Cross product with orthonormal basis Let ${\{u_1,u_2,\ldots,u_n}\}$ be an ortohonormal basis of $\mathbb{R^n}$ and $f,g:\mathbb{R}^n\to\mathbb{R}$ differtinable functions at $p\in\mathbb{R^n}$.
If $n=3$ does
$\nabla{f(p)}\times\nabla{g(p)}=(f_{u_1}(p),f_{u_1}(p),f_{u_3}(p))\times(g_{u_1}(p),g_{u_1}(p),g_{u_3}(p))$
?
 A: In $n=3$ if $a$ and $b$ are two vectors, $a\times b$ is a third vector which is orthogonal to both of them, such that $|a\times b|=|a||b|\sin\theta$ and $\{a,b,a\times b\}$ is an ordered basis compatible with the standard orientation of $\mathbb{R}^3$. Here $\theta$ is the angle between $a$ and $b$ such that $a\cdot b=|a||b|\cos\theta$. For any function $h$, we have that $\nabla h=\left(\dfrac{\partial h}{\partial x}, \dfrac{\partial h}{\partial y}, \dfrac{\partial h}{\partial z}\right)^t$, where $x,y,z$ are the standard coordinates of $\mathbb{R}^3$ and $t$ denotes the transpose. If $\{u_1, u_2, u_3\}$ is an orthonormal basis of $\mathbb{R}^3$ let denote by $U$ the matrix associated to this basis, that is the matrix whose columns are the vectors $u_i$. Then if your notation $f_{u_i}$ means the directional derivative of $f$ in the direction of $u_i$ then you are wrong, you are missing the change of basis matrix. Let us explain:
$\left(f_{u_1},f_{u_2}, f_{u_3}\right)^t=U^t\cdot\nabla f$, so $\nabla f=U\cdot\left(f_{u_1}, f_{u_2}, f_{u_3}\right)^t=U\cdot\left(f_{u_1}, f_{u_2}, f_{u_3}\right)^t$, this last equality is due to the fact that $U^tU=I$, that is, $\{u_1,u_2,u_3\}$ is an orthonormal basis. If we write $\partial_uf=\left(f_{u_1}, f_{u_2}, f_{u_3}\right)^t$ then
$$
\nabla f\times\nabla g=U\partial_uf\times U\partial_ug=\det(U)\left(U^{-1}\right)^t\partial_uf\times\partial_ug=U\partial_uf\times\partial_ug
$$
For the second equality check http://en.wikipedia.org/wiki/Cross_product#Algebraic_properties
