Question

The equation of plane whose perpendicular distance from origin is $\sqrt{17}\text{units}$ and normal vector to plane is $2\hat{i}-3\hat{j}-2\hat{k},$ is

• A. $2x-3y-2z=\sqrt{17}$
• B. $2x-3y-2z=17$
• C. $-2x+3y-2z=\sqrt{17}$
• D. $2x-3y+2z=17$

Answer 1

The correct option is B $2x-3y-2z=17$

Equation of plane in normal form : $\overrightarrow{r}\cdot \hat{n}=p$
where $p=$ perpendicular distance from origin, $\hat{n}$ is unit vector of normal to the plane.
$\Rightarrow \hat{n}=\frac{2\hat{i}-3\hat{j}-2\hat{k}}{\sqrt{17}}$
So, equation of plane is :
$(x\hat{i}+y\hat{j}+z\hat{k})\cdot \left(\frac{2\hat{i}-3\hat{j}-2\hat{k}}{\sqrt{17}}\right)=\sqrt{17}\Rightarrow 2x-3y-2z=17$