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Loesungen00001_und_00002.tex
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Loesungen00001_und_00002.tex
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\section{Aufgabe 1}
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\begin{enumerate}
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\item Gegeben sind: \hspace{10mm}$\vec a = \left( {3,4} \right)$,\hspace{5mm}$\vec b = \left( {10,5} \right)$,\hspace{5mm}$ \vec c =
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\vec a - \frac{1} {2}\vec b$
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\begin{enumerate}
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\item Bestimmen Sie $\vec c$ durch Zeichnung und Rechnung!\\ \\%
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Zeichnung:\\
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\includegraphics{Loesung0001}\\
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Rechnung:\\ \\%
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$\underline{\underline{\vec c}} = \vec a - \frac{1} {2}\vec b = \left( {3,4} \right) - \frac{1} {2}\left( {10,5}
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\right) = \left( {3,4} \right) - \left( {10 \cdot \frac{1} {2},5 \cdot \frac{1} {2}} \right) = \left( {3,4} \right) - \left(
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{5,\frac{5}
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{2}} \right) =$\\%
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$\left( {3 - 5,4 - \frac{5} {2}} \right) = \underline{\underline {\left( { - 2,\frac{3} {2}} \right)}}$\\%
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\item Bestimmen Sie $\left|\vec a\right|$, $\left|\vec b\right|$, $\left|\vec c\right|$.\\
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\\
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$\underline{\underline {\left| {\vec a} \right|}} = \left| {\left( {3,4} \right)} \right| = \sqrt {3^2 + 4^2 } = \sqrt {9 +
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16} = \sqrt {25} = \underline{\underline 5}$\\ \\%
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$\underline{\underline {\left| {\vec b} \right|}} = \left| {\left( {10,6} \right)} \right| = \sqrt {10^2 + 5^2 } = \sqrt
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{100 + 25} = \sqrt {125} = \sqrt {5 \cdot 25} = 5\sqrt 5 \\ \\%
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\cong \underline{\underline {11,18033988}}$
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\item Bestimmen Sie den Öffnungswinkel $\alpha \left( {\vec a,\vec c} \right)$ und $\alpha \left( {\vec a,\vec b} \right)$.
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\end{enumerate}
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\end{enumerate}
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