Question

Find the equation of plane which is at a distance $\frac{4}{\sqrt{14}}$ from the origin and is normal to the vector $2\hat{i}+\hat{j}-3\hat{k}$.

Answer 1

Vector equation of a plane at a distance 'd' from the origin

and unit vector to normal from origin $\hat{n}$ is

$\overrightarrow{r}.\hat{n}=d$

unit vector of $\overrightarrow{n}=\hat{n}=\frac{1}{\left|\overrightarrow{n}\right|}\left(\overrightarrow{n}\right)$

Now, distance from origin = $d=\frac{4}{\sqrt{14}}$

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

Magnitude of $\hat{n}=\sqrt{2^2+1^2+(-3{)}^2}=\sqrt{4+1+9}=\sqrt{14}$

$\left|\right(\overrightarrow{n}\left)\right|=\sqrt{14}$

Now, $\hat{n}=\frac{1}{\left|\overrightarrow{n}\right|}\left(\overrightarrow{n}\right)$

$=\frac{1}{\sqrt{14}}(2\hat{i}+\hat{j}-3\hat{k})$

$=\frac{2}{\sqrt{14}}\hat{i}+\frac{1}{\sqrt{14}}\hat{j}-\frac{3}{\sqrt{14}}\hat{k}$

vector equation is

$\overrightarrow{r}.\hat{n}=d$

$\overrightarrow{r}.(\frac{2}{\sqrt{14}}\hat{i}+\frac{1}{\sqrt{14}}\hat{j}-\frac{3}{\sqrt{14}}\hat{k})=\frac{4}{\sqrt{14}}$

Therefore, vector equation of the plane is

$\overrightarrow{r}.(\frac{2}{\sqrt{14}}\hat{i}+\frac{1}{\sqrt{14}}\hat{j}-\frac{3}{\sqrt{14}}\hat{k})=\frac{4}{\sqrt{14}}$