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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
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Auf Version 9.1
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... ... @@ -135,7 +135,7 @@
135 135  {{formula}}
136 136  \overrightarrow{OM}= \overrightarrow{OA}+\frac{1}{2} \cdot \overrightarrow{AC}=
137 137  \begin{pmatrix}2{,}5\\1\\3\end{pmatrix}, \
138 -\Bigl| \overrightarrow{AM} \Bigr|+ \Bigl| \overrightarrow{MB} \Bigr|+\Bigl| \overrightarrow{CM} \Bigr|=\sqrt{11{,}25}
138 +\Bigl| \overrightarrow{AM} \Bigr|= \Bigl| \overrightarrow{MB} \Bigr|=\Bigl| \overrightarrow{CM} \Bigr|=\sqrt{11{,}25}
139 139  {{/formula}}
140 140  <p>
141 141  Somit haben alle drei Punkte den gleichen Abstand vom Mittelpunkt der Hypotenuse {{formula}}AC{{/formula}}.
... ... @@ -147,5 +147,32 @@
147 147  
148 148  
149 149  {{detail summary="Erläuterung der Lösung"}}
150 -
150 +Skizze:
151 +<br>
152 +[[image:SkizzeKreis (1).svg||width="250"]]
153 +<br>
154 +Aus der vorherigen Teilaufgabe wissen wir, dass {{formula}}AC{{/formula}} die Hypotenuse des Dreieckes ist. Auf der Hypotenuse {{formula}}AC{{/formula}} hat nur ihr Mittelpunkt {{formula}}M{{/formula}} denselben Abstand von {{formula}}A{{/formula}} und {{formula}}C{{/formula}}.
155 +<br>
156 +Den Mittelpunkt {{formula}}M{{/formula}} der Strecke {{formula}}AC{{/formula}} erhalten wir durch
157 +<br>
158 +{{formula}}
159 +\overrightarrow{OM}= \overrightarrow{OA}+\frac{1}{2} \cdot \overrightarrow{AC}=\begin{pmatrix}5\\-1\\4\end{pmatrix}+\frac{1}{2}\cdot \begin{pmatrix}-5\\4\\-2\end{pmatrix} =
160 +\begin{pmatrix}2{,}5\\1\\3\end{pmatrix}
161 +{{/formula}}
162 +<p></p>
163 +Nun müssen wir prüfen, dass der berechnete Mittelpunkt von den Punkten {{formula}}A{{/formula}}, {{formula}}B{{/formula}} und {{formula}}C{{/formula}} jeweils den selben Abstand besitzt:
164 +<br>
165 +{{formula}}
166 +\begin{align*}
167 +&\Bigl| \overrightarrow{AM} \Bigr| &=\left| \begin{pmatrix}-2,5\\2\\-1\end{pmatrix}\right|=\sqrt{(-2,5)^2+2^2+(-1)^2} =\sqrt{11{,}25} \\
168 +&\Bigl| \overrightarrow{MB} \Bigr| &=\left| \begin{pmatrix}-1,5\\0\\3\end{pmatrix}\right|=\sqrt{(-1,5)^2+0^2+3^2} =\sqrt{11{,}25}\\
169 +&\Bigl| \overrightarrow{CM} \Bigr|&=\left| \begin{pmatrix}2,5\\-2\\1\end{pmatrix}\right|=\sqrt{2,5^2+(-2)^2+1^2} =\sqrt{11{,}25} \\
170 +\end{align*}
171 +{{/formula}}
172 +<p>
173 +Somit haben alle drei Punkte den gleichen Abstand vom Mittelpunkt der Hypotenuse {{formula}}AC{{/formula}}.
174 +Sie liegen deshalb auf einem Kreis mit diesem Punkt als Mittelpunkt.
175 +</p>
176 +Hinweis:
177 +Eine Argumentation mit dem Thaleskreis ist ebenso zulässig.
151 151  {{/detail}}
SkizzeKreis.svg
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