Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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

The magnitude of the normal vector is $|\vec{n}|=\sqrt{2^2+1^2+(-2{)}^2}=\sqrt{9}=3$.

The unit normal vector is $\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{2\hat{i}+\hat{j}-2\hat{k}}{3}$.

The vector equation of a plane at a perpendicular distance $d$ from the origin is given by $\vec{r}\cdot \hat{n}=d$.

Substituting $d=5$ and $\hat{n}=\frac{2\hat{i}+\hat{j}-2\hat{k}}{3}$, we get:

$\vec{r}\cdot \left(\frac{2\hat{i}+\hat{j}-2\hat{k}}{3}\right)=5$

$\vec{r}\cdot (2\hat{i}+\hat{j}-2\hat{k})=15$

Answer: $\vec{r}\cdot (2\hat{i}+\hat{j}-2\hat{k})=15$