Question

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

• A. $x-2y+3z-14=0$
• B. $x-2y+z+9=0$
• C. $x+2y-3z+5=0$
• D. $x+2y-3z-14=0$

Answer 1

The correct option is D $x+2y-3z-14=0$

Let $(1,2,-3)$ be point $P.$
Since $P(1,2,-3)$ is the foot of perpendicular from the origin to the plane, $OP$ is normal to the palne.
Thus, the direction ratios of normal to the plane are $1,2$ and $-3.$
Now, since the plane passes through $(1,2,-3),$ its equation is given by
$(x\hat{i}+y\hat{j}+z\hat{k})\cdot (\hat{i}+2\hat{j}-3\hat{k})=(\hat{i}+2\hat{j}-3\hat{k})\cdot (\hat{i}+2\hat{j}-3\hat{k})$
$\Rightarrow x+2y-3z=1+4+9$
$\Rightarrow x+2y-3z-14=0$