Der Punkt $P(1\mathrm{\mid }2\mathrm{\mid }5)P(1|2|5)$ wird am Punkt $Z(1\mathrm{\mid }3\mathrm{\mid }-4)Z(1|3|-4)$ gespiegelt.

$\overrightarrow{OP}=\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)\backslash overrightarrow\{OP\}=\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 5\backslash end\{pmatrix\}$und $\overrightarrow{OZ}=\left(\begin{array}{c}1\\ 3\\ -4\end{array}\right)\backslash overrightarrow\{OZ\}=\backslash begin\{pmatrix\}1\backslash \backslash 3\backslash \backslash -4\backslash end\{pmatrix\}$

Berechne den Vektor:

$\overrightarrow{PZ}=\overrightarrow{OZ}-\overrightarrow{OP}=\left(\begin{array}{c}1\\ 3\\ -4\end{array}\right)-\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ -9\end{array}\right)\backslash overrightarrow\{PZ\}=\backslash overrightarrow\{OZ\}-\backslash overrightarrow\{OP\}=\backslash begin\{pmatrix\}1\backslash \backslash 3\backslash \backslash -4\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 5\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash -9\backslash end\{pmatrix\}$.

Setze die Vektoren in $\overrightarrow{O{P}^{\mathrm{\prime }}}=\overrightarrow{OP}+2\cdot \overrightarrow{PZ}\backslash overrightarrow\{OP\text{'}\}=\backslash overrightarrow\{OP\}+2\backslash cdot\backslash overrightarrow\{PZ\}$ ein:

$\overrightarrow{O{P}^{\mathrm{\prime }}}=\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)+2\cdot \left(\begin{array}{c}0\\ 1\\ -9\end{array}\right)=\left(\begin{array}{c}1\\ 4\\ -13\end{array}\right)\backslash overrightarrow\{OP\text{'}\}=\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 5\backslash end\{pmatrix\}+2\backslash cdot\backslash begin\{pmatrix\}0\backslash \backslash 1\backslash \backslash -9\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}1\backslash \backslash 4\backslash \backslash -13\backslash end\{pmatrix\}$

Antwort: Der gespiegelte Punkt $P^{\mathrm{\prime }}P\text{'}$ hat die Koordinaten $P^{\mathrm{\prime }}(1\mathrm{\mid }4\mathrm{\mid }-13)P\text{'}(1|4|-13)$.

Der Punkt $P(1\mathrm{\mid }2\mathrm{\mid }5)P(1|2|5)$ wird am Punkt $Z(-1\mathrm{\mid }3\mathrm{\mid }1)Z(-1|3|1)$ gespiegelt.

$\overrightarrow{OP}=\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)\backslash overrightarrow\{OP\}=\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 5\backslash end\{pmatrix\}$und $\overrightarrow{OZ}=\left(\begin{array}{c}-1\\ 3\\ 1\end{array}\right)\backslash overrightarrow\{OZ\}=\backslash begin\{pmatrix\}-1\backslash \backslash 3\backslash \backslash 1\backslash end\{pmatrix\}$

Berechne den Vektor:

$\overrightarrow{PZ}=\overrightarrow{OZ}-\overrightarrow{OP}=\left(\begin{array}{c}-1\\ 3\\ 1\end{array}\right)-\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)=\left(\begin{array}{c}-2\\ 1\\ -4\end{array}\right)\backslash overrightarrow\{PZ\}=\backslash overrightarrow\{OZ\}-\backslash overrightarrow\{OP\}=\backslash begin\{pmatrix\}-1\backslash \backslash 3\backslash \backslash 1\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 5\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-2\backslash \backslash 1\backslash \backslash -4\backslash end\{pmatrix\}$.

Setze die Vektoren in $\overrightarrow{O{P}^{\mathrm{\prime }}}=\overrightarrow{OP}+2\cdot \overrightarrow{PZ}\backslash overrightarrow\{OP\text{'}\}=\backslash overrightarrow\{OP\}+2\backslash cdot\backslash overrightarrow\{PZ\}$ ein:

$\overrightarrow{O{P}^{\mathrm{\prime }}}=\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)+2\cdot \left(\begin{array}{c}-2\\ 1\\ -4\end{array}\right)=\left(\begin{array}{c}-3\\ 4\\ -3\end{array}\right)\backslash overrightarrow\{OP\text{'}\}=\backslash begin\{pmatrix\}1\backslash \backslash 2\backslash \backslash 5\backslash end\{pmatrix\}+2\backslash cdot\backslash begin\{pmatrix\}-2\backslash \backslash 1\backslash \backslash -4\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-3\backslash \backslash 4\backslash \backslash -3\backslash end\{pmatrix\}$

Antwort: Der gespiegelte Punkt $P^{\mathrm{\prime }}P\text{'}$ hat die Koordinaten $P^{\mathrm{\prime }}(-3\mathrm{\mid }4\mathrm{\mid }-3)P\text{'}(-3|4|-3)$.