Question

Find the vector and Catesian equation of the plane that passes through the point $(1,0,-2)$ and the normal to the plane is $\hat{i}+\hat{j}-\hat{k}$

Answer 1

The position vector of point $(1,0,-2)$ is $\overrightarrow{n}$ perpendicular to the plane is $\overrightarrow{n}=\hat{i}+\hat{j}-\hat{k}$.
The vector equation of the plane is given by, $(\overrightarrow{r}-\overrightarrow{a})\cdot \overrightarrow{n}=0$

$\Rightarrow$ $[\overrightarrow{r}-(\hat{i}-2\hat{k}\left)\right].(\hat{i}+\hat{j}-\hat{k})=\mathrm{0.......}\left(1\right)$

The position vector of any point $P(x,y,z)$ in the plane is given by,
$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Therefore, equation (1) becomes
$[(x\hat{i}+y\hat{j}-z\hat{k})-(\hat{i}-2\hat{k})].(\hat{i}+\hat{j}-\hat{k})=0$

$\Rightarrow$ $\left[\right(x-1)\hat{i}+y\hat{j}+(z+2\left)\hat{k}\right].(\hat{i}+\hat{j}-\hat{k})=0$

$\Rightarrow$ $(x-1)+y-(z+2)=0$

$\Rightarrow$ $x+y-z-3=0$

$\Rightarrow$ $x+y-z=3$