Why $u(x)\cdot ri(x)=0$?

Question

I was working on a paper, and I went into a problem with a simple inner product. Let r(x)=(x,f(x)), so we can write it as $$r(x)=\sum_{i=1}^n x_ie_i+f(x)e_{n+1}$$ then, $r_i(x)=e_i+f_i(x)e_{n+1}$ and $r_i(x).r_j(x)=\delta_{ij}+f_i(x)f_j(x)$ where $\delta_{ij}$ denotes the Kronecker delta. It said that the vector $u(x)=-\sum_{i=1}^n f_i(x)e_i+e_{n+1}=-\nabla f+e_{n+1}$ satisfies $u(x).r_i(x)=0$. The problem is that I can't prove that the vector u satisfies this equation. Any help would be appreciated.

Answer 1

Presumably, (if $r_i(x) = \frac{\partial}{\partial x_{i}} r(x)$), $$ u \cdot r_i = (-\nabla f + e_{n+1}) \cdot (e_i + f_i e_{n+1}) = -(e_i \cdot \nabla)f + e_{n+1} \cdot e_i - (f_i e_{n+1} \cdot \nabla) f + f_i. $$ The middle two terms are zero ($f$ only depends on $x_1,\dotsc, x_n$), and $e_i \cdot \nabla = \frac{\partial}{\partial x_i}$ means the other two terms cancel.