Transversality of Vector Fields Defined in terms of Diff. Forms and Open Books. All:
 Sorry for the length of the post, but I think it is necessary to set things up so that the post is understandable. I'm trying to understand how it is that the transversality (in this case , the transversality of the Reeb field $R_{\omega}$ associated with a given contact form $\omega$) is determined with respect to a certain differential form :
I'm going through an argument in which we intend to show that a given Reeb vector field $R_ω $   is positively-tangent to a  link ( as in a space whose connected components are $S^1$-knots) . This means the Reeb field  lives in the tangent space to the Link, along the positive direction) , and Rω   is positively- transverse to a surface S ( so that
 Rω   intersects S positively at points).
 The argument is made using properties of differential forms, in the context of open book decompositions of contact 3-manifolds.
 A needed definition is this: A contact structure ζ is supported by an open book decomposition (B, π ) if ζ can be isotoped thru contact structures so that there is a contact 1-form ω for ζ satisfying:
1)dω is a positive area form on each page $∑_θ$   of the open book, and:
2)ω>0 on the binding B ; both B and the pages are oriented. 
End of setup.

Actual Question:
To be more specific, I'm trying to understand the following arguments purporting to show 
 the equivalence between these two conditions:
(1) The contact manifold (M,ζ ) is supported by the open book (B,π)
(3) There is a Reeb field Rω   for a contact structure isotopic to ω , so that $R_ω$ is positively-tangent to the binding B, and positively-transverse to the pages of the open book.
Proof: 
 (3)->(1) : Since $R_ω$   is assumed positively -tangent to the binding B , we have ω>0 on oriented tangent vectors to B. Since the Reeb field $R_ω$   is positively-transverse to the pages of the OB (open book) , we have that $dω=i_{R_ω} (ω \wedge dω) >0$ on the pages of the OB (where i is --I am? -- the interior product , or contraction of the form $ω\wedge dω$ by the vector field $R_ω$   
Questions:
 i)How does $R_ω$   being positively-tangent to the ( knots in the ) binding imply $ω >0$ ? 
 I know this means the vector field being positively-tangent to the binding means that $R_ω$   lies along the tangent space ; a 1-d tangent space, to each of the knots, along the chosen positive direction orientation.
ii)Why is $dω$ equal to the contraction of $ω \wedge dω$ ? , and how does the positive transversality imply that $dω>0$?
(1)->(3): Assume (1), and let ω be the form with the given conditions, and let R ω   be the Reeb field associated with ω. Then "It is clear that $R_ω$   is positively-transverse to the pages of the OB, since dω is an area form on the pages of the open book"
I have no clue about the connection between the Reeb field being positively-transverse to the pages, and dω being an area form on the pages. I know if dw  is a positive area form on the pages, then $dω(X,Y)>0$ at any pair of positively-oriented tangent vectors. And I know a Reeb field associated to a contact structure is transverse to the planes in the contact structure .But I can't see how this relates to $dω being an area form for the pages of the open book.
Thanks for any suggestions, ideas. 
 A: Question i:
From the definition of Reeb vector field we know that $\omega(R_{\omega})=1$ and being positively-tangent to the binding implies that $T_\theta B = f(\theta) R_{\omega}$ for some positive function $f$. Hence $\omega(T_\theta B) = f(\theta) > 0$.
Question ii:
Again from the definition of Reeb vector field we know that $\iota_{R_\omega} d \omega = 0$. From the definition of interior product we know that $\iota_{R_\omega} (\omega \wedge d \omega) = (\iota_{R_\omega} \omega) \wedge d \omega + (-1)^{deg \omega} \omega \wedge (\iota_{R_\omega} d \omega) = 1 \wedge d \omega + (-1)^{deg \omega} \omega \wedge 0 = d \omega$, where $(\iota_{R_\omega} \omega) = 1$ is straightforward reformulation of $\omega(R_{\omega})=1$. Hence in fact $\iota_{R_\omega} (\omega \wedge d \omega) = d \omega$.
To see how how does the positive transversality imply that $d \omega > 0$ we must recall definition of contact structure, being more specific the condition $\omega \wedge d \omega > 0$. One may ask himself a question: why does this inequality doesn't give us $d\omega = \iota_{R_\omega} (\omega \wedge d \omega) > 0 $ directly? Because inequality $\omega \wedge d \omega > 0$ is associated with orientation on tangent space and positive transversality condition tells us that by "feeding" $\omega \wedge d \omega$ with $R_\omega$ we are dealing exactly with the case where the inequality holds.
I hope that the last paragraph answers to you the last question, too.
