Warning: this is an htmlized version!
The original is across this link,
and the conversion rules are here.
% (find-angg "LATEX/2016-1-GA-P1.tex")
% (find-angg "LATEX/2016-1-GA-P1.lua")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2016-1-GA-P1.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2016-1-GA-P1.pdf"))
% (defun e () (interactive) (find-LATEX "2016-1-GA-P1.tex"))
% (defun u () (interactive) (find-latex-upload-links "2016-1-GA-P1"))
% (defun z () (interactive) (find-zsh "flsfiles-tgz 2016-1-GA-P1.fls 2016-1-GA-P1.tgz"))
% (find-xpdfpage "~/LATEX/2016-1-GA-P1.pdf")
% (find-xdvipage "~/LATEX/2016-1-GA-P1.dvi")
% (find-sh0 "cp -v  ~/LATEX/2016-1-GA-P1.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2016-1-GA-P1.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2016-1-GA-P1.pdf
%               file:///tmp/2016-1-GA-P1.pdf
%           file:///tmp/pen/2016-1-GA-P1.pdf
% http://angg.twu.net/LATEX/2016-1-GA-P1.pdf
\documentclass[oneside]{book}
\usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref")
%\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\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")
%
\begin{document}

% (find-LATEXfile "2015-2-GA-P1-gab.tex" "setofxyst")
% (find-LATEXfile "2015-1-GA-P2-gabarito.tex" "setxyst")
\def\nip{\par\noindent}
\def\uu{{\vec u}}
\def\vv{{\vec v}}
\def\ww{{\vec w}}
\def\Vec#1{\overrightarrow{#1}}
\def\VEC#1{{\overrightarrow{(#1)}}}
\def\Pr{{\text{Pr}}}
\def\Pru{\Pr_\uu}
\def\Prv{\Pr_\vv}
\def\Prw{\Pr_\ww}
\def\setofxyst#1{\setofst{(x,y)∈\R^2}{#1}}
\def\setofexprt#1{\setofst{#1}{t∈\R}}
\def\setofexpru#1{\setofst{#1}{u∈\R}}




%   ____      _                    _ _           
%  / ___|__ _| |__   ___  ___ __ _| | |__   ___  
% | |   / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \ 
% | |__| (_| | |_) |  __/ (_| (_| | | | | | (_) |
%  \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/ 
%                                                

{\setlength{\parindent}{0em}
\footnotesize
\par Geometria Analítica
\par PURO-UFF - 2016.1
\par P1 - 8/jun/2016 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Proibido usar quaisquer aparelhos eletrÃ∧nicos.

% \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(B 0.45 pts) Bar



% (find-angg "LATEX/2015-2-GA-P2.tex")

1) \T(Total: 2.5 pts) Sejam
%
$$\begin{array}{rcl}
  r_1 &=& \setofxyst{5x-2y=0}, \\
  r_2 &=& \setofexprt{(2,2)+t\VEC{2,-1}}, \\
% r_3 &=& \setofexpru{(2,5)+u\VEC{3,-4}} \\
% r_3 &=& \setofexpru{(5,1)+u\VEC{3,-4}} \\
  r_3 &=& \setofexpru{(5+3u,1-4u)}. \\
  \end{array}
$$

a) \B(1.5 pts) Determine as coordenadas de $C∈r_1∩r_2$, $D∈r_1∩r_3$ e
$E∈r_2∩r_3$. Obs: se você conseguir encontrar as coordenadas de algum
ponto só pelo gráfico basta provar que ele pertence às retas
adequadas.

b) \B(1.0 pts) Determine a área do triângulo $CDE$.

\bsk



2) \T(Total: 2.5 pts) Sejam:
%
$$\begin{array}{ccl}
  C = \setofxyst{(x-3)^2 + (y-4)^2 = 25} \\
  C' = \setofxyst{(x-7)^2 + (y-3)^2 = 4} \\
  C∩C' = \{I,I'\} \\
  \end{array}
$$

a) \B(2.0 pts) Encontre os dois pontos de interseção $I$ e $I'$ dos
dois círculos. Obs: se você conseguir encontrar algum ponto só pelo
gráfico basta provar que ele pertence aos dois círculos.

b) \B(0.5 pts) Determine o ponto médio $M$ de $I$ e $I'$.

\bsk

3) \T(Total: 1.5 pts) Verdadeiro ou falso? Justifique.

Se $\uu$ e $\vv$ são ortogonais e não-nulos e $\ww = a\uu+b\vv$ então
$\ww = \Pru \ww + \Prv \ww$.


% a) \B(1.5 pts) 
% b) \B(1.0 pts)



\bsk

4) \T(Total: 1.0 pts) Determine a distância entre as retas com
equações $y=1-\frac x 3$ e $y=2 - \frac x 3$.

\bsk


\def\sen{\operatorname{sen}}
\def\ang{\operatorname{ang}}

5) \T(Total: 2.5 pts) Sejam $A=(2,5)$, $B=(1,3)$, $C=(2,3)$, $D=(2,1)$,
%
$$r=\setofxyst{x+y=3}.$$

a) \B(0.5 pts) Calcule $\cos(A \hat B C)$.

b) \B(2.0 pts) Encontre uma reta $s$ que passa por $D$ e que faz com
$r$ o mesmo ângulo que $A \hat B C$.


\newpage


%  ____           _             _    __   
% |  _ \  ___  __| |_ __   __ _| |_ / /_  
% | | | |/ _ \/ _` | '_ \ / _` | __| '_ \ 
% | |_| |  __/ (_| | | | | (_| | |_| (_) |
% |____/ \___|\__,_|_| |_|\__,_|\__|\___/ 
%                                         

\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")
% (find-dn6 "picture.lua" "V")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end
%L V.__div      = function (v, k) return v*(1/k) end
%L V.__index.tow = function (A, B, t) return A+(B-A)*t   end  -- towards
%L V.__index.mid = function (A, B)    return A+(B-A)*0.5 end  -- midpoint

\def\e{\expr}



%  _   _ _         
% | |_(_) | __ ____
% | __| | |/ /|_  /
% | |_| |   <  / / 
%  \__|_|_|\_\/___|
%                  

% \mygrid and \myaxes
% (find-es "tikz" "mygrid")
\tikzset{mycurve/.style=very thick}
\tikzset{axis/.style=semithick}
\tikzset{tick/.style=semithick}
\tikzset{grid/.style=gray!20,very thin}
\tikzset{anydot/.style={circle,inner sep=0pt,minimum size=1.2mm}}
\tikzset{opdot/.style={anydot, draw=black,fill=white}}
\tikzset{cldot/.style={anydot, draw=black,fill=black}}
%
\def\mygrid(#1,#2) (#3,#4){
  \clip              (#1-0.4, #2-0.4) rectangle (#3+0.4, #4+0.4);
  \draw[step=1,grid] (#1-0.2, #2-0.2) grid      (#3+0.2, #4+0.2);
  \draw[axis] (-10,0) -- (10,0);
  \draw[axis] (0,-10) -- (0,10);
  \foreach \x in {-10,...,10} \draw[tick] (\x,-0.2) -- (\x,0.2);
  \foreach \y in {-10,...,10} \draw[tick] (-0.2,\y) -- (0.2,\y);
}
\def\myaxes(#1,#2) (#3,#4){
  \clip              (#1-0.4, #2-0.4) rectangle (#3+0.4, #4+0.4);
 %\draw[step=1,grid] (#1-0.2, #2-0.2) grid      (#3+0.2, #4+0.2);
  \draw[axis] (-20,0) -- (20,0);
  \draw[axis] (0,-20) -- (0,20);
  \foreach \x in {-20,...,20} \draw[tick] (\x,-0.2) -- (\x,0.2);
  \foreach \y in {-20,...,20} \draw[tick] (-0.2,\y) -- (0.2,\y);
}

% Grid color
\tikzset{grid/.style=gray!50,very thin}

\def\tikzp#1{\mat{\begin{tikzpicture}#1\end{tikzpicture}}}

\def\mydraw       #1;{\draw [mycurve]  \expr{#1};}
\def\mydot        #1;{\node [cldot] at \expr{#1} [] {};}
\def\myldot #1 #2 #3;{\node [cldot] at \expr{#1} [label=#2:${#3}$] {};}
\def\myseg     #1 #2;{\draw [mycurve]  \expr{#1} -- \expr{#2};}
\def\mylabel #1 #2 #3;{\node []     at \expr{#1} [label=#2:${#3}$] {};}
\def\myseggrid  #1 #2;{\draw [grid]    \expr{#1} -- \expr{#2};}



%   ____       _                _ _        
%  / ___| __ _| |__   __ _ _ __(_) |_ ___  
% | |  _ / _` | '_ \ / _` | '__| | __/ _ \ 
% | |_| | (_| | |_) | (_| | |  | | || (_) |
%  \____|\__,_|_.__/ \__,_|_|  |_|\__\___/ 
%                                          

\def\Area{\operatorname{Area}}




Mini-gabarito:

\bsk



1a)

%L r1 = Line.new(v(0, 0), v(2,  5), -0.2, 1.2)
%L r2 = Line.new(v(2, 2), v(2, -1), -0.2, 1.2)
%L r3 = Line.new(v(5, 1), v(3, -4), -0.2, 1.2)
%L C  = v(1, 2.5)
%L D  = v(2, 5)
%L E  = v(5.6, 0.2)
\pu

$\tikzp{[scale=0.5,auto]
    \myaxes (-2,-2) (8,6);
    \myseg r1:t(-0.2) r1:t(1.2);  \mylabel r1:t(0.75) 180 r_1;
    \myseg r2:t(-1)   r2:t(3);    \mylabel r2:t(0.5)  270 r_2;
    \myseg r3:t(-1.2) r3:t(0.5);  \mylabel r3:t(-0.5) 0   r_3;
    % \myldot C 0 C;
    \myldot C 270 C;
    \myldot D   0 D;
    \myldot E  45 E;
    % \mypgrid 3;
  }
  \quad
\begin{tabular}{l}
Se $C=(1,2.5)$ então \\
$C ∈ r_1$ porque $5·1 - 2·2.5=0$, \\
$C ∈ r_2$ porque $C = (2,2) + (-0.5)\VEC{2,-1}$, \\
portanto $C∈r_1∩r_2$. \\[5pt]
Se $D=(2,5)$ então \\
$D ∈ r_1$ porque $5·2 - 2·5=0$, \\
$D ∈ r_3$ porque $D = (5+3·0, 1-4·0)$, \\
portanto $C∈r_1∩r_3$. \\
\\
\\
\end{tabular}
$

Se $E = (x,y) ∈ r_2∈r_3$ então

$E = (2+2t, 2-t) = (5+3u, 1-4u)$

$2+2t = 5+3u$

$2-t = 1-4u$

$2-1+4u = t$

$t=1+4u$

$2+2(1+4u) = 5+3u$

$2+2-5 + 8u-3u = 0$

$5u = 1$

$u = \frac 1 5$

$t = 1 + 4 \frac 1 5 = \frac 9 5$

$E = (x,y) = (2+2 \frac 9 5, 2 - \frac 9 5) = (5+3 \frac 1 5, 1-4 \frac 1 5) = (\frac {28} 5, \frac 1 5) = (5.6, 0.2)$

\msk

1b) $\Vec{CD} = \VEC{1, 2.5}$, $\Vec{CE} = \VEC{4.6, -2.3}$, $[\Vec{CD}, \Vec{CE}] = \psm{1 & 2.5 \\ 4.6 & -2.3}$

$|[\Vec{CD}, \Vec{CE}]| = \left|\sm{1 & 2.5 \\ 4.6 & -2.3}\right| = -2.3 - 4.6 · 2.5 = -2.3 - 11.5 = -13.8$

$\Area(CDE) = 13.8/2 = 6.9$

\bsk




\newpage

2)

%L C0  = v(3,4); R  = 5
%L CC0 = v(7,3); RR = 2
%L C  = Ellipse.newcircle(C0,  R)
%L CC = Ellipse.newcircle(CC0, RR)
%L I  = v(7,1)
%L uu = CC0 - C0
%L vv = I   - C0
%L M  = C0 + uu:proj(vv)
%L II = I:tow(M, 2)
\pu

$\tikzp{[scale=0.5,auto]
    \myaxes (-2,-1) (10,9);
    \mydraw C:draw();
    \mydraw CC:draw();
    \myseg  C0 CC0;
    \myldot C0  90 C_0;
    \myldot CC0 90 C'_0;
    \myseg  I II;
    \myldot I  270 I;
    \myldot II  45 I';
    \myldot M    0 \phantom{i}M;
  }
$

Seja $I = (7,1)$. Então

$I ∈ C$ porque $(7-3)^2 + (1-4)^2 = 25$ e

$I ∈ C'$ porque $(7-7)^2 + (1-3)^2 = 4$.

Sejam $\uu = \Vec{C_0 C'_0}$, $\vv = \Vec{C_0 I}$, $\ww = \Pru \vv$,
$M = C_0+\ww$, $I' = M + \Vec{IM}$.

Então $\uu = \VEC{4,-1}$, $\vv = \VEC{4,-3}$,

$\ww = \frac {\VEC{4,-1}·\VEC{4,-3}} {\VEC{4,-1}·\VEC{4,-1}}
\VEC{4,-1} = \frac{19}{17} \VEC{4,-1} = \VEC{\frac{76}{17},-\frac{19}{17}}$,

$M = (3,4) + \VEC{\frac{76}{17},-\frac{19}{17}} =
(\frac{51}{17},\frac{68}{17}) + \VEC{\frac{76}{17},-\frac{19}{17}} =
(\frac{127}{17},\frac{49}{17})$

$\Vec{IM} = (\frac{127}{17},\frac{49}{17}) - (7,1) =
(\frac{127}{17},\frac{49}{17}) - (\frac{119}{17},\frac{17}{17}) =
\VEC{\frac{8}{17},\frac{32}{17}}$

$I' = (\frac{127}{17},\frac{49}{17}) +
\VEC{\frac{8}{17},\frac{32}{17}} = (\frac{135}{17},\frac{81}{17})$

\bsk
\bsk




3) $\Pru\ww = \Pru(a\uu + b\vv) =$

$  \frac{\uu·(a\uu + b\vv)}{\uu·\uu} \uu
 = \frac{\uu·(a\uu) + \uu·(b\vv)}{\uu·\uu} \uu
 = \frac{a(\uu·\uu) + b(\uu·\vv)}{\uu·\uu} \uu
 = \frac{a(\uu·\uu) + b·0}{\uu·\uu} \uu
 = \frac{a(\uu·\uu)}{\uu·\uu} \uu
 = a \uu
$

$\Prv\ww = \Prv(a\uu + b\vv) =$

$  \frac{\vv·(a\uu + b\vv)}{\vv·\vv} \vv
 = \frac{\vv·(a\uu) + \vv·(b\vv)}{\vv·\vv} \vv
 = \frac{a(\vv·\uu) + b(\vv·\vv)}{\vv·\vv} \vv
 = \frac{a·0 + b(\vv·\vv)}{\vv·\vv} \vv
 = \frac{b(\vv·\vv)}{\vv·\vv} \vv
 = b \vv
$

$\Pru\ww + \Prv\ww = a \uu + b \vv= \ww$

\bsk
\bsk





4)
%L vv = v(1, -1/3)
%L A = v(0, 1); r = Line.new(A, vv, -0.2, 1.2)
%L B = v(0, 2); s = Line.new(B, vv, -0.2, 2.2)
%L C = B + vv:proj(A - B)
\pu
%
$\tikzp{[scale=0.5,auto]
    \myaxes (-1,-1) (7,3);
    \myseg r:t(-0.5) r:t(3.5);
    \myseg s:t(-0.5) s:t(6.5);
    \myseg A C;
    \myldot A 225 A;
    \myldot B 135 B;
    \myldot C 45 C;
    \mylabel r:t(2.5) 270 r;
    \mylabel s:t(3.5) 90 s;
    % \mydraw s:draw();
 }
$

Sejam:

$r = \setofxyst{y = 1-\frac x 3}$, $A = (0,1)$,

$s = \setofxyst{y = 2-\frac x 3}$, $B = (0,2)$.

Então $d(r,s) = d(A,s) = \frac{d(A,B)}{1 + (-1/3)^2} = \frac 1 {10/9} = \frac{9}{10}$. 

\newpage



5)
%L A = v(2, 5); B = v(1, 3); C = v(2, 3); D = v(2, 1)
%L r = Line.new(v(0,3), v(1,-1), 0, 3)
%L CC = v(3, 0); AA = v(5, 2)
\pu
%
$\tikzp{[scale=0.5,auto]
    \myaxes (-1,-1) (6,6);
    \myseg r:t(-0.5) r:t(3.5);
    \myseg A B;
    \myseg B C;
    \myldot A 90 A;
    \myldot B 90 B;
    \myldot C 45 C;
    %
    \myseg AA D;
    \myseg D CC;
    \myldot AA 90 A';
    \myldot D  90 D;
    \myldot CC 45 C';
    % \mydraw s:draw();
 }
$




% a) \B(0.5 pts) Calcule $\cos(A \hat B C)$.

% b) \B(2.0 pts) Encontre uma reta $s$ que passa por $D$ e que faz com
% $r$ o mesmo ângulo que $A \hat B C$.



5a) $\cos(A \hat B C)
     = \frac {\Vec{BA}·\Vec{BC}} {||\Vec{BA}|| \, ||\Vec{BC}||}
     = \frac 1 {\sqrt{5}}$

5b) $\VEC{1,-1} + 2\VEC{1,1} = \VEC{3,1}$

    $s = \setofexprt{(2,1) + t\VEC{3,1}}$




\end{document}



--[[
* (eepitch-lua51)
* (eepitch-kill)
* (eepitch-lua51)
dofile "edrxtikz.lua"
vv = v(1, -1/3)
A = v(0, 1); r = Line.new(A, vv, -0.2, 1.2)
B = v(0, 2); s = Line.new(B, vv, -0.2, 2.2)
C = B + vv:proj(A - B)
= A
= B
= r
= s
= C
= r:t(0)


V.__index.tow = function (A, B, t) return A+(B-A)*t   end  -- towards
C0  = v(3,4); R  = 5
CC0 = v(7,3); RR = 2
C  = Ellipse.newcircle(C0,  R)
CC = Ellipse.newcircle(CC0, RR)
I  = v(7,1)
uu = CC0 - C0
vv = I   - C0
ww = uu:proj(vv)
M  = C0 + ww
IM = M - I
II = M + IM
= ww * 17
= M * 17
= IM * 17
= II * 17

II = I:tow(M, 2)

= uu



r1 = Line.new(v(0, 0), v(2, 5), -0.2, 1.2)
r2 = Line.new(v(2, 2), v(2, -1), -0.2, 1.2)
r3 = Line.new(v(5, 1), v(3, -4), -0.2, 1.2)
= r1:t(0)
= r1:t(1)
= r2:t(0)
= r2:t(1)
= r3:t(0)
= r3:t(1)

= r
= r:draw()

= r:proj(v(0, 1))
= r:proj(v(-2, 4))
= r:sym(v(0, 1))
= r:sym(v(-2, 4))

--]]





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% coding: utf-8-unix
% End: