Evaluating differential forms.

Can someone please check my work? It's an exercise from Barret O'Neill's Elementary Differential Geometry. I want to be really sure that my understanding of this is right. I see that the forms $\mathrm{dx, dy, dz}$ work as projections of the vector part, and the functions multiplying these forms are evaluated in the point in question. I'm given the tangent vector $v_p = (1,2,-3)_p$, where $p = (0,-2,1)$ I must evaluate some forms on $v_p$. My work:

a) $y^2~ \mathrm{dx}$. $$\begin{align} (y^2 ~\mathrm{dx})(v_p) &= (-2)^2 \cdot \mathrm{dx}(1,2,-3) \\ &= 4 \cdot 1 \\ &= 4 \end{align}$$

b) $z~ \mathrm{dy} - y~ \mathrm{dz}$. $$\begin{align} (z~ \mathrm{dy} - y~ \mathrm{dz})(v_p) &= 1 \cdot \mathrm{dy}(1,2,-3) - (-2) \cdot \mathrm{dz}(1,2,-3) \\ &= 2 + 2 \cdot (-3) \\ &= -4 \end{align}$$

c) $ (z^2 - 1) ~\mathrm{dx} - \mathrm{dy} + x^2~ \mathrm{dz}$. $$\begin{align} \left((z^2 - 1)~ \mathrm{dx} - \mathrm{dy} + x^2 ~ \mathrm{dz} \right)(v_p) &= (1^2 - 1) \cdot \mathrm{dx}(1, 2,-3) - \mathrm{dy}(1,2,-3) + 0^2 \cdot \mathrm{dz}(1,2,-3) \\ &= - 2 \end{align}$$

Thank you.

Solution 1:

I've done the calculations again and convinced myself that aren't any mistakes.