\begin{exercice} \label{TSG3_E_orth_deux_cubes}
On considère deux cubes, disposés comme dans la figure associée. $M$ est le milieu de $[FG]$. On souhaite démontrer que $(AK)$ est orthogonale au plan $(MHC)$.
\begin{center}
 	\def \dx {-0.4}
 	\def \r {0.3}
	\begin{tikzpicture}[general, scale=2]
	%points
	\coordinate (A) at (0,0); \draw (A) node[above left] {$A$};
	\coordinate (B) at (\dx,-0.5); \draw (B) node[below left] {$B$};
	\coordinate (C) at (1+\dx,-0.5); \draw (C) node[below] {$C$};
	\coordinate (D) at (1,0); \draw (D) node[above right] {$D$};
	\coordinate (E) at (0,1); \draw (E) node[above left] {$E$};
	\coordinate (F) at (\dx,0.5); \draw (F) node[left] {$F$};
	\coordinate (G) at (1+\dx,0.5); \draw (G) node[above left] {$G$};
	\coordinate (H) at (1,1); \draw (H) node[above right] {$H$};
	\coordinate (I) at (2+\dx,-0.5); \draw (I) node[below] {$I$};
	\coordinate (J) at (2,0); \draw (J) node[right] {$J$};
	\coordinate (K) at (2+\dx,0.5); \draw (K) node[right] {$K$};
	\coordinate (L) at (2,1); \draw (L) node[above right] {$L$};
	\coordinate (M) at (0.5+\dx,0.5); \draw (M) node[above] {$M$};
	%arêtes
	\draw (E)--(F)--(K)--(L)--(E);
	\draw (C)--(G)--(H); \draw (I)--(B)--(F);
	\draw (L)--(J)--(I)--(K);
	\draw[dashed] (B)--(A)--(E); \draw[dashed] (C)--(D)--(H); \draw[dashed] (A)--(J);
	%droites
	\draw[thin, color=C1, dashed] (A)--(K);
	%plan
	\fill[color=A1, opacity=0.3] (M)--(H)--(C);
	\end{tikzpicture}
\end{center}
\begin{enumerate}
 \item Démontrer que $\vv{AK} \cdot \vv{CM} = \vv{BK} \cdot \vv{CM}$, puis en déduire la valeur de ce produit scalaire.
 \item En suivant cette méthode, calculer $\vv{AK} \cdot \vv{HM}$.
 \item Conclure.
\end{enumerate}
\end{exercice}
