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Homework Help: Simple vector help

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data
    vector a = (4i+3j-2k)
    vector b = (2i-3j+2k)

    1. a x b
    2. 3a x 2b
    3. |3a x 2b|

    3. The attempt at a solution

    1. -12j - 18k

    2. 6(a x b)
    6(-12j - 18k)
    -48j -108k

    3. |6(a x b)|
    sqrt(48^2 + 108^2)


    idk if im using identities wrong or if im way off, thx for confirmation on right/wrong and any help
  2. jcsd
  3. Sep 5, 2012 #2


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    Homework Helper

    Hi oreosama, welcome to PF. Your work is correct except a siliy mistake, How much is 12*6?

  4. Sep 5, 2012 #3
    thanks for that.

    using the same vectors


    the vectors a and b define a plane surface. determine a possible vector perp. to that surface. determine a possible vector parallel to that surface.

    perp: a x b= -12j -18k

    assuming that's right..

    parralel: I have no idea :(

    i know parallel vectors just have scaler multiplier but im working with 2 vectors(of which im assuming intersect to create a plane?). im pretty confused at this point so im going to sleep and hope someone provides insight by the time i wake up. thanks for any help
  5. Sep 5, 2012 #4


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    Gold Member

    If <a> and <b> are contained within the plane, then a vector parallel to <a> or <b> will also be parallel to the plane.
    If two vectors are parallel, then their cross product is the zero vector. Therefore compute:
    [tex] \vec{a}\, ×\,\vec{c} = 0[/tex] where [itex] \vec{c} [/itex] is the vector you want. You can solve a linear system of 3 variables, and get a dependence on the components of <c>.
    Note: <b> can be used in place of <a>
  6. Sep 5, 2012 #5


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    Homework Helper

    The vectors a and b lay in the surface, and so do their linear combinations. All are parallel with the plane. (you can shift a vector parallel with itself, it is the same vector.)

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