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So, I'm working in a real geometric / Clifford algebra generated by a set of 3 orthonormal vectors $\text{e}_1,\text{e}_2,\text{e}_3$ all with positive squares. I came to this rotor equation and I'm trying to solve it for the unknown rotor $\chi$, independent of the variable $E$ :

$ -\omega^2 \chi E \tilde{\chi} = \omega_p^2E + \frac{1}{2}\text{I}\omega \{\hat{\Omega} \chi E \tilde{\chi} - \chi E \tilde{\chi}\hat{\Omega} \} $

As for the notation here: $\chi$ is a rotor, so it has a scalar and simple bivector part. $\tilde{\chi} $ is the reverse of $\chi$. The values $\omega$ and $\omega_p$ are scalars, $\hat{\Omega}$ is a simple bivector that commutes with $\chi$ (i.e. $\hat{\Omega}\chi=\chi\hat{\Omega}$ ) and $\text{I}=\text{e}_1\text{e}_2\text{e}_3$ is the psuedoscalar of the algebra.

The variable $E$ has vector and bivector components.

As an update to this problem, I realized from the physics that $E$ must have the form $E = E_1 + E_1\wedge E_2$ where $E_1$ and $ E_2$ are vectors. That is to say that the vector and bivector parts of $E$ lie in a plane.

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This isn't a full answer, but perhaps enough to get you started. Because $ \hat{\Omega} $ commutes with $ \chi $, it also commutes with $ \tilde{\chi} $, as does the pseudoscalar $ I $ (which commutes with all grades). This means you can write

$$\begin{aligned}\frac{1}{2} I \omega\left( { \hat{\Omega} \chi E \tilde{\chi} -\chi E \tilde{\chi}\hat{\Omega} } \right)&=\frac{1}{2} I \omega\left( {\chi\hat{\Omega}E \tilde{\chi} -\chi E\hat{\Omega}\tilde{\chi}} \right) \\ &=\frac{1}{2} \omega\chi\left( {I \hat{\Omega}E-EI \hat{\Omega}} \right)\tilde{\chi} \\ &=\chi\omega\frac{1}{{2}}\left( {I \hat{\Omega}E-EI \hat{\Omega}} \right)\tilde{\chi}\end{aligned}$$

Note that $ I \hat{\Omega} $ is a vector, so

$$\frac{1}{{2}}\left( {I \hat{\Omega}E-EI \hat{\Omega}} \right)=\left( { I \hat{\Omega} } \right) \wedge E_1+\left( { I \hat{\Omega} } \right) \cdot E_2,$$

where $ E_1 = {\left\langle{{E}}\right\rangle}_{1} $, and $ E_2 = {\left\langle{{E}}\right\rangle}_{2} $, the grade one and two components of $ E $ respectively. This sum, like $ E $, therefore also has only vector and bivector grades, so your problem is now reduced to the form

$$\chi A \tilde{\chi}=B,$$

where $$\begin{aligned}A&= -\omega^2 E +\frac{\omega}{2}\left( {EI \hat{\Omega}-I \hat{\Omega}E} \right) \\ B &=\omega_p^2 E,\end{aligned}$$

are both multivectors with only vector and bivector grades.

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