Beispiel Nr: 05
$\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 \\ 3 \\ 2 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 1 \\ -2 \\ 1 \\ \end{array} \right) + \sigma \left( \begin{array}{c} 5 \\ 3 \\ 1 \\ \end{array} \right) \\ \text{Ebene2: } -5 x_1+4 x_2-13 x_3-28=0 \\ \\ \\ \textbf{Rechnung:} \\ \text{Ebene: } \vec{x} =\left( \begin{array}{c} 2 \\ 3 \\ 2 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 1 \\ -2 \\ 1 \\ \end{array} \right) + \sigma \left( \begin{array}{c} 5 \\ 3 \\ 1 \\ \end{array} \right) \\ \text{Ebene: } -5 x_1+4 x_2-13 x_3-28=0 \\ \begin{array}{cccc} x_1=& 2 &+1\lambda &+5\sigma \\ x_2=&3 &-2\lambda &+3\sigma \\ x_3=&2 &+1\lambda &+3\sigma\\ \end{array} \\ -5( 2+1\lambda+5\sigma) +4(3-2\lambda+3\sigma) -13 (2+1\lambda+1\sigma)-28=0 \\ -26\lambda-26\sigma-52=0 \\ \\ \sigma=\frac{+26 \lambda +52}{-26} \\ \sigma= -1 \lambda -2 \\ \vec{x} = \left( \begin{array}{c} 2 \\ 3 \\ 2 \\ \end{array} \right) +\lambda \cdot \left( \begin{array}{c} 1 \\ -2 \\ 1 \\ \end{array} \right) +(-1\lambda-2) \cdot \left( \begin{array}{c} 5 \\ 3 \\ 1 \\ \end{array} \right) \\ \text{Schnittgerade: } \vec{x} =\left( \begin{array}{c} -8 \\ -3 \\ 0 \\ \end{array} \right) + \lambda \left( \begin{array}{c} -4 \\ -5 \\ 0 \\ \end{array} \right) \\ \\ $