by $A\times B \times C = (A \cdot C)B-(A \cdot B)C$, i need to expand $n \times \bigtriangledown \times n$, where all of these are vectors.
Here is what i have right now
$n \times \bigtriangledown \times n = (n \cdot n) \bigtriangledown- (n \cdot \bigtriangledown) n$
I don't know how operator algebra works in order of operation, so i'm assuming right now this means
$\bigtriangledown (||n||^2)- n(\bigtriangledown \cdot n)$
Is this correct? This is part of a huge problem, i'm just trying to make sure how operators work in these operations
Ok, i recently edited, the problem i'm trying to solve is $(n \times \bigtriangledown \times n)^2$, maybe i will post the exact equations in another question.. This equation does not operate on anything, it becomes a scalar when squared
Cross product not associative, so you need to parenthesize as $$A \times (B \times C)=(A\cdot C)B -(A\cdot B)C.$$ Note that the "multiplications" on the right are scalar multiplications, since each of $A \cdot C$ and $A \cdot B$ are scalars.
The operator $\bigtriangledown$ makes sense only (in your three dimensional case) when applied to a function $f(x,y,z)$ of three variables, and then $\bigtriangledown f$ is the vector $(f_x,f_y,f_z)$ of the partial derivatives of $f$ with respect to the variables $x,y,z$ (in that order).
So if you are considering $n \times (\bigtriangledown \times n),$ it could be interpreted as another operator, which can be computed by the formula you suggest but without the extra step. Namely you can say $$[n \times (\bigtriangledown \times n)](f)=(n \cdot n)\bigtriangledown f -(n \cdot \bigtriangledown f) n,$$ but the initial "factor" $n \cdot n$ here remains as a scalar multiplier to the evaluated operator $\bigtriangledown f,$ which is a vector valued function of three variables. As mentioned above the latter is a vector of partials of the given function $f$ of three variables which the whole expression is acting on. The second scalar factor $u \cdot \bigtriangledown f$ would be then also a scalar function of the three variables $x,y,z$ since $ \bigtriangledown f$ is a 3-vector and $n$ is also.