Aus der Zeichnung kannst du zwei Punkt ablesen, die in der Ebene liegen:

$\mathrm{A}(1;0;0)\backslash mathrm\; A(1;\backslash ;0;\backslash ;0)$  

$\mathrm{B}(0;3;0)\backslash mathrm\; B(0;\backslash ;3;\backslash ;0)$

Der Vektor  $\overrightarrow{\mathrm{A}\mathrm{B}}\backslash overrightarrow\{\backslash mathrm\{AB\}\}$   ist einer der Richtungsvektoren der Ebene $\mathrm{E}\backslash mathrm\; E$ .

$\overrightarrow{\mathrm{A}\mathrm{B}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}=\left(\begin{array}{c}0\\ 3\\ 0\end{array}\right)-\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}-1\\ 3\\ 0\end{array}\right)\backslash overrightarrow\{\backslash mathrm\{AB\}\}=\backslash overrightarrow\{\backslash mathrm\; B\}-\backslash overrightarrow\{\backslash mathrm\; A\}=\backslash begin\{pmatrix\}0\backslash \backslash 3\backslash \backslash 0\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-1\backslash \backslash 3\backslash \backslash 0\backslash end\{pmatrix\}$

Der zweite Richtungsvektor ist parallel zur  ${\mathrm{x}}_3-\mathrm{A}\mathrm{c}\mathrm{h}\mathrm{s}\mathrm{e}\{\backslash mathrm\; x\}\_3-\backslash mathrm\{Achse\}$. Wir können also als zweiten Richtungsvektor $\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}$nehmen.

Wähle den Punkt  $\mathrm{A}\backslash mathrm\{A\}$  oder  $\mathrm{B}\backslash mathrm\; B$  als Aufpunkt und stelle die Gleichung der Ebene  $\mathrm{E}\backslash mathrm\; E$  in Parameter auf.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+\lambda \cdot \left(\begin{array}{c}-1\\ 3\\ 0\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}-1\backslash \backslash 3\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}$

Lese die gegebenen Punkte ab:

$\mathrm{A}(2;0;0)\backslash mathrm\; A(2;\backslash ;0;\backslash ;0)$   ,   $\mathrm{B}(0;0;3)\backslash mathrm\; B(0;\backslash ;0;\backslash ;3)$

Der Vektor  $\overrightarrow{\mathrm{A}\mathrm{B}}\backslash overrightarrow\{\backslash mathrm\{AB\}\}$   ist einer der Richtungsvektoren der Ebene $\mathrm{E}\backslash mathrm\; E$ .

$\overrightarrow{\mathrm{A}\mathrm{B}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}=\left(\begin{array}{c}0\\ 0\\ 3\end{array}\right)-\left(\begin{array}{c}2\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}-2\\ 0\\ 3\end{array}\right)\backslash overrightarrow\{\backslash mathrm\{AB\}\}=\backslash overrightarrow\{\backslash mathrm\; B\}-\backslash overrightarrow\{\backslash mathrm\; A\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}2\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-2\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}$

Der zweite Richtungsvektor  $\overrightarrow{\mathrm{v}}\backslash overrightarrow\{\backslash mathrm\; v\}$  ist parallel zur  ${\mathrm{x}}_2-\mathrm{A}\mathrm{c}\mathrm{h}\mathrm{s}\mathrm{e}\{\backslash mathrm\; x\}\_2-\backslash mathrm\{Achse\}$  .

$\Rightarrow \overrightarrow{\mathrm{v}}=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\backslash Rightarrow\backslash ;\backslash overrightarrow\{\backslash mathrm\; v\}=\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 0\backslash end\{pmatrix\}$

Wähle den Punkt  $\mathrm{A}\backslash mathrm\; A$  oder  $\mathrm{B}\backslash mathrm\; B$  als Aufpunkt und Stelle die Gleichung der Ebene  $\mathrm{E}\backslash mathrm\; E$  auf.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}2\\ 0\\ 0\end{array}\right)+\lambda \cdot \left(\begin{array}{c}-2\\ 0\\ 3\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}2\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}-2\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 0\backslash end\{pmatrix\}$

Lese die gegebenen Punkte ab:

$\mathrm{A}(1;0;0)\backslash mathrm\; A(1;\backslash ;0;\backslash ;0)$   ,   $\mathrm{B}(0;3;0)\backslash mathrm\; B(0;\backslash ;3;\backslash ;0)$   ,   $\mathrm{C}(0;0;2)\backslash mathrm\; C(0;\backslash ;0;\backslash ;2)$

Der Vektor  $\overrightarrow{\mathrm{A}\mathrm{B}}\backslash overrightarrow\{\backslash mathrm\{AB\}\}$   und $\overrightarrow{\mathrm{A}\mathrm{C}}\backslash overrightarrow\{\backslash mathrm\{AC\}\}$  sind die Richtungsvektoren der Ebene $\mathrm{E}\backslash mathrm\; E$ .

$\overrightarrow{\mathrm{A}\mathrm{B}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}=\left(\begin{array}{c}0\\ 3\\ 0\end{array}\right)-\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}-1\\ 3\\ 0\end{array}\right)\backslash overrightarrow\{\backslash mathrm\{AB\}\}=\backslash overrightarrow\{\backslash mathrm\; B\}-\backslash overrightarrow\{\backslash mathrm\; A\}=\backslash begin\{pmatrix\}0\backslash \backslash 3\backslash \backslash 0\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-1\backslash \backslash 3\backslash \backslash 0\backslash end\{pmatrix\}$

$\overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{\mathrm{A}}=\left(\begin{array}{c}0\\ 0\\ 2\end{array}\right)-\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}-1\\ 0\\ 2\end{array}\right)\backslash overrightarrow\{AC\}=\backslash overrightarrow\; C-\backslash overrightarrow\{\backslash mathrm\; A\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 2\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-1\backslash \backslash 0\backslash \backslash 2\backslash end\{pmatrix\}$

Wähle einen der  Punkte  $\mathrm{A}\backslash mathrm\; A$ ,  $\mathrm{B}\backslash mathrm\; B$  oder $\mathrm{C}\backslash mathrm\; C$  als Aufpunkt und Stelle die Gleichung der Ebene  $\mathrm{E}\backslash mathrm\; E$  auf.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+\lambda \cdot \left(\begin{array}{c}-1\\ 3\\ 0\end{array}\right)+\mu \cdot \left(\begin{array}{c}-1\\ 0\\ 2\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}-1\backslash \backslash 3\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}-1\backslash \backslash 0\backslash \backslash 2\backslash end\{pmatrix\}$

Lese die gegebenen Punkte ab:

$\mathrm{A}(0;0;3)\backslash mathrm\; A(0;\backslash ;0;\backslash ;3)$

Einer der Richtungsvektoren der Ebene  $\mathrm{E}\backslash mathrm\; E$  ist parallel zur  ${\mathrm{x}}_1-\mathrm{A}\mathrm{c}\mathrm{h}\mathrm{s}\mathrm{e}\{\backslash mathrm\; x\}\_1-\backslash mathrm\{Achse\}$ .

$\Rightarrow \overrightarrow{\mathrm{v}}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\backslash Rightarrow\backslash ;\backslash overrightarrow\{\backslash mathrm\; v\}=\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}$

Der andere Richtungsvektor ist parallel zur  ${\mathrm{x}}_2-\mathrm{A}\mathrm{c}\mathrm{h}\mathrm{s}\mathrm{e}\{\backslash mathrm\; x\}\_2-\backslash mathrm\{Achse\}$ .

$\Rightarrow \overrightarrow{\mathrm{u}}=\left(\begin{array}{c}0\\ 2\\ 0\end{array}\right)\backslash Rightarrow\backslash ;\backslash overrightarrow\{\backslash mathrm\; u\}=\backslash begin\{pmatrix\}0\backslash \backslash 2\backslash \backslash 0\backslash end\{pmatrix\}$

Wähle den Punkt  $\mathrm{A}\backslash mathrm\; A$  als Aufpunkt und Stelle die Gleichung Ebene  $\mathrm{E}\backslash mathrm\; E$  auf.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ 0\\ 3\end{array}\right)+\lambda \cdot \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 0\backslash end\{pmatrix\}$