I'm having some trouble reproducing Spivak's construction of the cross product (p.83 in his "Calculus on Manifolds").
He restricts it to $\mathbb{R}^n$, and does so by defining for $v_1,\dots,v_{n-1}\in\mathbb{R}^n$, with $\varphi$ defined by $$\varphi(w)=\det\begin{pmatrix}v_1\\ \vdots\\v_{n-1}\\w\end{pmatrix}$$ then $\varphi\in\Lambda^1(\mathbb{R}^n)$. Therefore, there is a unique $z\in\mathbb{R}^n$ s.t. $$\langle w,z\rangle=\varphi(w)=\det\begin{pmatrix}v_1\\ \vdots\\v_{n-1}\\w\end{pmatrix}$$
Now I've given this a go in $\mathbb{R}^3$, using the usual basis for $v$ (i.e. $v_1=e_1$, $v_2=e_2$ etc) and an arbitrary $w$ $$w=\begin{bmatrix}\frac{1}{\sqrt{2}}\\0\\\frac{1}{\sqrt{2}}\end{bmatrix}$$ so $\varphi(w)=\frac{1}{\sqrt{2}}$. However, the requirement for $z$ is $$\begin{bmatrix}\frac{1}{\sqrt{2}}\\0\\\frac{1}{\sqrt{2}}\end{bmatrix}\cdot z=\frac{1}{\sqrt{2}}$$ which to me seems fulfillable by $z=e_1$ or $z=e_3$, which would not be unique (and since this is the $\mathbb{R}^3$ case we know that only $e_3$ is the correct answer). What's wrong with my understanding?
The value of $\langle w,z\rangle$ should agree with the determinant for all $w$. The vector $z$ is unique, for, if we find some vector $z'$ such that $\langle w,z\rangle=\langle w,z'\rangle$ for all $w$, we must have $\langle w,z-z'\rangle=0$ for all $w$. In particular, $\langle z-z',z-z'\rangle=0$ and hence $z-z'=0$.
Since $\varphi$ is a linear form, we have $\varphi(w)=\varphi(\sum_iw_ie_i)=\sum_iw_i\varphi(e_i)$. So, we may explicitly express $z$ as $\big(\varphi(e_1),\varphi(e_2),\ldots,\varphi(e_n)\big)$.