Question

Find the distance of the point whose position vector is ${\displaystyle (2\hat{i}+\hat{j}-\hat{k})}$ from the plane ${\displaystyle \overrightarrow{r}\cdot (\hat{i}-2\hat{j}+4\hat{k})}$ = 9

Answer 1

Here ${\displaystyle \overrightarrow{a}=2\hat{i}+\hat{j}-\hat{k}}$

${\displaystyle \overrightarrow{n}=\hat{i}-2\hat{j}+4\hat{k}}$

And d = 9

So, the required distance is ${\displaystyle \frac{|(2\hat{i}+\hat{j}-\hat{k})\cdot (\hat{i}-2\hat{j}+4\hat{k})-9|}{\sqrt{1+4+16}}}$

= ${\displaystyle \frac{|2-2-4-9|}{\sqrt{21}}}$

= ${\displaystyle \frac{13}{\sqrt{21}}}$.