Du erhältst den Ortsvektor $\vec{C}\backslash vec\; C$ von C, indem du den Verbindungsvektor $\overrightarrow{AB}\backslash overrightarrow\{AB\}$ zum Ortsvektor $\overrightarrow{B}\backslash overrightarrow\{B\}$ addierst:

$\vec{C}=\vec{B}+\overrightarrow{AB}=\left(\begin{array}{c}0\\ 4\\ 3\end{array}\right)+\left(\begin{array}{c}0-(-2)\\ 4-1\\ 3-1\end{array}\right)=\left(\begin{array}{c}2\\ 7\\ 5\end{array}\right)\backslash vec\; C=\backslash vec\; B+\backslash overrightarrow\{AB\}=\backslash begin\{pmatrix\}\; 0\backslash \backslash 4\backslash \backslash 3\backslash end\{pmatrix\}+\backslash begin\{pmatrix\}\; 0-(-2)\backslash \backslash 4-1\backslash \backslash 3-1\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\backslash \backslash 7\backslash \backslash 5\backslash end\{pmatrix\}$

und somit $C(2\mathrm{\mid }7\mathrm{\mid }5)C(2|7|5)$.

Entweder kennst du die Formel $\overrightarrow{P}=\frac{1}{2}(\overrightarrow{A}+\overrightarrow{B})\backslash overrightarrow\{P\}=\backslash frac\; 1\; 2\; (\backslash overrightarrow\; A\; +\backslash overrightarrow\; B)$ für den Mittelpunkt der Strecke oder du verwendest eine Vektorkette $\overrightarrow{P}=\overrightarrow{A}+\frac{1}{2}\overrightarrow{AB}\backslash overrightarrow\; P=\backslash overrightarrow\; A\; +\backslash frac\; 1\; 2\; \backslash overrightarrow\{AB\}$ zur Bestimmung des Ortsvektors.

Möglichkeit 1

$\overrightarrow{P}=\frac{1}{2}(\overrightarrow{A}+\overrightarrow{B})=\frac{1}{2}(\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)+\left(\begin{array}{c}3\\ 1\\ 1\end{array}\right))=\frac{1}{2}\cdot \left(\begin{array}{c}5\\ 2\\ -3\end{array}\right)=\left(\begin{array}{c}\mathrm{2,5}\\ 1\\ -\mathrm{1,5}\end{array}\right)\backslash overrightarrow\; P=\backslash frac\; 1\; 2(\backslash overrightarrow\; A+\backslash overrightarrow\; B)=\backslash frac\; 1\; 2\; \backslash left(\backslash begin\{pmatrix\}\; 2\backslash \backslash 1\backslash \backslash -4\backslash end\{pmatrix\}+\backslash begin\{pmatrix\}3\backslash \backslash 1\backslash \backslash 1\backslash end\{pmatrix\}\backslash right)=\backslash frac\; 1\; 2\; \backslash cdot\; \backslash begin\{pmatrix\}5\backslash \backslash 2\backslash \backslash -3\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\{,\}5\backslash \backslash 1\backslash \backslash -1\{,\}5\backslash end\{pmatrix\}$

Möglichkeit 2

$\overrightarrow{P}=\overrightarrow{A}+\frac{1}{2}\overrightarrow{AB}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)+\frac{1}{2}\cdot \left(\begin{array}{c}3-2\\ 1-1\\ 1-(-4)\end{array}\right)=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)+\frac{1}{2}\cdot \left(\begin{array}{c}1\\ 0\\ 5\end{array}\right)=\left(\begin{array}{c}\mathrm{2,5}\\ 1\\ -\mathrm{1,5}\end{array}\right)\backslash overrightarrow\; P=\backslash overrightarrow\; A+\backslash frac\; 1\; 2\; \backslash overrightarrow\; \{AB\}=\; \backslash begin\{pmatrix\}2\backslash \backslash 1\backslash \backslash -4\backslash end\{pmatrix\}+\backslash frac\; 1\; 2\backslash cdot\backslash begin\{pmatrix\}3-2\backslash \backslash 1-1\backslash \backslash 1-(-4)\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\backslash \backslash 1\backslash \backslash -4\backslash end\{pmatrix\}+\backslash frac\; 1\; 2\backslash cdot\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 5\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}2\{,\}5\backslash \backslash 1\backslash \backslash -1\{,\}5\backslash end\{pmatrix\}$

Die Mitte der Strecke ist also P(2,5|1|-1,5).

Verwende die Formel $\vec{S}=\frac{1}{3}(\vec{A}+\vec{B}+\vec{C})=\frac{1}{3}(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ 3\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ 0\\ -4\end{array}\right))=\frac{1}{3}\cdot \left(\begin{array}{c}1\\ 3\\ -4\end{array}\right)=\left(\begin{array}{c}\frac{1}{3}\\ 1\\ -\frac{4}{3}\end{array}\right)\backslash vec\; S=\backslash frac\; 1\; 3(\backslash vec\; A+\backslash vec\; B+\backslash vec\; C)=\backslash frac\; 1\; 3\backslash left(\backslash begin\{pmatrix\}1\backslash \backslash 0\backslash \backslash 0\backslash end\{pmatrix\}+\backslash begin\{pmatrix\}0\backslash \backslash 3\backslash \backslash 0\backslash end\{pmatrix\}+\backslash begin\{pmatrix\}0\backslash \backslash 0\backslash \backslash -4\backslash end\{pmatrix\}\backslash right)=\backslash frac\; 1\; 3\backslash cdot\; \backslash begin\{pmatrix\}\; 1\backslash \backslash 3\backslash \backslash -4\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\backslash frac\; 1\; 3\backslash \backslash 1\backslash \backslash -\backslash frac\; 4\; 3\backslash end\{pmatrix\}$

Der Schwerpunkt liegt also bei $S(\frac{1}{3}\mathrm{\mid }1\mathrm{\mid }-\frac{4}{3})S(\backslash frac\; 1\; 3|1|-\backslash frac\; 4\; 3)$

In einem Parallelogramm sind gegenüberliegende Seiten parallel und gleich lang. Somit ist $\overrightarrow{CB}=\overrightarrow{DA}\backslash overrightarrow\{CB\}=\backslash overrightarrow\{DA\}$.

$\overrightarrow{CB}=\overrightarrow{DA}=\left(\begin{array}{c}2\\ 0\\ -2\end{array}\right)\backslash overrightarrow\{CB\}=\backslash overrightarrow\{DA\}=\backslash begin\{pmatrix\}2\backslash \backslash 0\backslash \backslash -2\backslash end\{pmatrix\}$

Somit kann man den Ortsvektor von A bestimmen, indem man $\overrightarrow{DA}\backslash overrightarrow\{DA\}$ zu $\overrightarrow{D}\backslash overrightarrow\; D$ addiert:

$\overrightarrow{A}=\overrightarrow{D}+\overrightarrow{DA}=\left(\begin{array}{c}-2\\ -3\\ 0\end{array}\right)+\left(\begin{array}{c}2\\ 0\\ -2\end{array}\right)=\left(\begin{array}{c}0\\ -3\\ -2\end{array}\right)\backslash overrightarrow\; A=\backslash overrightarrow\; D+\backslash overrightarrow\{DA\}=\backslash begin\{pmatrix\}-2\backslash \backslash -3\backslash \backslash 0\backslash end\{pmatrix\}+\backslash begin\{pmatrix\}2\backslash \backslash 0\backslash \backslash -2\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}0\backslash \backslash -3\backslash \backslash -2\backslash end\{pmatrix\}$

Da die beiden Seiten parallel aber nicht gleich lang sind, gilt:

$\overrightarrow{DC}=\mathrm{0,4}\cdot \overrightarrow{AB}\backslash overrightarrow\{DC\}=0\{,\}4\backslash cdot\backslash overrightarrow\{AB\}$

Eine Skizze hilft, den Term zu verstehen:

• Damit die Vektoren die gleiche Richtung haben, muss man zum Beispiel $\overrightarrow{AB}\backslash overrightarrow\{AB\}$ mit $\overrightarrow{DC}\backslash overrightarrow\{DC\}$ vergleichen (und nicht $\overrightarrow{CD}\backslash overrightarrow\{CD\}$)

• Die Strecke $\overline{DC}\backslash overline\{DC\}$ ist die kürzere, also muss $\overrightarrow{AB}\backslash overrightarrow\{AB\}$ verkürzt werden.

• Eine Verkürzung um 60% bedeutet, dass noch $100\mathrm{\%}-60\mathrm{\%}=40\mathrm{\%}100\backslash \%-60\backslash \%=40\backslash \%$ übrig sind.

Setze in die Gleichung ein:

Für den Ortsvektor $\overrightarrow{B}\backslash overrightarrow\{B\}$ kannst du nun verwenden, dass

$\overrightarrow{B}=\overrightarrow{A}+\overrightarrow{AB}\backslash overrightarrow\; B=\backslash overrightarrow\; A+\backslash overrightarrow\{AB\}$

Da A(0|0|0) der Ursprung des Koordinatensystems ist, ist $\overrightarrow{B}=\overrightarrow{AB}\backslash overrightarrow\; B\; =\; \backslash overrightarrow\{AB\}$ und somit

B(5|7,5|0)