Proposition 10

Suppose that the field $\mathbf{A}$ is maximal ordered, that the plane $\mathbf{E}$ is oriented, and that the form $\Phi$ is positive. For every pair of non-zero vectors $x, y$ of $\mathbf{E}$ one has

where $e$ denotes the positive bivector such that $\Phi_{(2)}(e, e) = 1$.

In fact, since the vectors $x' = x/|x|$ and $y' = y/|y|$ both have length 1, there exists one and only one rotation $\varphi$ such that $\varphi(x') = y'$ (No. 1, Cor. 2 of Prop. 1). If one writes $\varphi = a + b \omega$ ($a, b$ in $\mathbf{A}$), then by definition one has $a = \cos (\widehat{x, y})$ and $b = \sin (\widehat{x, y})$. The relation $y' = \varphi(x') = ax' + b \omega(x')$ gives $\Phi(x', y') = a \Phi(x', x') = a$ since $x'$ and $\omega(x')$ are orthogonal (Remark 2 of No. 1); this proves (19). On the other hand this relation also gives $x' \wedge y' = bx' \wedge \omega(x') = b.e$ by the definition of the extension $\Phi_{(2)}$ of $\Phi$ to $\bigwedge^2 \mathbf{E}$ ($§ 1,$ No. 9, formula (37)) and the choice of $\omega$; this proves (20).