We calculate the distance using the Hesse normal form.
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Definition 1. A normal form \(\left\{\mathbf{v}\in\mathbb{R}^3:\ \langle
\mathbf{v}-\mathbf{p},\mathbf{n} \rangle=0 \right\}\) of a plane
is called Hesse normal form if the length of \(\mathbf{n}\) equals 1. (\(\|n\|=1\))
Comment: A normal form can be transformed into a HNF by
dividing the given normal vector by its length.
Example 2. Consider the plane \[E:=\left\{\mathbf{v}\in\mathbb{R}^3:\
\left\langle
\mathbf{v} - {\begin{pmatrix} 1\\2\\3 \end{pmatrix}},
{\begin{pmatrix} 2\\-2\\1 \end{pmatrix}}\right\rangle =
0\right\}.\] The given representation is a normal form, but not a
Hesse normal form, since the normal vector \(\mathbf{n}={ \begin{pmatrix} 2\\-2\\1
\end{pmatrix}}\) has length \(\|\mathbf{n}\|=3\). A normal vector of
length 1 is \(\frac{1}{3}\mathbf{n} =
{\begin{pmatrix}
2/3 \\ -2/3 \\ 1/3
\end{pmatrix}}\). Hence, a Hesse normal form is \[E=\left\{\mathbf{v}\in\mathbb{R}^3:\ \left\langle
\mathbf{v} - {\begin{pmatrix} 1\\2\\3 \end{pmatrix}},
{ \begin{pmatrix} 2/3\\-2/3\\1/3 \end{pmatrix}}\right\rangle =
0\right\}.\]
Distance
point/plane
Let \(\mathbf{q}\) be a point
(position vector) and let \(E\) be a
plane in Hesse normal form \(E=\left\{ \mathbf{v} \in \mathbb{R}^3:\ \langle
\mathbf{v} - \mathbf{p}, \mathbf{n} \rangle= 0 \right\}\) then
the distance between \(\mathbf{q}\) and
\(E\) equals \[\left| \langle \mathbf{q} - \mathbf{p},
\mathbf{n} \rangle \right|.\]
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