Hesse normal form

Note: For simplicity, the following derivation discusses the 3D case. However, it is also applicable in 2D.

In the normal form,

${\displaystyle ({\vec {r}}-{\vec {a}})\cdot {\vec {n}}=0\,}$

a plane is given by a normal vector ${\displaystyle {\vec {n}}}$ as well as an arbitrary position vector ${\displaystyle {\vec {a}}}$ of a point ${\displaystyle A\in E}$. The direction of ${\displaystyle {\vec {n}}}$ is chosen to satisfy the following inequality

${\displaystyle {\vec {a}}\cdot {\vec {n}}\geq 0\,}$

By dividing the normal vector ${\displaystyle {\vec {n}}}$ by its magnitude ${\displaystyle |{\vec {n}}|}$, we obtain the unit (or normalized) normal vector

${\displaystyle {\vec {n}}_{0}={{\vec {n}} \over {|{\vec {n}}|}}\,}$

and the above equation can be rewritten as

${\displaystyle ({\vec {r}}-{\vec {a}})\cdot {\vec {n}}_{0}=0.\,}$

Substituting

${\displaystyle d={\vec {a}}\cdot {\vec {n}}_{0}\geq 0\,}$

we obtain the Hesse normal form

${\displaystyle {\vec {r}}\cdot {\vec {n}}_{0}-d=0.\,}$

In this diagram, d is the distance from the origin. Because ${\displaystyle {\vec {r}}\cdot {\vec {n}}_{0}=d}$ holds for every point in the plane, it is also true at point Q (the point where the vector from the origin meets the plane E), with ${\displaystyle {\vec {r}}={\vec {r}}_{s}}$, per the definition of the Scalar product

${\displaystyle d={\vec {r}}_{s}\cdot {\vec {n}}_{0}=|{\vec {r}}_{s}|\cdot |{\vec {n}}_{0}|\cdot \cos(0^{\circ })=|{\vec {r}}_{s}|\cdot 1=|{\vec {r}}_{s}|.\,}$

The magnitude ${\displaystyle |{\vec {r}}_{s}|}$ of ${\displaystyle {{\vec {r}}_{s}}}$ is the shortest distance from the origin to the plane.

• Weisstein, Eric W. "Hessian Normal Form". MathWorld.