Die Ebene $EE$ wird an der Ebene $HH$ gespiegelt.

Für die rechte Seite der gespiegelten Ebene $E{}^{\mathrm{\prime }}E\{\}\text{'}$ in Koordinatenform gilt die Gleichung:

$(\mathrm{I})d_3=2\cdot d_2-d_1\backslash mathrm\{(I)\}\backslash ;\backslash ;d\_3=2\backslash cdot\; d\_2-d\_1$

Lies $d_{1}d\_1$ aus der Ebenengleichung $EE$ ab: $d_1=7d\_1=7$

Lies $d_{2}d\_2$ aus der Ebenengleichung $HH$ ab: $d_2=12d\_2=12$

Setze $d_1=7d\_1=7$ und $d_2=12d\_2=12$ in Gleichung $(\mathrm{I})\backslash mathrm\{(I)\}$ ein:

$d_3=2\cdot 12-7=24-7=17d\_3=2\backslash cdot\; 12-7=24-7=17$

Antwort: Die Gleichung der Spiegelebene $E^{\mathrm{\prime }}E\text{'}$ lautet: $2\cdot x_1-3x_2+x_3=172\backslash cdot\; x\_1-\; 3x\_2+\; x\_3=17$

1. Die Schnittgerade $g_{S}g\_S$ ist gegeben: $g_S:\vec{x}=\overrightarrow{OA}+r\cdot \vec{u}=\left(\begin{array}{c}{\displaystyle \frac{18}{7}}\\ 0\\ {\displaystyle \frac{1}{14}}\end{array}\right)+r\cdot \left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)g\_S:\backslash ;\backslash vec\; x=\backslash overrightarrow\{OA\}+r\backslash cdot\; \backslash vec\; u\; =\backslash begin\{pmatrix\}\backslash dfrac\{18\}\{7\}\backslash \backslash [1ex]0\backslash \backslash [1ex]\backslash dfrac\{1\}\{14\}\backslash end\{pmatrix\}+r\backslash cdot\backslash begin\{pmatrix\}-1\backslash \backslash 1\backslash \backslash 1\backslash end\{pmatrix\}$.

2. Finde einen Punkt $PP$ auf der Ebene $EE$. Der Punkt $PP$ darf nicht auf der Schnittgeraden $g_{S}g\_S$ liegen.

$E:2\cdot x_1+4\cdot x_2-2\cdot x_3=5E:\backslash ;\; 2\backslash cdot\; x\_1+4\backslash cdot\; x\_2-2\backslash cdot\; x\_3=5$

Setze z. B. $x_1=1x\_1=1$ und $x_2=1x\_2=1$ und berechne $x_{3}x\_3$:

$2\cdot 1+4\cdot 1-2\cdot x_3=5\Rightarrow x_3=\mathrm{0,5}\Rightarrow P(1\mathrm{\mid }1\mathrm{\mid }\mathrm{0,5})2\backslash cdot\; 1+4\backslash cdot1-2\backslash cdot\; x\_3=5\backslash ;\backslash Rightarrow\backslash ;x\_3=0\{,\}5\backslash ;\backslash Rightarrow\backslash ;P(1|1|0\{,\}5)$

Liegt $PP$ auf $g_{S}g\_S$? Setze $PP$ in die Geradengleichung ein:

$\left(\begin{array}{c}1\\ 1\\ \mathrm{0,5}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{18}{7}}\\ 0\\ {\displaystyle \frac{1}{14}}\end{array}\right)+r\cdot \left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)\Rightarrow \left(\begin{array}{c}-{\displaystyle \frac{11}{7}}\\ 1\\ {\displaystyle \frac{3}{7}}\end{array}\right)=r\cdot \left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)\backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash 0\{,\}5\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{18\}\{7\}\backslash \backslash [1ex]0\backslash \backslash [1ex]\backslash dfrac\{1\}\{14\}\backslash end\{pmatrix\}+r\backslash cdot\backslash begin\{pmatrix\}-1\backslash \backslash 1\backslash \backslash 1\backslash end\{pmatrix\}\backslash ;\backslash Rightarrow\backslash ;\backslash begin\{pmatrix\}-\backslash dfrac\{11\}\{7\}\backslash \backslash 1\backslash \backslash \backslash dfrac\{3\}\{7\}\backslash end\{pmatrix\}=r\backslash cdot\backslash begin\{pmatrix\}-1\backslash \backslash 1\backslash \backslash 1\backslash end\{pmatrix\}$

Aus der ersten Zeile folgt $r={\displaystyle \frac{11}{7}}r=\backslash dfrac\{11\}\{7\}$ und aus der zweiten Zeile folgt $r=1r=1$. Das ist ein Widerspruch. Somit liegt $PP$ nicht auf $g_S\mathrm{.}g\_S.$

3. Wird ein Punkt $PP$ der Ebene $EE$ an der Ebene $HH$ gespiegelt, so liegen der Punkt $PP$ und der Spiegelpunkt $P{}^{\mathrm{\prime }}P\{\}\text{'}$ auf einer Geraden, die senkrecht auf der Ebene $HH$ steht. Diese Gerade ist die Lotgerade $l_{Lot}l\_\{Lot\}$. Sie wird benötigt, um den Lotfußpunkt $FF$ auf der Ebene $HH$ zu berechnen.

Erstelle nun eine Lotgerade $l_{Lot}l\_\{Lot\}$ mit dem gefundenen Punkt $P(1\mathrm{\mid }1\mathrm{\mid }\mathrm{0,5})P(1|1|0\{,\}5)$ als Aufpunkt und dem Normalenvektor ${\vec{n}}_H=\left(\begin{array}{c}3\\ -1\\ 4\end{array}\right)\backslash vec\; n\_H=\backslash begin\{pmatrix\}3\backslash \backslash -1\backslash \backslash 4\backslash end\{pmatrix\}$ der Ebene $HH$: $l_{Lot}:\vec{x}=\left(\begin{array}{c}1\\ 1\\ \mathrm{0,5}\end{array}\right)+r\cdot \left(\begin{array}{c}3\\ -1\\ 4\end{array}\right)l\_\{Lot\}:\backslash vec\; x=\backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash 0\{,\}5\backslash end\{pmatrix\}+r\backslash cdot\backslash begin\{pmatrix\}3\backslash \backslash -1\backslash \backslash 4\backslash end\{pmatrix\}$

4. Schneide die Lotgerade $l_{Lot}l\_\{Lot\}$ mit der Ebene $HH$, um den Lotfußpunkt $FF$ zu erhalten.

Setze $r={\displaystyle \frac{2}{13}}r=\backslash dfrac\{2\}\{13\}$ in die Lotgerade $l_{Lot}l\_\{Lot\}$ ein, um den Punkt $FF$ zu berechnen.

${\vec{x}}_F=\left(\begin{array}{c}1\\ 1\\ \mathrm{0,5}\end{array}\right)+(\frac{2}{13})\cdot \left(\begin{array}{c}3\\ -1\\ 4\end{array}\right)=\left(\begin{array}{c}1+{\displaystyle \frac{6}{13}}\\ 1-{\displaystyle \frac{2}{13}}\\ \mathrm{0,5}+{\displaystyle \frac{8}{13}}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{19}{13}}\\ {\displaystyle \frac{11}{13}}\\ {\displaystyle \frac{29}{26}}\end{array}\right)\Rightarrow \backslash vec\; x\_F=\; \backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash 0\{,\}5\backslash end\{pmatrix\}+(\backslash frac\{2\}\{13\})\backslash cdot\backslash begin\{pmatrix\}3\backslash \backslash -1\backslash \backslash 4\backslash end\{pmatrix\}\; =\backslash begin\{pmatrix\}1+\backslash dfrac\{6\}\{13\}\backslash \backslash [2ex]\; 1-\backslash dfrac\{2\}\{13\}\backslash \backslash [2ex]0\{,\}5+\backslash dfrac\{8\}\{13\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{19\}\{13\}\backslash \backslash [2ex]\; \backslash dfrac\{11\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{29\}\{26\}\backslash end\{pmatrix\}\backslash ;\backslash Rightarrow\backslash ;$$F({\displaystyle \frac{19}{13}}\mid {\displaystyle \frac{11}{13}}\mid {\displaystyle \frac{29}{26}})F\backslash left(\backslash dfrac\{19\}\{13\}\backslash Big\backslash vert\backslash dfrac\{11\}\{13\}\backslash Big\backslash vert\backslash dfrac\{29\}\{26\}\backslash right)$

5. Für den Spiegelpunkt $P{}^{\mathrm{\prime }}P\{\}\text{'}$ gilt immer die Gleichung $\overrightarrow{O{P}^{\mathrm{\prime }}}=\overrightarrow{OP}+2\cdot \overrightarrow{PF}\backslash overrightarrow\{OP\text{'}\}=\backslash overrightarrow\{OP\}+2\backslash cdot\backslash overrightarrow\{PF\}$

Berechne zunächst den Vektor $\overrightarrow{PF}\backslash overrightarrow\{PF\}$:

$\overrightarrow{PF}=\overrightarrow{OF}-\overrightarrow{OP}=\left(\begin{array}{c}{\displaystyle \frac{19}{13}}\\ {\displaystyle \frac{11}{13}}\\ {\displaystyle \frac{29}{26}}\end{array}\right)-\left(\begin{array}{c}1\\ 1\\ \mathrm{0,5}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{19}{13}}-1\\ {\displaystyle \frac{11}{13}}-1\\ {\displaystyle \frac{29}{26}}-\mathrm{0,5}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{6}{13}}\\ -{\displaystyle \frac{2}{13}}\\ {\displaystyle \frac{8}{13}}\end{array}\right)\backslash overrightarrow\{PF\}=\backslash overrightarrow\{OF\}-\backslash overrightarrow\{OP\}=\backslash begin\{pmatrix\}\backslash dfrac\{19\}\{13\}\backslash \backslash [2ex]\; \backslash dfrac\{11\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{29\}\{26\}\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash 0\{,\}5\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{19\}\{13\}-1\backslash \backslash [2ex]\; \backslash dfrac\{11\}\{13\}-1\backslash \backslash [2ex]\backslash dfrac\{29\}\{26\}-0\{,\}5\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{6\}\{13\}\backslash \backslash [2ex]\; -\backslash dfrac\{2\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{8\}\{13\}\backslash end\{pmatrix\}$

6. Setze dann $\overrightarrow{OP}\backslash overrightarrow\{OP\}$ und $\overrightarrow{PF}\backslash overrightarrow\{PF\}$ in die Vektorgleichung $\overrightarrow{O{P}^{\mathrm{\prime }}}=\overrightarrow{OP}+2\cdot \overrightarrow{PF}\backslash overrightarrow\{OP\text{'}\}=\backslash overrightarrow\{OP\}+2\backslash cdot\backslash overrightarrow\{PF\}$ ein:

$\overrightarrow{O{P}^{\mathrm{\prime }}}=\left(\begin{array}{c}1\\ 1\\ \mathrm{0,5}\end{array}\right)+2\cdot \left(\begin{array}{c}{\displaystyle \frac{6}{13}}\\ -{\displaystyle \frac{2}{13}}\\ {\displaystyle \frac{8}{13}}\end{array}\right)=\left(\begin{array}{c}1+{\displaystyle \frac{12}{13}}\\ 1-{\displaystyle \frac{4}{13}}\\ \mathrm{0,5}+{\displaystyle \frac{16}{13}}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{25}{13}}\\ {\displaystyle \frac{9}{13}}\\ {\displaystyle \frac{45}{26}}\end{array}\right)\backslash overrightarrow\{OP\text{'}\}=\backslash begin\{pmatrix\}1\backslash \backslash 1\backslash \backslash 0\{,\}5\backslash end\{pmatrix\}+2\backslash cdot\; \backslash begin\{pmatrix\}\backslash dfrac\{6\}\{13\}\backslash \backslash [2ex]\; -\backslash dfrac\{2\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{8\}\{13\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}1+\backslash dfrac\{12\}\{13\}\backslash \backslash [2ex]1-\backslash dfrac\{4\}\{13\}\backslash \backslash [2ex]0\{,\}5+\backslash dfrac\{16\}\{13\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{25\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{9\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{45\}\{26\}\backslash end\{pmatrix\}$

Der Spiegelpunkt hat die Koordinaten $P^{\mathrm{\prime }}({\displaystyle \frac{25}{13}}\mid {\displaystyle \frac{9}{13}}\mid {\displaystyle \frac{45}{26}})P\text{'}\backslash left(\backslash dfrac\{25\}\{13\}\backslash Big\backslash vert\backslash dfrac\{9\}\{13\}\backslash Big\backslash vert\backslash dfrac\{45\}\{26\}\backslash right)$.

7. Der berechnete Punkt $P^{\mathrm{\prime }}P\text{'}$ ist ein Punkt der Spiegelebene $E^{\mathrm{\prime }}E\text{'}$. Erstelle eine Parameterform für die Spiegelebene $E^{\mathrm{\prime }}E\text{'}$ mit der Schnittgeraden $g_{S}g\_S$ und einem weiteren Richtungsvektor $\vec{v}=\overrightarrow{A{P}^{\mathrm{\prime }}}\backslash vec\; v=\backslash overrightarrow\{AP\text{'}\}$ ($AA$ ist der Aufpunkt der Schnittgeraden)

$\Rightarrow \backslash ;\backslash Rightarrow\backslash ;$ $E^{\mathrm{\prime }}:\vec{x}=\overrightarrow{OA}+r\cdot \vec{u}+s\cdot \vec{v}E\text{'}:\backslash ;\; \backslash vec\; x=\backslash overrightarrow\{OA\}+r\backslash cdot\; \backslash vec\; u+s\backslash cdot\; \backslash vec\; v$.

Berechne den zweiten Richtungsvektor:

$\vec{v}=\overrightarrow{A{P}^{\mathrm{\prime }}}=\overrightarrow{O{P}^{\mathrm{\prime }}}-\overrightarrow{OA}=\left(\begin{array}{c}{\displaystyle \frac{25}{13}}\\ {\displaystyle \frac{9}{13}}\\ {\displaystyle \frac{45}{26}}\end{array}\right)-\left(\begin{array}{c}{\displaystyle \frac{18}{7}}\\ 0\\ {\displaystyle \frac{1}{14}}\end{array}\right)=\left(\begin{array}{c}{\displaystyle \frac{25}{13}}-{\displaystyle \frac{18}{7}}\\ {\displaystyle \frac{9}{13}}-0\\ {\displaystyle \frac{45}{26}}-{\displaystyle \frac{1}{14}}\end{array}\right)=\left(\begin{array}{c}-{\displaystyle \frac{59}{91}}\\ {\displaystyle \frac{9}{13}}\\ {\displaystyle \frac{151}{91}}\end{array}\right)={\displaystyle \frac{1}{91}}\cdot \left(\begin{array}{c}-59\\ 63\\ 151\end{array}\right)\backslash vec\; v=\backslash overrightarrow\{AP\text{'}\}=\backslash overrightarrow\{OP\text{'}\}-\backslash overrightarrow\{OA\}=\backslash begin\{pmatrix\}\backslash dfrac\{25\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{9\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{45\}\{26\}\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}\backslash dfrac\{18\}\{7\}\backslash \backslash [1ex]0\backslash \backslash [1ex]\backslash dfrac\{1\}\{14\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash dfrac\{25\}\{13\}-\backslash dfrac\{18\}\{7\}\backslash \backslash [2ex]\backslash dfrac\{9\}\{13\}-0\backslash \backslash [2ex]\backslash dfrac\{45\}\{26\}-\backslash dfrac\{1\}\{14\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-\backslash dfrac\{59\}\{91\}\backslash \backslash [2ex]\backslash dfrac\{9\}\{13\}\backslash \backslash [2ex]\backslash dfrac\{151\}\{91\}\backslash end\{pmatrix\}=\backslash dfrac\{1\}\{91\}\backslash cdot\backslash begin\{pmatrix\}-59\backslash \backslash 63\backslash \backslash 151\backslash end\{pmatrix\}$

Die Spiegelebene $E^{\mathrm{\prime }}E\text{'}$ kann dann als Parametergleichung geschrieben werden:

$E^{\mathrm{\prime }}:\vec{x}=\left(\begin{array}{c}{\displaystyle \frac{18}{7}}\\ 0\\ {\displaystyle \frac{1}{14}}\end{array}\right)+r\cdot \left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)+s\cdot \left(\begin{array}{c}-59\\ 63\\ 151\end{array}\right)E\text{'}:\backslash ;\backslash vec\; x=\backslash begin\{pmatrix\}\backslash dfrac\{18\}\{7\}\backslash \backslash [1ex]0\backslash \backslash [1ex]\backslash dfrac\{1\}\{14\}\backslash end\{pmatrix\}+r\backslash cdot\backslash begin\{pmatrix\}-1\backslash \backslash 1\backslash \backslash 1\backslash end\{pmatrix\}+s\backslash cdot\; \backslash begin\{pmatrix\}-59\backslash \backslash 63\backslash \backslash 151\backslash end\{pmatrix\}$