Write in vector form the plane 2x-y+z+3=0

Write in vector form the plane $2x-y+z+3=0$.

Step 1: Solution

The normal is $\vec n=2\vec i-\vec j+\vec k$, and $2x-y+z=\vec r\cdot\vec n$: $\vec r\cdot(2\vec i-\vec j+\vec k)=-3$.

Normal form: dividing by $|\vec n|=\sqrt6$ gives $\vec r\cdot\vec n_0=-\tfrac{3}{\sqrt6}$. The distance of the plane from $O$ is $\tfrac{3}{\sqrt6}=\tfrac{\sqrt6}{2}$.