Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cross product of a vector and its derivative

  1. Jul 8, 2009 #1
    Hello guys, this must be a very trivial question, but I just don't see it.

    Ihave a "vectorized quaternion" (quaternion with a null scalar part) :

    [tex]u(t) = (0, u_x(t), u_y(t), u_z(t) )[/tex]

    We also have it's time derivative :

    [tex] \frac{d}{dt}u(t) = \dot{u}(t) = (0, \dot{u}_x(t), \dot{u}_y(t), \dot{u}_z(t) )[/tex]

    If I'm not mistaken, the scalar product of them should be :

    [tex]\dot{u}(t)\cdot u(t) = 0[/tex]

    But what would the cross product look like? I know I can write it this way :

    [tex]\dot{u}(t)\times u(t) = (0, \dot{u}_y(t) u_z(t) - \dot{u}_z(t) u_y(t), \dot{u}_x(t) u_z(t) - \dot{u}_z(t) u_x(t), \dot{u}_x(t) u_y(t) - \dot{u}_y(t) u_x(t) )[/tex]

    Is there a simpler answer? Anything wrong?

    Thank you

    EDIT : argh, sorry, it's prolly in the wrong subforum
     
    Last edited: Jul 8, 2009
  2. jcsd
  3. Jul 8, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This may be where you are going wrong. The dot product of v and v' is 0 only if v has constant length.

     
  4. Jul 8, 2009 #3
    Oh yeah, I'm sorry, I forgot to say that those are rotation quaternions (rotation vector) on a unit sphere, so indeed, the length is constant
     
    Last edited: Jul 8, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cross product of a vector and its derivative
Loading...