Änderungen von Dokument Lösung Lineare Algebra
Zuletzt geändert von akukin am 2026/01/17 12:06
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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}}. ... ... @@ -147,32 +147,5 @@ 147 147 148 148 149 149 {{detail summary="Erläuterung der Lösung"}} 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. 150 + 178 178 {{/detail}}
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... ... @@ -1,1 +1,0 @@ 1 -XWiki.akukin - Größe
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