Question

A plane is such that the foot of perpendicular drawn from origin to it is $(2,-1,1)$. The distance of $(1,2,3)$ from the plane is

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

Answer 1

The correct option is B $\sqrt{\frac{3}{2}}$

foot of $\perp$ from origin $=(2,-1,1)$

$\therefore$ direction of normal $=(2,-1,1)=\overrightarrow{n}$

$\therefore$ perpendicular distance from origin $=\sqrt{{(2-0)}^2+{(-1-0)}^2+{(1-0)}^2}=\sqrt{6}=p$

$\therefore$Equation of plane,

$\overrightarrow{r}.\hat{n}=p\therefore \overrightarrow{r}.\frac{(2\hat{i}-\hat{j}+\hat{k})}{\sqrt{6}}=\sqrt{6}\therefore \overrightarrow{r}.(2\hat{i}-\hat{j}+\hat{k})=6$

In cartesian form,

$2x-y+z=6$

$\therefore$ distance from $(1,2,3)=\frac{|2\times 1-2+3-6|}{\sqrt{2^2+1^2+1^2}}=\sqrt{\frac{3}{2}}$.