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the vector a-5b and a-b are perpendiculat. If vector a and b are unit vectors, then determine axb

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Marie

  1. If this is in just two dimensions:
    A = ax i + ay j
    B = bx i + by j

    C = A-5B = (ax-5bx) i + (ay-5by)j
    D = A- B = (ax- bx) i + (ay- by)j

    if perpendicular C dot D = 0
    (ax-5bx)(ax-bx)+(ay-5by)(ay-by) = 0
    ax^2 -6ax bx + 5bx^2 +ay^2 -6ay by +5 by^2 = 0
    but unit vectors so ax^2+ay^2 = 1
    and
    5(bx^2+by^2) = 5
    so
    6 - 6 ax bx -6 ay by = 0
    so
    ax bx + ay by = 1

    Now we want to know A x B
    determinant of
    i j k
    ax ay 0
    bx by 0
    = (ax by - ay bx)k
    but
    ax bx + ay by = 1
    ay = (1-ax bx)/bx
    so cross product is
    (ax by - 1+ax bx)k
    = [ax(bx+by)-1] k

    Check carefully, I went pretty fast and there may be errors.

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    Damon

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