Distance from a Point to a Plane in 3D
In three-dimensional geometry, one of the most useful measurements is the shortest distance from a point
\(P_0(x_0,y_0,z_0)\) to a plane. The “shortest” path is always the segment drawn perpendicular to the plane,
so it aligns with the plane’s normal vector. If the plane is written in general form
\(ax+by+cz+d=0\), then the vector \(\mathbf{n}=(a,b,c)\) is perpendicular to the plane. Every point on the plane satisfies the equation,
and the expression \(ax+by+cz+d\) acts like a signed “height” function that tells you which side of the plane a point lies on.
The key idea behind the distance formula is to compare the point \(P_0\) to the plane using the normal direction.
Compute the signed value
\[
s = a x_0 + b y_0 + c z_0 + d.
\]
If \(s=0\), then \(P_0\) lies on the plane and the distance is zero. If \(s>0\), the point lies on the side toward which the normal points,
and if \(s<0\), it lies on the opposite side. However, \(s\) is not the distance yet because the normal vector may not have unit length.
To convert this signed value into a true geometric distance, you divide by the normal’s magnitude:
\[
\|\mathbf{n}\| = \sqrt{a^2+b^2+c^2}.
\]
The perpendicular distance is therefore
\[
\text{dist}(P_0,\Pi)=\frac{|s|}{\|\mathbf{n}\|}
= \frac{|a x_0 + b y_0 + c z_0 + d|}{\sqrt{a^2+b^2+c^2}}.
\]
This formula is widely used in physics and engineering because it is fast and numerically stable: it depends only on the plane coefficients
and the point coordinates.
Many applications need not only the distance, but also the closest point on the plane, called the foot of the perpendicular
(or the projection of \(P_0\) onto the plane). If the plane is \(\mathbf{n}\cdot\mathbf{x}+d=0\), the foot point is obtained by moving from \(P_0\)
in the direction of the normal:
\[
F = P_0 - \frac{\mathbf{n}\cdot P_0 + d}{\|\mathbf{n}\|^2}\,\mathbf{n}.
\]
The vector \(P_0F\) is parallel to \(\mathbf{n}\) and meets the plane at a right angle, which is exactly why it represents the shortest segment.
In real-world terms, this is like dropping a vertical line from a drone down to a sloped ground surface, or computing the altitude from a point to a
terrain plane approximation.
Sometimes you do not know the plane coefficients directly, but you do know three non-collinear points \(A\), \(B\), and \(C\) that lie on the plane.
In that case, you can compute two direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\), then take the cross product
\(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}\) to obtain a normal. With \(\mathbf{n}\) known, you can compute \(d\) using any point on the plane:
\(d=-\mathbf{n}\cdot A\). This calculator supports both input styles and visualizes the plane and the perpendicular segment so the algebra matches the geometry.
