Lösung Flächenberechnung Dreieck

Zuletzt geändert von Daniel Stocker am 2024/02/06 11:55

\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)

\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)

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}}
\Rightarrow \alpha \approx 22,89^{\circ}

Fläche = \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}