Finding an equation of a plane a certain distance from a given plane

Question

I just wanted to know the methodology of how to solve for the equation of a plane that is some distance from some given plane.

Thanks. Any help is appreciated

Answer 1

If the given plane equation is $$ax+by+cz=d$$ where $(a,b,c)$ is the unit normal then the required equation is $$ax+by+cz=d\pm e$$ where e is the distance between the planes

Added

The normal to both planes is $(a,b,c)$ and so the required plane equation is $$ax+by+cz=F$$ Then for any point $(x_0,y_0,z_0)$ on this plane we have $$\color{red}{ax_0+by_0+cz_0}=F$$ and the distance $e$ from the point $(x_0,y_0,z_0)$ to the given plane is $$e=\frac{|\color{red}{ax_0+by_0+cz_0}-d|}{\sqrt{a^2+b^2+c^2}}= \frac{|\color{red}F-d|}{\sqrt{a^2+b^2+c^2}} \\ \implies e\sqrt{a^2+b^2+c^2}=|F-d|\implies F=d\pm e\sqrt{a^2+b^2+c^2}$$ If the normal is unit normal then $$ax+by+cz=d\pm e$$ Note the distance between a point $(x_0,y_0,z_0)$ and a plane $ax+by+cd-d=0$ is given by the formula $$\color{blue}{\frac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}}$$

Answer 2

If the desired distance is $n$, then the required equation is $ax+by+cz+d\pm n\sqrt{a^2+b^2+c^2}=0$.