Equation of a Plane in Normal Form

Theorem 6.15 (Normal form). The plane at perpendicular distance $p\ge0$ from the origin, with unit normal vector $\hat d$, has vector equation

Proof. Let $A$ be the foot of the perpendicular from $O$ to the plane, so $\vec{OA}=p\hat d$. For any point $P$ (position vector $\vec r$) on the plane, $\vec{AP}=\vec r-p\hat d$ is perpendicular to $\vec{OA}=p\hat d$, so $(\vec r-p\hat d)\cdot p\hat d=0$, i.e. $(\vec r-p\hat d)\cdot\hat d=0$ (dividing by $p$, assuming $p\ne0$), giving $\vec r\cdot\hat d=p$.

(b) Cartesian normal form. Writing $\hat d=l\hat i+m\hat j+n\hat k$ (direction cosines $l,m,n$) and $\vec r=x\hat i+y\hat j+z\hat k$, the equation becomes

Remark.

(i) If the plane passes through the origin, $p=0$, giving $lx+my+nz=0$. …