Für diese Aufgabe benötigst Du folgendes Grundwissen: Lagebeziehungen zwischen 3 Ebenen
Untersuchung auf Parallelität oder Identität Dazu wird zuerst die Ebene E 3 E_3 E 3 Koordinatenform umgewandelt.
Berechne das Vektorprodukt der beiden Richtungsvektoren der Ebene E 3 E_3 E 3
n ⃗ = ( 1 2 0 ) × ( 0 1 2 ) = ( 4 − 2 1 ) \vec n=\begin{pmatrix}1\\2\\0\end{pmatrix}\times\begin{pmatrix}0\\1\\2\end{pmatrix}=\begin{pmatrix}4\\-2\\1\end{pmatrix} n = 1 2 0 × 0 1 2 = 4 − 2 1 Normalenform ein:
E : ( X → − A → ) ∘ n ⃗ \displaystyle E:\;\left(\overrightarrow{X}-\overrightarrow{A}\right)\circ\vec n E : ( X − A ) ∘ n = = = 0 \displaystyle 0 0 ↓ Setze A → = ( 1 0 0 ) \overrightarrow{A}=\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix} A = 1 0 0 n ⃗ = ( 4 − 2 1 ) \vec n=\begin{pmatrix}4\\-2\\1\end{pmatrix} n = 4 − 2 1
( X → − ( 1 0 0 ) ) ∘ ( 4 − 2 1 ) \displaystyle \left(\overrightarrow{X}-\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}\right)\circ\begin{pmatrix}4\\-2\\1\end{pmatrix} X − 1 0 0 ∘ 4 − 2 1 = = = 0 \displaystyle 0 0 ↓ Berechne das Skalarprodukt .
4 ⋅ x 1 − 2 ⋅ x 2 + x 3 − 4 \displaystyle 4\cdot x_1-2\cdot x_2+x_3-4 4 ⋅ x 1 − 2 ⋅ x 2 + x 3 − 4 = = = 0 \displaystyle 0 0 4 ⋅ x 1 − 2 ⋅ x 2 + x 3 \displaystyle 4\cdot x_1-2\cdot x_2+x_3 4 ⋅ x 1 − 2 ⋅ x 2 + x 3 = = = 4 \displaystyle 4 4
Die umgewandelte Ebenengleichung der Ebene E 3 E_3 E 3
E 3 : 4 ⋅ x 1 − 2 ⋅ x 2 + x 3 = 4 E_3:\;4\cdot x_1-2\cdot x_2+x_3=4 E 3 : 4 ⋅ x 1 − 2 ⋅ x 2 + x 3 = 4
Betrachte nun die Normalenvektoren der drei Ebenen:
n ⃗ 1 = ( 2 1 1 ) \vec n_1 =\begin{pmatrix}2\\1\\1\end{pmatrix} n 1 = 2 1 1 n ⃗ 2 = ( 3 4 1 ) \vec n_2=\begin{pmatrix}3\\4\\1\end{pmatrix} n 2 = 3 4 1 n ⃗ 3 = ( 4 − 2 1 ) \vec n_3 =\begin{pmatrix}4\\-2\\1\end{pmatrix} n 3 = 4 − 2 1
Die Normalenvektoren sind keine Vielfache voneinander, d.h. die Ebenen sind nicht parallel (und damit auch nicht identisch). Demnach müssen sich die Ebenen schneiden.
Berechnung der Schnittgeraden Erste Schnittgerade g 12 : E 1 ∩ E 2 g_{12}:\;E_1\cap E_2 g 12 : E 1 ∩ E 2
Betrachte die Ebenengleichungen E 1 E_1 E 1 E 2 E_2 E 2
I 2 ⋅ x 1 + 1 ⋅ x 2 + 1 ⋅ x 3 = 2 I I 3 ⋅ x 1 + 4 ⋅ x 2 + 1 ⋅ x 3 = 4 \def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&2\cdot x_1&+&1\cdot x_2&+&1\cdot x_3&=&2\\\mathrm{II}&3\cdot x_1&+&4\cdot x_2&+&1\cdot x_3&=&4\end{array} I II 2 ⋅ x 1 3 ⋅ x 1 + + 1 ⋅ x 2 4 ⋅ x 2 + + 1 ⋅ x 3 1 ⋅ x 3 = = 2 4
Rechne I I − I ⇒ x 1 + 3 ⋅ x 2 = 2 \mathrm{II}-\mathrm{I}\;\Rightarrow\;x_1+3\cdot x_2=2 II − I ⇒ x 1 + 3 ⋅ x 2 = 2 ⇒ x 1 = 2 − 3 ⋅ x 2 \;\Rightarrow\;x_1=2-3\cdot x_2 ⇒ x 1 = 2 − 3 ⋅ x 2
Eine Variable ist frei wählbar.
Setze x 2 = r ⇒ x 1 = 2 − 3 ⋅ r x_2=r\;\Rightarrow\;x_1=2-3\cdot r x 2 = r ⇒ x 1 = 2 − 3 ⋅ r
Löse Gleichung I I \mathrm{II} II x 3 x_3 x 3 x 1 = 2 − 3 ⋅ r x_1=2-3\cdot r x 1 = 2 − 3 ⋅ r x 2 = r x_2=r x 2 = r
3 ⋅ x 1 + 4 ⋅ x 2 + x 3 \displaystyle 3\cdot x_1+4\cdot x_2+x_3 3 ⋅ x 1 + 4 ⋅ x 2 + x 3 = = = 4 \displaystyle 4 4 − 3 ⋅ x 1 − 4 ⋅ x 2 \displaystyle -3\cdot x_1-4\cdot x_2 − 3 ⋅ x 1 − 4 ⋅ x 2 x 3 \displaystyle x_3 x 3 = = = 4 − 3 ⋅ x 1 − 4 ⋅ x 2 \displaystyle 4-3\cdot x_1-4\cdot x_2 4 − 3 ⋅ x 1 − 4 ⋅ x 2 ↓ Setze x 1 = 2 − 3 ⋅ r x_1=2-3\cdot r x 1 = 2 − 3 ⋅ r x 2 = r x_2=r x 2 = r
x 3 \displaystyle x_3 x 3 = = = 4 − 3 ⋅ ( 2 − 3 ⋅ r ) − 4 ⋅ r \displaystyle 4-3\cdot (2-3\cdot r)-4\cdot r 4 − 3 ⋅ ( 2 − 3 ⋅ r ) − 4 ⋅ r ↓ Löse die Klammer auf.
x 3 \displaystyle x_3 x 3 = = = 4 − 6 + 9 ⋅ r − 4 ⋅ r \displaystyle 4-6+9\cdot r-4\cdot r 4 − 6 + 9 ⋅ r − 4 ⋅ r ↓ Fasse zusammen.
x 3 \displaystyle x_3 x 3 = = = − 2 + 5 ⋅ r \displaystyle -2+5\cdot r − 2 + 5 ⋅ r
Untereinander geschrieben:
x 1 = 2 − 3 ⋅ r x 2 = 0 + 1 ⋅ r ⇒ ( x 1 x 2 x 3 ) = ( 2 0 − 2 ) + r ⋅ ( − 3 1 5 ) x 3 = − 2 + 5 ⋅ r x_1=2-3\cdot r\\
x_2=0+1\cdot r\;\Rightarrow\;\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}2\\0\\-2\end{pmatrix}+r\cdot\begin{pmatrix}-3\\1\\5\end{pmatrix}\\
x_3=-2+5\cdot r x 1 = 2 − 3 ⋅ r x 2 = 0 + 1 ⋅ r ⇒ x 1 x 2 x 3 = 2 0 − 2 + r ⋅ − 3 1 5 x 3 = − 2 + 5 ⋅ r
Die Schnittgerade g 12 g_{12} g 12
g 12 : X ⃗ = ( 2 0 − 2 ) + r ⋅ ( − 3 1 5 ) \displaystyle g_{12}:\;\vec X=\begin{pmatrix}2\\0\\-2\end{pmatrix}+r\cdot\begin{pmatrix}-3\\1\\5\end{pmatrix} g 12 : X = 2 0 − 2 + r ⋅ − 3 1 5 Zweite Schnittgerade g 23 : E 2 ∩ E 3 g_{23}:\;E_2\cap E_3 g 23 : E 2 ∩ E 3
Betrachte die Ebenengleichungen E 2 E_2 E 2 E 3 E_3 E 3
I I 3 ⋅ x 1 + 4 ⋅ x 2 + 1 ⋅ x 3 = 4 I I I 4 ⋅ x 1 − 2 ⋅ x 2 + 1 ⋅ x 3 = 4 \def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{II}&3\cdot x_1&+&4\cdot x_2&+&1\cdot x_3&=&4\\\mathrm{III}&4\cdot x_1&-&2\cdot x_2&+&1\cdot x_3&=&4\end{array} II III 3 ⋅ x 1 4 ⋅ x 1 + − 4 ⋅ x 2 2 ⋅ x 2 + + 1 ⋅ x 3 1 ⋅ x 3 = = 4 4
Rechne I I I − I I ⇒ x 1 − 6 ⋅ x 2 = 0 \mathrm{III}-\mathrm{II}\;\Rightarrow\;x_1-6\cdot x_2=0 III − II ⇒ x 1 − 6 ⋅ x 2 = 0 ⇒ x 1 = 6 ⋅ x 2 \;\Rightarrow\;x_1=6\cdot x_2 ⇒ x 1 = 6 ⋅ x 2
Eine Variable ist frei wählbar.
Setze x 2 = r ⇒ x 1 = 6 ⋅ r x_2=r\;\Rightarrow\;x_1=6\cdot r x 2 = r ⇒ x 1 = 6 ⋅ r
Löse Gleichung I I I \mathrm{III} III x 3 x_3 x 3 x 1 = 6 ⋅ r x_1=6\cdot r x 1 = 6 ⋅ r x 2 = r x_2=r x 2 = r
4 ⋅ x 1 − 2 ⋅ x 2 + x 3 \displaystyle 4\cdot x_1-2\cdot x_2+x_3 4 ⋅ x 1 − 2 ⋅ x 2 + x 3 = = = 4 \displaystyle 4 4 − 4 ⋅ x 1 + 2 ⋅ x 2 \displaystyle -4\cdot x_1+2\cdot x_2 − 4 ⋅ x 1 + 2 ⋅ x 2 x 3 \displaystyle x_3 x 3 = = = 4 − 4 ⋅ x 1 + 2 ⋅ x 2 \displaystyle 4-4\cdot x_1+2\cdot x_2 4 − 4 ⋅ x 1 + 2 ⋅ x 2 ↓ Setze x 1 = 6 ⋅ r x_1=6\cdot r x 1 = 6 ⋅ r x 2 = r x_2=r x 2 = r
x 3 \displaystyle x_3 x 3 = = = 4 − 4 ⋅ 6 ⋅ r + 2 ⋅ r \displaystyle 4-4\cdot 6\cdot r+2\cdot r 4 − 4 ⋅ 6 ⋅ r + 2 ⋅ r ↓ Fasse zusammen.
x 3 \displaystyle x_3 x 3 = = = 4 − 22 ⋅ r \displaystyle 4-22\cdot r 4 − 22 ⋅ r
Untereinander geschrieben:
x 1 = 0 + 6 ⋅ r x 2 = 0 + 1 ⋅ r ⇒ ( x 1 x 2 x 3 ) = ( 0 0 4 ) + r ⋅ ( 6 1 − 22 ) x 3 = 4 − 22 ⋅ r x_1=0+6\cdot r\\
x_2=0+1\cdot r\;\Rightarrow\;\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}0\\0\\4\end{pmatrix}+r\cdot\begin{pmatrix}6\\1\\-22\end{pmatrix}\\
x_3=4-22\cdot r x 1 = 0 + 6 ⋅ r x 2 = 0 + 1 ⋅ r ⇒ x 1 x 2 x 3 = 0 0 4 + r ⋅ 6 1 − 22 x 3 = 4 − 22 ⋅ r
Die Schnittgerade g 23 g_{23} g 23
g 23 : X ⃗ = ( 0 0 4 ) + r ⋅ ( 6 1 − 22 ) \displaystyle g_{23}:\;\vec X=\begin{pmatrix}0\\0\\4\end{pmatrix}+r\cdot\begin{pmatrix}6\\1\\-22\end{pmatrix} g 23 : X = 0 0 4 + r ⋅ 6 1 − 22 Dritte Schnittgerade g 13 : E 1 ∩ E 3 g_{13}:\;E_1\cap E_3 g 13 : E 1 ∩ E 3
Betrachte die Ebenengleichungen E 1 E_1 E 1 E 3 E_3 E 3
I 2 ⋅ x 1 + 1 ⋅ x 2 + 1 ⋅ x 3 = 2 I I I 4 ⋅ x 1 − 2 ⋅ x 2 + 1 ⋅ x 3 = 4 \def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&2\cdot x_1&+&1\cdot x_2&+&1\cdot x_3&=&2\\\mathrm{III}&4\cdot x_1&-&2\cdot x_2&+&1\cdot x_3&=&4\end{array} I III 2 ⋅ x 1 4 ⋅ x 1 + − 1 ⋅ x 2 2 ⋅ x 2 + + 1 ⋅ x 3 1 ⋅ x 3 = = 2 4
Rechne I I I − I I ⇒ 2 ⋅ x 1 − 3 ⋅ x 2 = 2 \mathrm{III}-\mathrm{II}\;\Rightarrow\;2\cdot x_1-3\cdot x_2=2 III − II ⇒ 2 ⋅ x 1 − 3 ⋅ x 2 = 2 ⇒ x 1 = 1 + 1 , 5 ⋅ x 2 \;\Rightarrow\;x_1=1+1{,}5\cdot x_2 ⇒ x 1 = 1 + 1 , 5 ⋅ x 2
Eine Variable ist frei wählbar.
Setze x 2 = r ⇒ x 1 = 1 + 1 , 5 ⋅ r x_2=r\;\Rightarrow\;x_1=1+1{,}5\cdot r x 2 = r ⇒ x 1 = 1 + 1 , 5 ⋅ r
Löse Gleichung I I I \mathrm{III} III x 3 x_3 x 3 x 1 = 1 + 1 , 5 ⋅ r x_1=1+1{,}5\cdot r x 1 = 1 + 1 , 5 ⋅ r x 2 = r x_2=r x 2 = r
4 ⋅ x 1 − 2 ⋅ x 2 + x 3 \displaystyle 4\cdot x_1-2\cdot x_2+x_3 4 ⋅ x 1 − 2 ⋅ x 2 + x 3 = = = 4 \displaystyle 4 4 − 4 ⋅ x 1 + 2 ⋅ x 2 \displaystyle -4\cdot x_1+2\cdot x_2 − 4 ⋅ x 1 + 2 ⋅ x 2 x 3 \displaystyle x_3 x 3 = = = 4 − 4 ⋅ x 1 + 2 ⋅ x 2 \displaystyle 4-4\cdot x_1+2\cdot x_2 4 − 4 ⋅ x 1 + 2 ⋅ x 2 ↓ Setze x 1 = 1 + 1 , 5 ⋅ r x_1=1+1{,}5\cdot r x 1 = 1 + 1 , 5 ⋅ r x 2 = r x_2=r x 2 = r
x 3 \displaystyle x_3 x 3 = = = 4 − 4 ⋅ ( 1 + 1 , 5 ⋅ r ) + 2 ⋅ r \displaystyle 4-4\cdot(1+1{,}5\cdot r)+2\cdot r 4 − 4 ⋅ ( 1 + 1 , 5 ⋅ r ) + 2 ⋅ r ↓ Löse die Klammer auf.
= = = 4 − 4 − 6 ⋅ r + 2 ⋅ r \displaystyle 4-4-6\cdot r+2\cdot r 4 − 4 − 6 ⋅ r + 2 ⋅ r ↓ Fasse zusammen.
x 3 \displaystyle x_3 x 3 = = = − 4 ⋅ r \displaystyle -4\cdot r − 4 ⋅ r
Untereinander geschrieben:
x 1 = 1 + 1 , 5 ⋅ r x 2 = 0 + 1 ⋅ r ⇒ ( x 1 x 2 x 3 ) = ( 1 0 0 ) + r ⋅ ( 1 , 5 1 − 4 ) x 3 = 0 − 4 ⋅ r x_1=1+1{,}5\cdot r\\
x_2=0+1\cdot r\;\Rightarrow\;\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}+r\cdot\begin{pmatrix}1{,}5\\1\\-4\end{pmatrix}\\
x_3=0-4\cdot r x 1 = 1 + 1 , 5 ⋅ r x 2 = 0 + 1 ⋅ r ⇒ x 1 x 2 x 3 = 1 0 0 + r ⋅ 1 , 5 1 − 4 x 3 = 0 − 4 ⋅ r
Die Schnittgerade g 13 g_{13} g 13
g 13 : X ⃗ = ( 1 0 0 ) + r ⋅ ( 1 , 5 1 − 4 ) \displaystyle g_{13}:\;\vec X=\begin{pmatrix}1\\0\\0\end{pmatrix}+r\cdot\begin{pmatrix}1{,}5\\1\\-4\end{pmatrix} g 13 : X = 1 0 0 + r ⋅ 1 , 5 1 − 4 Schneiden sich die 3 Ebenen eventuell in einem Punkt? Betrachte dazu das lineare Gleichungssystem, das aus den drei Ebenengleichungen besteht:
I 2 ⋅ x 1 + 1 ⋅ x 2 + 1 ⋅ x 3 = 2 I I 3 ⋅ x 1 + 4 ⋅ x 2 + 1 ⋅ x 3 = 4 I I I 4 ⋅ x 1 − 2 ⋅ x 2 + 1 ⋅ x 3 = 4 → ( 2 1 1 2 3 4 1 4 4 − 2 1 4 ) \def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&2\cdot x_1&+&1\cdot x_2&+&1\cdot x_3&=2\\\mathrm{II}&3\cdot x_1&+&4\cdot x_2&+&1\cdot x_3&=4\\\mathrm{III}&4\cdot x_1&-&2\cdot x_2&+&1\cdot x_3&=4\end{array}\quad\to\quad\left(\begin{array}{ccc|c}2&1&1&2\\3&4&1&4\\4&-2&1&4\end{array}\right) I II III 2 ⋅ x 1 3 ⋅ x 1 4 ⋅ x 1 + + − 1 ⋅ x 2 4 ⋅ x 2 2 ⋅ x 2 + + + 1 ⋅ x 3 1 ⋅ x 3 1 ⋅ x 3 = 2 = 4 = 4 → 2 3 4 1 4 − 2 1 1 1 2 4 4
Zur leichten Anwendung des Gaußverfahrens wird die 3. 3. 3. 1. 1. 1.
( 1 2 1 2 1 3 4 4 1 4 − 2 4 ) ⟶ 1 ⋅ I I I − 1 ⋅ I 1 ⋅ I I − 1 ⋅ I ( 1 2 1 2 0 1 3 2 0 2 − 3 2 ) ⟶ 1 ⋅ I I I − 2 ⋅ I I ( 1 2 1 2 0 1 3 2 0 0 − 9 − 2 ) \def\arraystretch{1.25} \left(\begin{array}{ccc|c}1&2&1&2\\1&3&4&4\\1&4&-2&4\end{array}\right)\overset{\mathrm{1 \cdot II - 1 \cdot I}}{\underset{\mathrm{1 \cdot III - 1 \cdot I}}\longrightarrow}\left(\begin{array}{ccc|c}1&2&1&2\\0&1&3&2\\0&2&-3&2\end{array}\right)\overset{\mathrm{1 \cdot III - 2 \cdot II}}\longrightarrow\left(\begin{array}{ccc|c}1&2&1&2\\0&1&3&2\\0&0&-9&-2\end{array}\right) 1 1 1 2 3 4 1 4 − 2 2 4 4 1 ⋅ III − 1 ⋅ I ⟶ 1 ⋅ II − 1 ⋅ I 1 0 0 2 1 2 1 3 − 3 2 2 2 ⟶ 1 ⋅ III − 2 ⋅ II 1 0 0 2 1 0 1 3 − 9 2 2 − 2
Dann folgt aus der letzten Zeile: − 9 ⋅ x 2 = − 2 ⇒ x 2 = 2 9 -9\cdot x_2=-2\;\Rightarrow\;x_2=\dfrac{2}{9} − 9 ⋅ x 2 = − 2 ⇒ x 2 = 9 2
In der letzten Matrix lautet die 2. 2. 2.
x 1 + 3 ⋅ x 2 \displaystyle x_1+3\cdot x_2 x 1 + 3 ⋅ x 2 = = = 2 \displaystyle 2 2 − 3 ⋅ x 2 \displaystyle -3\cdot x_2 − 3 ⋅ x 2 ↓ Löse nach x 1 x_1 x 1
x 1 \displaystyle x_1 x 1 = = = 2 − 3 ⋅ x 2 \displaystyle 2-3\cdot x_2 2 − 3 ⋅ x 2 ↓ Setze x 2 = 2 9 x_2=\frac{2}{9} x 2 = 9 2
x 1 \displaystyle x_1 x 1 = = = 2 − 3 ⋅ 2 9 \displaystyle 2-3\cdot \frac{2}{9} 2 − 3 ⋅ 9 2 x 1 \displaystyle x_1 x 1 = = = 4 3 \displaystyle \dfrac{4}{3} 3 4
In der letzten Matrix lautet die 1. 1. 1.
1 ⋅ x 3 + 2 ⋅ x 1 + 1 ⋅ x 2 \displaystyle 1\cdot x_3+2\cdot x_1+1\cdot x_2 1 ⋅ x 3 + 2 ⋅ x 1 + 1 ⋅ x 2 = = = 2 \displaystyle 2 2 − 2 ⋅ x 1 − 1 ⋅ x 2 \displaystyle -2\cdot x_1-1\cdot x_2 − 2 ⋅ x 1 − 1 ⋅ x 2 ↓ Löse nach x 3 x_3 x 3
x 3 \displaystyle x_3 x 3 = = = 2 − 2 ⋅ x 1 − 1 ⋅ x 2 \displaystyle 2-2\cdot x_1-1\cdot x_2 2 − 2 ⋅ x 1 − 1 ⋅ x 2 ↓ Setze x 1 = 4 3 x_1=\frac{4}{3} x 1 = 3 4 x 2 = 2 9 x_2=\frac{2}{9} x 2 = 9 2
x 3 \displaystyle x_3 x 3 = = = 2 − 2 ⋅ 4 3 − 1 ⋅ 2 9 \displaystyle 2-2\cdot \frac{4}{3}-1\cdot \frac{2}{9} 2 − 2 ⋅ 3 4 − 1 ⋅ 9 2 x 3 \displaystyle x_3 x 3 = = = 2 − 8 3 − 2 9 \displaystyle 2-\dfrac{8}{3}-\dfrac{2}{9} 2 − 3 8 − 9 2 ↓ Erweitere auf den Nenner 9 9 9
= = = 18 9 − 24 9 − 2 9 \displaystyle \dfrac{18}{9}-\dfrac{24}{9}-\dfrac{2}{9} 9 18 − 9 24 − 9 2 ↓ Fasse zusammen.
x 3 \displaystyle x_3 x 3 = = = − 8 9 \displaystyle -\dfrac{8}{9} − 9 8
Das lineare Gleichungssystem hat die Lösung:
L = { 4 3 ; 2 9 ; − 8 9 } \displaystyle \mathbb{L}=\{\frac{4}{3};\frac{2}{9};-\frac{8}{9}\} L = { 3 4 ; 9 2 ; − 9 8 } Die drei Ebenen haben einen gemeinsamen Schnittpunkt S ( 4 3 ∣ 2 9 ∣ − 8 9 ) S(\frac{4}{3}|\frac{2}{9}|-\frac{8}{9}) S ( 3 4 ∣ 9 2 ∣ − 9 8 )
▸ Zusätzliche graphische Veranschaulichung
Anmerkung: Wenn man die 3 3 3 S S S
