Analytische Geometrie-Lagebeziehung-Ebene - Ebene
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Beispiel Nr: 06
$\begin{array}{l} \text{Gegeben:} \\
\text{Ebene1: }
\vec{x} =\left(
\begin{array}{c}
a1 \\
a2 \\
a3 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
b1 \\
b2 \\
b3 \\
\end{array}
\right)
+ \sigma
\left(
\begin{array}{c}
c1 \\
c2 \\
c3 \\
\end{array}
\right) \\
\text{Ebene2: } n1 x_1+n2 x_2+n3 x_3+k1=0 \\
\\ \text{Gesucht:} \\\text{Lage der Ebenen zueinander}
\\ \\ \textbf{Gegeben:} \\
\text{Ebene1: }
\vec{x} =\left(
\begin{array}{c}
2 \\
3 \\
2 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
1 \\
-2 \\
1 \\
\end{array}
\right)
+ \sigma
\left(
\begin{array}{c}
5 \\
3 \\
1 \\
\end{array}
\right) \\
\text{Ebene2: } -5 x_1+4 x_2-13 x_3-28=0 \\
\\ \\ \textbf{Rechnung:} \\
\text{Ebene: }
\vec{x} =\left(
\begin{array}{c}
2 \\
3 \\
2 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
1 \\
-2 \\
1 \\
\end{array}
\right)
+ \sigma
\left(
\begin{array}{c}
5 \\
3 \\
1 \\
\end{array}
\right) \\
\text{Ebene: } -5 x_1+4 x_2-13 x_3-28=0 \\
\begin{array}{cccc}
x_1=& 2 &+1\lambda &+5\sigma \\
x_2=&3 &-2\lambda &+3\sigma \\
x_3=&2 &+1\lambda &+3\sigma\\
\end{array} \\
-5( 2+1\lambda+5\sigma) +4(3-2\lambda+3\sigma) -13 (2+1\lambda+1\sigma)-28=0 \\
-26\lambda-26\sigma-52=0 \\
\\
\sigma=\frac{+26 \lambda +52}{-26} \\
\sigma= -1 \lambda -2 \\
\vec{x} = \left(
\begin{array}{c}
2 \\
3 \\
2 \\
\end{array}
\right) +\lambda \cdot
\left(
\begin{array}{c}
1 \\
-2 \\
1 \\
\end{array}
\right) +(-1\lambda-2) \cdot
\left(
\begin{array}{c}
5 \\
3 \\
1 \\
\end{array}
\right) \\
\text{Schnittgerade: }
\vec{x} =\left(
\begin{array}{c}
-8 \\
-3 \\
0 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
-4 \\
-5 \\
0 \\
\end{array}
\right) \\
\\
\end{array}$