How do you find the equation of a plane which is parallel to another plane which is at a unit distance of from some point

Question

Ok I know how to solve the problem if you just assume the plane's equation is $x-2y+2z+k = 0$ ($k$ is some constant) and then use the distance formula to find the value of k and then finally obtaining the required equation ...

But is there some other way of doing this which involves something like: $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c1}{c2}=$constant (parallelism condition)?

Thank you!

Answer 1

Let's say you have a plane $$a x + b y + c z = d \tag{1}\label{NA1}$$

The normal of the plane is $$\vec{n} = ( a , b , c )$$

The signed distance from point $(x, y, z)$ to the plane is $$s(x, y, z) = \frac{ a x + b y + c z - d}{\sqrt{ a^2 + b^2 + c^2 }} \tag{2}\label{NA2}$$ Signed distance is positive in the same direction as the normal, negative in the opposite direction.

OP wants to find a plane with normal $\vec{n} = (a, b, c)$, that is at signed distance $\rho$ from point $(\chi, \gamma, \zeta)$, i.e. $s(\chi, \gamma, \zeta) = \rho$. (Planes are parallel if their normals are parallel, i.e. $\vec{n}_1 = k \, \vec{n}_2$, $k \in \mathbb{R}$, $k \ne 0$. OP can simply choose $k = 1$, and reuse the normal vector from the original plane.)

Substituting into $\eqref{NA2}$ yields $$\rho = \frac{a \chi + b \gamma + c \zeta - d}{\sqrt{ a^2 + b^2 + c^2 }}$$ which we can solve for d, getting $$\bbox[1em, border:3px double lightgray]{d = a \chi + b \gamma + c \zeta - \rho \sqrt{ a^2 + b^2 + c^2 } } \tag{3}\label{NA3}$$ Note that $\eqref{NA3}$ works for all planes having normal $(a, b, c)$, any points $(\chi, \gamma, \zeta)$, and all signed distances $\rho$.

Answer 2

There's a solution which involves the vector equation of the planes: if the given plane $H_0$ has equation $ax+by+cz=k$, let $\vec n=(a,b,c)$ be the normal vector to $H_0$, $M_0(_0,y_0,z_0)$ a point in $H_0$. Denote $H$ a plane parallel to $H0$ and at distance $1$.

This means the point $N_0=M_0\pm\dfrac{\vec n}{\bigl|\mkern2mu\vec n\mkern2mu\bigr|}$ must be in $H$ so that its vector equation is $$\vec n\cdot M=\vec n\cdot N_0=\vec n\cdot M_0 \pm\frac{\vec n\cdot\vec n}{\bigl|\mkern2mu\vec n\mkern2mu\bigr|}=\vec n\cdot M_0\pm\bigl|\mkern2mu\vec n\mkern2mu\bigr|$$ But as $M_0$ is in $H_0$, $\;\vec n\cdot M_0=ax_0+by_0+cz_0=k $, so an equation of $H$ is $$ax+by+cz=k\pm\bigl|\mkern2mu\vec n\mkern2mu\bigr|=k\pm\sqrt{a^2+b^2+c^2}.$$