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Vector products -Finding the direction of angular momentum

  1. Sep 21, 2008 #1
    How do you resolve a vector products when the two vectors lie in different planes, i.e. how do you find the vector product of a vector that lies in the zx plane and a vector that lies in the xy plane. Using the right hand rule for vectors is redundant as the two vectors lie in different planes.

    Can anyone help resolve this problem. Thanks
     
  2. jcsd
  3. Sep 21, 2008 #2
    The specific problem I have is finding out what the torque is, about the origin O, due to a force, of magnitude 2.0N on a Point A, given by the position vector r in the xz plane where r is equal to 3.0m and the angle it makes is 30 degrees. The force is directed along the y axis. What is the direction of the torque? I can't use the right hand rule because the vector products are in a different plane.
     
  4. Sep 21, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi ctpengage! Welcome to PF! :smile:

    Two vectors (from the same point) are always in a common plane. :smile:

    In this case, the common plane is sloping at 30º to the "horizontal", and goes through the whole y-axis. :wink:
     
  5. Sep 21, 2008 #4
    Re: Welcome to PF!

    How do you use the right hand rule in that case?
     
  6. Sep 21, 2008 #5

    tiny-tim

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    Hi ctpengage! :smile:

    I'm wondering what version of the right hand rule you've been taught. :confused:

    The one I know is where you point your right thumb along one vector, your right index finger along the other vector, and then the cross-product is where your middle finger is pointing.

    If you use that, there should be no difficulty. :wink:
     
  7. Sep 21, 2008 #6
    How do I use it to solve the problem given above? In that the force vector is directed perpendicular to the position vector? As said two vectors from a point lie in a single plane. But how do you figure it out with respects to the x, y and z axes? Hopefully you'll understand what I mean.
     
  8. Sep 21, 2008 #7

    tiny-tim

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    It's supposed to be!

    The easiest case is where they are perpendicular!
    Nooo … :confused:

    If you know a coordinate method, you just use that.

    If you just know the ordinary vector method, then just remember that the cross-product must be perpendicular to that sloping plane I mentioned earlier. :smile:

    (which automatically makes it perpendicular to the original two vectors :wink:)
     
  9. Sep 21, 2008 #8

    HallsofIvy

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    The force is along the y axis and its magnitude is 2.0N. Okay, write it as <0, 2.0, 0>.

    The position vector (assuming you mean "that angle it makes with the x-axis is 30 degrees) is <3.0cos(30), 0, 3.0sin(30)> = <[itex]\3\sqrt{3}/2, 0, 3/2>. The torque is the cross product of those two vectors.
     
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