Translation of a plane with known normal.

Question

How to translate a plane with equation $ax+by+cz=d$ by a distance vector $<i, j, k>$. ?

Eg. Plane $x + y + z = 1$ to position $<2, 2, 2>$. If you can use this example to explain.

Answer 1

I'll give a general answer: if $\vec n$ is a normal vector to an affine plane $\Pi$, the equation of this plane can be written in the form $$\overrightarrow{OM}\cdot\vec n= c$$ Suppose we translate plane $\Pi$ by vector $\vec v$, and set $\;\Pi'=\Pi+\vec v$. A point $M'\in \Pi'$ is the translated of point $M\in \Pi$, hence $\overrightarrow{OM}=\overrightarrow{OM'}-\vec v$. So an equation of $\Pi'$ is simply $$\overrightarrow{OM}\cdot\vec n=(\overrightarrow{OM'}-\vec v)\cdot\vec n=c\iff \color{red}{\overrightarrow{OM'}\cdot\vec n=c+\vec v\cdot\vec n}$$

In the present case, one obtains $$x+y+z=1+6=7.$$

Answer 2

Hint: A plane passing through a point $\mathbf{x}_0$ with normal vector $\mathbf{n}$ can be specified as $\mathbf{n}\cdot(\mathbf{x}-\mathbf{x}_0)=0.$