Evaluating a 2-form 
This is from Spivak Calculus on Manifolds, section 5.3
I have done part a, but I am stuck on part (b) and have been for a day now:
let $p = (p_1,p_2,p_3)$ then $w(p)(v_p,w_p) = \dfrac{p_1 dy \wedge dz (v_p,w_p) + p_2 dz \wedge dx (v_p,w_p) + p_3dx \wedge dy (v_p,w_p)}{|p|^{3/2}}$ but I am unsure on what to do here
 A: The first term in the denominator is $p_1(v_2w_3-v_3w_2),$ and similarly for the other terms. This gives you the mixed product.
A: Write $v_p = (v_1, v_2, v_3), w_p = (w_1, w_2, w_3)$ and $p = (x,y,z)$. Then
$$ w(p)(v_p, w_p) = \frac{x \cdot (dy \wedge dz)(v_p, w_p) + y \cdot (dz \wedge dx)(v_p, w_p) + z \cdot (dx \wedge dy)(v_p, w_p)}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} = \frac{x \cdot (v_2 w_3 - v_2 v_3) + y \cdot (v_3 w_1 - w_3 v_1) + z \cdot (v_1 w_2 - v_2 w_1)}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}.$$
Compare this to $\frac{ \left< v_p \times w_p, p \right>}{|p|^3}$.
A: You have a silly mistake in the denominator, and that's it. Write $v_p = (v_1,v_2,v_3)$ and $w_p = (w_1,w_2,w_3)$. Indeed we have: $$\begin{align} \omega(p)(v_p,w_p) &= \frac{p_1 ({\rm d}y \wedge {\rm d}z)(v_p,w_p)+p_2({\rm d}z \wedge {\rm d}x)(v_p,w_p) + p_3({\rm d}x \wedge {\rm d}y)(v_p,w_p)}{(p_1^2+p_2^2+p_3^2)^{3/2}} \\ &= \frac{p_1 ({\rm d}y \wedge {\rm d}z)(v_p,w_p)-p_2({\rm d}x \wedge {\rm d}z)(v_p,w_p) + p_3({\rm d}x \wedge {\rm d}y)(v_p,w_p)}{(|p|^2)^{3/2}}  \\ &= \frac{p_1 \begin{vmatrix}v_2 & w_2 \\ v_3 & w_3 \end{vmatrix}-p_2\begin{vmatrix} v_1 & w_1 \\ v_3 & w_3\end{vmatrix} + p_3\begin{vmatrix} v_1 & w_1 \\ v_2 & w_2 \end{vmatrix}}{|p|^3} \\ &= \frac{\begin{vmatrix} p_1 & p_2 & p_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 & \end{vmatrix}}{|p|^3} \\ &= \frac{\langle v_p \times w_p, p\rangle}{|p|^3} \end{align}$$
