Aufgabe:
Untersuchen Sie die gegenseitige Lage der Geraden g und h.
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8. a) \( g: \vec{x}=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)+t \cdot\left(\begin{array}{l}2 \\ 0 \\ 4\end{array}\right) \quad h: \vec{x}=\left(\begin{array}{l}1 \\ 2 \\ 6\end{array}\right)+s \cdot\left(\begin{array}{c}-2 \\ 1 \\ 1\end{array}\right) \quad \) (keine vielfache)
\( \begin{array}{l} \left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)+t \cdot\left(\begin{array}{l} 2 \\ 0 \\ 4 \end{array}\right)=\left(\begin{array}{l} 1 \\ 2 \\ 6 \end{array}\right)+s \cdot\left(\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right) \\ \begin{array}{ll} I & 1+2 t=1-2 s & -1+2 t=1-2 s \mid+2 s \\ \frac{11}{11} 1+0 t=2+1 s & 2 s+1+2 t=1 \quad 1-1 \\ \frac{11}{11} 1+4 t=6+1 s & 2 s+2 t=0 \end{array} \quad \begin{array}{l} -1 s+1+0 t=2 \quad 1-1 \\ -1 s+0 t=1 \end{array} \end{array} \)
\( \begin{array}{l} \begin{array}{l} \text { III } 1+4 t=6+1 s \mid-1 s \\ -1 s+1+4 t=6 \mid-1 \\ -1 s+4 t=3 \end{array} \\ h: \vec{x}=\left(\begin{array}{l} 1 \\ 2 \\ 6 \end{array}\right)+1 \cdot\left(\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right)+\left(\begin{array}{l} 1 \\ 2 \\ 6 \end{array}\right)=\left(\begin{array}{l} 1 \\ 3 \\ 7 \end{array}\right) \quad S=\left(\begin{array}{l} 1 \\ 3 \\ 7 \end{array}\right) \end{array} \)
Problem/Ansatz:
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8. a) \( g: \vec{x}=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)+t \cdot\left(\begin{array}{l}2 \\ 0 \\ 4\end{array}\right) \quad h: \vec{x}=\left(\begin{array}{l}1 \\ 2 \\ 6\end{array}\right)+s \cdot\left(\begin{array}{c}-2 \\ 1 \\ 1\end{array}\right) \quad \) (keine vielfache)
\( \begin{array}{l} \left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)+t \cdot\left(\begin{array}{l} 2 \\ 0 \\ 4 \end{array}\right)=\left(\begin{array}{l} 1 \\ 2 \\ 6 \end{array}\right)+s \cdot\left(\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right) \\ \begin{array}{ll} I & 1+2 t=1-2 s & -1+2 t=1-2 s \mid+2 s \\ \frac{11}{11} 1+0 t=2+1 s & 2 s+1+2 t=1 \quad 1-1 \\ \frac{11}{11} 1+4 t=6+1 s & 2 s+2 t=0 \end{array} \quad \begin{array}{l} -1 s+1+0 t=2 \quad 1-1 \\ -1 s+0 t=1 \end{array} \end{array} \)
\( \begin{array}{l} \begin{array}{l} \text { III } 1+4 t=6+1 s \mid-1 s \\ -1 s+1+4 t=6 \mid-1 \\ -1 s+4 t=3 \end{array} \\ h: \vec{x}=\left(\begin{array}{l} 1 \\ 2 \\ 6 \end{array}\right)+1 \cdot\left(\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right)+\left(\begin{array}{l} 1 \\ 2 \\ 6 \end{array}\right)=\left(\begin{array}{l} 1 \\ 3 \\ 7 \end{array}\right) \quad S=\left(\begin{array}{l} 1 \\ 3 \\ 7 \end{array}\right) \end{array} \)
Ich habe zu dieser Aufgabe den Schnittpunkt S=(1/3/7) raus, allerdings steht in den Lösungen der Schnittpunkt:(3/1/5). Wo liegt der Fehler in meiner Aufgabe?
