Question

The foot of the perpendicular drawn from the origin to a plane is $(-12,-4,3)$. Find the equation of the plane.

Answer 1

Let $P(-12,-4,3)$ is the foot of perpendicular from the origin to the plane.

$\therefore \overrightarrow{OP}=\overrightarrow{n}=-12\overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k}$

And $p=|-12\overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k}|=\sqrt{{(-12)}^2+16+9}$

$=\sqrt{144+16+9}$

$=\sqrt{144+25}$

$=\sqrt{169}=13$

We know that,

$\hat{n}=\frac{-12\overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k}}{13}$

Again, we know that

The vector equation of the plane is $\overrightarrow{r}.\hat{n}=p$

$(x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}).\frac{(-12\overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k})}{13}=13$

$-12x-4y+3z=169$

$12x+4y-3z+169=0$

Hence, this is the answer.