Question

Find the distance of point $(2,-1,3)$ from the plane $\overrightarrow{r}.(3\hat{i}+2\hat{j}-6\hat{k})+15=0$

Answer 1

Equation of the given plane
$=\overrightarrow{r}.(3\hat{i}+2\hat{j}-6\hat{k})+15=0$
$\overrightarrow{r}.\overrightarrow{n}=q$
$\overrightarrow{n}=3\hat{i}+2\hat{j}-6\hat{k}$
$q=-15$
Given point is $(2,-1,3)$
$\therefore \overrightarrow{a}=2\hat{i}-\hat{j}+3\hat{k}$
Distance of point from plane
$=\frac{|\overrightarrow{a}.\overrightarrow{n}-q|}{|\overrightarrow{n}|}$
$=\frac{|(2\hat{i}-\hat{j}+3\hat{k}).(3\hat{i}+2\hat{j}-6\hat{k})+15|}{|3\hat{i}+2\hat{j}-6\hat{k}|}$
$=\frac{|6-2-18+15|}{\sqrt{9+4+36}}$
$=\frac{1}{\sqrt{49}}=\frac{1}{7}$.