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% (find-angg "LATEX/2016-2-GA-P2.tex") % (defun c () (interactive) (find-LATEXsh "lualatex -record 2016-2-GA-P2.tex")) % (defun d () (interactive) (find-xpdfpage "~/LATEX/2016-2-GA-P2.pdf")) % (defun e () (interactive) (find-LATEX "2016-2-GA-P2.tex")) % (defun u () (interactive) (find-latex-upload-links "2016-2-GA-P2")) % (find-xpdfpage "~/LATEX/2016-2-GA-P2.pdf") % (find-sh0 "cp -v ~/LATEX/2016-2-GA-P2.pdf /tmp/") % (find-sh0 "cp -v ~/LATEX/2016-2-GA-P2.pdf /tmp/pen/") % file:///home/edrx/LATEX/2016-2-GA-P2.pdf % file:///tmp/2016-2-GA-P2.pdf % file:///tmp/pen/2016-2-GA-P2.pdf % http://angg.twu.net/LATEX/2016-2-GA-P2.pdf \documentclass[oneside]{book} \usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref") %\usepackage[latin1]{inputenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{pict2e} \usepackage{color} % (find-LATEX "edrx15.sty" "colors") \usepackage{colorweb} % (find-es "tex" "colorweb") %\usepackage{tikz} % \usepackage{edrx15} % (find-angg "LATEX/edrx15.sty") \input edrxaccents.tex % (find-angg "LATEX/edrxaccents.tex") \input edrxchars.tex % (find-LATEX "edrxchars.tex") \input edrxheadfoot.tex % (find-dn4ex "edrxheadfoot.tex") \input edrxgac2.tex % (find-LATEX "edrxgac2.tex") % \begin{document} \catcode`\^^J=10 \directlua{dednat6dir = "dednat6/"} \directlua{dofile(dednat6dir.."dednat6.lua")} \directlua{texfile(tex.jobname)} \directlua{verbose()} %\directlua{output(preamble1)} \def\expr#1{\directlua{output(tostring(#1))}} \def\eval#1{\directlua{#1}} \def\pu{\directlua{pu()}} \directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua") \directlua{dofile "edrxpict.lua"} % (find-LATEX "edrxpict.lua") %L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end \pu \def\V(#1){\VEC{#1}} % ____ _ _ _ % / ___|__ _| |__ ___ ___ __ _| | |__ ___ % | | / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \ % | |__| (_| | |_) | __/ (_| (_| | | | | | (_) | % \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/ % {\setlength{\parindent}{0em} \footnotesize \par Geometria Analítica \par PURO-UFF - 2016.2 \par P2 - 18/jan/2017 - Eduardo Ochs \par Respostas sem justificativas não serão aceitas. \par Proibido usar quaisquer aparelhos eletrônicos. % \par Versão: 14/mar/2016 % \par Links importantes: % \par \url{http://angg.twu.net/2015.2-C2.html} (página do curso) % \par \url{http://angg.twu.net/2015.2-C2/2015.2-C2.pdf} (quadros) % \par \url{http://angg.twu.net/LATEX/2015-2-C2-material.pdf} % \par {\tt eduardoochs@gmail.com} (meu e-mail) } \bsk \bsk \setlength{\parindent}{0em} \def\T(Total: #1 pts){{\bf(Total: #1 pts)}} \def\T(Total: #1 pts){{\bf(Total: #1)}} \def\B (#1 pts){{\bf(#1 pts)}} % Usage: % 1) \T(Total: 2.34 pts) Foo % a) \B(0.45 pts) Bar Lembre que uma equação de cônica é uma equação da forma $ax^2 + bxy + cy^2 + dy + ey + f = 0$; $4+(x+y)(x-y)=5y$ não é uma equação de cônica mas é equivalente a uma: $x^2-y^2-5y+4=0$. Nas questões 3 e 4 vamos usar a abreviação $[\text{equação}] = \setofxyzst{\text{equação}}$. \bsk 1) \T(Total: 4.0 pts) Faça esboços das cônicas com as equações abaixo. Algumas delas são degeneradas. % $$ \begin{tabular}[t]{rl} a) (0.1 pts) & $xy=0$ \\ b) (0.1 pts) & $(x+y)(x-y)=0$ \\ c) (0.1 pts) & $(y-1)(y-2)=0$ \\ d) (0.1 pts) & $x+y=0$ \\ e) (0.1 pts) & $x+y=1$ \\ f) (0.1 pts) & $(x+y)^2=1$ \\ g) (0.1 pts) & $y+2=0$ \\ h) (0.1 pts) & $y+2=1$ \\ \end{tabular} \quad \begin{tabular}[t]{rl} i) (0.4 pts) & $(x+y)^2+(y+2)^2=0$ \\ j) (1.0 pts) & $(x+y)^2+(y+2)^2=1$ \\ k) (0.4 pts) & $(x+y)(y+2)=0$ \\ l) (1.0 pts) & $(x+y)(y+2)=1$ \\ m) (0.4 pts) & $(\frac{x-4}{2})^2 + (\frac{y-6}{3})^2=1$ \\ \end{tabular} $$ \bsk 2) \T(Total: 1.0 pts) Encontre os focos das elipses cujos pontos óbvios são: a) \B(0.5 pts) $(\pm 3, 0)$, $(0, \pm 5)$ b) \B(0.5 pts) $(4 \pm 3, 2)$, $(4, 2 \pm 5)$ \bsk \bsk 3) \T(Total: 1.5 pts) Sejam $π_1=[y=3-2x]$, $π_2=[x+y+z=4]$, $r=π_1∩π_2$. a) \B(0.5 pts) Encontre o ponto de $r$ que tem $x=0$. b) \B(0.5 pts) Encontre o ponto de $r$ que tem $x=1$. c) \B(0.5 pts) Se $r'=\setofst{(x,ax+b,cx+d)}{x∈\R}$ e $r=r'$, quem são $a$, $b$, $c$ e $d$? \bsk \bsk 4) \T(Total: 3.5 pts) Sejam $A=(1,0,0)$, $B=(0,1,0)$, $\uu=\vec{AB}$, $C=(1,1,0)$, $D=(1,0,1)$, $\vv=\vec{CD}$, $r=\setofst{A+t\uu}{t∈\R}$, $r'=\setofst{C+t'\vv}{t'∈\R}$. a) \B(1.5 pts) Demonstre (algebricamente!) que $r$ e $r'$ são reversas, ou, equivalentemente, que $A$, $B$, $C$ e $D$ não são coplanares. b) \B(2.0 pts) Calcule a distância entre $r$ e $r'$. \newpage % ____ _ _ _ % / ___| __ _| |__ __ _ _ __(_) |_ ___ % | | _ / _` | '_ \ / _` | '__| | __/ _ \ % | |_| | (_| | |_) | (_| | | | | || (_) | % \____|\__,_|_.__/ \__,_|_| |_|\__\___/ % Gabarito: \msk % (find-LATEX "2016-2-GA-algebra.tex" "elipses") \unitlength=5pt \def\closeddot{\circle*{0.6}} \def\pictpoint#1{\put(#1){\closeddot}} \def\pictline#1{{\linethickness{1.0pt}\expr{Line.new(#1):pict()}}} \def\pictlinethin#1{{\linethickness{0.2pt}\expr{Line.new(#1):pict()}}} \def\pictLine(#1)(#2)#3{% \vcenter{\hbox{% \beginpicture(#1)(#2)% \pictaxes% \pictline{#3} \end{picture}% }}% } \def\pictellipse#1{{\linethickness{1.0pt}\expr{Ellipse.new(#1):pict()}}} \def\pictEllipse(#1)(#2)#3{% \vcenter{\hbox{% \beginpicture(#1)(#2)% \pictaxes% \pictellipse{#3} \end{picture}% }}% } \def\pictEllipseF(#1)(#2)#3(#4)(#5){% \vcenter{\hbox{% \beginpicture(#1)(#2)% \pictaxes% \pictellipse{#3} \put(#4){\closeddot} \put(#5){\closeddot} \end{picture}% }}% } \def\picthyperbole#1#2{{\linethickness{1.0pt}\expr{Hyperbole.new(#1):pict(#2)}}} 1a) $\vcenter{\hbox{% \beginpicture(-2,-2)(2,2)% \pictaxes% \pictline{v(0,0), v(1,0), -2, 2} \pictline{v(0,0), v(0,1), -2, 2} \end{picture}% }}$ \quad 1b) $\vcenter{\hbox{% \beginpicture(-2,-2)(2,2)% \pictaxes% \pictline{v(0,0), v(1,1), -2, 2} \pictline{v(0,0), v(1,-1), -2, 2} \end{picture}% }}$ \quad 1c) $\vcenter{\hbox{% \beginpicture(-2,-2)(2,2)% \pictaxes% \pictline{v(0,1), v(1,0), -2, 2} \pictline{v(0,2), v(1,0), -2, 2} \end{picture}% }}$ \quad 1d) $\vcenter{\hbox{% \beginpicture(-2,-2)(2,2)% \pictaxes% \pictline{v(0,0), v(-1,1), -2, 2} \end{picture}% }}$ \quad 1e) $\vcenter{\hbox{% \beginpicture(-2,-2)(2,2)% \pictaxes% \pictline{v(1,0), v(-1,1), -1, 2} \end{picture}% }}$ \quad 1f) $\vcenter{\hbox{% \beginpicture(-2,-2)(2,2)% \pictaxes% \pictline{v(1,0), v(-1,1), -1, 2} \pictline{v(-1,0), v(-1,1), -2, 1} \end{picture}% }}$ \msk 1g) $\vcenter{\hbox{% \beginpicture(-2,-2)(2,2)% \pictaxes% \pictline{v(0,-2), v(1,0), -2, 2} \end{picture}% }}$ \quad 1h) $\vcenter{\hbox{% \beginpicture(-2,-2)(2,2)% \pictaxes% \pictline{v(0,-1), v(1,0), -2, 2} \end{picture}% }}$ % \msk \def\mygrid{ \pictlinethin{v(0,-1), v(1,0), -1, 5} % y-2 = 1 \pictlinethin{v(0,-2), v(1,0), -1, 5} % y-2 = 0 \pictlinethin{v(0,-3), v(1,0), -1, 5} % y-2 = -1 \pictlinethin{v(0,-1), v(1,-1), -2, 3} % x+y = -1 \pictlinethin{v(0, 0), v(1,-1), -2, 4} % x+y = 0 \pictlinethin{v(1, 0), v(1,-1), -2, 4} % x+y = 1 } 1i) $\vcenter{\hbox{% \beginpicture(-2,-4)(5,2)% \pictaxes% \mygrid \put(2,-2){\closeddot} \end{picture}% }}$ \quad 1j) $\vcenter{\hbox{% \beginpicture(-2,-4)(5,2)% \pictaxes% \mygrid \pictellipse{v(2,-2), v(1,0), v(-1,1)} \end{picture}% }}$ \quad 1k) $\vcenter{\hbox{% \beginpicture(-2,-4)(5,2)% \pictaxes% \mygrid \pictline{v(0,-2), v(1,0), -1, 5} % y-2 = 0 \pictline{v(0, 0), v(1,-1), -2, 4} % x+y = 0 \end{picture}% }}$ \quad 1l) $\vcenter{\hbox{% \beginpicture(-2,-4)(5,2)% \pictaxes% \mygrid \picthyperbole{v(2,-2), v(1,0), v(-1,1), 1}{10, -4, -1/2, 1/4, 4} \end{picture}% }}$ \quad 1m) $\vcenter{\hbox{% \beginpicture(-1,-1)(7,10)% \pictaxes% \pictellipse{v(4,6), v(2,0), v(0,3)} \end{picture}% }}$ \bsk \unitlength=5pt \def\closeddot{\circle*{0.6}} 2a) $\pictEllipseF(-4,-6)(4,6){v(0,0), v(3,0), v(0,5)}(0,4)(0,-4)$ \quad 2b) $\pictEllipseF(-1,-4)(8,8){v(4,2), v(3,0), v(0,5)}(4,6)(4,-2)$ \bsk 3a) $x=0 \;⇒\; y=3 \;⇒\; z=1 \;⇒\; P=(0,3,1)$ 3b) $x=1 \;⇒\; y=1 \;⇒\; z=2 \;⇒\; P=(1,1,2)$ 3c) $r'=\setofst{(x,-2x+3,1x+1)}{x∈\R}$, $a=-2$, $b=3$, $c=1$ e $d=1$ \bsk 4a) $[\Vec{AB}, \Vec{AC}, \Vec{AD}] = [\V(-1,1,0), \V(0,1,0), \V(0,0,1)] = \vsm{-1&1&0 \\ 0&1&0 \\ 0&0&1} = -1 \neq 0$, portanto $A$, $B$, $C$ e $D$ não são coplanares. 4b) Seja $\ww=\Vec{AC}=\V(0,1,0)$. Então % $$\begin{array}{rcl} d(r,r') &=& [\Vec{AB}, \Vec{CD}, \Vec{AC}] / ||\Vec{AB} × \Vec{CD}|| \\ &=& [\V(-1,1,0), \V(0,-1,1), \V(0,1,0)] / ||\V(-1,1,0) × \V(0,-1,1)|| \\ &=& \vsm{-1&1&0 \\ 0&-1&1 \\ 0&1&0} / ||\V(1,1,1)|| \\ &=& 1/\sqrt{3} \\ &=& \sqrt{3}/3 \\ \end{array} $$ % \end{document} % Local Variables: % coding: utf-8-unix % ee-anchor-format: "«%s»" % End: