Question

The equation of plane whose perpendicular distance from origin is $2\text{units}$ and vector $\hat{i}-2\hat{j}-2\hat{k}$ is normal to the plane, is

• A. $x-2y-2z=2$
• B. $x-2y-2z=3$
• C. $x-2y-2z=6$
• D. $x+2y+2z=6$

Answer 1

The correct option is C $x-2y-2z=6$

Equation of plane in normal form : $\overrightarrow{r}\cdot \hat{n}=p$
where $p=$ perpendicular distance from origin, $\hat{n}$ is unit vector normal to plane.
$\Rightarrow \hat{n}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{9}}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{3}$
So, equation of plane is :
$(x\hat{i}+y\hat{j}+z\hat{k})\cdot \left(\frac{\hat{i}-2\hat{j}-2\hat{k}}{3}\right)=2\Rightarrow x-2y-2z=6$