Rotations in 4D

So the geometric algebra specifies a rotation on $\mathbb{R}^n$ in quite a nice form. Given unit vectors $a,b$ then defining $\sigma\colon v \mapsto b a v a b$ we get the $\sigma$ is a rotation that leaves everything perpendicular to the $a,b$ plan pointwise fixed, and within the $a,b$ plane $\sigma$ is the rotation from $a$ to $b$ applied twice (so that the rotation of $\sigma$ is twice the angle between $a$ and $b$). Now in the plane if we choose $a$ to be $e_1 = (1,0)$ and we choose $b = \cos(\theta/2) e_1 + \sin(\theta/2) e_2 = (\cos(\theta/2),\sin(\theta/2))$ so that applying the rotation from $a$ to $b$ twice is a counterclockwise rotation of the plane by angle $\theta$ then the geometric product $ab = \cos(\theta/2) + \sin(\theta/2) e_1 e_2$ and the geometric $ba = \cos(\theta/2) - \sin(\theta/2) e_1 e_2$. Now in 4 dimensions a rotation is made up of a plane, and its orthogonal plane, each of them being setwise invariant, and each undergoing a rotation. Thus we can see a model of a 4 dimensional rotation by taking the product $(\cos(\alpha/2) + \sin(\alpha/2) e_1 e_2) (\cos(\beta/2) + \sin(\beta/2) e_3 e_4)$. Then $v \mapsto q^{-1} v q$ is a rotation in $\mathbb{R}^4$ which is a rotation in the direction from $e_1$ to $e_2$ by angle $\alpha$ in the $e_1,e_2$ plane, and is a rotation in the $e_3,e_4$ plane rotating in the direction from $e_3$ to $e_4$ by angle $\beta$. More interestingly, given a general unit member $z = a + b_{12} e_1 e_2 + b_{13} e_1 e_3 + b_{14} e_1 e_4 + b_{23} e_2 e_3 + b_{24} e_2 e_4 + b_{34} e_3 e_4 + c e_1 e_2 e_3 e_4$ one wonders whether we can factor it as a product of two planar orthogonal rotations. That is, can you read off the planes of rotation and the corresponding angles of rotation from any unit even member of the geometric algebra. The answer is that yes one can. Assume that the two rotations are $g = \cos(\alpha/2) + \sin(\alpha/2) (r_{12} e_1 e_2 + r_{13} e_1 e_3 + r_{14} e_1 e_4 + r_{23} e_2 e_3 + r_{24} e_2 e_4 + r_{34} e_3 e_4)$ and by $h = \cos(\beta/2) + \sin(\beta/2) (s_{12} e_1 e_2 + s_{13} e_1 e_3 + s_{14} e_1 e_4 + s_{23} e_2 e_3 + s_{24} e_2 e_4 + s_{34} e_3 e_4)$. For a rotation to represent a plane it must have $r_{12} r_{34} + r_{13} r_{24} + r_{14} r_{23} = 0$ For the planes of these two members of the product to be orthogonal we must have $r_{12} = s_{34}$, $r_{13} = s_{24}$, $r_{14} = s_{23}$, $r_{23} = s_{14}$, $r_{24} = s_{13}$, $r_{34} = s_{12}$. Consequently the grade 4 part of the product $g h$ is $r_{12} s_{34} + r_{13} s_{24} + r_{14} s_{23} + r_{23} s_{14} + r_{24} s_{13} + r_{34} s_{14}$ which is equal to $r_{12}^2 + r_{13}^2 + r_{14}^2 + r{23}^2 + r_{24}^2 + r_{34}^2 = 1$. Consequently if $z = gh$ then $a = \cos(\alpha/2) \cos(\beta/2)$ and $c= \sin(\alpha/2) \sin(\beta/2)$. We then have $a - c = \cos((\alpha + \beta)/2)$ and $a + c = \cos((\alpha - \beta)/2)$. From this we can readily extract both $\alpha$ and $\beta$. Then since $b_{ij} = \cos(\beta/2)\sin(\alpha/2) r_{ij} + \sin(\beta/2) \cos(\alpha/2) s_{ij} = \cos(\beta/2)\sin(\alpha/2) r_{ij} + \sin(\beta/2) \cos(\alpha/2) r_{kl}$ and $b_{kl} = \cos(\beta/2)\sin(\alpha/2) r_{kl} + \sin(\beta/2) \cos(\alpha/2) s_{kl} = \cos(\beta/2)\sin(\alpha/2) r_{kl} + \sin(\beta/2) \cos(\alpha/2) r_{ij}$ where $kl$ are the two members different than $ij$ we have equations we can readily solve for $r_{ij}$ and $r_{kl}$ in terms of $b_{ij}$ and $b_{kl}$ (since $\beta$ and $\alpha$ are known). Thus one can decompose any even unit member of the geometric algebra of $\mathbb{R}^4$ into two orthogonal plane rotations with known angles and directions of rotation.