Analytische Geometrie-Lagebeziehung-Ebene - Ebene

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Beispiel Nr: 09
$\begin{array}{l} \text{Gegeben:} \\ \text{Ebene1: } \vec{x} =\left( \begin{array}{c} a1 \\ a2 \\ a3 \\ \end{array} \right) + \lambda \left( \begin{array}{c} b1 \\ b2 \\ b3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} c1 \\ c2 \\ c3 \\ \end{array} \right) \\ \text{Ebene2: } n1 x_1+n2 x_2+n3 x_3+k1=0 \\ \\ \text{Gesucht:} \\\text{Lage der Ebenen zueinander} \\ \\ \textbf{Gegeben:} \\ \text{Ebene1: } \vec{x} =\left( \begin{array}{c} -2 \\ -4 \\ 2 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 1 \\ 2 \\ 2 \\ \end{array} \right) + \sigma \left( \begin{array}{c} 0 \\ -1 \\ -2 \\ \end{array} \right) \\ \text{Ebene2: } 1 x_1+1 x_2+0 x_3+0=0 \\ \\ \\ \textbf{Rechnung:} \\ \text{Ebene: } \vec{x} =\left( \begin{array}{c} -2 \\ -4 \\ 2 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 1 \\ 2 \\ 2 \\ \end{array} \right) + \sigma \left( \begin{array}{c} 0 \\ -1 \\ -2 \\ \end{array} \right) \\ \text{Ebene: } 1 x_1+1 x_2+0 x_3+0=0 \\ \begin{array}{cccc} x_1=& -2 &+1\lambda &+0\sigma \\ x_2=&-4 &+2\lambda &-1\sigma \\ x_3=&2 &+2\lambda &-1\sigma\\ \end{array} \\ 1( -2+1\lambda+0\sigma) +1(-4+2\lambda-1\sigma) +0 (2+2\lambda-2\sigma)+0=0 \\ 3\lambda-1\sigma-6=0 \\ \\ \sigma=\frac{-3 \lambda +6}{-1} \\ \sigma= 3 \lambda -6 \\ \vec{x} = \left( \begin{array}{c} -2 \\ -4 \\ 2 \\ \end{array} \right) +\lambda \cdot \left( \begin{array}{c} 1 \\ 2 \\ 2 \\ \end{array} \right) +(3\lambda-6) \cdot \left( \begin{array}{c} 0 \\ -1 \\ -2 \\ \end{array} \right) \\ \text{Schnittgerade: } \vec{x} =\left( \begin{array}{c} -2 \\ 2 \\ 14 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 1 \\ -1 \\ -4 \\ \end{array} \right) \\ \\ \end{array}$