Übertrage in einen Spaltenvektor:

$\vec{A}=\left(\begin{array}{c}4\\ 1\\ -1\end{array}\right)\backslash vec\{A\}=\backslash begin\{pmatrix\}4\backslash \backslash 1\backslash \backslash -1\backslash end\{pmatrix\}$

Dieser Vektor verläuft vom Ursprung zum Punkt A.

Subtrahiere "Spitze minus Fuß":

$\overrightarrow{AB}=\left(\begin{array}{c}3-1\\ 3-0\\ 3-1\end{array}\right)=\left(\begin{array}{c}2\\ 3\\ 2\end{array}\right)\backslash overrightarrow\{AB\}=\backslash begin\{pmatrix\}3-1\backslash \backslash 3-0\backslash \backslash 3-1\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\backslash \backslash 3\backslash \backslash 2\backslash end\{pmatrix\}$

Drehe alle Vorzeichen um:

$-\vec{v}=\left(\begin{array}{c}1\\ -1\\ -2\end{array}\right)-\backslash vec\; v\; =\backslash begin\{pmatrix\}1\backslash \backslash -1\backslash \backslash -2\backslash end\{pmatrix\}$

Der Verbindungsvektor zwischen zwei Punkten $A(a_1\mathrm{\mid }{a}_2\mathrm{\mid }{a}_3)A(a\_1|a\_2|a\_3)$ und $B(b_1\mathrm{\mid }{b}_2\mathrm{\mid }{b}_3)B(b\_1|b\_2|b\_3)$ ist $\overrightarrow{AB}=\left(\begin{array}{c}{b}_1-a_1\\ b_2-a_2\\ b_3-a_3\end{array}\right)\backslash overrightarrow\{AB\}=\backslash begin\{pmatrix\}b\_1-a\_1\backslash \backslash b\_2-a\_2\backslash \backslash b\_3-a\_3\backslash end\{pmatrix\}$. Somit gilt für den Punkt B:

$\left(\begin{array}{c}1\\ 0\\ 3\end{array}\right)=\left(\begin{array}{c}{b}_1-(-2)\\ b_2-1\\ b_3-2\end{array}\right)\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 3\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}b\_1-(-2)\backslash \backslash b\_2-1\backslash \backslash b\_3-2\backslash end\{pmatrix\}$

also:

$1=b_1+2\iff b_1=-11=b\_1+2\; \backslash Leftrightarrow\; b\_1=-1$

$0=b_2-1\iff b_2=10=b\_2-1\backslash Leftrightarrow\; b\_2=1$

$3=b_3-2\iff b_3=53=b\_3-2\; \backslash Leftrightarrow\; b\_3=5$

der Ortsvektor ist also $\vec{B}=\left(\begin{array}{c}-1\\ 1\\ 5\end{array}\right)\backslash vec\; B=\backslash begin\{pmatrix\}-1\backslash \backslash 1\backslash \backslash 5\backslash end\{pmatrix\}$ und die Punktkoordinaten sind $B(-1\mathrm{\mid }1\mathrm{\mid }5)B(-1|1|5)$.