$i_X(dx\wedge dy\wedge dz) $ with $X=a\frac{{\partial }}{{\partial x}}+b\frac{{\partial }}{{\partial y}}+c\frac{{\partial }}{{\partial z}}$ I can not understand the practical way to calculate the following:
Let $X=a\frac{{\partial }}{{\partial x}}+b\frac{{\partial }}{{\partial y}}+c\frac{{\partial }}{{\partial z}}$
show that $dx\wedge dy\wedge dz(X,v_1,v_2)=ady\wedge dz-bdx\wedge dz+cdx\wedge dy.$
I have this:
 $i_X(dx\wedge dy\wedge dz)(v_1,v_2)=dx\wedge dy\wedge dz(X,v_1,v_2)=dx(X,v_1,v_2)\wedge (dy\wedge dz)(X,v_1,v_2)\\
=dx(X,v_1,v_2)\wedge (dy(X)dz(v_1)-dy(v_1)dz(X)+dy(X)dz(v_2)-dy(v_2)dz(X)+dy(v_1)dz(v_2)-dy(v_2)dz(v_1))\\
=[dx(v_2)dy(X)dz(v_1)-dx(v_2)dy(v_1)dz(X)]+[dx(v_1)dy(X)dz(v_2)-dx(v_1)dy(v_2)dz(X)+dx(v_2)dy(v_1)dz(v_2)-dx(X)dy(v_2)dz(v_1)]\\
=[bdx(v_2)dz(v_1)-cdx(v_2)dy(v_1)]+[bdx(v_1)dz(v_2)-cdx(v_1)dy(v_2)+ady(v_1)dz(v_2)-ady(v_2)dz(v_1)]$
(The only term that gives me is $ady \wedge dz$ the others do not fit me with the sign to form what I want.
How would it be? I do not handle this multiplication game well.
 A: Remember that
$$\begin{align}
dx \wedge dy \wedge dz 
& = dx \otimes dy \otimes dz
+ dy \otimes dz \otimes dx
+ dz \otimes dx \otimes dy \\
& - dx \otimes dz \otimes dy
- dy \otimes dx \otimes dz
- dz \otimes dy \otimes dx
\end{align}$$
Therefore,
$$\begin{align}
(dx \wedge dy \wedge dz)(X,Y,Z)
& = dx(X) \otimes dy(Y) \otimes dz(Z)
+ dy(X) \otimes dz(Y) \otimes dx(Z) \\
& + dz(X) \otimes dx(Y) \otimes dy(Z)
 - dx(X) \otimes dz(Y) \otimes dy(Z) \\
& - dy(X) \otimes dx(Y) \otimes dz(Z)
- dz(X) \otimes dy(Y) \otimes dx(Z)
\end{align}$$
In this case, $X = a \partial_x + b \partial_y + c \partial_z,$ so
$dx(X) = a, \ dy(X) = b, \ dz(X) = c,$ which gives
$$\begin{align}
(dx \wedge dy \wedge dz)(X,Y,Z)
& = a dy(Y) \otimes dz(Z)
+ b dz(Y) \otimes dx(Z) \\
& + c dx(Y) \otimes dy(Z)
 - a dz(Y) \otimes dy(Z) \\
& - b dx(Y) \otimes dz(Z)
- c dy(Y) \otimes dx(Z) \\
& = a (dy(Y) \otimes dz(Z) - dz(Y) \otimes dy(Z)) \\
& + b (dz(Y) \otimes dx(Z) - dx(Y) \otimes dz(Z)) \\
& + c (dx(Y) \otimes dy(Z) - dy(Y) \otimes dx(Z)) \\
& = a (dy \wedge dz)(Y, Z) + b (dz \wedge dx)(Y, Z) + c (dx \wedge dy)(Y, Z) \\
& = (a \, dy \wedge dz + b \, dz \wedge dx + c \, dx \wedge dy)(Y, Z)
\end{align}$$
