The Normal Form of Equation of Plane

Theorem 6.12 (normal form). If $p$ is the (non-negative) distance of a plane from the origin and $\hat n$ is the unit vector along the perpendicular dropped from the origin onto the plane, the plane's equation is simply $\vec r \cdot \hat n = p$.

Why this is true: let $N$ be the foot of that perpendicular, so $\overrightarrow{ON} = p$ in length; since $\hat n$ is the unit vector along $\overrightarrow{ON}$, this means $\overrightarrow{ON} = p\hat n$. For any point $P(\vec r)$ of the plane, $\overrightarrow{NP}$ lies in the plane and is therefore perpendicular to $\hat n$ (the perpendicular to the plane), so $\overrightarrow{NP}\cdot\hat n = 0$. Writing $\overrightarrow{NP} = \vec r - p\hat n$ turns this into $(\vec r - p\hat n)\cdot\hat n = 0$, i.e. $\vec r\cdot\hat n - p(\hat n\cdot\hat n) = 0$; since $\hat n$ is a unit vector, $\hat n\cdot\hat n = 1$, leaving $\vec r\cdot \hat n = p$ (Fig. 6.11 shows the axes, the foot $N$, the segment $ON$ of length $p$, the unit vector $\hat n$ along it, and a general point $P$ of the plane with position vector $\vec r$).

Several useful facts follow immediately. If $l, m, n$ are the direction cosines of the normal, then $\hat n = l\hat i + m\hat j + n\hat k$, and the coordinates of the foot $N$ itself are $(pl, pm, pn)$ — a shortcut used repeatedly below. In Cartesian coordinates the normal form reads $lx + my + nz = p$. Finally, because a unit normal can point either way, there are always two planes at a given distance $p$ from the origin with the same underlying normal direction, namely $\vec r\cdot\hat n = \pm p$.

Ex.(7): the plane at distance $6$ from the origin, normal to $2\hat i - \hat j + 2\hat k$. Here $|\vec n| = \sqrt{4+1+4} = 3$, so $\hat n = \tfrac{1}{3}(2\hat i - \hat j + 2\hat k)$, and $\vec r\cdot\hat n = p$ becomes $\vec r\cdot\tfrac{1}{3}(2\hat i-\hat j+2\hat k)=6$, i.e. $\vec r\cdot(2\hat i - \hat j + 2\hat k) = 18$.

Ex.(8): the perpendicular distance of the origin from $x - 3y + 4z - 6 = 0$. The normal has direction ratios $1, -3, 4$, so $|\vec n| = \sqrt{1+9+16}=\sqrt{26}$; dividing through converts $x-3y+4z=6$ to normal form $\tfrac{1}{\sqrt{26}}x - \tfrac{3}{\sqrt{26}}y + \tfrac{4}{\sqrt{26}}z = \tfrac{6}{\sqrt{26}}$, so the distance is $\tfrac{6}{\sqrt{26}}$. …