1. Berechne den Vektor $\overrightarrow{\mathrm{A}\mathrm{B}}\backslash overrightarrow\{\backslash mathrm\{AB\}\}$:

$\overrightarrow{\mathrm{A}\mathrm{B}}=\left(\begin{array}{c}4\\ 6\end{array}\right)-\left(\begin{array}{c}3\\ 1\end{array}\right)=\left(\begin{array}{c}1\\ 5\end{array}\right)\backslash overrightarrow\{\backslash mathrm\{AB\}\}=\backslash begin\{pmatrix\}4\backslash \backslash 6\backslash end\{pmatrix\}-\backslash begin\{pmatrix\}3\backslash \backslash 1\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}1\backslash \backslash 5\backslash end\{pmatrix\}$

Aus dem Vektor $\vec{v}=\left(\begin{array}{c}x\\ y\end{array}\right)\backslash vec\{v\}=\backslash begin\{pmatrix\}x\backslash \backslash y\backslash end\{pmatrix\}$ wird bei einer Drehung um $\alpha ={90}^{\circ }\backslash alpha=90^\backslash circ$der Bildvektor $\overrightarrow{v^{\mathrm{\prime }}}=\left(\begin{array}{c}-y\\ x\end{array}\right)\backslash vec\{v\text{'}\}=\backslash begin\{pmatrix\}-y\backslash \backslash x\backslash end\{pmatrix\}$

2. Drehe den Vektor $\overrightarrow{\mathrm{A}\mathrm{B}}\backslash overrightarrow\{\backslash mathrm\{AB\}\}$ um seinen Fußpunkt $AA$ um $\alpha ={90}^{\circ }\backslash alpha=90^\backslash circ$: $\overrightarrow{\mathrm{A}{\mathrm{B}}^{\mathrm{\prime }}}=\left(\begin{array}{c}-5\\ 1\end{array}\right)\backslash overrightarrow\{\backslash mathrm\{AB\text{'}\}\}=\backslash begin\{pmatrix\}-5\backslash \backslash 1\backslash end\{pmatrix\}$

3. Berechnung der Koordinaten von $B^{\mathrm{\prime }}\left(x^{\mathrm{\prime }}\mathrm{\mid }{y}^{\mathrm{\prime }}\right)B\text{'}\backslash left(x\text{'}\backslash vert\; y\text{'}\backslash right)$ durch Koordinatenvergleich:

$\overrightarrow{\mathrm{A}{\mathrm{B}}^{\mathrm{\prime }}}=\left(\begin{array}{c}{x}^{\mathrm{\prime }}-3\\ y^{\mathrm{\prime }}-1\end{array}\right)=\left(\begin{array}{c}-5\\ 1\end{array}\right)\backslash overrightarrow\{\backslash mathrm\{AB\text{'}\}\}=\backslash begin\{pmatrix\}x\text{'}-3\backslash \backslash y\text{'}-1\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-5\backslash \backslash 1\backslash end\{pmatrix\}$

Du erhältst zwei Gleichungen:

Antwort: Der Punkt $B^{\mathrm{\prime }}B\text{'}$ hat die Koordinaten $B^{\mathrm{\prime }}(-2\mathrm{\mid }2)B\text{'}\backslash left(-2\backslash vert\; 2\backslash right)$.

Andere Möglichkeit der Berechnung: Aufstellen einer Vektorkette