Änderungen von Dokument Lösung Lineare Algebra
Zuletzt geändert von akukin am 2026/01/17 12:06
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... ... @@ -7,7 +7,7 @@ 7 7 8 8 9 9 {{detail summary="Erläuterung der Lösung"}} 10 -Da die {{formula}}x_2{{/formula}}-Komponente des Richtungsvektors von {{formula}}g{{/formula}} 0ist, ist {{formula}}g{{/formula}} parallel zur10 +Da die {{formula}}x_2{{/formula}}-Komponente des Richtungsvektors von {{formula}}g{{/formula}} null ist, ist {{formula}}g{{/formula}} parallel zur {{formula}}x_1x_3{{/formula}}. Daher existieren nur zwei Spurpunkte. 11 11 <p></p> 12 12 Den Schnittpunkt mit der {{formula}}x_2x_3{{/formula}}-Ebene erhalten wir, indem wir die erste Zeile der Geradengleichung gleich 0 setzen ({{formula}}x_1=0{{/formula}}) und nach {{formula}}r{{/formula}} auflösen: 13 13 <br> ... ... @@ -49,7 +49,7 @@ 49 49 Die Gleichung der Geraden {{formula}}h{{/formula}} stellen wir auf, indem wir einen der beiden Punkte, hier {{formula}}A{{/formula}}, als Stützvektor verwenden und als Richtungsvektor den Verbindungsvektor der Punkte {{formula}}A{{/formula}} und {{formula}}B{{/formula}}. 50 50 <br> 51 51 {{formula}} 52 -\over line{AB}=\begin{pmatrix}1\\1\\0\end{pmatrix} - \begin{pmatrix} 5\\-1\\4\end{pmatrix}=\begin{pmatrix}-4\\2\\-4\end{pmatrix}52 +\overrightarrow{AB}=\begin{pmatrix}1\\1\\0\end{pmatrix} - \begin{pmatrix} 5\\-1\\4\end{pmatrix}=\begin{pmatrix}-4\\2\\-4\end{pmatrix} 53 53 {{/formula}} 54 54 <p></p> 55 55 Somit: ... ... @@ -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}}
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