Aufgaben zur Umwandlung der Ebenendarstellung

Berechne zuerst den Normalenvektor der Ebene als Kreuzprodukt der beiden Richtungsvektoren:

$\overrightarrow{\mathrm{n}}=\left(\begin{array}{c}-1\\ 2\\ -2\end{array}\right)\times \left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)=\left(\begin{array}{c}6\\ -1\\ -4\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; n\}=\backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -2\backslash end\{pmatrix\}\backslash times\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 1\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}6\backslash \backslash -1\backslash \backslash -4\backslash end\{pmatrix\}$

Wähle den Punkt $AA$ mit dem Ortsvektor  $\overrightarrow{\mathrm{a}}=\left(\begin{array}{c}1\\ 0\\ 3\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; a\}=\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}$  als Aufpunkt der Ebene.

$\mathrm{E}:\overrightarrow{\mathrm{n}}\circ [\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}]=0\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; n\}\backslash circ\backslash left[\backslash overrightarrow\{\backslash mathrm\; x\}-\backslash overrightarrow\{\backslash mathrm\; a\}\backslash right]=0$

Setze  $\overrightarrow{\mathrm{n}}\backslash overrightarrow\{\backslash mathrm\; n\}$  und  $\overrightarrow{\mathrm{a}}\backslash overrightarrow\{\backslash mathrm\; a\}$  ein.

$\mathrm{E}:\left(\begin{array}{c}6\\ -1\\ -4\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)-\left(\begin{array}{c}1\\ 0\\ 3\end{array}\right)]=0\backslash mathrm\; E:\backslash ;\backslash begin\{pmatrix\}6\backslash \backslash -1\backslash \backslash -4\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}\backslash right]=0$

Berechne zuerst den Normalenvektor der Ebene als Kreuzprodukt der beiden Richtungsvektoren:

$\overrightarrow{\mathrm{n}}=\left(\begin{array}{c}3\\ 2\\ 1\end{array}\right)\times \left(\begin{array}{c}-1\\ 0\\ 2\end{array}\right)=\left(\begin{array}{c}4\\ -7\\ 2\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; n\}=\backslash begin\{pmatrix\}3\backslash \backslash 2\backslash \backslash 1\backslash end\{pmatrix\}\backslash times\backslash begin\{pmatrix\}-1\backslash \backslash 0\backslash \backslash 2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}4\backslash \backslash -7\backslash \backslash 2\backslash end\{pmatrix\}$

Wähle den Punkt $AA$ mit dem Ortsvektor  $\overrightarrow{\mathrm{a}}=\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; a\}=\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}$  als Aufpunkt der Ebene.

$\mathrm{E}:\overrightarrow{\mathrm{n}}\circ [\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}]=0\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; n\}\backslash circ\backslash left[\backslash overrightarrow\{\backslash mathrm\; x\}-\backslash overrightarrow\{\backslash mathrm\; a\}\backslash right]=0$

$\mathrm{E}:\left(\begin{array}{c}4\\ -7\\ 2\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)-\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)]=0\backslash mathrm\; E:\backslash ;\backslash begin\{pmatrix\}4\backslash \backslash -7\backslash \backslash 2\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne zuerst den Normalenvektor der Ebene als Kreuzprodukt der beiden Richtungsvektoren:

$\overrightarrow{\mathrm{n}}=\left(\begin{array}{c}1\\ 0\\ 3\end{array}\right)\times \left(\begin{array}{c}-1\\ 2\\ 0\end{array}\right)=\left(\begin{array}{c}-6\\ -3\\ 2\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; n\}=\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}\backslash times\backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash 0\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-6\backslash \backslash -3\backslash \backslash 2\backslash end\{pmatrix\}$

Wähle den Punkt Koordinatenursprung als Aufpunkt der Ebene.

$\mathrm{E}:\overrightarrow{\mathrm{n}}\circ [\overrightarrow{\mathrm{x}}-\overrightarrow{0}]=0\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; n\}\backslash circ\backslash left[\backslash overrightarrow\{\backslash mathrm\; x\}-\backslash overrightarrow0\backslash right]=0$

$\mathrm{E}:\left(\begin{array}{c}-6\\ -3\\ 2\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)-\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)]=0\backslash mathrm\; E:\backslash ;\backslash begin\{pmatrix\}-6\backslash \backslash -3\backslash \backslash 2\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}\backslash right]=0$

$\iff \mathrm{E}:\left(\begin{array}{c}-6\\ -3\\ 2\end{array}\right)\circ \left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)=0\backslash Leftrightarrow\backslash ;\backslash mathrm\; E:\backslash ;\backslash begin\{pmatrix\}-6\backslash \backslash -3\backslash \backslash 2\backslash end\{pmatrix\}\backslash circ\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash end\{pmatrix\}=0$

Berechne zuerst den Normalenvektor der Ebene als  Kreuzprodukt der beiden Richtungsvektoren:

$\overrightarrow{\mathrm{n}}=\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)\times \left(\begin{array}{c}-1\\ 2\\ 0\end{array}\right)=\left(\begin{array}{c}-4\\ -2\\ 1\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; n\}=\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}\backslash times\backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash 0\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-4\backslash \backslash -2\backslash \backslash 1\backslash end\{pmatrix\}$

Wähle den Punkt $AA$ mit dem Ortsvektor  $\overrightarrow{\mathrm{a}}=\left(\begin{array}{c}1\\ 1\\ -1\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; a\}=\backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash -1\backslash end\{pmatrix\}$  als Aufpunkt der Ebene.

$\mathrm{E}:\overrightarrow{\mathrm{n}}\circ [\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}]=0\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; n\}\backslash circ\backslash left[\backslash overrightarrow\{\backslash mathrm\; x\}-\backslash overrightarrow\{\backslash mathrm\; a\}\backslash right]=0$

Setze  $\overrightarrow{\mathrm{n}}\backslash overrightarrow\{\backslash mathrm\; n\}$  und  $\overrightarrow{\mathrm{a}}\backslash overrightarrow\{\backslash mathrm\; a\}$  ein.

$\mathrm{E}:\left(\begin{array}{c}-4\\ -2\\ 1\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)-\left(\begin{array}{c}1\\ 1\\ -1\end{array}\right)]=0\backslash mathrm\; E:\backslash ;\backslash begin\{pmatrix\}-4\backslash \backslash -2\backslash \backslash 1\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash -1\backslash end\{pmatrix\}\backslash right]=0$

Berechne zuerst den Normalenvektor der Ebene als  Kreuzprodukt der beiden Richtungsvektoren:

$\overrightarrow{\mathrm{n}}=\left(\begin{array}{c}-1\\ 2\\ -1\end{array}\right)\times \left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)=\left(\begin{array}{c}4\\ 0\\ -4\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; n\}=\backslash begin\{pmatrix\}-1\backslash \backslash 2\backslash \backslash -1\backslash end\{pmatrix\}\backslash times\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 1\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}4\backslash \backslash 0\backslash \backslash -4\backslash end\{pmatrix\}$

Klammere  $\frac{1}{4}\backslash frac14$  aus, um einen möglichst einfachen Normalenvektor zu erhalten.

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

Wähle den Punkt $AA$ mit dem Ortsvektor  $\overrightarrow{\mathrm{a}}=\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; a\}=\backslash begin\{pmatrix\}3\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}$ als Aufpunkt der Ebene.

$\mathrm{E}:\overrightarrow{\mathrm{n}}\circ [\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}]=0\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; n\}\backslash circ\backslash left[\backslash overrightarrow\{\backslash mathrm\; x\}-\backslash overrightarrow\{\backslash mathrm\; a\}\backslash right]=0$

Setze  $\overrightarrow{\mathrm{n}}\backslash overrightarrow\{\backslash mathrm\; n\}$  und  $\overrightarrow{\mathrm{a}}\backslash overrightarrow\{\backslash mathrm\; a\}$  ein.

$\mathrm{E}:\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)-\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)]=0\backslash mathrm\; E:\backslash ;\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash -1\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}3\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne zuerst den Normalenvektor der Ebene als  Kreuzprodukt der beiden Richtungsvektoren:

(Hinweis: Die Ebene ist parallel zur  ${\mathrm{x}}_1{\mathrm{x}}_3-\mathrm{E}\mathrm{b}\mathrm{e}\mathrm{n}\mathrm{e}\{\backslash mathrm\; x\}\_1\{\backslash mathrm\; x\}\_3-\backslash mathrm\{Ebene\}$)

Klammere  $-7-7$  aus, um einen möglichst einfachen Normalenvektor zu erhalten.

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

Wähle den Punkt $AA$ mit dem Ortsvektor  $\overrightarrow{\mathrm{a}}=\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; a\}=\backslash begin\{pmatrix\}2\backslash \backslash 2\backslash \backslash 2\backslash end\{pmatrix\}$  als Aufpunkt der Ebene.

$\mathrm{E}:\overrightarrow{\mathrm{n}}\circ [\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}]=0\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; n\}\backslash circ\backslash left[\backslash overrightarrow\{\backslash mathrm\; x\}-\backslash overrightarrow\{\backslash mathrm\; a\}\backslash right]=0$

Setze  $\overrightarrow{\mathrm{n}}\backslash overrightarrow\{\backslash mathrm\; n\}$  und  $\overrightarrow{\mathrm{a}}\backslash overrightarrow\{\backslash mathrm\; a\}$  ein.

$\mathrm{E}:\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)-\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right)]=0\backslash mathrm\; E:\backslash ;\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 0\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}2\backslash \backslash 2\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne zuerst den Normalenvektor der Ebene als

Kreuzprodukt der beiden Richtungsvektoren:

$\overrightarrow{\mathrm{n}}=\left(\begin{array}{c}-20\\ -20\\ 10\end{array}\right)\times \left(\begin{array}{c}15\\ 10\\ 20\end{array}\right)=\left(\begin{array}{c}-500\\ 550\\ 100\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; n\}=\backslash begin\{pmatrix\}-20\backslash \backslash -20\backslash \backslash 10\backslash end\{pmatrix\}\backslash times\backslash begin\{pmatrix\}15\backslash \backslash 10\backslash \backslash 20\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-500\backslash \backslash 550\backslash \backslash 100\backslash end\{pmatrix\}$

Klammere  $5050$  aus, um einen möglichst einfachen Normalenvektor zu erhalten.

$\Rightarrow \overrightarrow{\mathrm{n}}=\left(\begin{array}{c}-10\\ 11\\ 2\end{array}\right)\backslash Rightarrow\backslash ;\backslash overrightarrow\{\backslash mathrm\; n\}=\backslash begin\{pmatrix\}-10\backslash \backslash 11\backslash \backslash 2\backslash end\{pmatrix\}$

Alternative: Meistens ist es einfacher, zuerst die

Richtungsvektoren der Ebene zu vereinfachen

und dann erst das Kreuzprodukt zu berechnen.

Klammere im ersten Richtungsvektor  $1010$  und im zweiten Richtungsvektor  $55$  aus.

$\overrightarrow{\mathrm{n}}=\left(\begin{array}{c}-2\\ -2\\ 1\end{array}\right)\times \left(\begin{array}{c}3\\ 2\\ 4\end{array}\right)=\left(\begin{array}{c}-10\\ 11\\ 2\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; n\}=\backslash begin\{pmatrix\}-2\backslash \backslash -2\backslash \backslash 1\backslash end\{pmatrix\}\backslash times\backslash begin\{pmatrix\}3\backslash \backslash 2\backslash \backslash 4\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-10\backslash \backslash 11\backslash \backslash 2\backslash end\{pmatrix\}$

Wähle den Punkt $AA$ mit dem Ortsvektor  $\overrightarrow{\mathrm{a}}=\left(\begin{array}{c}40\\ 80\\ 0\end{array}\right)\backslash overrightarrow\{\backslash mathrm\; a\}=\backslash begin\{pmatrix\}40\backslash \backslash 80\backslash \backslash 0\backslash end\{pmatrix\}$

als Aufpunkt der Ebene.

$\mathrm{E}:\overrightarrow{\mathrm{n}}\circ [\overrightarrow{\mathrm{x}}-\overrightarrow{\mathrm{a}}]=0\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; n\}\backslash circ\backslash left[\backslash overrightarrow\{\backslash mathrm\; x\}-\backslash overrightarrow\{\backslash mathrm\; a\}\backslash right]=0$

Setze  $\overrightarrow{\mathrm{n}}\backslash overrightarrow\{\backslash mathrm\; n\}$  und  $\overrightarrow{\mathrm{a}}\backslash overrightarrow\{\backslash mathrm\; a\}$  ein.

$\mathrm{E}:\left(\begin{array}{c}-10\\ 11\\ 2\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)-\left(\begin{array}{c}40\\ 80\\ 0\end{array}\right)]=0\backslash mathrm\; E:\backslash ;\backslash begin\{pmatrix\}-10\backslash \backslash 11\backslash \backslash 2\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}40\backslash \backslash 80\backslash \backslash 0\backslash end\{pmatrix\}\backslash right]=0$

$E:\left(\begin{array}{c}6\\ -1\\ -4\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}1\\ 0\\ 3\end{array}\right)]=0E:\backslash begin\{pmatrix\}6\backslash \backslash -1\backslash \backslash -4\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}x\_1\backslash \backslash x\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}\backslash right]=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus und berechne das Skalarprodukt.

$E:6x_1-6-x_2-4x_3+12=0E:\backslash ;6\{x\}\_1-6-\{x\}\_2-4\{x\}\_3+12=0$

Fasse zusammen.

$E:6x_1-x_2-4x_3+6=0E:6\{x\}\_1-\{x\}\_2-4\{x\}\_3+6=0$

$E:\left(\begin{array}{c}4\\ -7\\ 2\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)]=0E:\backslash begin\{pmatrix\}4\backslash \backslash -7\backslash \backslash 2\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne zunächst das Skalarprodukt :

$E:4\cdot x_1+(-7)\cdot (x_2-1)+2\cdot (x_3-2)=0E:\; 4\backslash cdot\{x\}\_1+\backslash left(-7\backslash right)\backslash cdot\backslash left(\{x\}\_2-1\backslash right)+2\backslash cdot\backslash left(\{x\}\_3-2\backslash right)=0$

Multipliziere nun die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:4x_1-7x_2+7+2x_3-4=0E:\backslash ;4\{x\}\_1-7\{x\}\_2+7+2\{x\}\_3-4=0$

Fasse zusammen:

$E:4x_1-7x_2+2x_3+3=0E:\; 4\{x\}\_1-7\{x\}\_2+2\{x\}\_3+3=0$

$E:\left(\begin{array}{c}-6\\ -3\\ 2\end{array}\right)\circ \left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=0E:\backslash ;\backslash begin\{pmatrix\}-6\backslash \backslash -3\backslash \backslash 2\backslash end\{pmatrix\}\backslash circ\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}=0$

Der zweite Faktor enthält keine Differenz. Berechne daher nur das Skalarprodukt :

$E:-6x_1-3x_2+2x_3=0\mathrm{\mid }\cdot (-1)E:-6\{x\}\_1-3\{x\}\_2+2\{x\}\_3=0\backslash ;\; |\; \backslash cdot\; (-1)$

$E:\phantom{-}6x_1+3x_2-2x_3=0E:\backslash phantom\{-\}6\{x\}\_1+3\{x\}\_2-2\{x\}\_3=0$

$E:\left(\begin{array}{c}-4\\ -2\\ 1\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}1\\ 1\\ -1\end{array}\right)]=0E:\; \backslash begin\{pmatrix\}-4\backslash \backslash -2\backslash \backslash 1\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash -1\backslash end\{pmatrix\}\backslash right]=0$

Berechne das Skalarprodukt :

$E:-4\cdot (x_1-1)+(-2)\cdot (x_2-1)+1\cdot (x_3+1)=0E:\; -4\backslash cdot\backslash left(\{x\}\_1-1\backslash right)+\backslash left(-2\backslash right)\backslash cdot\backslash left(\{x\}\_2-1\backslash right)+1\backslash cdot\backslash left(\{x\}\_3+1\backslash right)=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:-4x_1+4-2x_2+2+x_3+1=0E:\; -4\{x\}\_1+4-2\{x\}\_2+2+\{x\}\_3+1=0$

Fasse zusammen:

$E:-4x_1-2x_2+x_3+7=0E:\; -4\{x\}\_1-2\{x\}\_2+\{x\}\_3+7=0$

$E:\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)]=0E:\; \backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash -1\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}3\backslash \backslash 1\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne das Skalarprodukt :

$E:1\cdot (x_1-3)+(-1)\cdot (x_3-2)=0E:\; 1\backslash cdot\backslash left(\{x\}\_1-3\backslash right)+\backslash left(-1\backslash right)\backslash cdot\backslash left(\{x\}\_3-2\backslash right)=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:x_1-3-x_3+2=0E:\; x\_1-3-\{x\}\_3+2=0$

Fasse zusammen:

$E:x_1-x_3-1=0E:\; \{x\}\_1-\{x\}\_3-1=0$

$E:\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right)]=0E:\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 0\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}2\backslash \backslash 2\backslash \backslash 2\backslash end\{pmatrix\}\backslash right]=0$

Berechne das Skalarprodukt :

$E:1\cdot (x_2-2)=0E:\; 1\backslash cdot\backslash left(\{x\}\_2-2\backslash right)=0$

$E:x_2-2=0E:\; \{x\}\_2-2=0$

(Hinweis: Die Ebene ist parallel zur  $x_1-x_3-\text{Ebene}\{x\}\_1-\{x\}\_3-\backslash text\{Ebene\}$ )

$E:\left(\begin{array}{c}-500\\ 550\\ 100\end{array}\right)\circ [\left(\begin{array}{c}{\mathrm{x}}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}40\\ 80\\ 0\end{array}\right)]=0E:\; \backslash begin\{pmatrix\}-500\backslash \backslash 550\backslash \backslash 100\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{\backslash mathrm\; x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}40\backslash \backslash 80\backslash \backslash 0\backslash end\{pmatrix\}\backslash right]=0$

Berechne das Skalarprodukt .

$E:-500\cdot (x_1-40)+550\cdot (x_2-80)+100\cdot x_3=0E:\; -500\backslash cdot\backslash left(\{x\}\_1-40\backslash right)+550\backslash cdot\backslash left(\{\; x\}\_2-80\backslash right)+100\backslash cdot\{x\}\_3=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus.

$E:-500x_1+20000+550x_2-44000+100x_3=0E:\; -500\{x\}\_1+20000+550\; \{x\}\_2-44000+100\; \{x\}\_3=0$

Fasse zusammen.

$E:-500x_1+550x_2+100x_3-24000=0E:\; -500\{x\}\_1+550\{x\}\_2+100\{x\}\_3-24000=0$

Um die Gleichung noch zu vereinfachen, kann man sie auf beide Seiten durch  $5050$  dividieren.

$E:-10x_1+11x_2+2x_3-480=0E:\; -10\{x\}\_1+11\{x\}\_2+2\{x\}\_3-480=0$

Alternative Lösung: 

Meistens ist es einfacher, zuerst den Normalenvektor der Ebene zu vereinfachen und dann erst das Skalarprodukt zu berechnen.

Klammere  $5050$  im Normalenvektor aus, teile durch $5050$ und berechne dann das Skalarprodukt:

$E:\left(\begin{array}{c}-10\\ 11\\ 22\end{array}\right)\circ [\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)-\left(\begin{array}{c}40\\ 80\\ 0\end{array}\right)]=0E:\; \backslash begin\{pmatrix\}-10\backslash \backslash 11\backslash \backslash 22\backslash end\{pmatrix\}\backslash circ\backslash left[\backslash begin\{pmatrix\}\{x\}\_1\backslash \backslash \{x\}\_2\backslash \backslash \{x\}\_3\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}40\backslash \backslash 80\backslash \backslash 0\backslash end\{pmatrix\}\backslash right]=0$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:-10\cdot (x_1-40)+11\cdot (x_2-80)+2\cdot x_3=0_{}^{}E:-10\backslash cdot\backslash left(x\_1-40\backslash right)+11\backslash cdot\backslash left(x\_2-80\backslash right)+2\backslash cdot\; x\_3=0\_\{\; \}^\{\; \}$

Multipliziere die Klammern mit Hilfe des  Distributivgesetzes aus:

$E:-10x_1+400+11x_2-880+2x_3=0E:\; -10\{x\}\_1+400+11\{x\}\_2-880+2\{x\}\_3=0$

Fasse zusammen.

$E:-10x_1+11x_2+2x_3-480=0E:\; -10\{x\}\_1+11\{x\}\_2+2\{x\}\_3-480=0$

Löse die Ebenengleichung nach ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$ auf:

${\mathrm{x}}_3=-\frac{2}{3}{\mathrm{x}}_1+\frac{1}{3}{\mathrm{x}}_2+\frac{5}{3}\{\backslash mathrm\; x\}\_3=-\backslash frac23\{\backslash mathrm\; x\}\_1+\backslash frac13\{\backslash mathrm\; x\}\_2+\backslash frac53$

Ersetze  ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$  durch  $\lambda \backslash mathrm\backslash lambda$  und  ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$  durch  $\mu \backslash mathrm\backslash mu$  .

${\mathrm{x}}_3=-\frac{2}{3}\lambda +\frac{1}{3}\mu +\frac{5}{3}\{\backslash mathrm\; x\}\_3=-\backslash frac23\backslash mathrm\backslash lambda+\backslash frac13\backslash mathrm\backslash mu+\backslash frac53$

Schreibe ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$,  ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$ und  ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$ passend übereinander.

$\begin{array}{c}{\mathrm{x}}_1=0+1\cdot \lambda +0\cdot \mu \\ {\mathrm{x}}_2=0+0\cdot \lambda +1\cdot \mu \\ {\mathrm{x}}_3=\frac{5}{3}-\frac{2}{3}\cdot \lambda +\frac{1}{3}\cdot \mu \end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\{\backslash mathrm\; x\}\_1\backslash ;=\backslash ;0\backslash ;\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash ;=\backslash ;0\backslash ;\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_3=\backslash frac53\backslash ;\backslash ;\backslash ;-\backslash frac23\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;+\backslash frac13\backslash cdot\backslash mathrm\backslash mu\backslash end\{array\}$

Lese den Aufpunkt und die Richtungsvektoren ab.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ 0\\ \frac{5}{3}\end{array}\right)+\lambda \cdot \left(\begin{array}{c}1\\ 0\\ -\frac{2}{3}\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 1\\ \frac{1}{3}\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash \backslash frac53\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash -\backslash frac23\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash \backslash frac13\backslash end\{pmatrix\}$

Die beiden Richtungsvektoren können noch mit  $33$  multipliziert werden.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ 0\\ \frac{5}{3}\end{array}\right)+\lambda \cdot \left(\begin{array}{c}3\\ 0\\ -2\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 3\\ 1\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash \backslash frac53\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}3\backslash \backslash 0\backslash \backslash -2\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 3\backslash \backslash 1\backslash end\{pmatrix\}$

Löse die Ebenengleichung nach  ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$  auf:

${\mathrm{x}}_3=\frac{1}{4}{\mathrm{x}}_1-\frac{1}{2}{\mathrm{x}}_2\{\backslash mathrm\; x\}\_3=\backslash frac14\{\backslash mathrm\; x\}\_1-\backslash frac12\{\backslash mathrm\; x\}\_2$

Ersetze ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$ durch $\lambda \backslash mathrm\backslash lambda$ und ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$ durch $\mu \backslash mathrm\backslash mu$ .

${\mathrm{x}}_3=\frac{1}{4}\lambda -\frac{1}{2}\mu \{\backslash mathrm\; x\}\_3=\backslash frac14\backslash mathrm\backslash lambda-\backslash frac12\backslash mathrm\backslash mu$

Schreibe ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$,  ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$ und  ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$ passend übereinander.

$\begin{array}{c}{\mathrm{x}}_1=0+1\cdot \lambda +0\cdot \mu \\ {\mathrm{x}}_2=0+0\cdot \lambda +1\cdot \mu \\ {\mathrm{x}}_3=0+\frac{1}{4}\cdot \lambda -\frac{1}{2}\cdot \mu \end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\{\backslash mathrm\; x\}\_1\backslash ;=\backslash ;0\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash ;=\backslash ;0\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash ;=\backslash ;0\backslash ;\backslash ;\backslash ;+\backslash frac14\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;-\backslash frac12\backslash cdot\backslash mathrm\backslash mu\backslash end\{array\}$

Lese den Aufpunkt und die Richtungsvektoren ab.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)+\lambda \cdot \left(\begin{array}{c}1\\ 0\\ \frac{1}{4}\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 1\\ -\frac{1}{2}\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash \backslash frac14\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash -\backslash frac12\backslash end\{pmatrix\}$

Die beiden Richtungsvektoren können noch mit $44$  bzw. $22$ multipliziert werden.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)+\lambda \cdot \left(\begin{array}{c}4\\ 0\\ 1\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 2\\ -1\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash \backslash textstyle0\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}4\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 2\backslash \backslash -1\backslash end\{pmatrix\}$

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\lambda \cdot \left(\begin{array}{c}4\\ 0\\ 1\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 2\\ -1\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}4\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 2\backslash \backslash -1\backslash end\{pmatrix\}$

Löse die Ebenengleichung nach  ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$  auf:

${\mathrm{x}}_3=-\frac{3}{4}{\mathrm{x}}_1+\frac{5}{4}\{\backslash mathrm\; x\}\_3=-\backslash frac34\{\backslash mathrm\; x\}\_1+\backslash frac54$

Ersetze ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$ durch $\lambda \backslash mathrm\backslash lambda$ und setze ${\mathrm{x}}_2=\mu \{\backslash mathrm\; x\}\_2=\backslash mathrm\backslash mu$.

${\mathrm{x}}_3=-\frac{3}{4}\lambda +\frac{5}{4}\{\backslash mathrm\; x\}\_3=-\backslash frac34\backslash mathrm\backslash lambda+\backslash frac54$

Schreibe ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$, ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$ und ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$ passend übereinander.

$\begin{array}{c}{\mathrm{x}}_1=0+1\cdot \lambda +0\cdot \mu \\ {\mathrm{x}}_2=0+0\cdot \lambda +1\cdot \mu \\ {\mathrm{x}}_3=\frac{5}{4}-\frac{3}{4}\cdot \lambda +0\cdot \mu \end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\{\backslash mathrm\; x\}\_1=\backslash ;\backslash ;0\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_2=\backslash ;\backslash ;0\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_3=\backslash ;\backslash frac54\backslash ;\backslash ;\backslash ;-\backslash frac34\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash mu\backslash end\{array\}$

Lese den Aufpunkt und die Richtungsvektoren ab.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ 0\\ \frac{5}{4}\end{array}\right)+\lambda \cdot \left(\begin{array}{c}1\\ 0\\ -\frac{3}{4}\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash \backslash frac54\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash -\backslash frac34\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 0\backslash end\{pmatrix\}$

Die erste Richtungsvektor kann noch mit $44$ multipliziert werden.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ 0\\ \frac{5}{4}\end{array}\right)+\lambda \cdot \left(\begin{array}{c}4\\ 0\\ -3\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash \backslash frac54\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}4\backslash \backslash 0\backslash \backslash -3\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash 0\backslash end\{pmatrix\}$

Die Ebenengleichung kann nicht nach  ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$  aufgelöst werden. Löse deshalb nach  ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$  auf:

${\mathrm{x}}_2=-\frac{2}{3}{\mathrm{x}}_1+\frac{1}{3}\{\backslash mathrm\; x\}\_2=-\backslash frac23\{\backslash mathrm\; x\}\_1+\backslash frac13$

Ersetze ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$ durch $\lambda \backslash mathrm\backslash lambda$ und setze ${\mathrm{x}}_3=\mu \{\backslash mathrm\; x\}\_3=\backslash mathrm\backslash mu$.

${\mathrm{x}}_2=-\frac{2}{3}\lambda +\frac{1}{3}\{\backslash mathrm\; x\}\_2=-\backslash frac23\backslash mathrm\backslash lambda+\backslash frac13$

Schreibe ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$, ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$ und ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$ passend übereinander.

$\begin{array}{c}{\mathrm{x}}_1=0+1\cdot \lambda +0\cdot \mu \\ {\mathrm{x}}_2=\frac{1}{3}-\frac{2}{3}\lambda +0\cdot \mu \\ {\mathrm{x}}_3=0+0\cdot \lambda +1\cdot \mu \end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\{\backslash mathrm\; x\}\_1\backslash ;=\backslash ;\backslash ;0\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash ;=\backslash ;\backslash frac13\backslash ;\backslash ;\backslash ;-\backslash frac23\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash ;=\backslash ;\backslash ;0\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash mu\backslash end\{array\}$

Lese den Aufpunkt und die Richtungsvektoren ab.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ \frac{1}{3}\\ 0\end{array}\right)+\lambda \cdot \left(\begin{array}{c}1\\ -\frac{2}{3}\\ 0\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash \backslash frac13\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}1\backslash \backslash -\backslash frac23\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}$

Die erste Richtungsvektor kann noch mit $33$ multipliziert werden.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ \frac{1}{3}\\ 0\end{array}\right)+\lambda \cdot \left(\begin{array}{c}3\\ -2\\ 0\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash \backslash frac13\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}3\backslash \backslash -2\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}$

Die Ebenengleichung kann nicht nach ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$ aufgelöst werden. Löse deshalb nach ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$ auf:

${\mathrm{x}}_2=\frac{3}{2}\{\backslash mathrm\; x\}\_2=\backslash frac32$

Setze ${\mathrm{x}}_1=\lambda \{\backslash mathrm\; x\}\_1=\backslash mathrm\backslash lambda$ und ${\mathrm{x}}_3=\mu \{\backslash mathrm\; x\}\_3=\backslash mathrm\backslash mu$ und schreibe ${\mathrm{x}}_1\{\backslash mathrm\; x\}\_1$, ${\mathrm{x}}_2\{\backslash mathrm\; x\}\_2$ und ${\mathrm{x}}_3\{\backslash mathrm\; x\}\_3$ passend übereinander.

$\begin{array}{c}{\mathrm{x}}_1=0+1\cdot \lambda +0\cdot \mu \\ {\mathrm{x}}_2=\frac{3}{2}+0\cdot \lambda +0\cdot \mu \\ {\mathrm{x}}_3=0+0\cdot \lambda +1\cdot \mu \end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\{\backslash mathrm\; x\}\_1\backslash ;=\backslash ;\backslash ;0\backslash ;\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_2\backslash ;=\backslash ;\backslash frac32\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash mu\backslash \backslash \{\backslash mathrm\; x\}\_3\backslash ;=\backslash ;\backslash ;0\backslash ;\backslash ;\backslash ;\backslash ;+0\backslash cdot\backslash mathrm\backslash lambda\backslash ;\backslash ;\backslash ;+1\backslash cdot\backslash mathrm\backslash mu\backslash end\{array\}$

Lese den Aufpunkt und die Richtungsvektoren ab.

$\mathrm{E}:\overrightarrow{\mathrm{x}}=\left(\begin{array}{c}0\\ \frac{3}{2}\\ 0\end{array}\right)+\lambda \cdot \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+\mu \cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\backslash mathrm\; E:\backslash ;\backslash overrightarrow\{\backslash mathrm\; x\}=\backslash begin\{pmatrix\}0\backslash \backslash \backslash frac32\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash lambda\backslash cdot\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}+\backslash mathrm\backslash mu\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash 1\backslash end\{pmatrix\}$