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Imaginary vectors

  1. Jul 25, 2005 #1
    here's a questn:
    "2 vectors acting in same direction have resultant 20 whereas in perpendicular direction resultant is 10. find the vectors."

    pls. explain the imaginary solution.
  2. jcsd
  3. Jul 25, 2005 #2


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    Set up your system of equations.

    [tex] |\vec{a}| \hat{u_{1}} + |\vec{b}| \hat{u_{1}} = 20\hat{u_{1}} [/tex]

    [tex] |\vec{a}| \hat{u_{1}} + |\vec{b}| \hat{u_{2}} = 10 \hat{u_{3}} [/tex]


    [tex] \hat{u_{3}} = \cos \theta \hat{u_{1}} + \sin \theta \hat{u_{2}} [/tex]

    where [itex] \hat{u_{1}} [/itex], [itex] \hat{u_{2}} [/itex], [itex] \hat{u_{3}} [/itex] are unit vectors, and [itex] \theta [/itex] is the angle between [itex] \hat{u_{1}} [/itex] and [itex] \hat{u_{3}} [/itex].
    Last edited: Jul 25, 2005
  4. Jul 25, 2005 #3

    Meir Achuz

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    In your case, the two equations are
    This leads to a quadratic equation with (as you say) no real solution.
    This means that you have described an impossible situation.
  5. Jul 25, 2005 #4

    Not impossible, just imaginary. The solution lies in the field
    of complex numbers i.e. complex vectors.
  6. Jul 25, 2005 #5
    This equivalent to "construct a right triangle so that the sum of legs is twice that of the hypotenuse". This is equivalent to the equation a + b = 2c which would suggest that a^2 + b^2 =4c^2-2ab. The only way this can be consistent (given the pythagorean relation is if) c^2=4c^2 - 2ab or ab =3/2 c^2 which is a violation of the CS inequality, so there is no solution.
  7. Jul 25, 2005 #6
    I didn't want to solve this kid's problem, Crosson but you forced me to.

    [tex] \vec{v1} = \hat{x} [ \sqrt{50} (\sqrt{2} + i) ] [/tex]
    [tex] \vec{v2} = \hat{a} [ \sqrt{50} (\sqrt{2} - i) ] [/tex]

    [tex] \hat{a} = \hat{x} [/tex] in the parallel case and
    [tex] \hat{a} = \hat{y} [/tex] in the perpendicular case.

    Edit: The solution is not unique, I just picked one.
    Last edited: Jul 25, 2005
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