Analytische Geometrie-Lagebeziehung-Gerade - Gerade
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Beispiel Nr: 09
$\begin{array}{l} \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}
8 \\
4 \\
2 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
1 \\
3 \\
7 \\
\end{array}
\right) \\
\text{Gerade 2: }
\vec{x} =\left(
\begin{array}{c}
7 \\
7 \\
8 \\
\end{array}
\right) + \sigma
\left(
\begin{array}{c}
5 \\
1 \\
4 \\
\end{array}
\right) \\
\\ \\ \textbf{Rechnung:} \\
\text{Gerade 1: }
\vec{x} =\left(
\begin{array}{c}
8 \\
4 \\
2 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
1 \\
3 \\
7 \\
\end{array}
\right) \\
\text{Gerade 2: }
\vec{x} =\left(
\begin{array}{c}
7 \\
7 \\
8 \\
\end{array}
\right) + \sigma
\left(
\begin{array}{c}
5 \\
1 \\
4 \\
\end{array}
\right) \\
\text{Richtungsvektoren: } \\
\left(
\begin{array}{c}
1 \\
3 \\
7 \\
\end{array}
\right) =k \cdot
\left(
\begin{array}{c}
5 \\
1 \\
4 \\
\end{array}
\right) \\
\begin{array}{cccc}
1&=&+5 k& \quad /:5 \quad \Rightarrow k=\frac{1}{5} \\
3&=&+1 k & \quad /:1 \quad \Rightarrow k=3 \\
7&=&+4 k & \quad /:4 \quad \Rightarrow k=1\frac{3}{4} \\
\end{array} \\
\\ \Rightarrow \text{Geraden sind nicht parallel} \\ \left(
\begin{array}{c}
8 \\
4 \\
2 \\
\end{array}
\right) + \lambda
\left(
\begin{array}{c}
1 \\
3 \\
7 \\
\end{array}
\right) =
\left(
\begin{array}{c}
7 \\
7 \\
8 \\
\end{array}
\right) + \sigma
\left(
\begin{array}{c}
5 \\
1 \\
4 \\
\end{array}
\right) \\
\begin{array}{cccccc}
8& +1\lambda &=& 7& +5\sigma& \quad /-8 \quad /-5 \sigma\\
4& +3\lambda &=& 7& +1 \sigma& \quad /-4 \quad /-1 \sigma\\
2& +7\lambda &=& 8& +4 \sigma& \quad /-2 \quad /-4 \sigma\\
\end{array} \\
\\I \qquad 1 \lambda -5 \sigma =-1\\
II \qquad 3 \lambda -1 \sigma = 3 \\
III \qquad 7 \lambda +4 \sigma = 6 \\ \\
\text{Aus 2 Gleichungen }\lambda \text{ und } \sigma \text{ berechnen } \\
I \qquad 1 \lambda -5 \sigma =-1 \qquad / \cdot3\\
II \qquad 3 \lambda -1 \sigma = 3 \qquad / \cdot\left(-1\right)\\
I \qquad 3 \lambda -15 \sigma =-3\\
II \qquad -3 \lambda +1 \sigma = -3 \\
\text{I + II}\\
I \qquad 3 \lambda -3 \lambda-15 \sigma +1 \sigma =-3 -3\\ -14 \sigma = -6 \qquad /:\left(-14\right) \\
\sigma = \frac{-6}{-14} \\
\sigma=\frac{3}{7} \\
\sigma \text{ in I}\\
I \qquad 3 \lambda -15 \cdot \frac{3}{7} =-3 \\
3 \lambda -6\frac{3}{7} =-3 \qquad / +6\frac{3}{7} \\
3 \lambda =-3 +6\frac{3}{7} \\
3 \lambda =3\frac{3}{7} \qquad / :3 \\
\lambda = \frac{3\frac{3}{7}}{3} \\
\lambda=1\frac{1}{7} \\
\lambda \text{ und } \sigma \text{ in die verbleibende Gleichung einsetzen} \\
III \quad 2+1\frac{1}{7}\cdot7=8+\frac{3}{7}\cdot4 \\
10=9\frac{5}{7} \\ \text{Geraden sind windschief} \\
\end{array}$