
\begin{enumerate}
\item On souhaite construire le point M tel que $\vv{MA}+\vv{MB}=\vv{0}$.

\medskip

Utilisons la relation de Chasles pour écrire :
\begin{align*}
\vv{MA}+\vv{MB} & = \vv{MA} + \big(\vv{MA} + \vv{AB}\big)\\
& = 2\vv{MA} + \vv{AB}.
\end{align*}

Ainsi,
\begin{align*}
\vv{MA}+\vv{MB}=\vv{0} & \iff 2\vv{MA} + \vv{AB} = \vv{0}\\
& \iff 2\vv{MA} = -\vv{AB}\\
& \iff -2\vv{AM} = \vv{BA}\\
& \iff \vv{AM} = -\dfrac{1}{2}\vv{BA}\\
& \iff \vv{AM} = \dfrac{1}{2}\vv{AB}.
\end{align*}

\begin{center}
\begin{tikzpicture}
\draw[->,>=latex] (0,0) node[below] {A} -- (6,0) node[below] {B};
\draw[thick,red,>=latex,->] (0,0) -- (3,0) node[below] {M};
\end{tikzpicture}
\end{center}

Ainsi, M est le milieu de [AB].

\begin{rem}
Quand on souhaite construire un point inconnu défini par une égalité vectorielle (comme ici), on utilise la relation de Chasles de sorte à faire apparaître un seul vecteur qui dépend du point inconnu (ici, $\vv{MA}$).

Il faut arriver à une égalité vectorielle de la forme $\vv{AM} = k\vv{AB}$, où $k\in\R$. 
\end{rem}

\item On souhaite construire le point N tel que $2\vv{NA}-3\vv{NB}=\vv{0}$.

\begin{align*}
2\vv{NA}-3\vv{NB} & = 2\vv{NA}-3\big(\vv{NA}+\vv{AB}\big)\\
& = 2\vv{NA}-3\vv{NA}-3\vv{AB}\\
& = -\vv{NA}-3\vv{AB}\\
& = \vv{AN}-3\vv{AB}.
\end{align*}
Ainsi,
\begin{align*}
2\vv{NA}-3\vv{NB}=\vv{0} & \iff  \vv{AN}-3\vv{AB} = \vv{0}\\
& \iff  \vv{AN} = 3\vv{AB}
\end{align*}
\begin{center}
\begin{tikzpicture}
\draw[->,>=latex] (0,0) node[below] {A} -- (2,0) node[below] {B};
\draw[->,>=latex] (2,0) -- (4,0);
\draw[thick,red,>=latex,->] (0,0) -- (6,0) node[below] {N};
\end{tikzpicture}
\end{center}

\item On souhaite construire le point P tel que $-3\vv{PA}+5\vv{PB}=\vv{0}$.

\begin{align*}
-3\vv{PA}+5\vv{PB} & = -3\vv{PA}+5\big(\vv{PA}+\vv{AB}\big)\\
& = -3\vv{PA}+5\vv{PA}+5\vv{AB}\\
& = 2\vv{PA} + 5\vv{AB}
\end{align*}

Ainsi,
\begin{align*}
-3\vv{PA}+5\vv{PB}=\vv{0} & \iff 2\vv{PA} + 5\vv{AB} = \vv{0}\\
& \iff -2\vv{AP} = -5\vv{AB}\\
& \iff \vv{AP} = \dfrac{5}{2}\vv{AB}.
\end{align*}

On divise donc [AB] en deux et on reporte 5 fois cette moitié :
\begin{center}
\begin{tikzpicture}
\draw[->,>=latex] (0,0) node[below] {A} -- (4,0) node[below] {B};
\draw[thick,red,>=latex,->] (0,0) -- (10,0) node[below] {P};
\foreach\x in {0,2,4,6,8} {
\draw (\x,-0.1) -- (\x,0.1);
}
\end{tikzpicture}
\end{center}

\item On souhaite construire le point R tel que $2\vv{RA}+3\vv{RB}=\vv{0}$.

\begin{align*}
2\vv{RA}+3\vv{RB} & = 2\vv{RA}+3\big(\vv{RA}+\vv{AB}\big)=\vv{0}\\
& = 2\vv{RA}+3\vv{RB} + 3\vv{AB}\\
& = 5\vv{RA}+3\vv{AB}.
\end{align*}

Ainsi,

\begin{align*}
2\vv{RA}+3\vv{RB} = \vv{0} & \iff 5\vv{RA}+3\vv{AB} = \vv{0}\\
& \iff 5\vv{RA}=-3\vv{AB}\\
& \iff \vv{RA} = -\dfrac{3}{5}\vv{AB}\\
& \iff -\vv{AR} = -\dfrac{3}{5}\vv{AB}\\
& \iff \vv{AR} = \dfrac{3}{5}\vv{AB}.
\end{align*}

Il faut donc couper en 5 le segment [AB] :

\begin{center}
\begin{tikzpicture}
\draw[->,>=latex] (0,0) node[below] {A} -- (5,0) node[below] {B};
\draw[thick,red,>=latex,->] (0,0) -- (3,0) node[below] {R};
\foreach\x in {1,2,3,4} {
\draw (\x,-0.1) -- (\x,0.1);
}
\end{tikzpicture}
\end{center}

\item On souhaite construire le point S tel que $3\vv{SA}-2\vv{SB}=\vv{0}$.

\begin{align*}
3\vv{SA}-2\vv{SB} & = 3\vv{SA}-2\big(\vv{SA} + \vv{AB}\big)\\
& = 3\vv{SA}-2\vv{SA} -2\vv{AB}\\
& = \vv{SA}-2\vv{AB}.
\end{align*}

Ainsi,

\begin{align*}
3\vv{SA}-2\vv{SB}=\vv{0} & \iff \vv{SA}-2\vv{AB} = \vv{0}\\
& \iff \vv{SA}=2\vv{AB}\\
& \iff \vv{AS} = 2\vv{BA}.
\end{align*}

\begin{center}
\begin{tikzpicture}
\draw[->,>=latex] (3,0) node[below] {B} -- (0,0) node[below] {A};
\draw[thick,red,>=latex,->] (0,0) -- (-6,0) node[below] {S};
\draw (-3,-0.1) -- (-3,0.1);
\draw (0,-0.1) -- (0,0.1);
\end{tikzpicture}
\end{center}

\item On souhaite construire le point T tel que $5\vv{TA}-2\vv{TB}=\vv{0}$.

\begin{align*}
5\vv{TA}-2\vv{TB} & = 5\vv{TA}-2\big(\vv{TA}+\vv{AB}\big)\\
& = 5\vv{TA}-2\vv{TA}-2\vv{AB}\\
& = 3\vv{TA} - 2\vv{AB}.
\end{align*}

Ainsi,

\begin{align*}
5\vv{TA}-2\vv{TB}=\vv{0} & \iff 3\vv{TA} - 2\vv{AB}= \vv{0} \\
& \iff 3\vv{TA} = 2\vv{AB}\\
& \iff \vv{TA} = \dfrac{2}{3}\vv{AB}\\
& \iff \vv{AT} = \dfrac{2}{3}\vv{BA}.
\end{align*}

Il faut donc couper [AB] en 3 et reporter les deux tiers de l'autre côté du point A :

\begin{center}
\begin{tikzpicture}
\draw[->,>=latex] (3,0) node[below] {B} -- (0,0) node[below] {A};
\draw[thick,red,>=latex,->] (0,0) -- (-2,0) node[below] {T};
\foreach\x in {-1,0,1,2,3} {\draw (\x,-0.1) -- (\x,0.1); }
\end{tikzpicture}
\end{center}
\end{enumerate}
