63 lines
3.7 KiB
TeX
63 lines
3.7 KiB
TeX
\section{Aufgabe 1}\label{A001}
|
|
|
|
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$
|
|
|
|
\begin{list}{\ding{42}}
|
|
{\setlength{\topsep}{0.5cm}
|
|
\setlength{\itemsep}{0.5cm}
|
|
\setlength{\leftmargin}{6mm}
|
|
\setlength{\labelwidth}{4mm}
|
|
\setlength{\parsep}{2mm}
|
|
\setlength{\labelsep}{2mm}
|
|
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
|
\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 $\alpha \left( {\vec a,\vec b} \right)$.
|
|
\end{list}
|
|
|
|
\subsection{Lösung}
|
|
\subsubsection{Zeichnung}
|
|
\begin{list}{\ding{42}}
|
|
{\setlength{\topsep}{0.5cm}
|
|
\setlength{\itemsep}{0.5cm}
|
|
\setlength{\leftmargin}{6mm}
|
|
\setlength{\labelwidth}{4mm}
|
|
\setlength{\parsep}{2mm}
|
|
\setlength{\labelsep}{2mm}
|
|
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
|
\item[a.]\includegraphics{Loesung0001}
|
|
\end{list}
|
|
|
|
\subsubsection{Rechnung}
|
|
\begin{list}{\ding{42}}
|
|
{\setlength{\topsep}{0.5cm}
|
|
\setlength{\itemsep}{0.5cm}
|
|
\setlength{\leftmargin}{6mm}
|
|
\setlength{\labelwidth}{4mm}
|
|
\setlength{\parsep}{2mm}
|
|
\setlength{\labelsep}{2mm}
|
|
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
|
\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}$\cos \left( \alpha \right) = \frac{{\left\langle {\vec a,\vec c} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec c} \right|}}\mathop = \limits_{vgl.b.} \frac{{\left\langle {\left( {3,4} \right),\left( { - 2,\frac{3}{2}} \right)} \right\rangle }}{{5 \cdot \frac{5}{2}}} = \frac{{3 \cdot \left( { - 2} \right) + 4 \cdot \frac{3}{2}}}{{\frac{{25}}{2}}} = \frac{{ - 6 + 6}}{{\frac{{25}}{2}}} = 0 \Rightarrow \underline{\underline {\alpha = \frac{\pi }{2}}} \left( { \entspricht 90^ \circ } \right)$
|
|
\\
|
|
\vspace{5mm}
|
|
|
|
$\beta = \alpha \left( {\vec a,\vec b} \right)$, $0 \le \beta \le \pi$\\
|
|
\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:\\
|
|
$\beta = \cos^{-1} \left(\frac{2}{\sqrt 5}\right)= \arccos\left(\frac{2}{\sqrt 5}\right) \cong \underline{\underline{0.463647609}} \left( \entspricht 26.565051177078^\circ \right)$
|
|
|
|
|
|
|
|
\end{list}
|