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Änderungen von Dokument Lösung Lineare Algebra

Zuletzt geändert von akukin am 2026/01/17 12:06
Von Version 6.1
bearbeitet von akukin
am 2026/01/15 19:24
Änderungskommentar: Neues Bild SkizzeKreis.svg hochladen
Auf Version 4.1
bearbeitet von akukin
am 2026/01/14 16:43
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... ... @@ -95,13 +95,12 @@
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 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 Alternativ 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 +
101 101  <p></p>
102 -Somit ist eine mögliche Lösung
103 -<br>
104 -{{formula}}E: \ \vec x =
103 +Somit ist eine mögliche Lösung {{formula}}E: \ \vec x =
105 105  \begin{pmatrix}0\\1\\2\end{pmatrix}+ r \cdot
106 106  \begin{pmatrix}1\\0\\0\end{pmatrix} + s \cdot
107 107  \begin{pmatrix}0\\0\\1\end{pmatrix}; \ r, s \in \mathbb{R}
... ... @@ -108,7 +108,6 @@
108 108  {{/formula}}
109 109  <br>
110 110  Eine weitere Lösung wäre
111 -<br>
112 112  {{formula}}E: \ \vec x =
113 113  \begin{pmatrix}0\\5\\2\end{pmatrix}+ r \cdot
114 114  \begin{pmatrix}1\\0\\0\end{pmatrix} + s \cdot
... ... @@ -123,9 +123,7 @@
123 123  
124 124  
125 125  {{detail summary="Erläuterung der Lösung"}}
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.
124 +
129 129  {{/detail}}
130 130  
131 131  === Teilaufgabe e) ===
SkizzeKreis.svg
Author
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1 -XWiki.akukin
Größe
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1 -7.8 KB
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