Question

Let $P,Q,R$ be points with position vectors $\overrightarrow{r_1}=3\hat{i}-2\hat{j}-\hat{k},\overrightarrow{r_2}=\hat{i}+3\hat{j}+4\hat{k}$ and $\overrightarrow{r_3}=2\hat{i}+\hat{j}-2\hat{k}$ relative to an origin $O$. The distance of $P$ from the plane $OQR$ is

• A. $2$
• B. $3$
• C. $1$
• D. $\frac{11}{\sqrt{3}}$

Answer 1

The correct option is B $3$

Let $P,Q,R$ be points with the given position vectors

Equation of the plane $OQR$ is $r$ $=\lambda \overrightarrow{r_2}+\mu \overrightarrow{r_3},$
i.e. $\overrightarrow{r}\cdot (\overrightarrow{r_2}\times \overrightarrow{r_3})=0$
So, the distance of $P$ from the plane $OQR$ is $\left|\frac{\overrightarrow{r_1.}(\overrightarrow{r_2}\times \overrightarrow{r_3})}{|\overrightarrow{r_2}\times \overrightarrow{r_3}|}\right|.$

Since, $\overrightarrow{r_2}\times \overrightarrow{r_3}=-10\hat{i}+10\hat{j}-5\hat{k},$ so $|\overrightarrow{r_2}\times \overrightarrow{r_3}|=15$

And the perpendicular distance $=\left|\frac{-30-20+5}{15}\right|=3$

Hence, option B is correct.