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Lösung Vektoraddition

Version 3.1 von Daniel Stocker am 2023/11/30 15:42

\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)

\text{I:}