Wiki-Quellcode von Lösung Flächenberechnung Dreieck
Zuletzt geändert von Holger Engels am 2024/10/21 20:07
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| author | version | line-number | content |
|---|---|---|---|
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12.1 | 1 | {{formula}}\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA} = \left(\begin{array}{c} 0-2 \\ 9-(-1) \\ -3-4\end{array}\right) = \left(\begin{array}{c} -2 \\ 10 \\ -7\end{array}\right){{/formula}} |
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13.1 | 2 | |
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14.1 | 3 | {{formula}}\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA} = \left(\begin{array}{c} -2-2 \\ 5-(-1) \\ 1-4\end{array}\right) = \left(\begin{array}{c} -4 \\ 6 \\ -3\end{array}\right) |
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13.1 | 4 | {{/formula}} |
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15.1 | 5 | |
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15.2 | 6 | {{formula}} cos(\alpha) = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\mid\overrightarrow{AB} \mid \cdot \mid \overrightarrow{AC} \mid} = \frac{(-2)\cdot (-4)+10\cdot 6 + (-7) \cdot (-3) }{\sqrt{(-2)^2+10^2+(-7)^2}\cdot \sqrt{(-4)^2+6^2+(-3)^2}} = \frac{89}{\sqrt{153} \cdot \sqrt{61}} = \frac{89}{\sqrt{9333}} {{/formula}} |
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16.1 | 7 | {{formula}} \Rightarrow \alpha \approx 22,89^{\circ}{{/formula}} |
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15.1 | 8 | |
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23.1 | 9 | Fläche = {{formula}} \frac{1}{2} \cdot \mid \overrightarrow{AB} \mid \cdot \mid \overrightarrow{AC} \mid \cdot sin(\alpha) = \frac{1}{2} \cdot \sqrt{153} \cdot \sqrt{61}\cdot sin(22,89^{\circ}) \approx 18,79 \text{ FE}{{/formula}} |
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27.1 | 10 | |
| 11 | [[image:Dreieck.png]] |