Dados os pontos (-1,3) (2, 5) 3, -1) de o vetor: 3AB+4CB
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A = (-1,3)
B = (2,5)
C = (3,-1)
3AB + 4CB = 3(-1,3)(2,5) + 4(3,-1)(2,5) = (-3,9)(2,5) + (12,-4)(2,5) = (-6,45) + (24,-20) = (18,20)
3AB + 4CB = (18,20)
B = (2,5)
C = (3,-1)
3AB + 4CB = 3(-1,3)(2,5) + 4(3,-1)(2,5) = (-3,9)(2,5) + (12,-4)(2,5) = (-6,45) + (24,-20) = (18,20)
3AB + 4CB = (18,20)
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