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Vectors

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A=i –2j B=-3i+j How do I find the component of A that is perpendicular to B in vector form?

  • Vectors - ,

    A = i - 2 j so |A| = sqrt 5
    B = -3 i + j so |B| = sqrt 10

    The magnitude of the cross product A x B is |A| |B| sin T
    where T is the angle between the vectors.
    i j k
    1 -2 0
    -3 1 0 --> -5 k
    so the magnitude of the cross product is 5
    so
    sqrt 5 * sqrt 10 * sin T = 5
    5 sqrt 2 * sin T = 5
    sin T = 1/sqrt 2
    The component of A perpendicular to B, call it R, therefore has magnitude sqrt 5 /sqrt 2
    We know that it is perpendicular to B so
    B dot R has magnitude = 0
    so
    Rx Bx + RyBy = 0
    Rx(-3)+ Ry(1) = 0
    so Ry = 3 Rx
    so call it
    R = c Rx + 3c Ry where c is a constant
    magnitude = c sqrt (10)
    which has to be sqrt 5/sqrt 2
    so c sqrt 10 = sqrt 5/sqrt 2
    c = sqrt 5 /(sqrt 2*sqrt 2 * sqrt 5)
    c = 1/2 but it could be negative
    so our vector is
    R = (1/2) i + (3/2) j
    or
    R = -(1/2) i - (3/2) j
    looking at a sketch of the vectors, use the second one but both are valid.

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