Lösung Vektoraddition

Version 2.1 von Daniel Stocker am 2023/11/30 15:39

$\begin{align*}\overrightarrow{AB} &= \left(\begin{array}{c} 5-3 \\ 2-1 \\ 4-5\end{array}\right) &= \left(\begin{array}{c} 2 \\ 1 \\ -1\end{array}\right) \\ \overrightarrow{CA} &= \left(\begin{array}{c} 3-8 \\ 1-7 \\ 5-1 \end{array}\right) &= \left(\begin{array}{c} -5 \\ -6 \\ 4 \end{array}\right) \\ \overrightarrow{BC} &= \left(\begin{array}{c} 8-5 \\ 7-2 \\ 1-4 \end{array}\right) &= \left(\begin{array}{c} 3 \\ 5 \\ -3 \end{array}\right) \\ \overrightarrow{DA} &= \left(\begin{array}{c} 3-d_1 \\ 1-d_2 \\ 5-d_3 \end{array}\right) \end{align*}$

$\overrightarrow{AB}-\overrightarrow{CA}+\overrightarrow{BC}-\overrightarrow{DA}&=\overrightarrow{o}$

$\left(\begin{array}{c} 2 \\ 1 \\ -1\end{array}\right)-\left(\begin{array}{c} -5 \\ -6 \\ 4 \end{array}\right) + \left(\begin{array}{c} 3 \\ 5 \\ -3 \end{array}\right) - \left(\begin{array}{c} 3-d_1 \\ 1-d_2 \\ 5-d_3 \end{array}\right) &= \left(\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right)$

$\begin{align*} &\text{I: } &2-(-5)+3-(3-d_1) &= 0 \\ & &10-3+d_1 &= 0\\ & &d_1 &= -7 \\ &\text{II: } &1-(-6)+5-(1-d_2) &= 0 \\ & &12-1+d_2 &= 0\\ & &d_2 &= -11 \\ &\text{III: } &-1-4+(-3)-(5-d_3) &= 0 \\ & &-8-5+d_3 &= 0\\ & &d_3 &= 13 \end{align*}$

$\Rightarrow D(-7|-11|13)$