
\begin{enumerate}
\item Réponse \ref{prodscalexoqcm1a} En effet,
\begin{align*}
\prodscal{AB}{AC} & = AB \times AC \times \cos\angleor{AB}{AC}\\[6pt]
& = 5\times6\times\cos\dfrac{\pi}{4}\\[6pt]
& = 30\times\dfrac{\sqrt2}{2}\\[6pt]
& = 15\sqrt2.
\end{align*}

\item Réponse \ref{prodscalexoqcm2c} En effet, on a la figure suivante:
\begin{center}
\begin{tikzpicture}
\draw (0,0) node[below left] {A} -- (3,0) node[below right] {B} -- (3,3) node[above right] {C} -- (0,3) node[above left] {D} -- cycle;
\draw (0,0) -- (3,3);
\draw (0,3) -- (3,0);
\node[below] at (1.5,1.5) {O};
\draw[very thick,red,->,>=latex] (0,0) -- (3,0);
\draw[very thick,red,->,>=latex] (1.5,1.5) -- (0,3);
\draw[very thick,red,->,>=latex] (0,0) -- (-1.5,1.5) node[midway,above,rotate=-45] {$\vv{OD}$};
\draw[very thick,red,->,>=latex] (0.5,0) arc (0:135:0.5);
\node[red, right] at (0,0.5) {$\frac{3\pi}{2}$};
\end{tikzpicture}
\end{center}
Ainsi,
\begin{align*}
\angleor{AB}{OD} & = AB \times OD \times \cos\angleor{AB}{OD}\\
& = 1\times\dfrac{\sqrt2}{2}\times\left(-\dfrac{\sqrt2}{2}\right)\\
& = -\dfrac{1}{2}\\
& = -0,5.
\end{align*}

\item Réponse \ref{prodscalexoqcm3d} $\vv{u}$ et $\vv{v}$ sont deux vecteurs orthogonaux (donc $\prodscal{u}{v}=0$) tels que $\|\vv{u}\|=2$ et $\|\vv{v}\|=1$.

\begin{align*}
\big(\vv{u}+\vv{v}\big)\cdot\big(2\vv{u}-\vv{v}\big)&=2\vv{u}^2-\prodscal{u}{v}+2\prodscal{u}{v}-\vv{v}^2\\
& = 2\|\vv{u}\|^2+\prodscal{u}{v}-\|\vv{v}\|^2\\
& = 2\times2^2 + 0 - 1^2\\
& = 7.
\end{align*}

\newpage

\item Réponse \ref{prodscalexoqcm4d}

\begin{center}
\begin{tikzpicture}
\draw (0,0) node[below left]{A} -- (60:3) node[above left] {B}-- ($(4,0)+(60:3)$) node[above right] {C}-- (4,0) node[below right] {D} -- cycle;
\draw[->,>=latex] (0.5,0) arc(0:60:0.5);
\node[right] at (30:0.6) {$\frac{\pi}{3}$};
\end{tikzpicture}
\end{center}
\begin{align*}
\prodscal{DA}{DC} & = -\prodscal{AD}{AB}\\
& = -AD \times AB \times \cos\dfrac{\pi}{3}\\
& = -3\times4\times\dfrac{1}{2}\\
& = -6.
\end{align*}
\end{enumerate}
