Wiki-Quellcode von Lösung Lineare Algebra 4_2
Verstecke letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
 |
1.1 | 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"}} | ||
| |
1.2 | 20 | <p> |
| 21 | {{formula}}B{{/formula}} liegt auf {{formula}}g{{/formula}}, da der Stützvektor der Geraden dem Ortsvektor von Punkt {{formula}}B{{/formula}} entspricht. | ||
| 22 | </p><p> | ||
| 23 | {{formula}} | ||
| 24 | \overrightarrow{AB} =\begin{pmatrix}3-4\\0-2\\-1-(-3)\end{pmatrix}=\begin{pmatrix}-1\\-2\\2\end{pmatrix} | ||
| 25 | {{/formula}} | ||
| 26 | </p><p> | ||
| 27 | Es gilt: {{formula}} | ||
| 28 | \overrightarrow{AB} \cdot \begin{pmatrix} -2\\4\\3\end{pmatrix}=(-1)\cdot (-2)+(-2)\cdot 4+2\cdot 3=2-8+6=0 | ||
| 29 | {{/formula}} | ||
| 30 | </p> | ||
| 31 | Da das Skalarprodukt von {{formula}}\overrightarrow{AB}{{/formula}} und des Richtungsvektors der Geraden {{formula}}g{{/formula}} null ist, ist {{formula}}\overrightarrow{AB}{{/formula}} senkrecht zu {{formula}}g{{/formula}}. Damit entspricht {{formula}}\Bigl| \overrightarrow{AB} \Bigr|{{/formula}} dem Abstand von {{formula}}A{{/formula}} zu {{formula}}g{{/formula}}. | ||
| |
1.1 | 32 | {{/detail}} |
| 33 | |||
| 34 | === Teilaufgabe b) === | ||
| 35 | {{detail summary="Erwartungshorizont"}} | ||
| 36 | <p> | ||
| 37 | Mögliche Lösung: | ||
| 38 | </p> | ||
| 39 | {{formula}} | ||
| |
1.2 | 40 | \overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{AB}= |
| |
1.1 | 41 | \begin{pmatrix}3\\0\\-1\end{pmatrix} |
| 42 | + | ||
| 43 | \begin{pmatrix}-1\\-2\\2\end{pmatrix} | ||
| 44 | = | ||
| 45 | \begin{pmatrix}2\\-2\\1\end{pmatrix},{{/formula}} damit {{formula}}\ C(2 \mid -2 \mid 1) | ||
| 46 | {{/formula}} | ||
| 47 | {{/detail}} | ||
| 48 | |||
| 49 | |||
| 50 | {{detail summary="Erläuterung der Lösung"}} | ||
| |
2.1 | 51 | Einen weiteren Punkt {{formula}}C{{/formula}} erhalten wir, indem wir Punkt {{formula}}A{{/formula}} am Punkt {{formula}}B{{/formula}} spiegeln durch: |
| 52 | <br> | ||
| 53 | {{formula}} | ||
| 54 | \overrightarrow{OC} = \overrightarrow{OA} + 2\cdot \overrightarrow{AB}= | ||
| 55 | \begin{pmatrix}4\\2\\-3\end{pmatrix} | ||
| 56 | + | ||
| 57 | 2\cdot \begin{pmatrix}-1\\-2\\2\end{pmatrix} | ||
| 58 | = | ||
| 59 | \begin{pmatrix}2\\-2\\1\end{pmatrix},{{/formula}} damit {{formula}}\ C(2 \mid -2 \mid 1) | ||
| 60 | {{/formula}} | ||
| 61 | <br> | ||
| 62 | oder alternativ: | ||
| 63 | <br> | ||
| 64 | {{formula}} | ||
| 65 | \overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{AB}= | ||
| 66 | \begin{pmatrix}3\\0\\-1\end{pmatrix} | ||
| 67 | + | ||
| 68 | \begin{pmatrix}-1\\-2\\2\end{pmatrix} | ||
| 69 | = | ||
| 70 | \begin{pmatrix}2\\-2\\1\end{pmatrix},{{/formula}} damit {{formula}}\ C(2 \mid -2 \mid 1) | ||
| 71 | {{/formula}} | ||
| |
1.1 | 72 | {{/detail}} |