Berechne die Koordinaten des Schnittpunktes D DD der Geraden g gg mit der Ebene E EE Setze g = E g=Eg = E :
( 2 1 − 1 ) + r ⋅ ( 1 1 − 2 ) = ( 0 1 4 ) + s ⋅ ( 4 2 3 ) + t ⋅ ( 2 0 1 ) \def\arraystretch{1.25} \left(\begin{array}{c}2 \\ 1 \\ -1\end{array}\right)+r \cdot\left(\begin{array}{c}1 \\ 1 \\ -2\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 4\end{array}\right)+s \cdot\left(\begin{array}{l}4 \\ 2 \\ 3\end{array}\right)+t \cdot\left(\begin{array}{l}2 \\ 0 \\ 1\end{array}\right) 2 1 − 1 + r ⋅ 1 1 − 2 = 0 1 4 + s ⋅ 4 2 3 + t ⋅ 2 0 1
⇒ ( 2 1 − 1 ) − ( 0 1 4 ) = r ⋅ ( − 1 − 1 2 ) + s ⋅ ( 4 2 3 ) + t ⋅ ( 2 0 1 ) \def\arraystretch{1.25} \Rightarrow\;\begin{pmatrix}2 \\ 1 \\ -1\end{pmatrix}-\begin{pmatrix}0 \\ 1 \\ 4\end{pmatrix}=r\cdot \left(\begin{array}{l}-1 \\ -1 \\ 2\end{array}\right)+s \cdot\left(\begin{array}{l}4 \\ 2 \\ 3\end{array}\right)+t \cdot\left(\begin{array}{l}2 \\ 0 \\ 1\end{array}\right)⇒ 2 1 − 1 − 0 1 4 = r ⋅ − 1 − 1 2 + s ⋅ 4 2 3 + t ⋅ 2 0 1
⇒ ( 2 0 − 5 ) = r ⋅ ( − 1 − 1 2 ) + s ⋅ ( 4 2 3 ) + t ⋅ ( 2 0 1 ) \def\arraystretch{1.25} \Rightarrow\;\begin{pmatrix}2\\0\\-5\end{pmatrix}=r\cdot \left(\begin{array}{l}-1 \\ -1 \\ 2\end{array}\right)+s \cdot\left(\begin{array}{l}4 \\ 2 \\ 3\end{array}\right)+t \cdot\left(\begin{array}{l}2 \\ 0 \\ 1\end{array}\right)⇒ 2 0 − 5 = r ⋅ − 1 − 1 2 + s ⋅ 4 2 3 + t ⋅ 2 0 1
Man erhält dann das folgende Gleichungssystem:
I − 1 ⋅ r + 4 ⋅ s + 2 ⋅ t = 2 I I − 1 ⋅ r + 2 ⋅ s + 0 ⋅ t = 0 I I I 2 ⋅ r + 3 ⋅ s + 1 ⋅ t = − 5 \def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&-1\cdot r&+&4\cdot s&+&2\cdot t&=2\\\mathrm{II}&-1\cdot r&+&2\cdot s&+&0\cdot t&=0\\\mathrm{III}&2\cdot r&+&3\cdot s&+&1\cdot t&=-5\end{array}I II III − 1 ⋅ r − 1 ⋅ r 2 ⋅ r + + + 4 ⋅ s 2 ⋅ s 3 ⋅ s + + + 2 ⋅ t 0 ⋅ t 1 ⋅ t = 2 = 0 = − 5
Aus Gleichung I I \mathrm{II}II folgt: r = 2 s r=2sr = 2 s
Setze r = 2 s r=2sr = 2 s in Gleichung I \mathrm{I}I und I I I \mathrm{III}III ein:
I − 1 ⋅ 2 s + 4 ⋅ s + 2 ⋅ t = 2 I I I 2 ⋅ 2 s + 3 ⋅ s + 1 ⋅ t = − 5 \def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I}&-1\cdot 2s&+&4\cdot s&+&2\cdot t&=2\\\mathrm{III}&2\cdot 2s&+&3\cdot s&+&1\cdot t&=-5\end{array}I III − 1 ⋅ 2 s 2 ⋅ 2 s + + 4 ⋅ s 3 ⋅ s + + 2 ⋅ t 1 ⋅ t = 2 = − 5
Zusammengefasst folgt:
I ′ 2 ⋅ s + 2 ⋅ t = 2 I I I ′ 7 ⋅ s + 1 ⋅ t = − 5 \def\arraystretch{1.25} \begin{array}{ccccc}\mathrm{I'}&2\cdot s&+&2\cdot t&=2\\\mathrm{III'}&7\cdot s&+&1\cdot t&=-5 \end{array}I ′ II I ′ 2 ⋅ s 7 ⋅ s + + 2 ⋅ t 1 ⋅ t = 2 = − 5
Rechne I ′ + ( − 2 ) ⋅ I I I ′ : \mathrm{I'}+(-2)\cdot\mathrm{III'}:I ′ + ( − 2 ) ⋅ II I ′ :
I ′ : 2 ⋅ s + 2 ⋅ t = 2 + ( − 2 ) ⋅ I I I ′ : − 14 ⋅ s − 2 ⋅ t = 10 − 12 ⋅ s + 0 = 12 \def\arraystretch{1.25} \begin{array}{rrrrrcrr}&\mathrm{I'}:&2\cdot s&+&2\cdot t&=&2\\&+(-2)\cdot\mathrm{III'}:&-14\cdot s&-&2\cdot t&=&10 \\ \hline &&-12\cdot s&+&0&=&12&\end{array} I ′ : + ( − 2 ) ⋅ II I ′ : 2 ⋅ s − 14 ⋅ s − 12 ⋅ s + − + 2 ⋅ t 2 ⋅ t 0 = = = 2 10 12
Damit ist s = 12 − 12 = − 1 s=\dfrac{12}{-12}=-1s = − 12 12 = − 1 . Dann folgt r = 2 ⋅ s = 2 ⋅ ( − 1 ) = − 2 r=2\cdot s=2\cdot (-1)=-2r = 2 ⋅ s = 2 ⋅ ( − 1 ) = − 2 und t = 1 − s = 1 − ( − 1 ) = 2 t=1- s=1-(-1)=2t = 1 − s = 1 − ( − 1 ) = 2
Setze r = − 2 r=-2r = − 2 in g gg ein:
O D → = ( 2 1 − 1 ) + ( − 2 ) ⋅ ( 1 1 − 2 ) = ( 0 − 1 3 ) ⇒ D ( 0 ∣ − 1 ∣ 3 ) \overrightarrow{OD}=\begin{pmatrix}2 \\ 1 \\ -1\end{pmatrix}+(-2) \cdot\begin{pmatrix}1 \\ 1 \\ -2\end{pmatrix}=\begin{pmatrix}0\\-1\\3\end{pmatrix}\;\Rightarrow\;D(0|-1|3)O D = 2 1 − 1 + ( − 2 ) ⋅ 1 1 − 2 = 0 − 1 3 ⇒ D ( 0∣ − 1∣3 )
Die Koordinaten des Schnittpunktes D DD der Geraden g gg mit der Ebene E EE lauten D ( 0 ∣ − 1 ∣ 3 ) D(0|-1|3)D ( 0∣ − 1∣3 ) .