Solving Vector Field for Independence of z

[Matt atkinson]

Homework Statement

A vector field $$ \vec{u}=(u_1,u_2,u_3) $$
satisfies the equations;
$$ \Omega\hat{z} \times \vec{u}=-\nabla p , \nabla \bullet \vec{u}=0$$
where p is a scalar variable, \Omega is a scalar constant. Show that \vec{u} is independant of z.
Hint ; how can we remove p from the equations

Homework Equations

Included above in question.

The Attempt at a Solution

I know that it means that \vec{u} doesn't have a z component and therefore is only described by x,y but I have no idea where to begin.
I tried removing p but I can't.

[edit]
I have made some progress I took the curl of the longer equation and got rid of \nabla p using the curl of a scalar gradient = 0, but from then I just have ;
\nabla \times ( \vec{u} \times \Omega \hat{z})=0

 

[vela]

How did you try removing p?

 

[Matt atkinson]

How did you try removing p?

I just updated it then because I realized I hadn't included what I'd done, kinda jumped the gun a little :) sorry, not sure where to go from there now though.

 

[vela]

Your textbook should have a list of vector identities. (If not, google it.) You can rewrite the curl of a cross product in a different form that'll let you use the fact that the divergence of u is 0.

 

[Matt atkinson]

\nabla \times ( \vec{u} \times \Omega \hat{z}) = (\Omega \hat{z} \bullet \nabla)\vec{u} - (\vec{u} \bullet \nabla)\Omega \hat{z} +\vec{u}(\nabla \bullet \Omega \hat{z}) - \Omega \hat{z}(\nabla \bullet \vec{u})
This one? and then the turns with \nabla \bullet \vec{u} cancel?
thanks for your help I think I Know what to do from here. If that's right ;D

 

[vela]

Yup, looks good.

 

[Matt atkinson]

Thanks a lot vela