Question

Find the distance of the point $2\hat{i}-\hat{j}-4\hat{k}$ from the plane $\overrightarrow{r}·\left(3\hat{i}-4\hat{j}+12\hat{k}\right)-9=0.$

Answer 1

$$\begin{gathered} \text{We know that the perpendicular distance of a point}\mathit{\text{P}}\text{of position vector}\overrightarrow{a}\text{from the plane}\overrightarrow{r}\text{.}\overrightarrow{n}\text{=}\mathit{\text{d}}\text{is given by} \\ p=\frac{\left|\overrightarrow{a}.\overrightarrow{n}-d\right|}{\left|\overrightarrow{n}\right|} \\ \text{Here,}\overrightarrow{a}=2\hat{i}-\hat{j}-4\hat{k};\overrightarrow{n}=3\hat{i}-4\hat{j}+12\hat{k};d=9 \\ \text{So, the required distance,}\mathit{\text{p}} \\ =\frac{\left|\left(2\hat{i}-\hat{j}-4\hat{k}\right).\left(3\hat{i}-4\hat{j}+12\hat{k}\right)-9\right|}{\left|3\hat{i}-4\hat{j}+12\hat{k}\right|} \\ =\frac{\left|6+4-48-9\right|}{\sqrt{9+16+144}} \\ =\frac{\left|-47\right|}{13} \\ =\frac{47}{13}\text{units} \end{gathered}$$