Math Distance calculations Hesse normal form

# Hesse normal form

A special type of normal equation is the Hesse normal form:

$\text{E: } (\vec{x} - \vec{a}) \cdot \vec{n_0}=0$
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### Remember

$\vec{n_0}$ is called the normalized or unit normal vector. It has the magnitude/length of 1.

$|\vec{n_0}|=1$

For the unit normal vector, the normal vector is divided by its magnitude.

$\vec{n_0}= \frac{\vec{n}}{|\vec{n}|}$

The Hesse normal form is obtained by dividing the vector equation of the plane by the magnitude of the normal vector.

$\text{E: } (\vec{x} - \vec{a}) \cdot \frac{\vec{n}}{|\vec{n}|}=0$

### Example

$\text{E: } \left(\vec{x} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}=0$

1. #### Absolute value of the normal vector

$\vec{n}=\begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$

$|\vec{n}|=\sqrt{2^2+(-2)^2+4^2}$ $=\sqrt{24}$

2. #### Unit normal vector

$\vec{n_0}= \frac{\vec{n}}{|\vec{n}|}$

$\vec{n_0}= \frac{ \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}}{\sqrt{24}}$ $=\frac{1}{\sqrt{24}}\cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$ $=\begin{pmatrix} 2/\sqrt{24} \\ -2/\sqrt{24} \\ 4/\sqrt{24} \end{pmatrix}$

3. #### Hesse normal form

$\text{E: } (\vec{x} - \vec{a}) \cdot \vec{n_0}=0$

$\text{E: } \left(\vec{x} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 2/\sqrt{24} \\ -2/\sqrt{24} \\ 4/\sqrt{24} \end{pmatrix}=0$

Alternatively, the following notation is also possible:

$\text{E: } \left(\vec{x} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \frac{1}{\sqrt{24}}\cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}=0$