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.

The quadrance (distance squared) from a line ${\displaystyle ax+by+c=0}$  to a point ${\displaystyle (x,y)}$  is

${\displaystyle {\frac {(ax+by+c)^{2}}{a^{2}+b^{2}}}.}$ 

If ${\displaystyle (a,b)}$  has unit length then this becomes ${\displaystyle (ax+by+c)^{2}.}$ 

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