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Find the vector

  1. Sep 1, 2007 #1

    mrjeffy321

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    I am given the following information,

    A x X = B
    AX = phi

    Where A, B, and X are vectors and phi is a scalar. A cross product is indicated by “x” and “●” indicates a dot product.

    I am told that A, B, and phi are ‘known’ to me by the question and my goal is to find the vector X. I am told that my answer should be in terms of A, B, phi, and the magnitude of A.


    It seems like the most obvious place to start would be to use the information given in the problem and find out what B and phi are equal to in terms of A and X.
    By performing the cross and dot products, I obtain the following,

    Bx = (Ay * Xz) – (Az * Xy)
    By = - (Ax * Xz) + (Az * Xx)
    Bz = (Ax * Xy) – (Ay * Xx)

    B = <Bx, By, Bz>

    Phi = (Ax * Xx) + (Ay * Xy) + (Az * Xz)

    Although the above is true, it does not look very useful, so it might also be beneficial to find B and phi using idical / summation notion,

    Bi = Σ εijk * Aj * Xk
    Where the summation is taken over the summing indices of j and k and I refers to the component of B.

    Phi = Σ Ai * Xi
    Where i is the summing index.


    But after doing this, I am at a loss as to how to proceed in order to solve for X.
     
  2. jcsd
  3. Sep 1, 2007 #2

    Hurkyl

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    It's four linear equations in three unknown variables, isn't it? Haven't you studied how to solve such systems of equations?


    I do wonder if a more direct calculation can be made using a triple product identity:
    http://mathworld.wolfram.com/VectorTripleProduct.html
    http://mathworld.wolfram.com/ScalarTripleProduct.html

    Linear algebra is so much nicer when you manipulate vectors directly, rather than components of coordinate representations.
     
    Last edited: Sep 1, 2007
  4. Sep 1, 2007 #3

    mrjeffy321

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    Yes, I have 4 equations and only 3 unknowns, so I should indeed be able to solve this algebraically as you suggested. Doing this is how I first thought to solve the problem.
    But what got me a little confused was why I needed to know the information about the dot product and phi, and why I would need to express my answer in terms of phi and the magnitude of A. Theoretically I could do it all using only A and B.
    I only assumed that there was something perhaps, a more elegant / cleaner solution if I used a different approach, since that is what the question wording seems to imply.
     
  5. Sep 1, 2007 #4

    Hurkyl

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    Because the problem asks you to do so. :smile: Basically, it's often easier to understand things when expressed in terms of 'geometric' quantities. To say the same thing differently, coordinates tend to obscure meaning.


    If you grind out the solution by solving the system of equations, isn't it plausible that you might be able to see how to arrange the answer in the desired form? You won't know unless you try. :smile:


    Incidentally, my rambling wasn't idle speculation... :wink:
     
    Last edited: Sep 1, 2007
  6. Sep 1, 2007 #5

    mrjeffy321

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    I have a solution for X…it is not pretty of course, but it works.

    If you are hinting at anything in particular with the triple scalar product, I don’t think I am seeing it.
     
  7. Sep 1, 2007 #6

    Hurkyl

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    You only really have two tools in your vector algebra toolbox: dot products, and cross products. You should experiment and see what you can produce by using your tools, and the triple product identities are useful for looking at the results, which is why I pointed at them.

    I hate giving spoilers, but my solution only involved the vector triple product. (Of course, some other solution might use the scalar triple product)
     
  8. Sep 2, 2007 #7

    HallsofIvy

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    A x X = B
    A ● X = phi
    It is also true that A ● X= |X||A|cos(phi) and that |A x X|= |X||A|sin(phi). Can you use those to find |A|, |X|, and phi?
     
  9. Sep 2, 2007 #8

    mrjeffy321

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    But phi is just some scalar in the problem, not the angle in between the vectors A and X, which I will call θ.

    A x X = B,
    |B| = |A| * |X| * sin (θ)
    A ● X = Phi
    Phi = |A| * |X| * cos (θ)


    Using the Vector triple product I got an interesting result,

    (A x B) x X = -A(B ● X) + B(A ● X)
    B ● X = 0 since B and X are perpendicular (A x X = B), and A ● X = phi, so we re-write the equation to be,
    (A x B) x X = B * phi
    But A x B should be X, or at least in the same direction as X, so this cross X will be zero, so now we have,
    0 = B * phi
    Which will be true when either B, Phi, or both are zero.

    In order for B to be the zero vector, A and X would need to be parallel, or one of them also the zero vector.
    In order for Phi to be zero, A and X would need to be perpendicular, or one of them be the zero vector.
    Lots of possibilities, but I don’t have a lot of confidence in this result.
     
  10. Sep 2, 2007 #9

    Hurkyl

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    Not true! Clearly X is perpendicular to B, but why should it be perpendicular to A? But maybe A x B is still interesting; have you tried computing that?



    Incidentally, if you didn't realize geometrically that B.X=0, you still could have worked that out from the scalar triple product:

    B . X = (A x X) . X = (X x X) . A = 0 x A = 0
     
    Last edited: Sep 2, 2007
  11. Sep 2, 2007 #10
    I got:

    [tex]\left| X\right| =\frac{\sqrt{{B}^{2}+{\phi}^{2}}}{\left| A\right| }[/tex]

    and theta = atan(mag (B)/phi)

    using this I can get vector X.
    but please check this for me.

    thanks a lot!

    oops, a minor mistake
     
    Last edited: Sep 2, 2007
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