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%%%
\begin{document}
\title{Produit scalaire}
\author{Première, enseignement de spécialité\\\url{mathweb.fr}}
\date{\today}
\maketitle
\thispagestyle{fancy}
\begin{consigne}{Réponses}
Vous trouverez ci-dessous les réponses correctes.
\end{consigne}
\begin{enumerate}
\item \bgroup\bfseries\boldmath Si $\|\vv{u}\|=3$, $\|\vv{v}\|=6$  et $\cos(\vv{u},\vv{v})=\frac{1}{2}$, alors:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $\prodscal{u}{v}=36$
\item[\checkbox{red}] $\prodscal{u}{v}=9$
\item[$\square$] $\prodscal{u}{v}=-36$
\item[$\square$] $\prodscal{u}{v}=-9$
\end{multicols}
\end{enumerate} % fin des propositions de la question 1

\bigskip

\item \bgroup\bfseries\boldmath Si $\|\vv{u}\|=2$, $\|\vv{v}\|=5$  et $\big( \vv{u},\vv{v} \big)=\frac{2\pi}{3}$, alors:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[\checkbox{red}] $\prodscal{u}{v}=-5$
\item[$\square$] $\prodscal{u}{v}=5$
\item[$\square$] $\prodscal{u}{v}=-5\sqrt3$
\item[$\square$] $\prodscal{u}{v}=5\sqrt3$
\end{multicols}
\end{enumerate} % fin des propositions de la question 2

\bigskip

\item \bgroup\bfseries\boldmath Si \(\|\vv{u}\| = 4\), \(\|\vv{v}\| = 5\) et \(\cos(\vv{u}, \vv{v}) = \frac{1}{2}\), alors:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $\prodscal{u}{v} = 20$
\item[\checkbox{red}] $\prodscal{u}{v} = 10$
\item[$\square$] $\prodscal{u}{v} = -20$
\item[$\square$] $\prodscal{u}{v} = -10$
\end{multicols}
\end{enumerate} % fin des propositions de la question 3

\bigskip

\item \bgroup\bfseries\boldmath Si \(\|\vv{u}\| = 7\), \(\|\vv{v}\| = 3\) et \(\big( \vv{u},\vv{v} \big) = \frac{\pi}{3}\), alors:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $\prodscal{u}{v} = 21$
\item[\checkbox{red}] $\prodscal{u}{v} = \frac{21}{2}$
\item[$\square$] $\prodscal{u}{v} = -\frac{21}{2}$
\item[$\square$] $\prodscal{u}{v} = -21$
\end{multicols}
\end{enumerate} % fin des propositions de la question 4

\bigskip

\item \bgroup\bfseries\boldmath Si \(\|\vv{u}\| = 2\), \(\|\vv{v}\| = 3\) et \(\big( \vv{u},\vv{v} \big) = \frac{\pi}{4}\), alors quelle est la valeur de \(\|\vv{u} + \vv{v}\|\) ?\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[\checkbox{red}] $\sqrt{13 + 6\sqrt{2}}$
\item[$\square$] $\sqrt{13 - 6\sqrt{2}}$
\item[$\square$] $\sqrt{13 + 4\sqrt{2}}$
\item[$\square$] $\sqrt{13 - 4\sqrt{2}}$
\end{multicols}
\begin{tcolorbox}[info]
On utilise la formule: \( \|\vv{u}+\vv{v}\|^2 = \|\vv{u}\|^2 + 2\vv{u}\cdot\vv{v} + \|\vv{v}\|^2 \).
\end{tcolorbox}
\end{enumerate} % fin des propositions de la question 5


\newpage
\item \bgroup\bfseries\boldmath Si \(\|\vv{u}\| = 6\), \(\|\vv{v}\| = 4\) et \(\big( \vv{u},\vv{v} \big) = \frac{2\pi}{3}\), alors:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $\prodscal{u}{v} = 24$
\item[\checkbox{red}] $\prodscal{u}{v} = -12$
\item[$\square$] $\prodscal{u}{v} = 12$
\item[$\square$] $\prodscal{u}{v} = -24\sqrt{3}$
\end{multicols}
\end{enumerate} % fin des propositions de la question 6

\bigskip

\item \bgroup\bfseries\boldmath Si \(\|\vv{u}\| = 5\), \(\|\vv{v}\| = 2\) et \(\cos(\vv{u}, \vv{v}) = -\frac{1}{\sqrt{2}}\), alors:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $\prodscal{u}{v} = -5\sqrt{2}$
\item[$\square$] $\prodscal{u}{v} = 5\sqrt{2}$
\item[\checkbox{red}] $\prodscal{u}{v} = -5$
\item[$\square$] $\prodscal{u}{v} = 5$
\end{multicols}
\end{enumerate} % fin des propositions de la question 7

\bigskip

\item \bgroup\bfseries\boldmath Si \(\|\vv{u}\| = 3\), \(\|\vv{v}\| = 4\) et \(\big( \vv{u},\vv{v} \big) = \frac{2\pi}{3}\), alors quelle est la valeur de \(\|\vv{u} - \vv{v}\|\) ?\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] 5
\item[$\square$] 13
\item[$\square$] 0
\item[\checkbox{red}] 37
\end{multicols}
\begin{tcolorbox}[info]
On utilise la formule: \( \|\vv{u}-\vv{v}\|^2 = \|\vv{u}\|^2 - 2\vv{u}\cdot\vv{v} + \|\vv{v}\|^2 \).
\end{tcolorbox}
\end{enumerate} % fin des propositions de la question 8

\bigskip

\item \bgroup\bfseries\boldmath Si \(\|\vv{u}\| = 8\), \(\|\vv{v}\| = 5\) et \(\big( \vv{u},\vv{v} \big) = \frac{5\pi}{6}\), alors:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $\prodscal{u}{v} = -20\sqrt{3}$
\item[$\square$] $\prodscal{u}{v} = 20\sqrt{3}$
\item[\checkbox{red}] $\prodscal{u}{v} = -20$
\item[$\square$] $\prodscal{u}{v} = 20$
\end{multicols}
\end{enumerate} % fin des propositions de la question 9

\bigskip

\item \bgroup\bfseries\boldmath Si $\vv{u}\cdot\vv{v}=5$, que vaut $\big(-2\vv{u}\big)\cdot\big(3\vv{v}\big)$ ?\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $-5$
\item[$\square$] $5$
\item[\checkbox{red}] $-30$
\item[$\square$] $30$
\end{multicols}
\begin{tcolorbox}[info]
On utilise ici la propriété de bilinéarité du produit scalaire: $\big(k\vv{u}\big)\cdot\big(k'\vv{v}\big)=kk'\vv{u}\cdot\vv{v}$.
\end{tcolorbox}
\end{enumerate} % fin des propositions de la question 10

\bigskip

\item \bgroup\bfseries\boldmath Si $\vv{u}\cdot\vv{v}=-2$, que vaut $\big(5\vv{u}\big)\cdot\big(-2\vv{v}\big)$ ?\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $6$
\item[$\square$] $-6$
\item[$\square$] $20$
\item[\checkbox{red}] $20$
\end{multicols}
\end{enumerate} % fin des propositions de la question 11

\bigskip

\item \bgroup\bfseries\boldmath Si $ABCD$ est un carré de côté 1 et de centre $O$, alors $\prodscal{AB}{AO}=\ldots$:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[\checkbox{red}] $\frac{1}{2}$
\item[$\square$] 1
\item[$\square$] $-\frac{1}{2}$
\item[$\square$] $-1$
\end{multicols}
\begin{tcolorbox}[info]
On peut utiliser le projeté orthogonal $H$ de $O$ sur $[AB]$, qui est en son milieu.\\Ainsi, $\prodscal{AB}{AO}=AB \times AH=1\times\frac{1}{2}=\frac12$.
\end{tcolorbox}
\end{enumerate} % fin des propositions de la question 12


\newpage
\item \bgroup\bfseries\boldmath Si $A\coord{-1}{2}$ et $B\coord{3}{1}$ dans un repère orthonormé d'origine $O$, alors $\prodscal{OA}{OB}=\ldots$:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] 1
\item[$\square$] 0
\item[\checkbox{red}] $-1$
\item[$\square$] $-2$
\end{multicols}
\end{enumerate} % fin des propositions de la question 13

\bigskip

\item \bgroup\bfseries\boldmath Si $A\coord{-1}{2}$, $B\coord{3}{1}$ et $C\coord{1}{-2}$ dans un repère orthonormé, alors $\prodscal{CA}{CB}=\ldots$:\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[\checkbox{red}] 8
\item[$\square$] 7
\item[$\square$] $-8$
\item[$\square$] $-16$
\end{multicols}
\end{enumerate} % fin des propositions de la question 14

\bigskip

\item \bgroup\bfseries\boldmath Si $\vv{u}\dbinom{x}{3}$ et $\vv{v}\dbinom{-5}{2}$, quelle doit être la valeur de $x$ pour que ces vecteurs soient orthogonaux?\egroup
\begin{enumerate}
\begin{multicols}{2}
\item[$\square$] $x=\frac56$
\item[\checkbox{red}] $x=1,2$
\item[$\square$] $x=-1,2$
\item[$\square$] $x=-\frac56$
\end{multicols}
\end{enumerate} % fin de la question 15
\end{enumerate} % fin du qcm
\end{document}