TITLE: Interior product between differential forms and vector fields
QUESTION [8 upvotes]: I don't understand what is meant when someone writes that forms (or form fields) "eat" vectors (or vector fields). For example when I have a one form field ω=3dx+5dy+3xdz and a vector field X=3x∂x+5y∂y+3∂z, then their interior product $\iota_{X}\omega=\omega(X)= (3dx+5dy+3xdz)(3x\partial_{x}+5y\partial_{y}+3\partial_{z})=6x(\partial_{x}dx)+25y(\partial_{y}dy)+6x(\partial_{z}dz)=9x+25y+6x=15x+25y$
I think this is right. So in this case ω(X) doesn’t mean that ω is function of X, it just means multiplication. Now how does this work with two forms? For example X=y∂x+2z∂y+3xy∂z and the two form ω=3dx∧dy−(14zx+2)dx∧dz
$\iota_{X}\omega=\omega(X,V)=(3dx\wedge dy-(14zx+2)dx\wedge dz)(y\partial_{x}+2z\partial_{y}+3xy\partial_{z}, V)$
Now how do I proceed from here?

REPLY [4 votes]: Thank you for your answers and Phillip Andreae has the best answer so far. But I want to answer the question myself because I found an easier and more intuitive way to look at this problem. I just think of differential forms as alternating multilinear functions, for example the determinant !
Let's take a simpler example:
$\omega = 3x dx\wedge dy$
$X = 2y\partial_{x}+4xy\partial_{y}$ 
Now the I can write the two form as a determinant with a factor $\omega = 3x dx\wedge dy(X,V) = 3x\begin{vmatrix} dx(X) & dx(V) \\ dy(X) & dy(V) \end{vmatrix}$
This equals $\omega = 3x[dx(X)dy(V)-dx(V)dy(X)]$
$dx(X)= 2y$ and $dy(X)=4xy$. 
So $\iota_{X}\omega= 6xydy-12x^{2}ydx$
The same goes with three forms and so on. 
Any objections?