I was trying to understand the generalization of he usual vector cross product of $\mathbb{R}^3$. I found a paper by Rost, where he says that it is easy to prove the equivalence between these two definitions.
DEFINITION A. Given a vector space $V$, endowed with a scalar product $\langle,\rangle$, a cross product is a bilinear map $\times : V\times V \rightarrow V$ such that $$\langle x\times y, z\rangle \text{ is alternating in }x,y,z$$ and $$(x\times y)\times x = \langle x,x\rangle y - \langle x,y\rangle x $$
DEFINITION B. Given a vector space $V$, endowed with a scalar product $\langle,\rangle$, a cross product is a bilinear map $\times : V\times V \rightarrow V$ such that $$\langle x\times y, x\text{ or }y \rangle =0$$ and $$\langle x\times y , x\times y\rangle = \langle x,x\rangle \langle y,y\rangle - \langle x,y\rangle ^2$$
I proved Definition A implies Definition B. But I'm having troubles showing the other way. Assuming Definition B, I proved the first condition of definition A. However I have no idea how to show condition 2 of definition A. The only thing that came to my mind was to compute $$ \langle (x \times y)\times x,(x \times y)\times x\rangle $$ using the second condition of definition B, but I got nothing clear.
Any help will be very much appreciated. Thank you.
\begin{align*} &\langle(x\times y)\times x - \langle x,x\rangle y + \langle x,y\rangle x,(x\times y)\times x - \langle x,x\rangle y + \langle x,y\rangle x\rangle\\ &=\langle(x\times y)\times x,(x\times y)\times x\rangle-\langle x,x\rangle\langle(x\times y)\times x,y\rangle-\langle x,x\rangle\langle y,(x\times y)\times x\rangle+\langle x,x\rangle^2\langle y,y\rangle-\langle x,x\rangle\langle x,y\rangle^2-\langle x,x\rangle\langle x,y\rangle^2+\langle x,x\rangle\langle x,y\rangle^2\\ &=\langle(x\times y)\times x,(x\times y)\times x\rangle-2\langle x,x\rangle\langle y,(x\times y)\times x\rangle+\langle x,x\rangle^2\langle y,y\rangle-\langle x,x\rangle\langle x,y\rangle^2\\ &=\langle(x\times y),(x\times y)\rangle\langle x,x\rangle+\langle x,x\rangle^2\langle y,y\rangle-\langle x,x\rangle\langle x,y\rangle^2-2\langle x,x\rangle\langle y,(x\times y)\times x\rangle\\ &=2\langle(x\times y),(x\times y)\rangle\langle x,x\rangle-2\langle x,x\rangle\langle (x\times y)\times x,y\rangle\\ &=2\langle x,x\rangle\langle(x\times y),(x\times y)\rangle-2\langle x,x\rangle\langle(x\times y),(x\times y)\rangle=0. \end{align*}
If $\langle x,x\rangle=0$, we have $x=0$. Also, note that an alternating multilinear map is always skew-symmetric(a fact that is used in the last step).