Distance Between a Point and a Plane

Suppose that we have $2$ parallel planes: $$P_1: Ax+By+Cz+D_1=0$$ $$P_2: Ax+By+Cz+D_2=0$$ We wish to find the distance between these $2$ planes. Note that we want to find the shortest possible distance between the two planes, i.e. the distance if one traveled from $P_1$ to $P_2$ along a segment parallel to the normal vector.

A normal vector to both of these planes is $\vec{n}=\left<A,B,C\right>$. Note that $|\vec{n}|=\sqrt{A^2+B^2+C^2}$.

Suppose you move along the normal vector such that your displacement is $k\vec{n}=\left<kA,kB,kC\right>$, for some scalar $k$. For all points $P=(x,y,z)\in\mathbb{R}^3$, define $f(P)=Ax+By+Cz$. Then: $$f(P+k\vec{n})=A(x+kA)+B(y+kB)+C(z+kC)$$ $$f(P+k\vec{n})=(Ax+By+Cz)+k(A^2+B^2+C^2)$$ $$f(P+k\vec{n})-f(P)=k|\vec{n}|^2$$ Therefore, a movement of $k|\vec{n}|$ in the normal direction produces a change of $k|\vec{n}|^2$ in $f(P)$. Suppose that we have $2$ points $M_1\in P_1$ and $M_2\in P_2$ such that $M_1M_2\perp P_1,P_2$. Then: $$f(M_2)-f(M_1)=(-D_2)-(-D_1)=D_1-D_2$$ $$D_1-D_2=k|\vec{n}|^2$$ $$k|\vec{n}|=\frac{D_1-D_2}{\sqrt{A^2+B^2+C^2}}$$ Note that $k|\vec{n}|$ is just the displacement in the normal direction! Thus, since geometric distances are always nonnegative, we can say: $$M_1M_2=\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$$ This means that the distance $d$ between $2$ parallel planes of the above form is: $$d=\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$$ This argument can easily be generalized to any number of dimensions. Suppose we have $2$ parallel $(n-1)$-dimensional hyperplanes in $n$-dimensional hyperspace of the form: $$P_1: A_1x_1+A_2x_2+...+A_nx_n+B_1=0$$ $$P_2: A_1x_1+A_2x_2+...+A_nx_n+B_2=0$$ Then, the distance $d$ between them is: $$d=\frac{|B_1-B_2|}{\sqrt{A_1^2+A_2^2+...+A_n^2}}$$

The perpendicular distance between a point and a plane is always the shortest distance between this point and the plane. So if a perpendicular line is drawn through the plane to the point, the only relevant information is that there are 2 points of intersection for this line. This distance can be calculated easily as the length of this line segment, having the 2 intersections mentioned above as its endpoints. If you wanted to know the distance from the point to a non-perpendicular point on the plane, the Pythagorean theorem will come in handy.

(Viewers, any suggestions as to what to add?)

If you have read the previous explanation, you would have seen that Neelesh gives a very good intuition of the distince between a point and a plane. However, he did not provide a formula, so here is the distance from a point $(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$:

$Distance = \dfrac{Ax_0 + By_0 + Cz_0 + D}{\sqrt{A^2 + B^2 + C^2}}$

Here you go! Good luck!