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Änderungen von Dokument Lösung Lineare Algebra

Zuletzt geändert von akukin am 2026/01/17 12:06
Von Version 4.1
bearbeitet von akukin
am 2026/01/14 16:43
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Auf Version 6.1
bearbeitet von akukin
am 2026/01/15 19:24
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... ... @@ -95,12 +95,13 @@
95 95  {{detail summary="Erläuterung der Lösung"}}
96 96  Damit die Ebene parallel zur {{formula}}x_1x_3{{/formula}}- Ebene ist, muss ein Richtungsvektor parallel zur {{formula}}x_1{{/formula}}-Achse sein und der andere zur {{formula}}x_3{{/formula}}-Achse.
97 97  <br>
98 -Das heißt, wir verwenden die Richtungsvektoren {{formula}}\begin{pmatrix}1\\0\\0\end{pmatrix}{{/formula}} und {{formula}}\begin{pmatrix}0\\0\\1\end{pmatrix}{{/formula}} (oder Alternativ Vielfache davon).
98 +Das heißt, wir verwenden die Richtungsvektoren {{formula}}\begin{pmatrix}1\\0\\0\end{pmatrix}{{/formula}} und {{formula}}\begin{pmatrix}0\\0\\1\end{pmatrix}{{/formula}} (oder Vielfache davon).
99 99  <p></p>
100 100  Damit die Ebene von {{formula}}C{{/formula}} den Abstand {{formula}}2{{/formula}} hat, wählen wir, ausgehend vom Punkt {{formula}}C(0|3|2){{/formula}}, als Stützvektor entweder {{formula}}\begin{pmatrix}0\\1\\2\end{pmatrix}{{/formula}} oder {{formula}}\begin{pmatrix}0\\5\\2\end{pmatrix}{{/formula}}, damit die Differenz der {{formula}}x_2{{/formula}}-Koordinaten {{formula}}2{{/formula}} ist.
101 -
102 102  <p></p>
103 -Somit ist eine mögliche Lösung {{formula}}E: \ \vec x =
102 +Somit ist eine mögliche Lösung
103 +<br>
104 +{{formula}}E: \ \vec x =
104 104  \begin{pmatrix}0\\1\\2\end{pmatrix}+ r \cdot
105 105  \begin{pmatrix}1\\0\\0\end{pmatrix} + s \cdot
106 106  \begin{pmatrix}0\\0\\1\end{pmatrix}; \ r, s \in \mathbb{R}
... ... @@ -107,6 +107,7 @@
107 107  {{/formula}}
108 108  <br>
109 109  Eine weitere Lösung wäre
111 +<br>
110 110  {{formula}}E: \ \vec x =
111 111  \begin{pmatrix}0\\5\\2\end{pmatrix}+ r \cdot
112 112  \begin{pmatrix}1\\0\\0\end{pmatrix} + s \cdot
... ... @@ -121,7 +121,9 @@
121 121  
122 122  
123 123  {{detail summary="Erläuterung der Lösung"}}
124 -
126 +Da {{formula}}\overrightarrow{AB}\cdot \overrightarrow{BC}=0{{/formula}} gilt, wissen wir, dass {{formula}}\overrightarrow{AB}{{/formula}} und {{formula}}\overrightarrow{BC}{{/formula}} senkrecht auf einander stehen. Die Punkte {{formula}}A, B{{/formula}} und {{formula}}C{{/formula}} bilden somit ein rechtwinkliges Dreieck. Die Strecken {{formula}}AB{{/formula}} und {{formula}}BC{{/formula}} sind dabei die Katheten des Dreiecks.
127 +<br>
128 +Da die Längen der Vektoren {{formula}}\overrightarrow{AB}{{/formula}} und {{formula}}\overrightarrow{BC}{{/formula}} den Längen der Katheten des Dreiecks {{formula}}ABC{{/formula}} entsprechen, wird durch den Term {{formula}} \frac{1}{2} \cdot \Bigl|\overrightarrow{AB}\Bigr| \cdot \Bigl|\overrightarrow{BC}\Bigr| {{/formula}} der Flächeninhalt des Dreiecks {{formula}}ABC{{/formula}} berechnet.
125 125  {{/detail}}
126 126  
127 127  === Teilaufgabe e) ===
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1 +XWiki.akukin
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