I think it's silly but, you could tell me how this operator $i_{v}$ is evaluated with some example. Is a 3-form $\omega$ in a vector field of dimension 3 $V$, something like that
$\omega=f dx$ ^ $dy$ ^ $dz$, $V= V_{1}\partial x+V_{2}\partial y+V_{3}\partial z$
$i_{V}\omega=¿?$
Let $X_{1} = g_{1}\partial x + g_{2}\partial y + g_{3}\partial z$, $X_{2} = h_{1}\partial x + h_{2}\partial y + h_{3}\partial z$, $\iota_{V}\omega = W_{1}dy\wedge dz + W_{2} dz\wedge dx + W_{3}dx\wedge dy$. Then by definition, we should have $(\iota_{V}\omega)(X_{1}, X_{2}) = \omega(V, X_{1}, X_{2})$, i.e. $$ \omega(V, X_{1}, X_{2}) = \sum_{i,j, k} V_{i}g_{j}h_{k}\omega(\partial x_{i}, \partial x_{j}, \partial x_{k}) = f\sum_{\sigma\in S_{3}}\mathrm{sgn}(\sigma)V_{\sigma(1)}g_{\sigma(2)}h_{\sigma(3)} = f \Bigg|\begin{matrix}V_{1} & g_{1} & h_{1} \\ V_{2} & g_{2} & h_{2} \\ V_{3} & g_{3} & h_{3}\end{matrix}\Bigg| \\ (\iota_{V}\omega)(X_{1}, X_{2}) = ... = \Bigg|\begin{matrix}W_{1} & g_{1} & h_{1} \\ W_{2} & g_{2} & h_{2} \\ W_{3} & g_{3} & h_{3} \end{matrix}\Bigg| $$ for any $X_{1}$ and $X_{2}$. Hence we get $W_{i} = fV_{i}$ and $$ \iota_{V}\omega = f(V_{1}dy\wedge dz + V_{2} dz\wedge dx + V_{3} dx\wedge dy) $$