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Wiki-Quellcode von Lösung In Summe Null

Zuletzt geändert von akukin am 2025/08/14 15:55
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akukin 5.1 1 {{formula}}
2 \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) \\
3 \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) \\
4 \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) \\
Daniel Stocker 1.1 5 \overrightarrow{DA} &= \left(\begin{array}{c} 3-d_1 \\ 1-d_2 \\ 5-d_3 \end{array}\right)
6 \end{align*}
7 {{/formula}}
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akukin 5.1 9 {{formula}}\overrightarrow{AB}-\overrightarrow{CA}+\overrightarrow{BC}-\overrightarrow{DA}=\overrightarrow{o}{{/formula}}
Daniel Stocker 1.1 10
akukin 5.1 11 {{formula}}\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){{/formula}}
Daniel Stocker 1.1 12
13 {{formula}}
Daniel Stocker 4.1 14 \begin{align*}
15 &\text{I: } &2-(-5)+3-(3-d_1) &= 0 \\
16 & &10-3+d_1 &= 0\\
17 & &d_1 &= -7 \\
18 &\text{II: } &1-(-6)+5-(1-d_2) &= 0 \\
19 & &12-1+d_2 &= 0\\
20 & &d_2 &= -11 \\
21 &\text{III: } &-1-4+(-3)-(5-d_3) &= 0 \\
22 & &-8-5+d_3 &= 0\\
23 & &d_3 &= 13
24 \end{align*}
Daniel Stocker 1.1 25 {{/formula}}
Daniel Stocker 4.1 26
27 {{formula}}
28 \Rightarrow D(-7|-11|13)
29 {{/formula}}
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