\documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.50.0.2890} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %TCIDATA{Created=Saturday, April 30, 2011 16:26:47} %TCIDATA{LastRevised=Saturday, April 30, 2011 16:44:31} %TCIDATA{} %TCIDATA{} %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 Flächeninhalt $F$ 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}