We know that:
$\nabla (Α\cdotΒ)=Α\times (\nabla \times Β)+Β\times (\nabla \times Α)+(Α\cdot\nabla )Β+(Β\cdot\nabla )Α$
Where $A$ and $B$ are two vectors.
Now, suppose that the curl and divergence of vector $A$ equals to zero. That is- $\nabla\times A=0$ and $A\cdot\nabla=0$. Also, let $\nabla\times B = 0$ and $B\cdot\nabla=n$, where $n$ is a positive number.
This means that vector $A$ does not changes with direction, while vector $B$ has only some positive divergence but no curl.
So, our original identity becomes: $\nabla (Α\cdotΒ)=(Β\cdot\nabla )Α$
We now notice that $\nabla (Α\cdotΒ)$, a vector, has its direction in the direction of vector $A$. It is counter intuitive to me because it is $B$ which is changing, not $A$.
So, my question is- is the last equation correct? Will the direction of $\nabla (Α\cdotΒ)$ be in the direction of $A$, in the conditions I have previously described? Or I am interpreting something wrongly?