I am trying to show that the following identity is true.
$$\Omega\times\omega=(\omega\cdot\nabla)(\Omega\times r)$$
I've tried using vector algebra and index notation to show that this is true with no luck.
I expanded the right side using the identity $$\nabla(A\cdot B)=(A\cdot\nabla)B+(B\cdot\nabla)A+A\times(\nabla\times B)+B\times(\nabla\times A)$$ and I got this.
$$\nabla(\omega\cdot (\Omega\times r))=(\omega\cdot\nabla)(\Omega\times r)+((\Omega\times r)\cdot\nabla)\omega+\omega\times(\nabla\times (\Omega\times r))+(\Omega\times r)\times(\nabla\times \omega)$$
Then I expanded the third term on the right using the identity $$\nabla\times(A\times B)=A(\nabla\cdot B)+(B\cdot\nabla)A-B(\nabla\cdot A)-(A\cdot\nabla)B$$
and I got this.
$$\nabla(\omega\cdot (\Omega\times r))=(\omega\cdot\nabla)(\Omega\times r)+((\Omega\times r)\cdot\nabla)\omega+(\omega\times\Omega)(\nabla\cdot r)+\omega\times(( r\cdot\nabla)\Omega)-(\omega\times r)(\nabla\cdot\Omega)-\omega\times((\Omega\cdot\nabla) r)+(\Omega\times r)\times(\nabla\times \omega)$$
In my problem $\Omega$ is a constant vector so any derivatives of it should go to zero.
$$\nabla(\omega\cdot (\Omega\times r))=(\omega\cdot\nabla)(\Omega\times r)+((\Omega\times r)\cdot\nabla)\omega+(\omega\times\Omega)(\nabla\cdot r)-\omega\times((\Omega\cdot\nabla) r)+(\Omega\times r)\times(\nabla\times \omega)$$
At this point I'm stuck and not sure whether I'm doing something wrong or if I haven't gone far enough. Any help would be appreciated.
Since this is a Fluid Dynamics problem, why not just do the following: $$\Omega\times r=\begin{pmatrix}\Omega_yz-\Omega_z y\\ \Omega_zx-\Omega_x z\\ \Omega_xy-\Omega_y x\\ \end{pmatrix}$$ And: $$\omega \cdot \nabla=\omega_x\frac \partial{\partial x}+\omega_y\frac \partial{\partial y}+\omega_z\frac \partial{\partial z}$$
Since $\Omega_x,\Omega_y,\Omega_z$ are constant, we get: $$(\omega \cdot \nabla)(\Omega\times r)=\begin{pmatrix}\Omega_y\omega_z-\Omega_z \omega_y\\ \Omega_z\omega_x-\Omega_x\omega_ z\\ \Omega_x\omega_y-\Omega_y \omega_x\\ \end{pmatrix}$$
Which is obviously $\Omega \times \omega$
P.S. : I am fully aware that this is a very particular and simplistic way of proving the identity, but given the context I assume this is appropriate...