Hesse normal form

Distance from the origin O to the line E calculated with the Hesse normal form. Normal vector in red, line in green, point O shown in blue.

In analytic geometry, the Hesse normal form (named after Otto Hesse) is an equation used to describe a line in the Euclidean plane R 2 {\displaystyle \mathbb {R} ^{2}} , a plane in Euclidean space R 3 {\displaystyle \mathbb {R} ^{3}} , or a hyperplane in higher dimensions.[1][2] It is primarily used for calculating distances (see point-plane distance and point-line distance).

It is written in vector notation as

r → ⋅ n → 0 − d = 0. {\displaystyle {\vec {r}}\cdot {\vec {n}}_{0}-d=0.\,}

The dot  ⋅ {\displaystyle \cdot } indicates the dot product (or scalar product). Vector  r → {\displaystyle {\vec {r}}} points from the origin of the coordinate system, O, to any point P that lies precisely in plane or on line E. The vector  n → 0 {\displaystyle {\vec {n}}_{0}} represents the unit normal vector of plane or line E. The distance  d ≥ 0 {\displaystyle d\geq 0} is the shortest distance from the origin O to the plane or line.

## Derivation/Calculation from the normal form

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

In the normal form,

( r → − a → ) ⋅ n → = 0 {\displaystyle ({\vec {r}}-{\vec {a}})\cdot {\vec {n}}=0\,}

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

a → ⋅ n → ≥ 0 {\displaystyle {\vec {a}}\cdot {\vec {n}}\geq 0\,}

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

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

and the above equation can be rewritten as

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

Substituting

d = a → ⋅ n → 0 ≥ 0 {\displaystyle d={\vec {a}}\cdot {\vec {n}}_{0}\geq 0\,}

we obtain the Hesse normal form

r → ⋅ n → 0 − d = 0. {\displaystyle {\vec {r}}\cdot {\vec {n}}_{0}-d=0.\,}

In this diagram, d is the distance from the origin. Because  r → ⋅ n → 0 = d {\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  r → = r → s {\displaystyle {\vec {r}}={\vec {r}}_{s}} , per the definition of the Scalar product

d = r → s ⋅ n → 0 = | r → s | ⋅ | n → 0 | ⋅ cos ⁡ ( 0 ∘ ) = | r → s | ⋅ 1 = | r → s | . {\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  | r → s | {\displaystyle |{\vec {r}}_{s}|} of  r → s {\displaystyle {{\vec {r}}_{s}}} is the shortest distance from the origin to the plane.

## Distance to a line

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

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

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

## References

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1. ^ Bôcher, Maxime (1915), Plane Analytic Geometry: With Introductory Chapters on the Differential Calculus, H. Holt, p. 44.
2. ^ John Vince: Geometry for Computer Graphics. Springer, 2005, ISBN 9781852338343, pp. 42, 58, 135, 273

## External links

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