Analytische Geometrie-Lagebeziehung-Ebene - Ebene
1
2
3
4
5
6
7
8
9
Beispiel Nr: 09
$\begin{array}{l} \text{Gegeben:} \\
\text{Ebene1: }
\vec{x} =\left(
\begin{array}{c}
a1 \\
a2 \\
a3 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
b1 \\
b2 \\
b3 \\
\end{array}
\right)
+ \sigma
\left(
\begin{array}{c}
c1 \\
c2 \\
c3 \\
\end{array}
\right) \\
\text{Ebene2: } n1 x_1+n2 x_2+n3 x_3+k1=0 \\
\\ \text{Gesucht:} \\\text{Lage der Ebenen zueinander}
\\ \\ \textbf{Gegeben:} \\
\text{Ebene1: }
\vec{x} =\left(
\begin{array}{c}
-2 \\
-4 \\
2 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
1 \\
2 \\
2 \\
\end{array}
\right)
+ \sigma
\left(
\begin{array}{c}
0 \\
-1 \\
-2 \\
\end{array}
\right) \\
\text{Ebene2: } 1 x_1+1 x_2+0 x_3+0=0 \\
\\ \\ \textbf{Rechnung:} \\
\text{Ebene: }
\vec{x} =\left(
\begin{array}{c}
-2 \\
-4 \\
2 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
1 \\
2 \\
2 \\
\end{array}
\right)
+ \sigma
\left(
\begin{array}{c}
0 \\
-1 \\
-2 \\
\end{array}
\right) \\
\text{Ebene: } 1 x_1+1 x_2+0 x_3+0=0 \\
\begin{array}{cccc}
x_1=& -2 &+1\lambda &+0\sigma \\
x_2=&-4 &+2\lambda &-1\sigma \\
x_3=&2 &+2\lambda &-1\sigma\\
\end{array} \\
1( -2+1\lambda+0\sigma) +1(-4+2\lambda-1\sigma) +0 (2+2\lambda-2\sigma)+0=0 \\
3\lambda-1\sigma-6=0 \\
\\
\sigma=\frac{-3 \lambda +6}{-1} \\
\sigma= 3 \lambda -6 \\
\vec{x} = \left(
\begin{array}{c}
-2 \\
-4 \\
2 \\
\end{array}
\right) +\lambda \cdot
\left(
\begin{array}{c}
1 \\
2 \\
2 \\
\end{array}
\right) +(3\lambda-6) \cdot
\left(
\begin{array}{c}
0 \\
-1 \\
-2 \\
\end{array}
\right) \\
\text{Schnittgerade: }
\vec{x} =\left(
\begin{array}{c}
-2 \\
2 \\
14 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
1 \\
-1 \\
-4 \\
\end{array}
\right) \\
\\
\end{array}$