Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Let $O(0,0,0)$ be the origin and $M(2,1,-2)$ be the foot of the perpendicular drawn from the origin to the plane.

The normal vector to the plane is $\vec{n}=\overrightarrow{OM}=2\hat{i}+\hat{j}-2\hat{k}$.

The plane passes through the point $M$, so its position vector is $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$.

The vector equation of the plane is given by $\vec{r}\cdot \vec{n}=\vec{a}\cdot \vec{n}$.

Substituting the values, we get:

$\vec{r}\cdot (2\hat{i}+\hat{j}-2\hat{k})=(2\hat{i}+\hat{j}-2\hat{k})\cdot (2\hat{i}+\hat{j}-2\hat{k})$

$\vec{r}\cdot (2\hat{i}+\hat{j}-2\hat{k})=(2)(2)+(1)(1)+(-2)(-2)$

$\vec{r}\cdot (2\hat{i}+\hat{j}-2\hat{k})=4+1+4$

$\vec{r}\cdot (2\hat{i}+\hat{j}-2\hat{k})=9$

Answer: $\bar{r}\cdot (2\hat{i}+\hat{j}-2\hat{k})=9$