Question

Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is $2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+6\hat{\mathrm{k}}$.

Answer 1

Given:
$$\begin{gathered} \mathrm{Normal}\mathrm{vector},\hat{n}=2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+6\hat{\mathrm{k}} \\ \mathrm{Perpendicular}\mathrm{distance},d=5\mathrm{units} \end{gathered}$$

The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector $\hat{n}=2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+6\hat{\mathrm{k}}$ is as follows:

$\overrightarrow{r.}\hat{n}=d$
$\overrightarrow{r.}(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+6\hat{\mathrm{k}})=5$