Cross product of a vector and its derivative

[nburo]

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) :

$$u(t) = (0, u_x(t), u_y(t), u_z(t) )$$

We also have it's time derivative :

$$\frac{d}{dt}u(t) = \dot{u}(t) = (0, \dot{u}_x(t), \dot{u}_y(t), \dot{u}_z(t) )$$

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

$$\dot{u}(t)\cdot u(t) = 0$$

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

$$\dot{u}(t)\times u(t) = (0, \dot{u}_y(t) u_z(t) - \dot{u}_z(t) u_y(t), \dot{u}_x(t) u_z(t) - \dot{u}_z(t) u_x(t), \dot{u}_x(t) u_y(t) - \dot{u}_y(t) u_x(t) )$$

Is there a simpler answer? Anything wrong?

Thank you

EDIT : argh, sorry, it's prolly in the wrong subforum

 

[HallsofIvy]

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) :

$$u(t) = (0, u_x(t), u_y(t), u_z(t) )$$

We also have it's time derivative :

$$\frac{d}{dt}u(t) = \dot{u}(t) = (0, \dot{u}_x(t), \dot{u}_y(t), \dot{u}_z(t) )$$

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

$$\dot{u}(t)\cdot u(t) = 0$$

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

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

$$\dot{u}(t)\times u(t) = (0, \dot{u}_y(t) u_z(t) - \dot{u}_z(t) u_y(t), \dot{u}_x(t) u_z(t) - \dot{u}_z(t) u_x(t), \dot{u}_x(t) u_y(t) - \dot{u}_y(t) u_x(t) )$$

Is there a simpler answer? Anything wrong?

Thank you

EDIT : argh, sorry, it's prolly in the wrong subforum

 

[nburo]

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

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