On considère le rectangle ABCD ci-dessous, ainsi que les points H et K, projetés orthogonaux respectivement des points A et C sur [BD]. On sait que $DC = 7$ et $AD = 3$.

\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (7,0) -- (7,3) -- (0,3) -- cycle;
\draw (0.974789693645493,2.274509285172817) -- (1.2347631670933656,2.1630920822665853) -- (1.346180369999597,2.4230655557144583) -- (1.0862068965517242,2.5344827586206895) -- cycle; 
\draw (6.173766576896148,0.3541000384730792) -- (6.28518377980238,0.6140735119209519) -- (6.025210306354507,0.725490714827183) -- (5.913793103448276,0.46551724137931033) -- cycle; 
\draw (0,0)-- (7,0);
\draw (7,0)-- (7,3);
\draw (7,3)-- (0,3);
\draw (0,3)-- (0,0);
\draw (0,3)-- (7,0);
\draw (1.0862068965517242,2.5344827586206895) -- (0,0);
\draw (7,3)-- (5.913793103448276,0.46551724137931033);
\draw (0,0) circle (0.5pt);
\draw (-0.25,-0.19) node {$A$};
\draw (7,0) circle (0.5pt);
\draw (7.19,-0.15) node {$B$};
\draw   (7,3) circle (0.5pt);
\draw  (7.15,3.27) node {$C$};
\draw  (0,3) circle (0.5pt);
\draw  (-0.25,3.31) node {$D$};
\draw  (1.25,2.7) node {$H$};
\draw  (5.65,0.3) node {$K$};
\end{tikzpicture}
\end{center}

\begin{enumerate}
\item Exprimer en fonction de HK le produit scalaire \prodscal{AC}{DB} en utilisant les points H et K.
\item En utilisant la relation $\vv{DB}=\vv{DC}+\vv{CB}$, calculer d'une autre manière \prodscal{AC}{DB}.
\item En déduire la valeur exacte de HK.
\end{enumerate}
\vspace*{-1.7em}

