Wiki-Quellcode von Lösung Lineare Algebra 4_2
Zuletzt geändert von akukin am 2025/12/29 17:28
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| author | version | line-number | content |
|---|---|---|---|
| 1 | === Teilaufgabe a) === | ||
| 2 | {{detail summary="Erwartungshorizont"}} | ||
| 3 | <p> | ||
| 4 | {{formula}}B{{/formula}} liegt auf {{formula}}g{{/formula}}. | ||
| 5 | </p><p> | ||
| 6 | {{formula}} | ||
| 7 | \overrightarrow{AB} =\begin{pmatrix}-1\\-2\\2\end{pmatrix} | ||
| 8 | {{/formula}} | ||
| 9 | </p><p> | ||
| 10 | Es gilt: {{formula}} | ||
| 11 | \overrightarrow{AB} \cdot \begin{pmatrix} -2\\4\\3\end{pmatrix}= 0 | ||
| 12 | {{/formula}} | ||
| 13 | </p> | ||
| 14 | Somit ist {{formula}}\overrightarrow{AB}{{/formula}} senkrecht zu {{formula}}g{{/formula}} | ||
| 15 | und damit entspricht {{formula}}\Bigl| \overrightarrow{AB} \Bigr|{{/formula}} dem Abstand von {{formula}}A{{/formula}} zu {{formula}}g{{/formula}}. | ||
| 16 | {{/detail}} | ||
| 17 | |||
| 18 | |||
| 19 | {{detail summary="Erläuterung der Lösung"}} | ||
| 20 | |||
| 21 | {{/detail}} | ||
| 22 | |||
| 23 | === Teilaufgabe b) === | ||
| 24 | {{detail summary="Erwartungshorizont"}} | ||
| 25 | <p> | ||
| 26 | Mögliche Lösung: | ||
| 27 | </p> | ||
| 28 | {{formula}} | ||
| 29 | \overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{AB} | ||
| 30 | \begin{pmatrix}3\\0\\-1\end{pmatrix} | ||
| 31 | + | ||
| 32 | \begin{pmatrix}-1\\-2\\2\end{pmatrix} | ||
| 33 | = | ||
| 34 | \begin{pmatrix}2\\-2\\1\end{pmatrix},{{/formula}} damit {{formula}}\ C(2 \mid -2 \mid 1) | ||
| 35 | {{/formula}} | ||
| 36 | {{/detail}} | ||
| 37 | |||
| 38 | |||
| 39 | {{detail summary="Erläuterung der Lösung"}} | ||
| 40 | |||
| 41 | {{/detail}} |