www.mathsenligne.com 2G1 - VECTEURS EXERCICES 3B EXERCICE 3B.1
A l’aide de la relation de Chasles, écrire sous forme d’un seul vecteur… si c’est possible : 1. \s\up10(¾® +
\s\up10(¾® = 2. \s\up10(¾® +
\s\up10(¾® = 3. \s\up10(¾® –
\s\up10(¾® = 4. \s\up10(¾® –
\s\up10(¾® = 5. \s\up10(¾® +
\s\up10(¾® = 6. \s\up10(¾® +
\s\up10(¾® = 7. \s\up10(¾® –
\s\up10(¾® = 8. -\s\up10(¾® –
\s\up10(¾® = EXERCICE 3B.2
Ecrire plus simplement les vecteurs suivants, en utilisant la relation de Chasles :
\s\up10(® =
\s\up10(¾® +
\s\up10(¾® +
\s\up10(¾®
\s\up10(® =
\s\up10(¾® +
\s\up10(¾® +
\s\up10(¾®
\s\up10(® =
\s\up10(¾® +
\s\up10(¾® +
\s\up10(¾®
\s\up10(® = \s\up10(¾® +
\s\up10(¾® + \s\up10(¾® +
\s\up10(¾®
EXERCICE 3B.3
Ecrire plus simplement les vecteurs suivants, en transformant les soustractions en addition de l’opposé, puis en utilisant la relation de Chasles :
\s\up10(® =
\s\up10(¾® –
\s\up10(¾®
\s\up10(® =
\s\up10(¾® –
\s\up10(¾® +
\s\up10(¾®
\s\up10(® = \s\up10(¾® +
\s\up10(¾® – \s\up10(¾®
+ \s\up10(¾®
\s\up10(® = 2\s\up10(¾®
– \s\up10(¾® –
\s\up10(¾® + \s\up10(¾®
EXERCICE 3B.4
Compléter les égalités vectorielles : 1. \s\up10(¾® = \s\up10(¾® +
\s\up10(¾® 2.\s\up10(¾® = \s\up10(¾® +
\s\up10(¾® 3.\s\up10(¾® = \s\up10(¾® +
\s\up10(¾®
4. \s\up10(¾® = \s\up10(¾® +
\s\up10(¾® 5.\s\up10(¾® = \s\up10(¾® +
\s\up10(¾® 6.\s\up10(¾® = \s\up10(¾® +
\s\up10(¾® + \s\up10(¾®
7. \s\up10(¾® = \s\up10(¾® +
\s\up10(¾® + \s\up10(¾® 8. \s\up10(¾® = \s\up10(¾® +
\s\up10(¾® + \s\up10(¾® +
\s\up10(¾®
9.\s\up10(¾® = \s\up10(¾® +
\s\up10(¾®+ \s\up10(¾®
EXERCICE 3B.5
a. Exprimer le vecteur \s\up10(® en fonction de \s\up10(¾® et \s\up10(¾®.
1. \s\up10(® = \s\up10(¾® 2.\s\up10(® = 2\s\up10(¾® +
\s\up10(¾® 3.\s\up10(® = 2\s\up10(¾® + 3\s\up10(¾® + \s\up10(¾®
b.Exprimer le vecteur \s\up10(® en fonction de \s\up10(¾® et \s\up10(¾®.
www.mathsenligne.com 2G1 - VECTEURS EXERCICES 3B 1. \s\up10(® = \s\up10(¾® +
\s\up10(¾® 2.\s\up10(® = \s\up10(¾® –
3\s\up10(¾® + \s\up10(¾® 3.\s\up10(® = 2\s\up10(¾® + 3\s\up10(¾® + \s\up10(¾®
