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%           file:///tmp/pen/2015-2-GA-P2.pdf
% http://angg.twu.net/LATEX/2015-2-GA-P2.pdf
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%
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%
\begin{document}

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

{\setlength{\parindent}{0em}
\footnotesize
\par Geometria Analítica
\par PURO-UFF - 2015.2
\par P2 - 21/mar/2016 - Eduardo Ochs
% \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)}}


1) \T(Total: 3.0 pts) Sejam $F=(9,0)$ e $r: x=25$.

a) \B(0.5 pts) Encontre 4 pontos de $\R^2$ tais que $d(P,F) = \frac35
d(P,r)$.

b) \B(2.0 pts) Seja $E$ o conjunto dos pontos que obedecem $d(P,F) =
\frac35 d(P,r)$. Dê a equação reduzida da elipse $E$, as coordenadas
dos seus focos, os comprimentos dos seus semi-eixos maior e menor e a
sua excentricidade.

c) \B(0.5 pts) Represente graficamente tudo o que você obteve nos
itens anteriores.

\bsk

2) \T(Total: 1.0 pts) Represente graficamente a hipérbole $(2(x-3) +
(y-4)) (2(x-3) - (y-4)) = 4$ e suas assíntotas.

\bsk

3) \T(Total: 3.0 pts) Seja $π_1$ o plano com equação $2x+3y+4z=6$.

a) \B(0.5 pts) Dê uma parametrização para a reta $r$ que passa pelo
ponto $A=(1,1,0)$ e é ortogonal a $π_1$.

b) \B(0.5 pts) Dê as coordenadas do ponto $B∈π_1∩r$.

c) \B(1.0 pts) Dê as coordenadas de um ponto $C∈r$ tal que
$d(C,π_1)=1$.

d) \B(1.0 pts) Dê a equação de um plano $π_2 \parallel π_1$ tal que
$d(π_1,π_2)=2$.

\bsk

4) \T(Total: 3.0 pts) Sejam $A=(2,0,0)$, $B=(0,2,0)$, $C=(0,0,5)$,
$D=(0,5,0)$. Sejam $r$ uma reta que passa por $A$ e $B$, $r'$ uma reta
que passa por $C$ e $D$, e $s$ uma reta ortogonal a $r$ e $r'$ que
corta ambas.

a) \B(0.5 pts) Calcule a distância entre $r$ e $r'$.

b) \B(1.5 pts) Dê uma parametrização para $s$.

c) \B(1.0 pts) Dê a equação de um plano $π$ paralelo a $r$ e $r'$ e
equidistante de ambas.

 \bsk

\newpage

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

% Dots, labels, vectors
%
\def\uu{\vec u}
\def\vv{\vec v}
\def\ww{\vec w}
\def\nn{\vec n}
\def\VEC#1{{\overrightarrow{(#1)}}}

\def\nm#1{\|#1\|}
\def\Reg#1{(#1)}

\def\setofxyst #1{\setofst{(x,y)  ∈\R^2}{#1}}
\def\setofxyzst#1{\setofst{(x,y,z)∈\R^3}{#1}}
\def\setofet  #1{\setofst{#1}{t∈\R}}
\def\setofeu  #1{\setofst{#1}{u∈\R}}
\def\setofpt  #1 #2 #3 #4 {\setofet{(#1,#2)+t\VEC{#3,#4}}}
\def\setofpu  #1 #2 #3 #4 {\setofeu{(#1,#2)+u\VEC{#3,#4}}}

\def\setofeaa  #1{\setofst{#1}{α∈\R}}
\def\setofebb  #1{\setofst{#1}{β∈\R}}

% \mygrid and \myaxes
% (find-es "tikz" "mygrid")
\tikzset{mycurve/.style=very thick}
\tikzset{axis/.style=semithick}
\tikzset{tick/.style=semithick}
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\def\mydraw       #1;{\draw [mycurve]  \expr{#1};}
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\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};}

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%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

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


Gabarito:

(Primeira versão, não revisada)

\bsk

%L e = Ellipse.new(v(0,0), v(15,0), v(0,12))
\pu

1) \quad
$\tikzp{[scale=0.2,auto]
  \myaxes(-23,-15) (32,15)
  \draw [mycurve] (-9,0) -- (0,12) -- (9,0);
  \draw [mycurve] (0,12) -- (25,12);
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  \myldot v(0,-12) 315 (0,-12);
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  \myldot v(-15,0) 225 (-15,0);
  \myldot v(15,0) 315 (15,0);
  \myldot v(25,0) 315 (25,0);
  \myldot v(25,12) 0 (25,12);
  \mydraw e:draw();
 }
$

\bsk
\bsk

%L h = Hyperbole.new(v(3,4), v(1,-2)/2, v(1,2)/2, 8)
\pu

2) Sejam $u = 2(x-3)$, $v=y-4$. Então

$\begin{array}[t]{rcl}
       H &=& \setofxyst{(2(x-3) + (y-4)) (2(x-3) - (y-4)) = 4=4} \\
         &=& \setofxyst{(u+v)(u-v)=4} \\
       \end{array}
      $.

$H$ tem assíntotas em $u+v=0$ e em $u-v=0$, e

$H$ passa pelo ponto que tem $(u+v)=(u-v)=2$, ou seja, $u=2$ e $v=0$.

\msk

$\def\s#1{{\tiny#1}}
 \def\s#1{{\scriptscriptstyle#1}}
 \def\s#1{{\scriptstyle#1}}
 %
 \tikzp{[scale=0.4,auto]
  \myaxes(-2,-4) (14,12)
  \draw [grid] (2,-10) -- (2,11);   \mylabel v(2,11)   90 \s{u{=}-2};
  \draw [grid] (3,-10) -- (3,10.5); \mylabel v(3,10.5) 90 \s{u{=}0};
  \draw [grid] (4,-10) -- (4,10);   \mylabel v(4,10)   90 \s{u{=}2};
  \draw [grid] (-10,2) -- (10,2);  \mylabel v(10,2)   0 \s{v{=}-2};
  \draw [grid] (-10,4) -- (10,4);  \mylabel v(10,4)   0 \s{v{=}0};
  \draw [grid] (-10,6) -- (10,6);  \mylabel v(10,6)   0 \s{v{=}2};
  \myaxes(-2,-4) (14,12)
  \mydraw h:draw(10);
  \mydraw h:drawau(-8,8);
  \mydraw h:drawav(-8,8);
 }
$

\newpage


3) $\begin{array}[t]{rcl}
    π_1 &=& \setofxyzst{2x + 3y + 4z = 6} \\
    \nn &=& \VEC{2,3,4} \\
    A &=& (1,1,0) \\
    \end{array}
   $

3a) $\begin{array}[t]{rcl}
     r &=& \setofet{A+t\nn} \\
       &=& \setofet{(1+2t, 1+3t, 4t)} \\
     \end{array}
    $

3b) Se $B∈π_1∩r$ então

    $\begin{array}[t]{rcl}
      6 &=& 2(1+2t) + 3(1+3t) + 4(4t) \\
        &=& 2+4t+3+9t+16t \\
        &=& 29t+5 \\
      % 29t &=& 6 - 5 = 1 \\
      t &=& \frac{1}{29} \\
      B &=& (1+\frac{2}{29}, 1+\frac{3}{29}, \frac{4}{29})
            = (\frac{31}{29}, \frac{32}{29}, \frac{4}{29})
     \end{array}
    $

3c) Seja $\nn' = \frac{\nn}{||\nn||} = \VEC{2,3,4}/\sqrt{4+9+16} = \VEC{2,3,4}/\sqrt{29}$.

    Seja $C = B+\nn'$.

3d) Seja $D = B+2\nn'$.

Seja $π_2 = \setofxyzst{2x + 3y + 4z = a}$, onde $a$ é tal que $D∈π_2$.

Então $a = 2D_1 + 3D_2 + 4D_3$.

\bsk
\bsk


4) Sejam

$\begin{array}[t]{rcl}
 \uu &=& \Vec{AB} = \VEC{-2,2,0}, \\
 \vv &=& \Vec{CD} = \VEC{0,5,-5}, \\
 r &=& \setofeaa{A+α\uu} \\
   &=& \setofeaa{(2,0,0)+α\VEC{-2,2,0}} \\
   &=& \setofeaa{(2-2α,2α,0)}, \\
 r' &=& \setofebb{C+β\vv} \\
   &=& \setofebb{(0,0,5)+β\VEC{0,5,-5}} \\
   &=& \setofebb{(0,5β,5-5β)}, \\
 P_α &=& (2-2α,2α,0), \\
 Q_β &=& (0,5β,5-5β), \\
 \end{array}
$

Então $\uu×\vv = \VEC{-10,-10,-10}$.

4a) $\begin{array}[t]{rcl}
     \uu×\vv &=& \VEC{-10,-10,-10} \\
     \Vec{AC} &=& \VEC{-2,0,5} \\
     \end{array}
    $

$   \frac{[\uu,\vv,\Vec{AC}]}{||\uu×\vv||}
  = \frac{(\uu×\vv)·\Vec{AC}]}{||\uu×\vv||}
  = \frac{\VEC{-10,-10,-10}·\VEC{-2,0,5}}{10\sqrt{3}}
  = \frac{20-50}{10\sqrt{3}}
  = - \sqrt{3}
$

$d(r,r') = \left| \frac{[\uu,\vv,\Vec{AC}]}{||\uu×\vv||} \right | = \sqrt{3}$

\newpage

\def\PaQb{\Vec{P_αQ_β}}

4b) Queremos que $s$ passe por $P_α$ e $Q_β$ e que $\PaQb⊥\uu$ e $\PaQb⊥\vv$.

$\begin{array}[t]{rcl}
 \PaQb &=& \VEC{2α-2, 5β-2α, 5-5β} \\
 \PaQb·\uu &=& (4-4α)+(10β-4α) = 4 - 8α + 10β = 0 \\
 \PaQb·\vv &=& (25β-10α)+(25β-25) = 25 - 10α + 50β = 0\\
 10α &=& 50β-25 \\
   α &=& 5β-\frac52 \\
 4 - 8α + 10β &=& 4 - 8(5β-\frac52) + 10β \\
              &=& 4 - 40β + 20 + 10β \\
              &=& 24 - 30β = 0 \\
 β &=& \frac{24}{30} = \frac45 \\
 α &=& 5\frac45 - \frac52 = \frac82 - \frac52 = \frac32 \\
 P_α &=& (2-2\frac32, 2\frac32, 0) = (-1, 3, 0) \\
 Q_β &=& (0, 5\frac45, 5 - 5\frac45, 0) = (0, 4, 1) \\
 \PaQb &=& \VEC{1,1,1} \\
 s &=& \setofet{P_α + t \PaQb} \\
   &=& \setofet{(-1,3,0) + t\VEC{1,1,1}} \\
 \end{array}
$

\msk

4c) Seja $M = \frac{P_α+Q_β}{2} = (-\frac12, \frac72, \frac12).$

Seja $π = \setofxyzst{x+y+z=a}$, onde $a$ é tal que $M∈π$.

Temos $-\frac12 + \frac72 + \frac12 = a = \frac72$,

$π = \setofxyzst{x+y+z=\frac72}$.


$\begin{array}[t]{l}
     \end{array}
    $

\end{document}




\end{document}


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