\section{Aufgabe 1}

Gegeben sind: \hspace{10mm}$\vec a = \left( {3,4} \right)$,\hspace{5mm}$\vec b = \left( {10,5} \right)$,\hspace{5mm}$ \vec c =
\vec a - \frac{1} {2}\vec b$

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\item[a.]Bestimmen Sie $\vec c$ durch Zeichnung und Rechnung!
\item[b.]Bestimmen Sie $\left|\vec a\right|$, $\left|\vec b\right|$, $\left|\vec c\right|$.
\item[c.]Bestimmen Sie den Öffnungswinkel $\alpha \left( {\vec a,\vec c} \right)$ und $\alpha \left( {\vec a,\vec b} \right)$.
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\subsection{Lösung}
\subsubsection{Zeichnung}
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\item[a.]\includegraphics{Loesung001}
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\subsubsection{Rechnung}
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\item[a.]$\underline{\underline{\vec c}} = \vec a - \frac{1} {2}\vec b = \left( {3,4} \right) - \frac{1} {2}\left( {10,5}
\right) = \left( {3,4} \right) - \left( {10 \cdot \frac{1} {2},5 \cdot \frac{1} {2}} \right) = \left( {3,4} \right) - \left({5,\frac{5}{2}} \right) =\left( {3 - 5,4 - \frac{5} {2}} \right) = \underline{\underline {\left( { - 2,\frac{3} {2}} \right)}}$
\item[b.]$\underline{\underline {\left| {\vec a} \right|}}  = \left| {\left( {3,4} \right)} \right| = \sqrt {3^2  + 4^2 }  = \sqrt {9 +
16}  = \sqrt {25}  = \underline{\underline 5}$\\ \\%
$\underline{\underline {\left| {\vec b} \right|}}  = \left| {\left( {10,6} \right)} \right| = \sqrt {10^2  + 5^2 }  = \sqrt
{100 + 25}  = \sqrt {125}  = \sqrt {5 \cdot 25}  = 5\sqrt 5  \cong \underline{\underline {11.18033988}}$\\%
$\underline{\underline {\left| {\vec c} \right|}} = \left| {\left( { - 2,\frac{3}{2}} \right)} \right| = \sqrt {\left( { - 2} \right)^2  + \left( {\frac{3}{2}} \right)^2 }  = \sqrt {4 + \frac{9}{4}}  = \sqrt {\frac{{16 + 9}}{4}}  = \sqrt {\frac{{25}}{4}}  = \underline{\underline {\frac{5}{2}}} $
\item[c.]$\alpha  = \alpha \left( {\vec a,\vec c} \right)\text{, }0 \le \alpha  \le \pi$\\
\vspace{2mm}
\hspace{-1.5mm}$\cos \left( \alpha  \right) = \frac{{\left\langle {\vec a,\vec c} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec c} \right|}}\mathop  = \limits_{vgl.b.} \frac{{\left\langle {\left( {3,4} \right),\left( { - 2,\frac{3}{2}} \right)} \right\rangle }}{{5 \cdot \frac{5}{2}}} = \frac{{3 \cdot \left( { - 2} \right) + 4 \cdot \frac{3}{2}}}{{\frac{{25}}{2}}} = \frac{{ - 6 + 6}}{{\frac{{25}}{2}}} = 0 \Rightarrow \underline{\underline {\alpha  = \frac{\pi }{2}}} \left( { \entspricht 90^ \circ  } \right)$

$\beta  = \alpha \left( {\vec a,\vec b} \right)$, $0 \le \beta  \le \pi$\\
\vspace{2mm}
\hspace{-1.5mm}$\cos \left( \beta  \right) = \frac{{\left\langle {\vec a,\vec b} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec b} \right|}}\mathop  = \limits_{vgl.b.}\frac{{\left\langle {\left( {3,4} \right),\left( {10,5} \right)} \right\rangle }}{{5 \cdot 5 \cdot \sqrt 5 }} = \frac{{3 \cdot 10 + 4 \cdot 5}}{{25 \cdot \sqrt 5 }} = \frac{{50}}{{25 \cdot \sqrt 5 }} = \frac{2}{{\sqrt 5 }} $.\\
Es folgt:\\
$\beta = \cos^{-1} \left(\frac{2}{\sqrt 5}\right)= \arccos\left(\frac{2}{\sqrt 5}\right) \cong  \underline{\underline{0.463647609}} \left( \entspricht 26.565051177078^\circ \right)$

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