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Show that this set of equations defines a unique vector

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that, given vectors A and B:

    A * X = |a|
    and
    A x X = B

    defines X as a unique vector

    2. Relevant equations



    3. The attempt at a solution

    No idea.

    My only conclusion thus far is that the magnitude of x is equal to 1/cos(theta) where theta is the angle it makes with vector A.

    Any ideas?
     
  2. jcsd
  3. Sep 9, 2012 #2

    SammyS

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    It appears that you're being a bit careless with upper/lower case.

    If you mean AX = |A|,

    then, yes, the magnitude of vector, X, is given by |X| = 1/cos(θ) .

    ******************

    So, it looks like you used CD = |C| |D| cos(θ) .

    Now use that the magnitude of the vector product is:
    |C × D| = |C| |D| sin(θ)

     
  4. Sep 9, 2012 #3
    Sorry! Yes, I tried to make all vectors upper case afterwards to prevent confusion with the cross product.

    I have also noted the magnitude of the cross product to be |A||X|sin(theta) = |B| but I am not sure how this relates to finding anything out about X, since I already know its magnitude based on theta above, and I know nothing about B other than it being perpendicular to both A and X.
     
  5. Sep 9, 2012 #4

    SammyS

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    And what is |X| ?
     
  6. Sep 9, 2012 #5
    Ah!

    So then that results in
    Tan(theta) = |B|/|A|

    So it's direction relative to A is
    arctan(|B|/|A|)

    And it's magnitude is
    1/cos(theta)
    Which is of course:

    1/cos(arctan(|B|/|A|)

    Geometrically, I figure that cos(arctan(x)) = (x^2 + 1)^(-1/2)
    So 1/cos(arctan(x)) = (x^2 + 1)^(1/2)

    Then letting x = |B|/|A|,

    The magnitude must be
    [itex]\sqrt{\frac{|B|^{2}}{|A|^{2}} + 1}[/itex]

    And its direction relative to A must be
    [itex]arctan(\frac{|B|}{|A|})[/itex]


    Hm... but arctan will give the same thing for some angle theta, and 180 - theta. So it's still not unique.. this must be resolved some way?


    Thanks for the help!
     
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