Determining a plane by a point in the plane and a orthogonal vector.
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Idea: A plane \(E\)
in \(\mathbb{R}^3\) is determined
uniquely if we know a point in the plane \(E\) and a vector (different from \(\mathbf{o}\)) which is orthogonal to \(E\).
Definition 1. Every plane \(E\) in \(\mathbb{R}^3\) can be described in the
following form: \[E=\left\{\mathbf{v}\in\mathbb{R}^3:\ \langle
\mathbf{v}-\mathbf{p},\mathbf{n}\rangle = 0 \right\}
= \left\{\mathbf{v}\in\mathbb{R}^3:\ \langle
\mathbf{v},\mathbf{n}\rangle = \langle \mathbf{p},\mathbf{n}\rangle
\right\}.\] Here, \(\mathbf{p}\)
represents any point of \(E\). The
vector \(\mathbf{n}\neq \mathbf{o}\) is
called normal vector of \(E\), i.e. it is orthogonal to \(E\). This representation is called
normal form.
Definition 2. Every plane \(E\) in \(\mathbb{R}^3\) can be written in the form
\[E=\left\{
\mathbf{x}={\scriptsize \begin{pmatrix} x_1\\x_2\\x_3
\end{pmatrix}}\in\mathbb{R}^3:~
a_1x_1+a_2x_2+a_3x_3=d
\right\}\] with \(a_1,a_2,a_3,d\in\mathbb{R}\). This
representation is called coordinate form.
The equation \(a_1x_1+a_2x_2+a_3x_3=d\) describes which
condition a point \((x_1,x_2,x_3)\)
must satisfy in order to part of the plane.
Example 3. Let a plane \(E\) be given in normal form \(E=\left\{\mathbf{v}\in\mathbb{R}^3:\ \langle
\mathbf{v}-\mathbf{p},\mathbf{n}\rangle = 0 \right\}\) then a
coordinate equation can be found as follows:
Set \(\mathbf{v}=\begin{pmatrix} x_1\\x_2\\x_3
\end{pmatrix}\) and rearrange \(\langle
\mathbf{v}-\mathbf{p},\mathbf{n}\rangle = 0\) to get an equation
of the form \(a_1x_1+a_2x_2+a_3x_3=d\).
Example 4. In \(\mathbb{R}^3\) there exists a unique plane
\(E\) which contains the points \(R:=(1,1,1)\), \(S:=(1,3,2)\) and \(T:=(-1,4,3)\). A parameter form of \(E\) is: \[\begin{aligned}
E = \left\{\begin{pmatrix} 1\\1\\1 \end{pmatrix} + \lambda
{\color{red}\begin{pmatrix} 0\\2\\1 \end{pmatrix}}
+ \mu {\color{red}\begin{pmatrix} -2\\3\\2 \end{pmatrix}}:\
\lambda,\mu\in\mathbb{R} \right\}.
\end{aligned}\] For a normal form of \(E\) we need a point \(\mathbf{p}\) and a normal vector \(\mathbf{n}\). (Hint: Use cross
product!)
point: \(\mathbf{p} = \begin{pmatrix}
1\\1\\1 \end{pmatrix}\)
normal vector: \({\color{blue}\mathbf{n}} =
{\color{red}\begin{pmatrix} 0\\2\\1 \end{pmatrix}}
\times {\color{red}\begin{pmatrix} -2\\3\\2 \end{pmatrix}} =
\begin{pmatrix} 1\\-2\\4 \end{pmatrix}\)
A normal form of \(E\) is: \[E=\left\{ \mathbf{v}\in\mathbb{R}^3:\
\left\langle \mathbf{v} - \begin{pmatrix} 1\\1\\1 \end{pmatrix},
\begin{pmatrix} 1\\-2\\4 \end{pmatrix} \right\rangle = 0 \right\}~
.\] To obtain a coordinate form, we substitute \(\mathbf{v}=\begin{pmatrix}
x_1\\x_2\\x_3 \end{pmatrix}\) into the equation of the
normal form: \[\left\langle \begin{pmatrix}
x_1\\x_2\\x_3 \end{pmatrix} - \begin{pmatrix} 1\\1\\1
\end{pmatrix}, \begin{pmatrix} 1\\-2\\4 \end{pmatrix} \right\rangle = 0
~~~
\Leftrightarrow
~~~
1x_1-2x_2+4x_3 = 3~ .\] Hence, \[E= \left\{ \begin{pmatrix}
x_1\\x_2\\x_3 \end{pmatrix}\in\mathbb{R}^3:\ x_1-2x_2+4x_3 = 3
\right\}~ .\]
Overview:
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