Wiki-Quellcode von Lösung Drachen begründen
Version 10.1 von Daniel Stocker am 2024/02/05 14:25
Zeige letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
| 1 | {{formula}}\overrightarrow{AC} = \left(\begin{array}{c} -3,5-8,5 \\ 8-5 \\ 2,5-(-3,5)\end{array}\right) = \left(\begin{array}{c} -12 \\ 3 \\ 6\end{array}\right){{/formula}} | ||
| 2 | |||
| 3 | {{formula}}\overrightarrow{BD} = \left(\begin{array}{c} 5-4 \\ 7-5 \\ 1-(-2)\end{array}\right) = \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right){{/formula}} | ||
| 4 | |||
| 5 | {{formula}}\overrightarrow{AC} \cdot \overrightarrow{BD} = -12 \cdot 1 + 3 \cdot 2 + 6 \cdot 1 = 0{{/formula}} | ||
| 6 | |||
| 7 | {{formula}}\mid \overrightarrow{AB} \mid = \( \Biggl| \left(\begin{array}{c} 4-8,5 \\ 5-5 \\ -2-(-3,5)\end{array}\right) \Biggr| \)= \( \Biggl| \left(\begin{array}{c} -4,5 \\ 0 \\ 1,5\end{array}\right) \Biggr| \)= \sqrt{(-4,5)^2 + 0^2 +1,5^2} = \sqrt{22,5}{{/formula}} | ||
| 8 | |||
| 9 | {{formula}}\mid \overrightarrow{AC} \mid = \mid \left(\begin{array}{c} -12 \\ 3 \\ 6\end{array}\right) \mid = \sqrt{(-12)^2+3^2+6^2} = \sqrt{22,5}{{/formula}} |