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.

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