Lösung Vektoraddition
Version 3.1 von Daniel Stocker am 2023/11/30 14:42
\[\begin{align*}\overrightarrow{AB} &= \left(\begin{array}{c} 5-3 \\ 2-1 \\ 4-5\end{array}\right) &= \left(\begin{array}{c} 2 \\ 1 \\ -1\end{array}\right) \\
\overrightarrow{CA} &= \left(\begin{array}{c} 3-8 \\ 1-7 \\ 5-1 \end{array}\right) &= \left(\begin{array}{c} -5 \\ -6 \\ -4 \end{array}\right) \\
\overrightarrow{BC} &= \left(\begin{array}{c} 8-5 \\ 7-2 \\ 1-4 \end{array}\right) &= \left(\begin{array}{c} 3 \\ 5 \\ -3 \end{array}\right) \\
\overrightarrow{DA} &= \left(\begin{array}{c} 3-d_1 \\ 1-d_2 \\ 5-d_3 \end{array}\right)
\end{align*}\]
\[\overrightarrow{AB}-\overrightarrow{CA}+\overrightarrow{BC}-\overrightarrow{DA}&=\overrightarrow{o}\]
\[\left(\begin{array}{c} 2 \\ 1 \\ -1\end{array}\right)-\left(\begin{array}{c} -5 \\ -6 \\ -4 \end{array}\right) + \left(\begin{array}{c} 3 \\ 5 \\ -3 \end{array}\right) - \left(\begin{array}{c} 3-d_1 \\ 1-d_2 \\ 5-d_3 \end{array}\right) &= \left(\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right)\]
\[\text{I:} \]