Question

Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ units from the origin and its normal vector from the origin is $2\hat{i}-3\hat{j}-4\hat{k}$

Answer 1

$d=\frac{6}{\sqrt{29}}$

Let $\overrightarrow{N}=2\hat{i}-3\hat{j}-4\hat{k}$

So, $\hat{n}=\frac{\overrightarrow{N}}{|\overrightarrow{N}|}=\frac{2\hat{i}-3\hat{j}-4\hat{k}}{\sqrt{29}}$

Therefore, the vector equation of the plane is

$\begin{array}{c}\overrightarrow{r}.\hat{n}=d\\ \overrightarrow{r}.\frac{2\hat{i}-3\hat{j}-4\hat{k}}{\sqrt{29}}=\frac{6}{\sqrt{29}}\\ \overrightarrow{r}.\left(2\hat{i}-3\hat{j}-4\hat{k}\right)=6\end{array}$