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Änderungen von Dokument Lösung Lineare Algebra

Zuletzt geändert von Anna Kukin am 2026/02/17 14:32
Von Version 11.1
bearbeitet von Anna Kukin
am 2026/02/17 14:32
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Auf Version 3.1
bearbeitet von Anna Kukin
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... ... @@ -4,7 +4,7 @@
4 4  \begin{align*}
5 5  &A_1(3|3|0), \ A_2(-3|3|0), \\
6 6  &A_3(-3|-3|0), \ A_4(3|-3|0) \\
7 -& C'_1(3|1|0), \ C'_2(1|3|0)
7 +& C'_1(3|1|0), C'_2(1|3|0)
8 8  \end{align*}
9 9  {{/formula}}
10 10  [[image:Lösunga).png||width="180"]]
... ... @@ -13,24 +13,12 @@
13 13  
14 14  {{detail summary="Erläuterung der Lösung"}}
15 15  //Aufgabenstellung//
16 -<br>
17 -Zeichne das Quadrat {{formula}} A_{1}A_{2}A_{3}A_{4} {{/formula}} in ein zweidimensionales {{formula}} x_{1}x_{2} {{/formula}}-Koordinatensystem ein.
18 -<br>
19 -Zeichne in dasselbe Koordinatensystem die orthogonalen Projektionen der Punkte {{formula}} C_{1} {{/formula}} und {{formula}} C_{2} {{/formula}} ein.
20 -<p></p>
16 +<br><p>
17 +
18 +</p>
21 21  //Lösung//
22 22  <br>
23 -Gegeben sind die Punkte {{formula}}
24 -A_1(3|3|0), \ A_2(-3|3|0)
25 -{{/formula}}.
26 -<br>
27 -Da der Grundriss nach Aufgabenstellung ein Quadrat mit einer Seitenlänge von 6 Metern ist, erhalten wir die fehlenden Punkte durch Verschiebung von {{formula}}A_1{{/formula}} und {{formula}}A_2{{/formula}} um 6 in negative x-Richtung: {{formula}}A_3(-3|-3|0), \ A_4(3|-3|0){{/formula}}.
28 -<p></p>
29 -Die orthogonalen Projektionen der Punkte {{formula}}C_1{{/formula}} und {{formula}} C_{2} {{/formula}} erhalten wir, indem wir die {{formula}}x_3{{/formula}} Koordinate gleich null setzen: {{formula}}
30 -C'_1(3|1|0), \ C'_2(1|3|0)
31 -{{/formula}}
32 -<br>
33 -[[image:Lösunga).png||width="180"]]
21 +
34 34  {{/detail}}
35 35  
36 36  === Teilaufgabe b) ===
... ... @@ -39,7 +39,7 @@
39 39  \Bigl|\overrightarrow{C_1C_2}\Bigl|=\Bigl|\overrightarrow{B_1C_1}\Bigl|=\Bigl|\overrightarrow{B_1C_2}\Bigl|=2\sqrt2
40 40  {{/formula}}
41 41  <br>
42 -Damit ist das Dreieck {{formula}}C_1B_1C_2{{/formula}} gleichseitig.
30 +Damit ist das Dreieck {{formula}}B_2C_1C_2{{/formula}} gleichseitig.
43 43  <p></p>
44 44  Mittelpunkt der Strecke {{formula}}C_1C_2{{/formula}}:
45 45  {{formula}}
... ... @@ -58,43 +58,10 @@
58 58  {{detail summary="Erläuterung der Lösung"}}
59 59  //Aufgabenstellung//
60 60  <br><p>
61 -Zeige, dass das Dreieck {{formula}} C_{1}B_{1}C_{2} {{/formula}} gleichseitig ist. Berechne den Flächeninhalt des Dreiecks {{formula}} C_{1}B_{1}C_{2} {{/formula}}.
49 +
62 62  </p>
63 63  //Lösung//
64 64  <br>
65 -Wir berechnen die Seitenlängen des Dreiecks {{formula}} C_{1}B_{1}C_{2} {{/formula}}:
66 -{{formula}}
67 -\begin{align*}
68 -\Bigl|\overrightarrow{C_1C_2}\Bigl|&=\left|\begin{pmatrix}1-3\\3-1\\4-4\end{pmatrix}\right|=\left|\begin{pmatrix}-2\\2\\0\end{pmatrix}\right|=\sqrt{(-2)^2+2^2+0^2}=2\sqrt{2} \\
69 -\Bigl|\overrightarrow{B_1C_1}\Bigl|&=\left|\begin{pmatrix}3-3\\1-3\\4-2\end{pmatrix}\right|=\left|\begin{pmatrix}0\\-2\\2\end{pmatrix}\right|=\sqrt{0^2+(-2)^2+2^2}=2\sqrt{2} \\
70 -\Bigl|\overrightarrow{B_1C_2}\Bigl|&=\left|\begin{pmatrix}1-3\\3-3\\4-2\end{pmatrix}\right|=\left|\begin{pmatrix}-2\\0\\2\end{pmatrix}\right|=\sqrt{(-2)^2+0^2+2^2}=2\sqrt{2}
71 -\end{align*}
72 -{{/formula}}
73 -<p></p>
74 -Somit gilt
75 -{{formula}}
76 -\Bigl|\overrightarrow{C_1C_2}\Bigl|=\Bigl|\overrightarrow{B_1C_1}\Bigl|=\Bigl|\overrightarrow{B_1C_2}\Bigl|=2\sqrt2
77 -{{/formula}}.
78 -<br>
79 -Damit ist das Dreieck {{formula}}C_1B_1C_2{{/formula}} gleichseitig.
80 -<p></p>
81 -[[image:DreieckSkizze.svg||width="180" style="float: right"]]
82 -Der Flächeninhalt des Dreiecks berechnet sich durch {{formula}}A=\frac{1}{2}\cdot g\cdot h{{/formula}}. Um die Höhe {{formula}}h{{/formula}} des Dreieckes zu bestimmen, benötigen wir zunächst den Mittelpunkt der Strecke {{formula}}C_1C_2{{/formula}} (siehe Skizze).
83 -<br>
84 -Diesen berechnen wir durch
85 -{{formula}}
86 -M\left(\frac{3+1}{2}\Bigl| \frac{1+3}{2} \Bigl|\frac{4+4}{2}\right)=M(2|2|4)
87 -{{/formula}}.
88 -<br>
89 -Die Höhe {{formula}}h{{/formula}} ergibt sich durch
90 -{{formula}}
91 -\Bigl|\overrightarrow{MB_1}\Bigl|=\left|\begin{pmatrix}3-2\\3-2\\2-4\end{pmatrix}\right|=\left|\begin{pmatrix}1\\1\\-2\end{pmatrix}\right|=\sqrt{1^2+1^2+(-2)^2}=\sqrt{6}
92 -{{/formula}}.
93 -<br>
94 -Damit:
95 -{{formula}}
96 -A=\frac12\cdot \Bigl|\overrightarrow{C_1C_2}\Bigl| \cdot \Bigl|\overrightarrow{MB_1}\Bigl|=2\sqrt{2}\cdot \sqrt{6}=2\sqrt3
97 -{{/formula}}
98 98  {{/detail}}
99 99  
100 100  === Teilaufgabe c) ===
... ... @@ -113,36 +113,11 @@
113 113  
114 114  {{detail summary="Erläuterung der Lösung"}}
115 115  //Aufgabenstellung//
116 -<br>
117 -Vier der acht Dreiecksflächen des Daches sind parallel zu den jeweils unterhalb liegenden Dreiecksflächen.
118 -<br>
119 -Ermittle die Koordinaten der Spitze {{formula}} S {{/formula}}.
120 -<p></p>
71 +<br><p>
72 +
73 +</p>
121 121  //Lösung//
122 122  <br>
123 -Die Punkte {{formula}}B_1{{/formula}}, {{formula}}C_1{{/formula}} und {{formula}}C_2{{/formula}} liegen in einer Ebene {{formula}}E{{/formula}}. Die Punkte {{formula}}D_1{{/formula}}, {{formula}}D_2{{/formula}} und {{formula}}S{{/formula}}
124 -liegen in einer Ebene {{formula}}F{{/formula}}. Aus der Skizze des Kirchturmes wird ersichtlich, dass die beiden Ebenen {{formula}}E{{/formula}} und {{formula}}F{{/formula}} parallel sein müssen. Das heißt, die beiden Ebenen besitzen den selben Normalenvektor.
125 -<br>
126 -Um also eine Ebenengleichung der Ebene {{formula}}F{{/formula}} zu bestimmen und so anschließend den Punkt {{formula}}S{{/formula}} zu bestimmen, bestimmen wir zunächst einen Normalenvektor der Ebene {{formula}}E{{/formula}}:
127 -<br>
128 -{{formula}}
129 -\begin{align*}
130 - \vec{n}_E=\Bigl|\overrightarrow{B_1C_1}\Bigl|\times \Bigl|\overrightarrow{B_1C_2}\Bigl|&=\begin{pmatrix}0\\-2\\2\end{pmatrix}\times \begin{pmatrix}-2\\0\\2\end{pmatrix} \\
131 -&=\begin{pmatrix}(-2)\cdot 2-2\cdot 0\\2\cdot (-2)-0\cdot 2\\0\cdot 0-(-2)\cdot (-2)\end{pmatrix} \\
132 -&=\begin{pmatrix}-4\\-4\\-4\end{pmatrix}=(-4)\cdot \begin{pmatrix}1\\1\\1\end{pmatrix}
133 -\end{align*}
134 -{{/formula}}
135 -<p></p>
136 -Somit ist {{formula}}
137 -\vec{n}_E=\vec{n}_F=\begin{pmatrix}1\\1\\1\end{pmatrix}
138 -{{/formula}}.
139 -<br>
140 -{{formula}}F{{/formula}} hat damit die Form {{formula}}F: \ x_1+x_2+x_3=b{{/formula}}.
141 -<p></p>
142 -Punktprobe mit einem der Punkte der Ebene {{formula}}F{{/formula}}, z.B. {{formula}}D_1{{/formula}}, liefert:
143 -{{formula}}3+1+8= 12= b{{/formula}}
144 -<br>
145 -Da der Punkt {{formula}}S{{/formula}} auf der {{formula}}x_3{{/formula}}-Achse liegt, sind die {{formula}}x_1{{/formula}}- und {{formula}}x_2{{/formula}}-Koordinate von {{formula}}S{{/formula}} null. Somit ergibt sich für die {{formula}}x_3{{/formula}}-Koordinate {{formula}}x_3=12{{/formula}} und der Punkt lautet {{formula}}S(0|0|12){{/formula}}.
146 146  {{/detail}}
147 147  
148 148  === Teilaufgabe d) ===
... ... @@ -154,20 +154,10 @@
154 154  {{detail summary="Erläuterung der Lösung"}}
155 155  //Aufgabenstellung//
156 156  <br><p>
157 -Der Mittelpunkt der Strecke {{formula}} D_{1}D_{2} {{/formula}} ist {{formula}} M {{/formula}}. Der Mittelpunkt der Strecke {{formula}} D_{2}D_{3} {{/formula}} ist {{formula}} N {{/formula}}.
158 -<br>
159 -Begründe, dass die Strecken {{formula}} MS {{/formula}} und {{formula}} NS {{/formula}} unterschiedliche Neigungswinkel haben.
87 +
160 160  </p>
161 161  //Lösung//
162 162  <br>
163 -Der Abstand der Kante {{formula}}D_2D_3{{/formula}} von der {{formula}}x_3{{/formula}}-Achse beträgt 3m. Der Abstand der Kante {{formula}}D_1D_2{{/formula}} von der {{formula}}x_3{{/formula}}-Achse ist kleiner, da {{formula}}D_1D_2{{/formula}} die Ecke des Quadrats abschneidet. Deshalb sind die Neigungen der Strecken {{formula}}MS{{/formula}} und {{formula}}NS{{/formula}} bei gleicher Spitze {{formula}}S{{/formula}} unterschiedlich.
164 -<p></p>
165 -//Anmerkung: Falls nicht ersichtlich ist, wieso der Abstand der Kante {{formula}}D_1D_2{{/formula}} von der {{formula}}x_3{{/formula}}-Achse kleiner ist als 3, kann man den Abstand auch berechnen:
166 -<br>
167 -{{formula}}
168 -\begin{align*}
169 -M\left(\frac{3+1}{2}\biggl|\frac{1+3}{2}\biggl|\frac{8+8}{2}\right)=M(2|2|8) \\ \rightarrow \ \text{Abstand zur} \ x_3\text{-Achse:} \sqrt{2^2+2^2}=\sqrt{8}<3
170 -\end{align*}{{/formula}}//
171 171  {{/detail}}
172 172  
173 173  === Teilaufgabe e) ===
... ... @@ -177,12 +177,12 @@
177 177  {{formula}}
178 178  \cos(\varphi)=\frac{\begin{pmatrix}-3\\-2\\4\end{pmatrix}\cdot
179 179  \begin{pmatrix}0\\0\\1\end{pmatrix}}{\sqrt{29}}\approx0{,}743; \
180 -\varphi\approx 42{,}0^\circ
100 +\varphi\approx42{,}0^\circ
181 181  {{/formula}}
182 -<br>
102 +
183 183  {{formula}}
184 184  \cos(\varphi)=\frac{\text{Kirchturmhöhe} \ h}{\text{gesuchter Abstand}\ a} \ \Leftrightarrow \
185 -a=\frac{h}{\cos(\varphi)}\approx\frac{30}{\cos(42{,}0^\circ)}\approx40{,}4
105 +a=\frac{h}{\cos(\varphi)}\approx\frac{30}{\cos(42^\circ)}\approx40{,}4
186 186  {{/formula}}
187 187  <p></p>
188 188  Der Abstand der Turmspitze und ihrem Schattenpunkt beträgt ca. 40,4 Meter.
... ... @@ -192,42 +192,10 @@
192 192  {{detail summary="Erläuterung der Lösung"}}
193 193  //Aufgabenstellung//
194 194  <br><p>
195 -Der Kirchplatz liegt in einer zur {{formula}} x_{1}x_{2} {{/formula}}-Ebene parallelen Ebene. Die Spitze {{formula}} S {{/formula}} befindet sich 30 m über dem Kirchplatz.
196 -<br>
197 -An einem Sommertag scheint die Sonne in der Richtung {{formula}} \vec{v}=\begin{pmatrix}3\\ 2\\ -4\end{pmatrix} {{/formula}}. Dadurch wirft sie einen Schatten von {{formula}} S {{/formula}} auf den Kirchplatz.
198 -<br>
199 -Berechne, wie groß der Abstand der Spitze {{formula}} S {{/formula}} von deren Schattenpunkt ist.
115 +
200 200  </p>
201 201  //Lösung//
202 202  <br>
203 -Skizze:
204 -<br>
205 -[[image:Skizzee).svg||width="350"]]
206 -<br>
207 -Gesucht: Abstand {{formula}}a{{/formula}}
208 -<p></p>
209 -Mit Hilfe der Formel in der Merkhilfe berechnen den Winkel des Sonnenlichts zur Vertikalen durch
210 -<br>
211 -{{formula}}
212 -\begin{align*}
213 -\cos(\varphi)&=\frac{\begin{pmatrix}-3\\-2\\4\end{pmatrix}\cdot
214 -\begin{pmatrix}0\\0\\1\end{pmatrix}}{\sqrt{(-3)^2+(-2)^2+4^2}\cdot \sqrt{0^2+0^2+1^2}}\\
215 - &=\frac{(-3)\cdot 0+(-2)\cdot 0+4\cdot 1}{\sqrt{29}\cdot \sqrt{1}}=\frac{4}{\sqrt{29}} \\
216 -\Leftrightarrow \varphi&=\cos^{-1}\left(\frac{4}{\sqrt{29}}\right)\approx42{,}0^\circ
217 -\end{align*}
218 -{{/formula}}
219 -<p></p>
220 -Über trignometrische Beziehungen erhalten wir:
221 -<br>
222 -{{formula}}
223 -\begin{align*}
224 -\cos(\varphi)&=\frac{\text{Ankathete}}{\text{Hypotenuse}}=\frac{\text{Kirchturmhöhe} \ h}{\text{gesuchter Abstand}\ a} \\
225 -\Leftrightarrow \
226 -a &=\frac{h}{\cos(\varphi)}\approx\frac{30}{\cos(42{,}0^\circ)}\approx40{,}4
227 -\end{align*}
228 -{{/formula}}
229 -<p></p>
230 -Der Abstand der Turmspitze und ihrem Schattenpunkt beträgt ca. 40,4 Meter.
231 231  {{/detail}}
232 232  
233 233  
DreieckSkizze.svg
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Skizzee).svg
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1 -XWiki.akukin
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