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Mathematics-Online course: Prepcourse Mathematics - Linear Algebra and Geometry - Lines and Planes

Hessian Normal Form of Planes

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The position vector $ \vec{x}$ of a given point $ X$ in a plane through point $ P$ orthogonal to a normal vector $ \vec{n}$ satisfies

$\displaystyle \vec{x}\cdot\vec{n} = d,\,\quad d=\vec{p}\cdot\vec{n}\, . $

If $ \vert\vec{n}\vert=1$ and $ d\ge0$ , then we obtain the so called Hessian normal form and $ d$ is the distance between the plane and the origin.

\includegraphics[width=10cm]{hesse_form}
Given a plane $ E$ with unit normal vector $ \vec{n} = \dfrac{1}{3} \begin{pmatrix}2\\ 2\\ 1\\ \end{pmatrix}$ throught point $ P=(1, 2, 3)$ .

We have

$\displaystyle d = \begin{pmatrix}1\\ 2\\ 3\\ \end{pmatrix}\cdot \frac{1}{3} \begin{pmatrix}2\\ 2\\ 1\\ \end{pmatrix} = 3\, , $

i.e. the Hessian normal form of the plane ist given by

$\displaystyle E: \quad \frac{2}{3}x_1+\frac{2}{3}x_2+\frac{1}{3}x_3=3\,. $

Then $ X=(4,0,1)$ lies in this plane, because we have

$\displaystyle \vec{x} \cdot \vec{n} = \frac{1}{3}(8+0+1) = d\, . $

On the other hand, $ X = (0,0,0)$ does not lie in this plane, because we have

$\displaystyle \vec{x} \cdot \vec{n} = 0 \neq d\,.$

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  automatically generated 9/18/2007