Wiki-Quellcode von Lösung gleichschenkliges Dreieck
Version 2.1 von Holger Engels am 2024/10/21 19:28
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| author | version | line-number | content |
|---|---|---|---|
| 1 | 1. {{formula}}\overrightarrow{AB}=\left(\begin{array}{c} 0 \\ 10 \\ 0 \end{array}\right) \qquad \overrightarrow{AC}=\left(\begin{array}{c} -10 \\ 10 \\ 0 \end{array}\right) \qquad \overrightarrow{BC}=\left(\begin{array}{c} -10 \\ 0 \\ 0 \end{array}\right){{/formula}} | ||
| 2 | Also: {{formula}}\left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|=10{{/formula}} | ||
| 3 | 1. {{formula}}\overrightarrow{AB}{{/formula}} und {{formula}}\overrightarrow{BC}{{/formula}} sind gleich lang und es gilt {{formula}}\overrightarrow{AB}\circ\overrightarrow{BC}=0{{/formula}} | ||
| 4 | |||
| 5 | {{formula}}D(-5|-5|12){{/formula}} |