A(-5 \ 0,5) hat den Ortsvektor $\overrightarrow{OA}=\left(\begin{array}{c}-5\\ \mathrm{0,5}\end{array}\right)\backslash overrightarrow\{OA\}\backslash \; =\backslash \; \backslash begin\{pmatrix\}-5\backslash \backslash 0\{,\}5\backslash end\{pmatrix\}$ und B(-2,5 \ 1,5) den Ortsvektor $\overrightarrow{OB}=\left(\begin{array}{c}-\mathrm{2,5}\\ \mathrm{1,5}\end{array}\right)\backslash overrightarrow\{OB\}\; =\; \backslash begin\{pmatrix\}-2\{,\}5\backslash \backslash 1\{,\}5\backslash end\{pmatrix\}$

$\overrightarrow{v}=\overrightarrow{AB}\backslash overrightarrow\{v\}\; =\; \backslash overrightarrow\{AB\}$ berechnest du dann wie folgt:

$\overrightarrow{v}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(\begin{array}{c}-\mathrm{2,5}-(-5)\\ \mathrm{1,5}-\mathrm{0,5}\end{array}\right)=\left(\begin{array}{c}\mathrm{2,5}\\ 1\end{array}\right)\backslash overrightarrow\{v\}\; =\; \backslash overrightarrow\{AB\}\; =\; \backslash overrightarrow\{OB\}\; -\; \backslash overrightarrow\{OA\}\; =\; \backslash begin\{pmatrix\}-2\{,\}5-(-5)\backslash \backslash 1\{,\}5-0\{,\}5\backslash end\{pmatrix\}=\; \backslash begin\{pmatrix\}2\{,\}5\backslash \backslash 1\backslash end\{pmatrix\}$

${\displaystyle \overrightarrow{v}=\left(\begin{array}{c}\mathrm{2,5}\\ 1\end{array}\right)}\backslash displaystyle\; \backslash overrightarrow\{v\}=\backslash begin\{pmatrix\}2\{,\}5\backslash \backslash 1\backslash end\{pmatrix\}$ ist also die Lösung

Ein weiterer Repräsentant des Vektors ${\displaystyle \overrightarrow{AB}}\backslash displaystyle\; \backslash overrightarrow\; \{AB\}$ ist der Vektor ${\displaystyle \overrightarrow{GH}}\backslash displaystyle\; \backslash overrightarrow\{GH\}$ .

Alle Pfeile, die gleich lang und parallel sind, und die in die gleiche Richtung zeigen, sind Repräsentanten desselben Vektors.

Der Gegenvektor von $\overrightarrow{v}=\left(\begin{array}{c}\mathrm{2,5}\\ 1\end{array}\right)\backslash overrightarrow\{v\}\backslash \; =\; \backslash begin\{pmatrix\}2\{,\}5\backslash \backslash 1\backslash end\{pmatrix\}$ ist $-\overrightarrow{v}-\backslash overrightarrow\{v\}$, also $-(1)\cdot \left(\begin{array}{c}\mathrm{2,5}\\ 1\end{array}\right)=\left(\begin{array}{c}-\mathrm{2,5}\\ -1\end{array}\right)-(1)\; \backslash cdot\; \backslash begin\{pmatrix\}\; 2\{,\}5\; \backslash \backslash \; 1\; \backslash end\{pmatrix\}\; =\; \backslash begin\{pmatrix\}\; -2\{,\}5\; \backslash \backslash \; -1\; \backslash end\{pmatrix\}$

Es gilt :${\displaystyle \overrightarrow{a}+(-\overrightarrow{a})=\overrightarrow{a}-\overrightarrow{a}}\backslash displaystyle\; \backslash overrightarrow\; \{a\}\; +(-\backslash overrightarrow\; \{a\}\; )=\backslash overrightarrow\{a\}\; -\backslash overrightarrow\; \{a\}$

Der Gegenvektor von ${\displaystyle \overrightarrow{v}}\backslash displaystyle\; \backslash overrightarrow\{v\}$ ist also $-\overrightarrow{v}=-\backslash overrightarrow\{v\}=$$\left(\begin{array}{c}-\mathrm{2,5}\\ -1\end{array}\right)\backslash begin\{pmatrix\}-2\{,\}5\backslash \backslash -1\backslash end\{pmatrix\}$