Proof of point-plane distance formula.

Question

I have been wondering about the derivation of the following formula $$d = \frac{Ax_1 + By_1 + Cz_1 - D}{\sqrt{A^2 + B^2 + C^2}} $$ which is commonly used to find the perpendicular/shortest distance from a point ($$x_1, y_1, z_1$$) to a plane. After some consideration, I managed to condense this to $$d = \frac{|p \cdot \vec{n} - D|}{|\vec{n}|}$$ which may or may not be of use. $p$ is the desired point from which distance is to be found whereas $D$ is $a.n$ where $A$ is any fixed(known) point on the line and $n$ is the normal vector of the plane. While I managed to obtain this form, I was unable to conclude the logic that would have lead to this reduced form. Any and all clarifications explaining the derivation are much appreciated.

Answer 1

Looks like the the plane equation you're using is $Ax+By+Cz-D=0$, but the usual one is $Ax+By+Cz+D=0$ (it's either this or you made a mistake somewhere). Anyways I'll suppose you're using the first one, since they are in a way equivalent it's just that $D$ will be different for two planes that are the same.

Let's give a name to your point $(x_1,y_1,z_1)$, let's call it $T_1$, and let's denote the plane as $\pi$.From point $T_1$ there is a perpendicular line through the plane $\pi$, that will intersect it at some point, which we will name $N$. Now the distance from point $T_1$ to point $N$ is the distance from $T_1$ to plane $\pi$. So $\displaystyle d(T_1,\pi)=d(T_1,N)=|[\overrightarrow{T_1 N}]|$. You can see the following: $\displaystyle |[\overrightarrow{T_1 N}]|=|[\overrightarrow{T_1 N}]|\left\lvert\frac{\vec n}{|\vec{n}|}\right\rvert=\left\lvert\frac{[\overrightarrow{T_1 N}]\vec n}{|\vec n|}\right\rvert$.
We know that $[\overrightarrow{OT_1}]+[\overrightarrow{T_1 N}]=[\overrightarrow{O N}]$ and from there $[\overrightarrow{T_1 N}]=[\overrightarrow{O N}]-[\overrightarrow{O T_1}]$. We also know that $\vec n=(A,B,C)$ and $[\overrightarrow{O T_1}]=(x_1,y_1,z_1)$. Finally, because $N$ is a point on the plane, the equation $[\overrightarrow{O N}]\cdot \vec n -D=0\implies[\overrightarrow{O N}]\cdot \vec n=D$ holds. With all this in mind we have $\displaystyle d(T_1,\pi)=\left\lvert\frac{[\overrightarrow{T_1 N}]\vec n}{|\vec n|}\right\rvert=\left\lvert\frac{[\overrightarrow{O T_1}]\cdot\vec n-[\overrightarrow{O N}]\cdot\vec n}{|\vec n|}\right\rvert=\frac{|Ax_1+By_1+Cz_1-D|}{\sqrt{A^2+B^2+C^2}}$.

Seems like you had a good idea, just couldn't execute it. Keep in mind your $p$ needs to be a vector, which would be equal to my $[\overrightarrow{O T_1}]$.