Beispiel Nr: 04
$ \text{Gegeben:} \text{Gerade 1: } \vec{x} =\left( \begin{array}{c} a1 \\ a2 \\ a3 \\ \end{array} \right) + \lambda \left( \begin{array}{c} b1 \\ b2 \\ b3 \\ \end{array} \right) \\ \text{Gerade 2: } \vec{x} =\left( \begin{array}{c} c1 \\ c2 \\ c3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} d1 \\ d2 \\ d3 \\ \end{array} \right) \\ \text{Gesucht:} \text{Die Lage der Geraden zueinander.} \\ \\ \textbf{Gegeben:} \\ \text{Gerade 1: } \vec{x} =\left( \begin{array}{c} 8 \\ 4 \\ 2 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 1 \\ 3 \\ 7 \\ \end{array} \right) \\ \text{Gerade 2: } \vec{x} =\left( \begin{array}{c} 7 \\ 7 \\ 8 \\ \end{array} \right) + \sigma \left( \begin{array}{c} 5 \\ 1 \\ 4 \\ \end{array} \right) \\ \\ \\ \textbf{Rechnung:} \\ \text{Gerade 1: } \vec{x} =\left( \begin{array}{c} 8 \\ 4 \\ 2 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 1 \\ 3 \\ 7 \\ \end{array} \right) \\ \text{Gerade 2: } \vec{x} =\left( \begin{array}{c} 7 \\ 7 \\ 8 \\ \end{array} \right) + \sigma \left( \begin{array}{c} 5 \\ 1 \\ 4 \\ \end{array} \right) \\ \text{Richtungsvektoren: } \\ \left( \begin{array}{c} 1 \\ 3 \\ 7 \\ \end{array} \right) =k \cdot \left( \begin{array}{c} 5 \\ 1 \\ 4 \\ \end{array} \right) \\ \begin{array}{cccc} 1&=&+5 k& \quad /:5 \quad \Rightarrow k=\frac{1}{5} \\ 3&=&+1 k & \quad /:1 \quad \Rightarrow k=3 \\ 7&=&+4 k & \quad /:4 \quad \Rightarrow k=1\frac{3}{4} \\ \end{array} \\ \\ \Rightarrow \text{Geraden sind nicht parallel} \\ \left( \begin{array}{c} 8 \\ 4 \\ 2 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 1 \\ 3 \\ 7 \\ \end{array} \right) = \left( \begin{array}{c} 7 \\ 7 \\ 8 \\ \end{array} \right) + \sigma \left( \begin{array}{c} 5 \\ 1 \\ 4 \\ \end{array} \right) \\ \begin{array}{cccccc} 8& +1\lambda &=& 7& +5\sigma& \quad /-8 \quad /-5 \sigma\\ 4& +3\lambda &=& 7& +1 \sigma& \quad /-4 \quad /-1 \sigma\\ 2& +7\lambda &=& 8& +4 \sigma& \quad /-2 \quad /-4 \sigma\\ \end{array} \\ \\I \qquad 1 \lambda -5 \sigma =-1\\ II \qquad 3 \lambda -1 \sigma = 3 \\ III \qquad 7 \lambda +4 \sigma = 6 \\ \\ \text{Aus 2 Gleichungen }\lambda \text{ und } \sigma \text{ berechnen } \\ I \qquad 1 \lambda -5 \sigma =-1 \qquad / \cdot3\\ II \qquad 3 \lambda -1 \sigma = 3 \qquad / \cdot\left(-1\right)\\ I \qquad 3 \lambda -15 \sigma =-3\\ II \qquad -3 \lambda +1 \sigma = -3 \\ \text{I + II}\\ I \qquad 3 \lambda -3 \lambda-15 \sigma +1 \sigma =-3 -3\\ -14 \sigma = -6 \qquad /:\left(-14\right) \\ \sigma = \frac{-6}{-14} \\ \sigma=\frac{3}{7} \\ \sigma \text{ in I}\\ I \qquad 3 \lambda -15 \cdot \frac{3}{7} =-3 \\ 3 \lambda -6\frac{3}{7} =-3 \qquad / +6\frac{3}{7} \\ 3 \lambda =-3 +6\frac{3}{7} \\ 3 \lambda =3\frac{3}{7} \qquad / :3 \\ \lambda = \frac{3\frac{3}{7}}{3} \\ \lambda=1\frac{1}{7} \\ \lambda \text{ und } \sigma \text{ in die verbleibende Gleichung einsetzen} \\ III \quad 2+1\frac{1}{7}\cdot7=8+\frac{3}{7}\cdot4 \\ 10=9\frac{5}{7} \\ \text{Geraden sind windschief} \\ $