R^3/Cross product/Oriented orthonormal basis/Fact/Proof

Proof

Let

${}x=c_{1}u_{1}+c_{2}u_{2}+c_{3}u_{3}\,$

and

${}y=d_{1}u_{1}+d_{2}u_{2}+d_{3}u_{3}\,.$

Due to fact  (2), we have

${}x\times y={\left(c_{1}u_{1}+c_{2}u_{2}+c_{3}u_{3}\right)}\times {\left(d_{1}u_{1}+d_{2}u_{2}+d_{3}u_{3}\right)}=\sum _{1\leq i,j\leq 3}c_{i}d_{j}{\left(u_{i}\times u_{j}\right)}\,.$

Due to fact  (3), we have

${}u_{i}\times u_{i}=0\,,$

and, because of fact  (1), we have

${}u_{i}\times u_{j}=-u_{j}\times u_{i}\,.$

According to fact  (6), ${}u_{1}\times u_{2}$ is perpendicular to ${}u_{1}$ and to ${}u_{2}$; therefore,

${}u_{1}\times u_{2}=\lambda u_{3}\,$

with some ${}\lambda \in \mathbb {R}$, as this orthogonality condition defines a line. Because of fact  (5) and the condition, we get

${}\lambda =\left\langle \lambda u_{3},u_{3}\right\rangle =\left\langle u_{1}\times u_{2},u_{3}\right\rangle =\det {\left(u_{1},\,u_{2},\,u_{3}\right)}=1\,;$

hence,

${}u_{1}\times u_{2}=u_{3}\,.$

Using fact  (3), we obtain ${}u_{1}\times u_{3}=-u_{2}$ and ${}u_{2}\times u_{3}=u_{1}$. Altogether we get

${}{\begin{aligned}x\times y&=\sum _{1\leq i,j\leq 3}c_{i}d_{j}{\left(u_{i}\times u_{j}\right)}\\&=\sum _{i<j}(c_{i}d_{j}-c_{j}d_{i}){\left(u_{i}\times u_{j}\right)}\\&=(c_{1}d_{2}-c_{2}d_{1})u_{3}-(c_{1}d_{3}-c_{3}d_{1})u_{2}+(c_{2}d_{3}-c_{3}d_{2})u_{1},\end{aligned}}$

and this is the claim.

To fact