Distance from point to plane

Here's a quick sketch of how to calculate the distance from a point P = (x₁, y₁, z₁) to a plane determined by normal vector N = (A, B, C) and point Q = (x₀, y₀, z₀).

The equation for this plane is A(x - x₀) + B(y - y₀) + C(z - z₀) = 0. (Or we might write the equation as Ax + By + Cz + D = 0, where D = - Ax₀ - By₀ - Cz₀.) In the below CVT, I've drawn P in red and Q in yellow. To simplify the figure, I've fixed the plane. Although you can move Q around, it is confined to lie in the plane.

To calculate the distance from point P to the plane, I've dropped a perpendicular from the point P to the plane, drawing it as a gray line. The point R where the perpendicular hits the plane is drawn in green. R is the point on the plane closest to P. Hence the distance from P to the plane is the distance from P to R, or the length of the gray line. (You cannot move R directly by dragging on it, as its position is determined by P. It will move as you move P.)

Next, we calculate an expression for a unit normal vector, i.e., a normal vector of length one. It is simply N divided by its length. We'll use the notation n for the unit normal vector:

$${\frac{{\mathbf{N}}}{{\vert\vert\mathbf{N}\vert\vert}}} {\frac{{(A,B,C)}}{{\sqrt{A^2+B^2+C^2}}}}$$

I've draw this unit normal vector n in green, shifted so its tail is fixed at the point R. (In this case, I haven't drawn a ball at the end of n so the figure doesn't get too cluttered.) The unit normal vector n looks short because I've draw the figure so that the x, y, and z axes each extend from -5 to 5.

Let v be the vector from Q to P (shown in blue). Since P = (x₁, y₁, z₁) and Q = (x₀, y₀, z₀), we calculate that v = (x₁ - x₀, y₁ - y₀, z₁ - z₀). The length of the gray line, i.e., the distance from P to the plane, is simply the length of the projection of v onto the unit normal vector n. Since n is length one, this distance is simply the absolute value of v ^. n. We'll label the distance d; it is

d = |v ^. n| = |(x₁ - x₀, y₁ - y₀, z₁ - z₀) ^. n|

$${\frac{{\vert A(x_1-x_0) +B(y_1-y_0) +C(z_1-z_0)\vert}}{{\sqrt{A^2+B^2+C^2}}}}$$ (1)

This distance is shown on the cyan slider labeled by d to the right of the figure.

Recall that we can also write the equation for the plane as Ax + By + Cz + D = 0, with D = - Ax₀ - By₀ - Cz₀. We'll substitute into the above formula, to arrive at the following expression for the distance from P = (x₁, y₁, z₁) to the plane Ax + By + Cz + D = 0:

$${\frac{{\vert Ax_1+By_1+Cz_1 +D\vert}}{{\sqrt{A^2+B^2+C^2}}}}$$ (2)

From this final formula, you can see that the distance didn't depend on the point Q = (x₀, y₀, z₀). As long as Q is in the plane Ax + By + Cz + D = 0, then we know that D = - Ax₀ - By₀ - Cz₀. So, formula (1) is equivalent to formula (2) no matter where in the plane Q is. It's clear from the figure how the distance d shouldn't change as you move Q around in the plane. The vector v changes, but its projection onto n is constant.