Beispiel Nr: 06
$\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} 1 \\ -2 \\ 8 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) \\ \text{Gerade 2: } \vec{x} =\left( \begin{array}{c} 9 \\ -5 \\ 3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} -4 \\ -4 \\ -3 \\ \end{array} \right) \\ \\ \\ \textbf{Rechnung:} \\ \text{Gerade 1: } \vec{x} =\left( \begin{array}{c} 1 \\ -2 \\ 8 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) \\ \text{Gerade 2: } \vec{x} =\left( \begin{array}{c} 9 \\ -5 \\ 3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} -4 \\ -4 \\ -3 \\ \end{array} \right) \\ \text{Richtungsvektoren: } \\ \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) =k \cdot \left( \begin{array}{c} -4 \\ -4 \\ -3 \\ \end{array} \right) \\ \begin{array}{cccc} 4&=&-4 k& \quad /:-4 \quad \Rightarrow k=-1 \\ -7&=&-4 k & \quad /:-4 \quad \Rightarrow k=1\frac{3}{4} \\ -8&=&-3 k & \quad /:-3 \quad \Rightarrow k=2\frac{2}{3} \\ \end{array} \\ \\ \Rightarrow \text{Geraden sind nicht parallel} \\ \left( \begin{array}{c} 1 \\ -2 \\ 8 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) = \left( \begin{array}{c} 9 \\ -5 \\ 3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} -4 \\ -4 \\ -3 \\ \end{array} \right) \\ \begin{array}{cccccc} 1& +4\lambda &=& 9& -4\sigma& \quad /-1 \quad /+4 \sigma\\ -2& -7\lambda &=& -5& -4 \sigma& \quad /+2 \quad /+4 \sigma\\ 8& -8\lambda &=& 3& -3 \sigma& \quad /-8 \quad /+3 \sigma\\ \end{array} \\ \\I \qquad 4 \lambda +4 \sigma =8\\ II \qquad -7 \lambda +4 \sigma = -3 \\ III \qquad -8 \lambda -3 \sigma = -5 \\ \\ \text{Aus 2 Gleichungen }\lambda \text{ und } \sigma \text{ berechnen } \\ I \qquad 4 \lambda +4 \sigma =8 \qquad / \cdot\left(-7\right)\\ II \qquad -7 \lambda +4 \sigma = -3 \qquad / \cdot\left(-4\right)\\ I \qquad -28 \lambda -28 \sigma =-56\\ II \qquad 28 \lambda -16 \sigma = 12 \\ \text{I + II}\\ I \qquad -28 \lambda +28 \lambda-28 \sigma -16 \sigma =-56 +12\\ -44 \sigma = -44 \qquad /:\left(-44\right) \\ \sigma = \frac{-44}{-44} \\ \sigma=1 \\ \sigma \text{ in I}\\ I \qquad -28 \lambda -28 \cdot 1 =-56 \\ -28 \lambda -28 =-56 \qquad / +28 \\ -28 \lambda =-56 +28 \\ -28 \lambda =-28 \qquad / :\left(-28\right) \\ \lambda = \frac{-28}{-28} \\ \lambda=1 \\ \lambda \text{ und } \sigma \text{ in die verbleibende Gleichung einsetzen} \\ III \quad 8+1\cdot\left(-8\right)=3+1\cdot\left(-3\right) \\ 0=0 \\ \lambda \text{ oder } \sigma \text{ in die Geradengleichung einsetzen} \\ \\ \vec{x} = \left( \begin{array}{c} 1 \\ -2 \\ 8 \\ \end{array} \right) +1 \cdot \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) \\ \text{Schnittpunkt: }S(5,-9,0) \\ \\$
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