Hello guys, this must be a very trivial question, but I just don't see it.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Cross product of a vector and its derivative

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