Question

The foot of the perpendicular drawn from the origin to the plane is $(4,-2,-5)$, then find the vector equation of plane.

Answer 1

Let equation of the given point lying in the plane be $\overrightarrow{r_0}$.

Let equation of the normal be $\overrightarrow{n}$.

$\therefore \overrightarrow{n}=\overrightarrow{0}-(4\hat{i}-2\hat{j}-5\hat{k})=-4\hat{i}+2\hat{j}+5\hat{k}$

Now, equation of plane with normal $\overrightarrow{n}$ and point $\overrightarrow{n_0}$ is $\overrightarrow{n}\cdot (\overrightarrow{r}-\overrightarrow{r_0})=0$,

where $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ is any general point in the plane.

$⟹(-4\hat{i}+2\hat{j}+5\hat{k})\cdot (\overrightarrow{r}-(4\hat{i}-2\hat{j}-5\hat{k}\left)\right)=0$

$⟹-4x+2y+5z+16+4+25=0$

$⟹4x-2y-5z=45$

This is the required equation of the plane.