\documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Thursday, February 21, 2002 14:53:39} %TCIDATA{LastRevised=Monday, October 03, 2005 10:02:18} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{AllPages= %F=36,\PARA{038
Scientific Notebook: On Line Mathematics\dotfill \thepage} %} \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]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Line\'{a}rne \'{u}tvary v $\mathbf{E}_{3}$} \subsection{D\^{o}kaz vety} H\v{l}adajme spolo\v{c}n\'{e} body priamok $p,q$, ktor\'{y}ch parametrick% \'{e} rovnice s\'{u} \[ p:\left\{ \begin{array}{l} x=a_{1}+tu_{1} \\ y=a_{2}+tu_{2} \\ z=a_{3}+tu_{3}% \end{array}% \right. \hspace{2cm}q:\left\{ \begin{array}{l} x=b_{1}+sv_{1} \\ y=b_{2}+sv_{2} \\ z=b_{3}+sv_{3}% \end{array}% \right. \]% Spolo\v{c}n\'{e} body odpovedaj\'{u} t\'{y}m hodnot\'{a}m parametrov $t,s$, pre ktor\'{e} plat\'{\i} \[ \begin{array}{rl@{\ =\ }l} a_{1}+tu_{1} & = & b_{1}+sv_{1} \\ a_{2}+tu_{2} & = & b_{2}+sv_{2} \\ a_{2}+tu_{2} & = & b_{2}+sv_{2}% \end{array}% \]% Po jednoduchej \'{u}prave dostaneme pre nezn\'{a}me hodnoty $t,s$ s\'{u}% stavu line\'{a}rnych rovn\'{\i}c: \[ \begin{array}{rl@{\ =\ }l} tu_{1}-sv_{1} & = & b_{1}-a_{1} \\ tu_{2}-sv_{2} & = & b_{2}-a_{2} \\ tu_{3}-sv_{3} & = & b_{3}-a_{3}% \end{array}% \]% Ozna\v{c}me $h$ hodnos\v{t} matice tejto s\'{u}stavy a $h^{\prime }$ hodnos% \v{t} roz\v{s}\'{\i}renej matice tejto s\'{u}stavy. Pre tieto hodnosti nastane pr\'{a}ve jedna z mo\v{z}nost\'{\i}: \begin{enumerate} \item[a)] $h=2,h^{\prime }=3$. To nast\'{a}va pr\'{a}ve vtedy, ke\v{d} st% \'{l}pce roz\v{s}\'{\i}renej matice s\'{u}stavy, teda vektory $\overline{u},% \overline{v},B-A$, s\'{u} line\'{a}rne nez\'{a}visl\'{e}, t.j. $[B-A,\,% \overline{u},\,\overline{v}]\neq 0$. V tomto pr\'{\i}pade s\'{u}stava nem% \'{a} rie\v{s}enie, t.j. $p\cap q=\emptyset $. Ke\v{d}\v{z}e vektory $% \overline{u},\overline{v}$ nie s\'{u} koline\'{a}rne, priamky nem\^{o}\v{z}u by\v{t} rovnobe\v{z}n\'{e}, s\'{u} teda mimobe\v{z}n\'{e}. \item[b)] $h=h^{\prime }=2$. To plat\'{\i} pr\'{a}ve vtedy, ke\v{d} vektory $% \overline{u},\overline{v}$ s\'{u} line\'{a}rne nez\'{a}visl\'{e} a vektory $% \overline{u},\overline{v},B-A$ s\'{u} line\'{a}rne z\'{a}visl\'{e}, t.j. $% \overline{u}\times \overline{v}\neq \overline{0},[B-A,\,\overline{u},\,% \overline{v}]=0$. V tomto pr\'{\i}pade m\'{a} s\'{u}stava pr\'{a}ve jedno rie% \v{s}enie. Priamky $p,q$ maj\'{u} spolo\v{c}n\'{y} jedin\'{y} bod, s\'{u} teda r\^{o}znobe\v{z}n\'{e}. \item[c)] $h=1,h^{\prime }=2$. To plat\'{\i} pr\'{a}ve vtedy, ke\v{d} vektory $\overline{u},\overline{v}$ s\'{u} line\'{a}rne z\'{a}visl\'{e} a vektory $\overline{u},B-A$ s\'{u} lin\'{a}rne nez\'{a}visl\'{e}, t.j. $% \overline{u}\times \overline{v}=\overline{0},(B-A)\times \overline{u}\neq \overline{0}$. V tomto pr\'{\i}pade s\'{u}stava nem\'{a} rie\v{s}enie, teda $% p\cap q=\emptyset $, a ke\v{d}\v{z}e smerov\'{e} vektory priamok $p,:q$ s% \'{u} koline\'{a}rne, tak tieto priamky s\'{u} rovnobe\v{z}n\'{e} a r\^{o}% zne. \item[d)] $h=h^{\prime }=1$. To nastane pr\'{a}ve vtedy, ke\v{d} vektory $% \overline{u},\overline{v}$ a tie\v{z} $\overline{u},B-A$ s\'{u} line\'{a}rne z\'{a}visl\'{e}, t.j. $\overline{u}\times \overline{v}=\overline{0},$ $% (B-A)\times \overline{u}=\overline{0}$. V tomto pr\'{\i}pade s\'{u}stava m% \'{a} rie\v{s}enie. Jednu nezn\'{a}mu, bu\v{d} $t$ alebo $s$, m\^{o}\v{z}eme voli\v{t} \v{l}ubovo\v{l}ne, a teda priamky $p,:q$ maj\'{u} v\v{s}etky svoje body spolo\v{c}n\'{e}, \v{c}i\v{z}e $p=q$.$\;\blacksquare $ \end{enumerate} \begin{center} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{Sp\"{a}\v{t}}{}{}{A63.tex#4}}% %BeginExpansion \msihyperref{Sp\"{a}\v{t}}{}{}{A63.tex#4}% %EndExpansion } \\ \hline \end{tabular} \end{center} \end{document}