Question

Let $P,Q,R$ be points with position vectors ${\overrightarrow{r}}_1=3\overrightarrow{i}-2\overrightarrow{j}-\overrightarrow{k},{\overrightarrow{r}}_2=\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}$ and ${\overrightarrow{r}}_3=2\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$ relative to an origin $O$. The distance of $P$ from the plane $OQR$ is (magnitude)

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

Answer 1

The correct option is B $3$

$P,Q,R$ be points with position vectors $r_1=3\overrightarrow{i}-2\overrightarrow{j}-\overrightarrow{k},r_2=\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}$ and $r_3=2\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$
Equation of a plane passing through the points $O,Q,R$ is $\overrightarrow{r}=s\overrightarrow{r_2}+t\overrightarrow{r_3}$
Since equation of plane through points $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ is $\overrightarrow{r}=(1-s-t)\overrightarrow{a}+s\overrightarrow{b}+t\overrightarrow{c}$
$\Rightarrow \overrightarrow{r}=s(\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k})+t(2\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k})$
Taking $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$ and comparing terms gives
$x=s+2t;y=3s+t;z=4s-2t$
Eliminating $s$ and $t$, we get
$4x-4y+2z=0$
Distance of the plane from $P$ is $\frac{18}{\sqrt{16+16+4}}$
$\therefore$ required distance is $3$