Normal Form of a Plane

Plane at perpendicular distance p from the origin, with unit normal (l, m, n) where l²+m²+n²=1 and p > 0. To convert ax+by+cz+d=0: divide by ±√(a²+b²+c²) choosing sign so p > 0.

Let $ON = p$ be the perpendicular from the origin $O$ to the plane, with $\hat{n} = (l,m,n)$ being the unit normal ($l^2+m^2+n^2=1$) in the direction of $ON$.

The foot of perpendicular is $N = p\hat{n} = (pl, pm, pn)$.

For any point $P(\mathbf{r})$ on the plane, $\overrightarrow{NP} \perp \hat{n}$, so:

$$(\mathbf{r}-p\hat{n})\cdot\hat{n} = 0 \implies \mathbf{r}\cdot\hat{n} = p$$ $$lx+my+nz = p$$

Converting from general to normal form: Given $ax+by+cz+d=0$:

Divide by $\pm\sqrt{a^2+b^2+c^2}$ choosing the sign so that $p > 0$:

$$\frac{ax+by+cz}{\pm\sqrt{a^2+b^2+c^2}} = \frac{-d}{\pm\sqrt{a^2+b^2+c^2}}$$

So $l = a/\sqrt{a^2+b^2+c^2}$, $m = b/\sqrt{a^2+b^2+c^2}$, $n = c/\sqrt{a^2+b^2+c^2}$, and $p = |d|/\sqrt{a^2+b^2+c^2}$ — which is exactly the distance-from-origin formula.