\documentclass[mathserif]{beamer}
%\documentclass[mathserif, handout]{beamer}

\input{/home/janine/Dropbox/Travail_S/Divers/Preambule/Preambule_Cours}

\begin{document}

\titreCours{1}{Les vecteurs}{Géométrie vectorielle}

\begin{frame}{1. La notion de vecteur}

\underline{Définition 1.1} On appelle \bl{vecteur} l'ensemble de \ro{toutes} les flèches qui définissent \ro{la même translation}. \\[0.2cm]

\begin{minipage}{0.5\textwidth}


\centering
\begin{tikzpicture}[scale=1.2]
\def\al{60}
\coordinate (A) at (0,0);
\coordinate (B) at (1,2);
\coordinate (A3) at (0.5,0);
\coordinate (B3) at (1.5,2);
\coordinate (A2) at (-0.25,-0.5);
\coordinate (B2) at (1.25,2.5);
\coordinate (S1) at (-0.25,0.5);
\coordinate (S2) at (0.25,1.5);

\draw[lblue] (-1.5,-1) grid[xstep=0.5cm, ystep=0.5cm] (3,3);

\only<beamer>{

\uncover<5->{\draw[dorange, thick] (A2) node[right]{- direction} -- (B2);}
\uncover<7->{\draw[dred, -latex, thick] (S1) -- node[left]{sens -} (S2);}
\uncover<9->{\draw[dgreen, *-*, thick] (A3) -- node[right]{- norme} (B3);}

\uncover<2->{
\fill[dblue] (A) circle (0.05cm) node[below]{A} node[left]{origine -} ;
\fill[dblue] (B) circle (0.05cm) node[above]{B} node[left]{extrémité -};}

\uncover<3->{
\draw[dgreen, very thick,
    decoration={markings,mark=at position 1 with {\arrow[scale=2,dred]{latex}}},
    postaction={decorate},
    shorten >=0.4pt
    ]
(A) -- (B);}}


\end{tikzpicture}


\end{minipage} \;\begin{minipage}{0.47\textwidth}

\uncover<4->{
Un vecteur se caractérise par 
\begin{itemize}
\item<4-> sa \ora{direction}
\item<6-> son \ro{sens}
\item<8-> sa longueur, appelée \ve{norme} 
\end{itemize}}


\end{minipage}

\vspace{0.2cm}
\uncover<10->{
\underline{Notation 1.1}
\begin{itemize}
\item $\bl{\Vecs{AB}}$: le vecteur de $A$ à $B$
\item<11-> $\ve{\Norm{AB}}$: la norme du vecteur $\Vecs{AB}$
\end{itemize}
}

\end{frame}



\begin{frame}
\underline{Exemple 1.1} Donner toutes les flèches représentant le vecteur $\Vecs{AC}$\\[0.2cm]

\centering
\begin{tikzpicture}
\coordinate (A) at (1,5);
\coordinate (B) at (3,5);
\coordinate (C) at (5,5);
\coordinate (D) at (7,5);
\coordinate (J) at (1,3);
\coordinate (K) at (3,3);
\coordinate (L) at (5,3);
\coordinate (E) at (7,3);
\coordinate (I) at (1,1);
\coordinate (H) at (3,1);
\coordinate (G) at (5,1);
\coordinate (F) at (7,1);

\draw[lblue] (0,0) grid[xstep=0.5cm, ystep=0.5cm] (8cm,6cm);
\draw[thick] (A) \nodec{above left}{A}-- (B) \nodec{above left}{B} --(C) \nodec{above right}{C} -- (D) \nodec{above right}{D};
\draw[thick] (D) -- (E) \nodec{above right}{E} -- (F) \nodec{below right}{F}; 
\draw[thick] (F) -- (G) \nodec{below right}{G} --(H) \nodec{below left}{H} -- (I) \nodec{below left}{I}; 
\draw[thick] (I) -- (J) \nodec{above left}{J} -- (A); 
\draw[thick] (J) -- (K) \nodec{above left}{K} --(L) \nodec{above right}{L} -- (E); 
\draw[thick] (B) -- (K) -- (H); 
\draw[thick] (C) -- (L) -- (G); 

\only<beamer>{
\uncover<3->{\fill[dred] (A) circle (0.1cm);}
\uncover<4->{\fill[dred] (C) circle (0.1cm);}
\uncover<5->{\draw[-latex,very thick, dred] (A) -- node[below left]{$\Vecs{AC}$} (C);}
\uncover<6->{\fill[dblue] (J) circle (0.1cm);}
\uncover<7->{\fill[dblue] (L) circle (0.1cm);}
\uncover<8->{\draw[|-latex,very thick, dblue] (J) -- node[below left]{$\Vecs{JL}$} (L);}
\uncover<10->{\fill[dgreen] (I) circle (0.1cm);}
\uncover<11->{\fill[dgreen] (G) circle (0.1cm);}
\uncover<12->{\draw[-latex,very thick, dgreen] (I) -- node[above left]{$\Vecs{IG}$} (G);}
\uncover<14->{\fill[dorange] (K) circle (0.1cm);}
\uncover<15->{\fill[dorange] (E) circle (0.1cm);}
\uncover<16->{\draw[-latex,very thick, dorange] (K) -- node[below right]{$\Vecs{KE}$}(E);}
\uncover<18->{\fill[dviolet] (H) circle (0.1cm);}
\uncover<19->{\fill[dviolet] (F) circle (0.1cm);}
\uncover<20->{\draw[-latex,very thick, dviolet] (H) -- node[above right]{$\Vecs{HF}$}(F);}
\uncover<22->{\fill[dbrown] (B) circle (0.1cm);}
\uncover<23->{\fill[dbrown] (D) circle (0.1cm);}
\uncover<24->{\draw[-latex,very thick, dbrown] (B) --node[below right]{$\Vecs{BD}$}(D);}}

\end{tikzpicture}

\only<beamer>{
\uncover<9->{
\vspace{0.2cm}
\centering
$\ro{\Vecs{AC}} = \bl{\Vecs{JL}} 
\uncover<13->{= \ve{\Vecs{IG}}}  
\uncover<17->{= \ora{\Vecs{KE}}}
\uncover<21->{= \vi{\Vecs{HF}}}
\uncover<25->{= \br{\Vecs{BD}}}$\\}

\uncover<26->{Toutes les flèches sont \bl{équivalentes}, elles représentent donc le \vi{même} vecteur}}
\end{frame}

\begin{frame}
\underline{Exercice 1.1} Donner toutes les flèches représentant le vecteur $\Vecs{BI}$\\[0.2cm]

\centering
\begin{tikzpicture}
\coordinate (A) at (1,5);
\coordinate (B) at (3,5);
\coordinate (C) at (5,5);
\coordinate (D) at (7,5);
\coordinate (J) at (1,3);
\coordinate (K) at (3,3);
\coordinate (L) at (5,3);
\coordinate (E) at (7,3);
\coordinate (I) at (1,1);
\coordinate (H) at (3,1);
\coordinate (G) at (5,1);
\coordinate (F) at (7,1);

\draw[lblue] (0,0) grid[xstep=0.5cm, ystep=0.5cm] (8cm,6cm);
\draw[thick] (A) \nodec{above left}{A}-- (B) \nodec{above left}{B} --(C) \nodec{above right}{C} -- (D) \nodec{above right}{D};
\draw[thick] (D) -- (E) \nodec{above right}{E} -- (F) \nodec{below right}{F}; 
\draw[thick] (F) -- (G) \nodec{below right}{G} --(H) \nodec{below left}{H} -- (I) \nodec{below left}{I}; 
\draw[thick] (I) -- (J) \nodec{above left}{J} -- (A); 
\draw[thick] (J) -- (K) \nodec{above left}{K} --(L) \nodec{above right}{L} -- (E); 
\draw[thick] (B) -- (K) -- (H); 
\draw[thick] (C) -- (L) -- (G); 

\only<beamer>{
\uncover<2->{\draw[-latex,very thick, dred] (B) -- node[left, yshift=0.5cm]{$\Vecs{BI}$} (I);}
\uncover<3->{\draw[-latex,very thick, dblue] (C) -- node[left, yshift=0.5cm]{$\Vecs{CH}$} (H);}
\uncover<4->{\draw[-latex,very thick, dgreen] (D) -- node[left, yshift=0.5cm]{$\Vecs{DG}$} (G);}
}

\end{tikzpicture}

\only<beamer>{
\uncover<5->{
$$\ro{\Vecs{BI}} = \bl{\Vecs{CH}} = \ve{\Vecs{DG}}$$}
}
\end{frame}
 
\begin{frame}

\underline{Définition 1.2} Deux vecteurs qui ont la \ro{même direction} sont dits \bl{colinéaires}. \\[0.2cm]

\begin{minipage}{0.5\textwidth}

\begin{tikzpicture}[scale=0.7]

\small

\coordinate (A) at (1,5);
\coordinate (B) at (3,5);
\coordinate (C) at (5,5);
\coordinate (D) at (7,5);
\coordinate (J) at (1,3);
\coordinate (K) at (3,3);
\coordinate (L) at (5,3);
\coordinate (E) at (7,3);
\coordinate (I) at (1,1);
\coordinate (H) at (3,1);
\coordinate (G) at (5,1);
\coordinate (F) at (7,1);

\draw[lblue] (0,0) grid[xstep=0.5cm, ystep=0.5cm] (8cm,6cm);
\draw[thick] (A) \nodec{above left}{A}-- (B) \nodec{above right}{B} --(C) \nodec{above right}{C} -- (D) \nodec{above right}{D};
\draw[thick] (D) -- (E) \nodec{above right}{E} -- (F) \nodec{below right}{F}; 
\draw[thick] (F) -- (G) \nodec{below right}{G} --(H) \nodec{below left}{H} -- (I) \nodec{below left}{I}; 
\draw[thick] (I) -- (J) \nodec{above left}{J} -- (A); 
\draw[thick] (J) -- (K) \nodec{above left}{K} --(L) \nodec{above right}{L} -- (E); 
\draw[thick] (B) -- (K) -- (H); 
\draw[thick] (C) -- (L) -- (G); 

\draw[-latex,very thick] (A) -- node[right, yshift=0.2cm]{$\Vecs{AG}$} (G);
\draw[-latex,very thick] (C) -- node[right, yshift=0.1cm]{$\Vecs{CE}$} (E);
\draw[-latex,very thick] (A) -- node[above]{$\Vecs{AB}$} (B);
\draw[-latex,very thick] (G) -- node[below]{$\Vecs{GH}$} (H);
\draw[-latex,very thick] (A) -- node[left]{$\Vecs{AJ}$} (J);

\only<beamer>{
\uncover<2->{
\draw[-latex,very thick, dblue] (A) -- node[right, yshift=0.2cm]{$\Vecs{AG}$} (G);
\draw[-latex,very thick, dblue] (C) -- node[right, yshift=0.1cm]{$\Vecs{CE}$} (E);}
\uncover<4->{
\draw[-latex,very thick, dgreen] (A) -- node[above]{$\Vecs{AB}$} (B);
\draw[-latex,very thick, dgreen] (G) -- node[below]{$\Vecs{GH}$} (H);}
\uncover<6->{
\draw[-latex,very thick, dred] (A) -- node[left]{$\Vecs{AJ}$} (J);}
}


\end{tikzpicture}

\end{minipage}\quad\begin{minipage}{0.45\textwidth}

\only<beamer>{
\underline{Exemple 1.2} \\

\uncover<3->{$\bl{\Vecs{AG}}$ et $\bl{\Vecs{CE}}$ sont colinéaires.}\\

\uncover<5->{$\ve{\Vecs{AB}}$ et $\ve{\Vecs{GH}}$ aussi.}\\

\uncover<7->{$\Norm{\ve{AB}} = \Norm{\ro{AJ}}$ mais les vecteurs ne sont pas colinéaires.}
}
\end{minipage}

\vspace{0.2cm}

\uncover<8->{
\begin{minipage}{0.5\textwidth}
\underline{Remarque 1.1}
\begin{enumerate}
\item $\Vecs{UV}=\Vecs{O} \Leftrightarrow U=V$
\item<9-> $-\Vecs{XY}=\Vecs{YX}$
\end{enumerate}
\end{minipage}\quad\begin{minipage}{0.45\textwidth}
\uncover<9->{
\begin{tikzpicture}[scale=0.7]

\coordinate (X1) at (0.5, 0.5);
\coordinate (Y1) at (2.5, 3);
\coordinate (X2) at ($(X1)+(4,0)$);
\coordinate (Y2) at ($(Y1)+(4,0)$);

\draw[lblue] (0,0) grid[xstep=0.5cm, ystep=0.5cm] (3cm,4cm);
\draw[lblue] (4-0.001,0) grid[xstep=0.5cm, ystep=0.5cm] (7cm,4cm);

\node[dot=X] at (X1){};
\node[dot=Y] at (Y1){};
\node[dot=X] at (X2){};
\node[dot=Y] at (Y2){};

\only<beamer>{
\uncover<10->{\draw[-latex,very thick, dblue] (X1) -- node[above left]{$\Vecs{XY}$} (Y1);}
\uncover<11->{\draw[-latex,very thick, dred] (Y2) -- node[above left]{$-\Vecs{XY}$} (X2);}
}

\end{tikzpicture}}
\end{minipage}}
\end{frame} 

\begin{frame}
\underline{Exercice 1.3} Soit le vecteur $\Vecs{v}=\Vecs{FA}$.
\begin{enumerate}
\item Donner toutes les flèches représentant $\Vecs{v}$: \uncover<2-|handout:0>{$\bl{\Vecs{FA} = \Vecs{OB} = \Vecs{EO} = \Vecs{DC}}$} 
\item Donner tous les vecteurs \ve{colinéaires} à $\Vecs{v}$: \\[0.2cm]
\only<beamer>{
\ve{\uncover<3->{$\Vecs{FA}$, $\Vecs{OB}$, $\Vecs{EO}$, $\Vecs{DC}$,} 
\uncover<4->{$\Vecs{AF}$, $\Vecs{BO}$, $\Vecs{OE}$, $\Vecs{CD}$,} 
\uncover<5->{$\Vecs{EB}$, $\Vecs{BE}$}} }
\only<handout>{
\vspace{1.5cm}
}
\end{enumerate}
\begin{center}
\begin{tikzpicture}
	
\def\rad{2cm};
\coordinate (O) at (0,0);

\draw[lblue] (-2.5,-2.5) grid[xstep=0.5cm, ystep=0.5cm] (2.5,2.5);
\node[dot=O] at (O) {};
\node[dot=A] at (120:\rad) {};
\node[dot=B] at (60:\rad) {};
\node[dot=C] at (0:\rad) {};
\node[dot=D] at (-60:\rad) {};
\node[dot=E] at (-120:\rad) {};
\node[dot=F] at (180:\rad) {};
\draw[hline] (A) -- (B) -- (C) -- (D) -- (E) -- (F) -- (A) -- cycle;
\foreach \som in {A,B,C,D,E,F}{
\draw[hline] (O) -- (\som);}

\draw[bline, -latex] (F) -- node[left]{\Vect{FA}} (A);  
	
\end{tikzpicture}	
\end{center} 
\end{frame}

\begin{frame}{2. L'addition de vecteurs}

L'addition est simplement la composition des translations.\\[0.5cm]

\begin{minipage}{0.5\textwidth}
\begin{tikzpicture}[scale=0.9]

\draw[lblue] (-1,-1) grid[xstep=0.5cm, ystep=0.5cm] (5,5);

\coordinate (A1) at (0,0);
\coordinate (A2) at (1,2);

\coordinate (B1) at (0.5,4);
\coordinate (B2) at (3,2.5);
\coordinate (B3) at ($(A2)+(B2)-(B1)$);
\coordinate (A3) at ($(B2)+(A2)-(A1)$);

\only<beamer>{
\uncover<2->{\draw[dblue] (A1) circle (0.1cm) node[above left]{1};}
\uncover<3->{\draw[dblue] (A2) circle (0.1cm) node[above]{2};}
\uncover<5->{\draw[dblue] (B3) circle (0.1cm) node[above right]{3};}
\uncover<4->{\draw[rline, -latex] (A2) -- node[above]{$\Vecs{b}$} (B3); }
\uncover<6->{\draw[bline, -latex] (A1) -- node[below]{$\Vecs{a} + \Vecs{b}$} (B3); }
}

\only<beamer>{
\uncover<8->{\draw[gline, -latex] (B2) -- node[right]{$\Vecs{a}$} (A3); }
\uncover<9->{\draw[bline, -latex] (B1) -- node[above]{$\Vecs{b} + \Vecs{a}$} (A3); }
}
 
\draw[gline, -latex] (A1) -- node[left]{$\Vecs{a}$} (A2); 
\draw[rline, -latex] (B1) -- node[above]{$\Vecs{b}$} (B2); 
	
\end{tikzpicture}	
\end{minipage}\quad\begin{minipage}{0.45\textwidth}

\uncover<7->{
\underline{Propriété 2.1}\\

Commutativité de l'addition\\
$$\boxed{\bl{\Vecs{a}+\Vecs{b} = \Vecs{b} + \Vecs{a}}}$$\\

%\uncover<10->{$\Norm{\ve{a}} + \Norm{\ro{b}} \cancel{=} ||\bl{\Vecs{a}+\Vecs{b}}||$}\\

\uncover<10->{
Inégalité triangulaire\\
$$\boxed{\Norm{\ve{a}} + \Norm{\ro{b}} \geq ||\bl{\Vecs{a}+\Vecs{b}}||}$$}

}
\end{minipage}

\end{frame}

\begin{frame}

\begin{minipage}{0.5\textwidth}
\begin{tikzpicture}[scale=0.9]

\draw[lblue] (-1,-1) grid[xstep=0.5cm, ystep=0.5cm] (5,5);
\node[dot=A] at (0,0){};
\node[dot=B] at (3,4){};
\node[dot=C] at (4,2){};

\only<beamer>{
\uncover<1->{\draw[gline, -latex] (A) -- node[left]{$\Vecs{AB}$}(B);}
\uncover<2->{\draw[rline, -latex] (B) -- node[right]{$\Vecs{BC}$}(C);}
\uncover<3->{\draw[bline, -latex] (A) -- node[below right, xshift=-0.2cm]{$\ve{\Vecs{AB}} + \ro{\Vecs{BC}} = \Vecs{AC}$}(C);}
}
		
\end{tikzpicture}	
\end{minipage}\quad\begin{minipage}{0.45\textwidth}

\uncover<4->{
\underline{Première relation de Chasles}
$$\boxed{\ve{\Vecs{AB}} + \ro{\Vecs{BC}} = \bl{\Vecs{AC}}}$$}

\uncover<5->{
\underline{Exemple 2.1} Réduire le vecteur suivant au maximum
$$\Vecs{AB} + \Vecs{BC} + \Vecs{CD} + \Vecs{DE} \only<beamer>{\uncover<6->{= \Vecs{AE}}}$$
}

\end{minipage}

\vspace{0.2cm}

\uncover<7->{
\underline{Exercice 2.1} Réduire les vecteurs suivants\\[0.2cm]
\begin{enumerate}
	\item<7-> \only<7>{\Vect{AB} + \Vect{DE} + \Vect{BD}}\only<beamer>{\only<8->{\Vect{A\ro{B}} + \Vect{\ve{D}E} + \Vect{\ro{B}\ve{D}}}
	\uncover<9->{= \Vect{A\ro{B}} + \Vect{\ro{B}\ve{D}} + \Vect{\ve{D}E} }
	\uncover<10->{= \Vect{A\ve{D}} + \Vect{\ve{D}E} }
	\uncover<11->{= \Vect{AE} }}\\[0.5cm]	
    \item<12-> \only<12>{\Vect{BC} - \Vect{DE} - \Vect{BD}}\only<beamer>{\only<13->{\Vect{BC} \ora{- \Vect{DE} - \Vect{BD}}}
    \only<14->{= \Vect{BC} \ora{+ \Vect{ED}} + \Vect{DB}}
	\uncover<15->{= \Vect{\ro{B}C} + \Vect{E\ve{D}} + \Vect{\ve{D}\ro{B}}}\\
	\qquad \qquad \qquad \,\, \uncover<16->{ = \Vect{E\ve{D}} + \Vect{\ve{D}\ro{B}} + \Vect{\ro{B}C}}
	\uncover<17->{= \Vect{E\ro{B}} + \Vect{\ro{B}C}}
	\uncover<18->{= \Vect{EC}}}
\end{enumerate}}

 
\end{frame}

\begin{frame}{3. La multiplication d'un vecteur par un scalaire}

\underline{Définition 3.1} On appelle \bl{scalaire} un nombre qui multiplie un vecteur. 

\begin{center}
\begin{tikzpicture}

\coordinate (A1) at (0.5, 3);
\coordinate (A2) at (2.5, 3.5);

\coordinate (2A1) at (1,2);
\coordinate (2A2) at ($(2A1)+3*(A2)-3*(A1)$);

\coordinate (3A1) at (3, 1);
\coordinate (3A2) at ($(3A1)+2*(A2)-2*(A1)$);


\draw[lblue] (0,0.5) grid[xstep=0.5cm, ystep=0.5cm] (8,4);

\draw[very thick, -latex] (A1) -- node[above]{$\Vecs{a}$} (A2);

\only<beamer>{
\uncover<2->{\draw[gray,thick, -latex] (2A1) -- node[below]{$\Vecs{a}$} ($(2A1) + (A2) - (A1)$);}
\uncover<3->{\draw[gray,thick, -latex] ($(2A1) + (A2) - (A1)$) -- node[below]{$\Vecs{a}$} ($(2A1) + 2*(A2) - 2*(A1)$);}
\uncover<4->{\draw[gray,thick, -latex] ($(2A1) + 2*(A2) - 2*(A1)$) -- node[below]{$\Vecs{a}$} (2A2);}
\uncover<5->{\draw[very thick,-latex] (2A1) -- node[above]{$\bl{3}\Vecs{a}$} (2A2);}
\uncover<6->{\draw[gray,thick, -latex] (3A2) -- node[below]{$-\Vecs{a}$} ($(3A2) - (A2) + (A1)$);}
\uncover<7->{\draw[gray,thick, -latex] ($(3A2) - (A2) + (A1)$) -- node[below]{$-\Vecs{a}$} (3A1);}
\uncover<8->{\draw[very thick,-latex] (3A2) -- node[above]{$\bl{-2}\Vecs{a}$} (3A1);}
}
\end{tikzpicture}
\end{center}
\uncover<9->{
\underline{Remarque 3.1}\\[0.2cm]

Pour tout \bl{scalaire} $\bl{k}$ , les vecteurs $\Vecs{a}$ et $\ve{k}\Vecs{a}$ ont \ve{la même direction}. \uncover<10->{Ils sont donc \ve{colinéaires}.}\\[0.2cm]

\uncover<11->{ Si \bl{k} est \ro{négatif}, $\Vecs{a}$ et $\bl{k}\Vecs{a}$  sont de \ro{sens contraire}.}}

\end{frame}

\begin{frame}

\underline{Vocabulaire 3.1}\\[0.2cm]

Si, par exemple, on a $$\Vecs{a} = 4\Vecs{u} + 3 \Vecs{v} - 5 \Vecs{w},$$ 
on dit que $\Vecs{a}$ est une \bl{combinaison linéaire} des vecteurs $\Vecs{u}$, $\Vecs{v}$ et $\Vecs{w}$. \\[0.2cm]

\uncover<2->{
Dans ce cas, on dit que $\Vecs{a}$, $\Vecs{u}$, $\Vecs{v}$ et $\Vecs{w}$ sont \bl{linéairement dépendants}. } \\[0.2cm]

\end{frame}
 
\end{document}
