2,2)$ and is perpendicular to the plane $6x-2y+2z=9$.

Answer 1

Vector equation of the line joining the points ($1,—1,2$) and ($2,—2,2$) is

$\begin{array}{c}\overrightarrow{a}=\left(2-1\right)\hat{i}+\left(-2-1\right)\hat{j}+\left(2-2\right)\hat{k}\\ =\hat{i}-\hat{j}\end{array}$

Given equation of the plane is

$\begin{array}{c}6x-2y+2z=0\\ \overrightarrow{n}=6\hat{i}-2\hat{j}+2\hat{k}\end{array}$

Equation of the plane passing through $\overrightarrow{a}$ and normal to $\overrightarrow{n}$ is $\left(\overrightarrow{r}-\overrightarrow{a}\right)\cdot \overrightarrow{n}=0$ i.e. $\overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{a}\cdot \overrightarrow{n}$

Hence the vector equation of the plane $\overrightarrow{r}\cdot \left(6\hat{i}-2\hat{j}+2\hat{k}\right)=6+2=8$

The Cartesian form of equation is $\left(x\hat{i}+y\hat{j}+z\hat{k}\right)\cdot \left(6\hat{i}-2\hat{j}+2\hat{k}\right)=8$

$6x-2y+2z=8$ is the required equation of the plane.