
Avant tout, un schéma peut aider. Pour cela, il faudra savoir comment construire les points $E$ et $F$ : $\begin{aligned}[t]
-2\vv{EA}+3\vv{EB}=\vv{0} & \iff -2\big(\vv{EB}+\vv{BA}\big)+3\vv{EB}=\vv{0} \\
& \iff \vv{EB}-2\vv{BA}=\vv{0} \\
& \iff \vv{BE} = 2\vv{AB}\\
-2\vv{FA}+3\vv{FC}=\vv{0}  & \iff -2\big(\vv{FC}+\vv{CA}\big)+3\vv{FC}=\vv{0}\\
& \iff \vv{FC}-2\vv{CA}=\vv{0}\\
& \iff \vv{CF} = 2\vv{AC}.
\end{aligned}$
%On a alors la figure page suivante.
\begin{center}
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\begin{tikzpicture}[scale=0.55]
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\draw [color=qqqqff] (8,0)-- ++(-0.5pt,-0.5pt) -- ++(1pt,1pt) ++(-1pt,0) -- ++(1pt,-1pt);
\draw[color=qqqqff] (8.226236097560978,0.3776048780487831) node {$B$};
\draw [color=qqqqff] (6,3)-- ++(-0.5pt,-0.5pt) -- ++(1pt,1pt) ++(-1pt,0) -- ++(1pt,-1pt);
\draw[color=qqqqff] (5.597998048780489,3.52006341463415) node {$C$};
\draw [color=qqqqff] (7.96,7.16)-- ++(-0.5pt,-0.5pt) -- ++(1pt,1pt) ++(-1pt,0) -- ++(1pt,-1pt);
\draw[color=qqqqff] (8.197668292682929,7.548123902439029) node {$D$};
\draw [color=ffqqqq] (7.98,3.58)-- ++(-1.5pt,-1.5pt) -- ++(3pt,3pt) ++(-3pt,0) -- ++(3pt,-3pt);
\draw[color=ffqqqq] (8.45477853658537,3.7486058536585403) node {$I$};
\draw [color=ffqqqq] (6.98,5.08)-- ++(-1.5pt,-1.5pt) -- ++(3pt,3pt) ++(-3pt,0) -- ++(3pt,-3pt);
\draw[color=ffqqqq] (6.826413658536587,5.605513170731712) node {$J$};
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\draw[color=xdxdff] (24.2242068292683,0.4918760975609783) node {$E$};
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\draw[color=qqqqff] (18.224967804878055,9.376463414634152) node {$F$};
\end{tikzpicture}
\end{center}
\begin{enumerate}
\item Des égalités vectorielles précédentes, on peut conclure:
\begin{align*}
\vv{EF} & = \vv{EB} + \vv{BC} + \vv{CF}\\
& = 2\vv{BA} + \vv{BC} + 2\vv{AC}\\
& = 2\big(\vv{BA} + \vv{AC}\big) + \vv{BC}\\
& = 2\vv{BC} + \vv{BC}\\
& = 3\vv{BC}.
\end{align*}

Ainsi, les vecteurs $\vv{EF}$ et $\vv{BC}$ sont colinéaires; les quatre points $E$, $F$, $B$ et $C$ sont donc coplanaires.

\item
\begin{enumerate}
\item Complétons la figure :
\begin{center}
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\begin{tikzpicture}[scale=0.55]
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\draw[color=qqqqff] (8.226236097560978,0.37760487804878473) node {$B$};
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\draw [color=ffqqqq] (6.98,5.08)-- ++(-1.5pt,-1.5pt) -- ++(3pt,3pt) ++(-3pt,0) -- ++(3pt,-3pt);
\draw[color=ffqqqq] (6.826413658536587,5.605513170731713) node {$J$};
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\draw[color=qqqqff] (18.224967804878055,9.376463414634154) node {$F$};
\draw [color=ttttff] (4.776,4.296)-- ++(-1.5pt,-1.5pt) -- ++(3pt,3pt) ++(-3pt,0) -- ++(3pt,-3pt);
\draw[color=ttttff] (4.6552604878048784,4.919885853658542) node {$K$};
\end{tikzpicture}
\end{center}
$K$ est défini comme le point d'intersection de la droite $(AD)$ et du plan $(EFI)$; par conséquent, toutes droites du plan $(EFI)$ coupant $(AD)$ passe par $K$.

$(AD)$ appartient au plan $(ADC)$ et la droite $(FJ)$ aussi; n'étant pas parallèles, elles sont sécantes : en $K$. Les points $F$, $J$ et $K$ sont donc alignés.

$(AD)$ appartient au plan $(ABD)$ et la droite $(EI)$ aussi; n'étant pas parallèles, elles sont aussi sécantes : en $K$.

\item Plaçons-nous dans le repère $\big(A;\vv{AB},\vv{AC},\vv{AD}\big)$, où:
\[
B(1;0;0)\quad;\quad
D(0;0;1)\quad;\quad
C(0;1;0)\quad;\quad
E(3;0;0)\quad;\quad
F(0;3;0)
\]
\[
I\left(\frac{x_B+x_D}{2};\frac{y_B+y_D}{2};\frac{z_B+z_D}{2}\right)=\left(\frac{1}{2};0;\frac{1}{2}\right)\quad;\quad
J\left(0;\frac{1}{2};\frac{1}{2}\right)
\]
\begin{itemize}
\item \textbf{\boldmath La droite $(JF)$.}

Un coefficient directeur est $\vv{JF}\begin{pmatrix}0-0\\3-1/2\\0-1/2\end{pmatrix}=\begin{pmatrix}0\\5/2\\-1/2\end{pmatrix}$.

Une représentation paramétrique est donc:
\[
(JF)\quad:\quad\begin{cases}
x = 0 + t \times 0\\
y = 3 + \frac{5}{2}t\\
z = 0 - \frac{1}{2}t
\end{cases}\quad,\quad t\in\R
\]
soit :
\[
(JF)\quad:\quad\begin{cases}
x = 0 \\
y = 3 + \frac{5}{2}t\\
z = - \frac{1}{2}t
\end{cases}\quad,\quad t\in\R
\]

\item \textbf{\boldmath La droite $(EI)$.}

Un coefficient directeur est $\vv{IE}\begin{pmatrix}3-1/2\\0-0\\0-1/2\end{pmatrix}=\begin{pmatrix}5/2\\0\\-1/2\end{pmatrix}$.

Une représentation paramétrique est donc:
\[
(EI)\quad:\quad\begin{cases}
x = 3 + \frac{5}{2}t'\\
y = 0 + t' \times 0\\
z = 0 - \frac{1}{2}t'
\end{cases}\quad,\quad t\in\R
\]
soit :
\[
(EI)\quad:\quad\begin{cases}
x = 3 + \frac{5}{2}t' \\
y = 0\\
z = - \frac{1}{2}t'
\end{cases}\quad,\quad t'\in\R
\]
\end{itemize}
$K\in(EI)$ et $K\in(JF)$ donc :
\[
\begin{cases}
x_K = 0 = 3+\frac{5}{2}t' \\
y_K = 3 + \frac{5}{2}t = 0\\
z_K = - \frac{1}{2}t = - \frac{1}{2}t'
\end{cases} \iff t' = -\frac{6}{5} = t
\]
On en déduit alors les coordonnées de $K$ en utilisant par exemple la représentation paramétrique de $(EI)$ :
\[
K\left( 0;0;\frac{3}{5} \right).
\]
Ce qui signifie que $\vv{AK}=\dfrac{3}{5}\vv{AD}$.
\end{enumerate}
\end{enumerate}
