Resolução de vetores e geometria analitica

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Resolução de vetores e geometria analitica

  1. 1. Resolu¸˜o das quest˜es de Vetores e Geometria Anal´ ca o ıtica1. ( Aula 03 : Quest˜o 3)Prove que − + − , − − − = |− |2 − |− |2 para quaisquer vetores a → → → → u v u v → u → v 2 u, v ∈ R . Prova: Sejam u, v ∈ R2 , ent˜o a → → → → − + − ,− − − u v u v = − ,− − − + − ,− − − → → → u u v → → → v u v → → − ,− − − ,− + − ,− − − ,− = u u → → u v → → v u → → v v = − ,− − − ,− + − ,− − − ,− → → u u → → u v → → u v → → v v =|u→ − |2 − |− |2 . → v √ →1. (Aula 03 : Quest˜o 5) Sabendo que |− | = 2, |− | = 3 e que − e − formam um ˆngulo de a → u v → → u v a 3 π, determine: 4 (a) | 2− − − , − − 2− | → → → u v u →v Solu¸˜o: Temos que − , − = |− ||− | cos θ. Assim, ca → → u v → → √ v u − , − = √2.3. cos 3 π = −√2.3. 2 = −3. → → u v 4 2 − → → 2− − − , − − 2− = 2− , − − 2− + −v, − − 2− → → → u v u → v → → u u → v u →v → → − ,− − 4 − ,− − − ,− + 2 − ,− =2 u u → → u v → → v u → → v v = 2|− |2 − 5 − , − + 2|− |2 → u √ → → u v → v = 2( 2)2 − 5.(−3) + 2.32 = 4 + 15 + 18 = 37. (b) |− − 2− | → u →v Solu¸˜o: Note que, ca u v → − → u → |− − 2− |2 = − − 2v, − − 2− → → u → v = − , − − 2− + −2− , − − 2− → → u u → v → → v u → v → → − ,− − 2 − ,− − 2 − ,− + 4 − ,− = u u → → u v → → v u → → v v → − |2 − 4 − , − + 4|− |2 = |√ u → → u v → v = ( 2)2 − 4.(−3) + 4.32 = 2 + 12 + 36 = 50 √ Portanto |− − 2− | = 50. → u → v Tutora: Thamires dos Santos 1

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