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Lavabo
@Lavabo
May 2019
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Bonjour, voilà un exercice court de 2nde sur la colinéarité de vecteurs. Si quelqu'un peut m'aider j'accepte volontiers merci :)
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Bonsoir
AI = 1/3 AB donc AB = 3AI et on inverse BA= -3 AI
d'apres la relation de chasles : BJ = BC + CJ, Alors BJ= BI + IC + CJ
(AI =1/3 AB dc BI = 2/3 BA et AJ= 3 AC donc CJ = 2AC)
ALors: BJ= 2/3BA + IC + 2AC
Comme prouvé precedemment BA = -3AI donc:
BJ= 2/3 (-3AI) + IC + 2AC
BJ= -2 AI + IC + 2(AI+IC)
BJ= -2AI + IC + 2AI +2IC
BJ= 3IC
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AI = 1/3 AB donc AB = 3AI et on inverse BA= -3 AI
d'apres la relation de chasles : BJ = BC + CJ, Alors BJ= BI + IC + CJ
(AI =1/3 AB dc BI = 2/3 BA et AJ= 3 AC donc CJ = 2AC)
ALors: BJ= 2/3BA + IC + 2AC
Comme prouvé precedemment BA = -3AI donc:
BJ= 2/3 (-3AI) + IC + 2AC
BJ= -2 AI + IC + 2(AI+IC)
BJ= -2AI + IC + 2AI +2IC
BJ= 3IC