Proof of the formula for the distance from a point to a plane

Proof

We find the orthogonal projection of $X$ on the plane and compare every other point of the plane with it.

1. Let $n = (a, b, c)$. It is a normal vector of $P$. Since $n \neq 0$, its length $\norm{n} = \sqrt{a^2 + b^2 + c^2}$ is positive.
2. Let $k = a x_0 + b y_0 + c z_0 - d$, let $t = k / \norm{n}^2$, and let $Q_0 = X - t n$.
3. With the dot product, the plane is $P = \set{Q \in \R^3 : n \cdot Q = d}$. Also $n \cdot X = k + d$.
4. The point $Q_0$ lies on $P$. Indeed, $n \cdot Q_0 = n \cdot X - t \, (n \cdot n) = (k + d) - k = d$.
5. Let $Q$ be any point of $P$. Then $n \cdot Q = d = n \cdot Q_0$, so $n \cdot (Q_0 - Q) = 0$.
6. We have $X - Q = (X - Q_0) + (Q_0 - Q) = t n + (Q_0 - Q)$. The two summands are orthogonal by step 5.
7. By the Pythagorean theorem for orthogonal vectors, $\norm{X - Q}^2 = t^2 \norm{n}^2 + \norm{Q_0 - Q}^2 \geq t^2 \norm{n}^2$.
8. The Euclidean distance is $d(X, Q) = \norm{X - Q}$. Taking square roots in step 7 gives $d(X, Q) \geq \abs{t} \, \norm{n}$ for each $Q \in P$.
9. For $Q = Q_0$, we get $d(X, Q_0) = \norm{t n} = \abs{t} \, \norm{n}$. So the bound of step 8 is reached at the point $Q_0$ of $P$.
10. Hence the smallest distance exists and equals $\abs{t} \, \norm{n} = \dfrac{\abs{k}}{\norm{n}} = \dfrac{\abs{a x_0 + b y_0 + c z_0 - d}}{\sqrt{a^2 + b^2 + c^2}}$.