Question

The foot of the perpendicular drawn from the origin to a plane is $(1,2,-3)$. Find the equation of the plane.

• A. $x-2y-3z=14$
• B. $x-2y+3z=14$
• C. $x+2y-3z=14$
• D. $x+2y+3z=14$

Answer 1

The correct option is C $x+2y-3z=14$

Given:

Let the equation of plane passing through $(1,2,3)$ and perpendicular to $OA$.

$A=\hat{i}+2\hat{j}-3\hat{k}$

$\overrightarrow{n}=\overrightarrow{OA}=\hat{i}+2\hat{j}-3\hat{k}$

$a=|\hat{i}+2\hat{j}-3\hat{k}|=\sqrt{1+4+9}=\sqrt{14}$

$\hat{n}=\frac{\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{14}}$

The equation of the required plane is $\overrightarrow{r}\cdot \overrightarrow{n}=a$

$\Rightarrow \overrightarrow{r}\cdot \left(\frac{\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{14}}\right)=\sqrt{14}$

$\Rightarrow \overrightarrow{r}\cdot (\hat{i}+2\hat{j}-3\hat{k})=14$

$\Rightarrow x+2y-3z=14$

Thus, the equation of the plane is $x+2y-3z=14$.