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Complex Geometry - Vectors and Paralellogram

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Lay down vector a, take its endpoint as the initial point of vector b, and complete the parallelogram. The area of this paralellogram taken with a certain sign is I(a conjugate)b. On what geometrical feature does the sign depend?



    2. Relevant equations



    3. The attempt at a solution
    My initial thought is that is has something to do with angles and arguements, but my problem is figuring out how the sign changes.
     
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  3. Feb 8, 2009 #2

    Office_Shredder

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    What is I(a conjugate)?
     
  4. Feb 8, 2009 #3
    I stands for imaginary like how in a+ib a is the Real part of z and b is the Imaginary part of z. a conjugate is like how the conjugate of a+ib is a-ib.
     
  5. Feb 8, 2009 #4

    Office_Shredder

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    Ah, I see now. Try putting a and b into polar form: rei*theta

    and the result will fall out pretty fast
     
  6. Feb 9, 2009 #5

    HallsofIvy

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    So you are thinking of these vectors as in the complex plane? That is, the vector <x, y> is represented by x+ iy?

    In three dimensions, the area of the parallelogram formed by vectors [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] is [itex]|\vec{u}\times\vec{v}|[/itex] and, of course, you need the absolute value because the sign of the cross product depends on the order of the vectors NOT on any property of the parallelogram. Looks to me like the same thing happens here: How does (conjugate a)b is differ from (conjugate b)a?
     
  7. Feb 9, 2009 #6
    I realized my problem. If I considered the angle [tex]\beta[/tex]-[tex]\alpha[/tex], then I was fine.
     
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