physics [vectors]

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The concept i get, but somehow i just can't execute this problem, please help me!

You are given vectors A = 5.5 6.2 and B = - 3.7 7.4 . A third vector C lies in the xy-plane. Vector C is perpendicular to vector A and the scalar product of C with B is 19.0.

I need x and y components for vector C

  • physics [vectors] -

    A = 5.5 i + 6.2 j
    B = -3.7i + 7.4 j

    C = x i + y j

    A dot C = |A| |C| cos theta
    if perpendicular theta is 90 degrees and cos theta = 0
    so
    5.5 x + 6.2 y = 0

    B dot C = 19
    so
    -3.7 x + 7.4 y = 19.0

    solve those two equations for x and y

  • physics [vectors] -

    5.5(_3.7x+7.4y=19)
    3.7(5.5X+2.2y=0)
    (20.35x+22.94y=0)
    + (_20.35x+40.70y=384.65)
    Y=384.68/63.64
    X=_(22.94(384.68/63.64)/20.35)

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