83 lines
5.5 KiB
TeX
83 lines
5.5 KiB
TeX
|
||
\documentclass{article}
|
||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||
%TCIDATA{OutputFilter=LATEX.DLL}
|
||
%TCIDATA{Version=5.50.0.2890}
|
||
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
|
||
%TCIDATA{BibliographyScheme=Manual}
|
||
%TCIDATA{Created=Saturday, April 30, 2011 16:26:47}
|
||
%TCIDATA{LastRevised=Saturday, April 30, 2011 16:44:31}
|
||
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
|
||
%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
|
||
%TCIDATA{CSTFile=40 LaTeX article.cst}
|
||
|
||
\newtheorem{theorem}{Theorem}
|
||
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
|
||
\newtheorem{algorithm}[theorem]{Algorithm}
|
||
\newtheorem{axiom}[theorem]{Axiom}
|
||
\newtheorem{case}[theorem]{Case}
|
||
\newtheorem{claim}[theorem]{Claim}
|
||
\newtheorem{conclusion}[theorem]{Conclusion}
|
||
\newtheorem{condition}[theorem]{Condition}
|
||
\newtheorem{conjecture}[theorem]{Conjecture}
|
||
\newtheorem{corollary}[theorem]{Corollary}
|
||
\newtheorem{criterion}[theorem]{Criterion}
|
||
\newtheorem{definition}[theorem]{Definition}
|
||
\newtheorem{example}[theorem]{Example}
|
||
\newtheorem{exercise}[theorem]{Exercise}
|
||
\newtheorem{lemma}[theorem]{Lemma}
|
||
\newtheorem{notation}[theorem]{Notation}
|
||
\newtheorem{problem}[theorem]{Problem}
|
||
\newtheorem{proposition}[theorem]{Proposition}
|
||
\newtheorem{remark}[theorem]{Remark}
|
||
\newtheorem{solution}[theorem]{Solution}
|
||
\newtheorem{summary}[theorem]{Summary}
|
||
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
|
||
\input{tcilatex}
|
||
\begin{document}
|
||
|
||
|
||
\section{Aufgabe 3}
|
||
|
||
Gegeben sind: $\vec{a}=\left(3,-1,2\right)\text{, }\vec{b}=\left(1,2,4\right)%
|
||
\text{, }\vec{c}=\left(1,1,1\right)$.
|
||
|
||
Berechnen Sie:
|
||
|
||
\begin{description}
|
||
\item[a.] ds\"{o}lfkj\"{o}dsak
|
||
|
||
\item[b.] fsdfsd
|
||
|
||
\item[c.]
|
||
\end{description}
|
||
|
||
\subsection{L\"{o}sung}
|
||
|
||
\begin{list}{\ding{42}}
|
||
{\setlength{\topsep}{0.3cm}
|
||
\setlength{\itemsep}{0.3cm}
|
||
\setlength{\leftmargin}{6mm}
|
||
\setlength{\labelwidth}{4mm}
|
||
\setlength{\parsep}{1mm}
|
||
\setlength{\labelsep}{2mm}
|
||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||
\item[a.]$\underline{\underline{\vec{a}+\vec{b}}}=\left(3,-1,2\right)+\left(1,2,4\right)=\left(3+1,-1+2,2+4\right)=\underline{\underline{\left(4,1,6\right)}}$ \\
|
||
$\underline{\underline{\vec{a}-\vec{b}}}=\left(3,-1,2\right)-\left(1,2,4\right)=\left(3-1,-1-2,2-4\right)=\underline{\underline{\left(2,-3,-2\right)}}$ \\
|
||
$\underline{\underline {4 \cdot \vec a}} = 4 \cdot \left( {3. - 1,2} \right) = \left( {4 \cdot 3,4 \cdot \left( { - 1} \right),4 \cdot 2} \right) = \underline{\underline {\left( {12, - 4,8} \right)}} $\\
|
||
$ \underline{\underline { - \frac{1}{4} \cdot \vec b}} = - \frac{1}{4} \cdot \left( {1,2,4} \right) = \left( {\left( { - \frac{1}{4}} \right) \cdot 1,\left( { - \frac{1}{4}} \right) \cdot 2,\left( { - \frac{1}{4}} \right) \cdot 4} \right) = \underline{\underline {\left( { - \frac{1}{4}, - \frac{1}{2}, - 1} \right)}} $ \\
|
||
$ \underline{\underline { - 5 \cdot \vec c}} = - 5 \cdot \left( {1,1,1} \right) = \underline{\underline {\left( { - 5, - 5, - 5} \right)}} $
|
||
\item[b.]$ \underline{\underline {\left| {\vec a} \right|}} = \left| {\left( {3, - 1,4} \right)} \right| = \sqrt {3^2 + \left( { - 1} \right)^2 + 2^2 } = \sqrt {9 + 1 + 4} = \underline{\underline {\sqrt {14} }} \cong 3.741657387 \\
|
||
\underline{\underline {\left| {\vec b} \right|}} = \left| {\left( {1,2,4} \right)} \right| = \sqrt {1^2 + 2^2 + 4^2 } = \sqrt {1 + 4 + 16} = \underline{\underline {\sqrt {21} }} \cong 4.582575695 \\
|
||
\underline{\underline {\left| {\vec c} \right|}} = \left| {\left( {1,1,1} \right)} \right| = \sqrt {1^2 + 1^2 + 1^2 } = \underline{\underline {\sqrt 3 }} \cong 1.732050808 $
|
||
\item[c.] Sei $ \alpha = \alpha \left( {\vec a,\vec b} \right)$, dann gilt $0 \le \alpha \le \pi$ und
|
||
$\cos \left( \alpha \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, - 1,2} \right),\left( {1,2,4} \right)} \right\rangle }}{{\sqrt {14} \cdot \sqrt {21} }} = \frac{{3 \cdot 1 + \left( { - 1} \right) \cdot 2 + 2 \cdot 4}}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} = \frac{{3 - 2 + 8}}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} = \frac{9}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} $\\
|
||
Der Taschenrechner liefert : $\underline{\underline {\alpha \cong 1.018209678}}\entspricht 58.33911721^ \circ $
|
||
\item[d.]$ \underline{\underline{\vec a \times \vec b }}= \left( {3, - 1,2} \right) \times \left( {1,2,4} \right) = \left( {\left( { - 1} \right) \cdot 4 - 2 \cdot 2,2 \cdot 1 - 4 \cdot 3,3 \cdot 2 - 1 \cdot \left( { - 1} \right)} \right) = \left( { - 4 - 4,2 - 12,6 + 1} \right) = \underline{\underline {\left( { - 8, - 10,7} \right)}}$ \\
|
||
Der gesuchte Fl<46>cheninhalt $F$ ist $F = \left| {\vec a \times \vec b} \right| = \sqrt {\left( { - 8} \right)^2 + \left( { - 10} \right)^2 + 7^2 } = \sqrt {64 + 100 + 49} = \underline{\underline {\sqrt {213} }}$ $ \left( { \cong 14.59451952} \right)$
|
||
\item[e.]$ \left[ {\vec a,\vec b,\vec c} \right] = \left\langle {\vec a \times \vec b,\vec c} \right\rangle \mathop = \limits_{vgl.d.} \left\langle {\left( { - 8, - 10,7} \right),\left( {1,1,1} \right)} \right\rangle = - 8 - 10 + 7 = \underline{\underline { - 11}} $ \\
|
||
Das gesuchte Volumen ist $\underline{\underline V} = \left| {\left[ {\vec a,\vec b,\vec c} \right]} \right| = \left| { - 11} \right| = \underline{\underline {11}}$
|
||
\end{list}
|
||
|
||
\end{document}
|