<jupyter_start><jupyter_text>$\color{red}{\langle Rotor \rangle} = R$>  Rotor $=:$ geometric product of a number of unit vectors>  reverse Rotor $=: R^{\dagger}$ reverses order of Rotor>>  inverse Rotor $=: R^{-1} \iff R^{\dagger}$ inverses of Rotor>>  $|v|=|w|=1, R = \vec{w}\vec{v} \iff R^{\dagger} = \vec{v}\vec{w} \\\because vwwv = wvvw = R^{\dagger}R = RR^{\dagger} = 1\quad \therefore R^{\dagger} = R^{-1}$>  $\therefore {^{R=e^{-\frac{\theta}{2}B}}_{R^{\dagger}=e^{\frac{\theta}{2}B}}{\Large\color{red}{g}}^{RuR^{\dagger}}_{(wv)^{-1}u\,(vw)}} \ni B^2 = -1, |B|^2 = - B^2$  $\color{red}{\langle Dagger  \rangle} = R^{\dagger}$>  conjugate Rotor>>  $ R = \vec{w}\vec{v} = cos(\frac{\theta}{2}) - B\,sin(\frac{\theta}{2}) = e^{-\frac{\theta}{2}B} \iff R^{\dagger} = \bar{R} =R^{\star} = cos() - B\,sin()$>  inverse Rotor>>  $ R = \vec{w}\vec{v} = cos(\frac{\theta}{2}) - B sin(\frac{\theta}{2}) = e^{-\frac{\theta}{2}B} \iff R^{\dagger} = R^{-1} = e^{B\frac{\theta}{2}}$ --->  $R = e^{\theta\hat{B}}\\R^{\dagger} = e^{-\theta\hat{B}} \\R^s = e^{s\theta\hat{B}}\\R^{s\dagger} = e^{-s\theta\hat{B}} \\R^{\dagger\dagger} = R \\ (R_1R_2)^{\dagger} = R_2^{\dagger}R_1^{\dagger} \\\quad R_1 = ab, R_2 = cd, (abcd)^{\dagger} = dcba = R_2^{\dagger}R_1^{\dagger}$ >  $  Rot(\alpha x + b\beta y) = R(\alpha x + \beta y)R^{\dagger}$ >>  $ \quad  = R(\alpha x)R^{\dagger} + R(\beta y)R^{\dagger}$ >>  $ \quad  = \alpha R(x)R^{\dagger} + \beta R(y)R^{\dagger}$ >  $ ^{u=\alpha x + \beta y}_{R(\alpha x + \beta y)R^{\dagger}}{\Large\color{red}{\dagger}}^{R(\alpha x)R^{\dagger} + R(\beta y)R^{\dagger}}_{\alpha R(x)R^{\dagger} + \beta R(y)R^{\dagger}}$>  $ ^{R=mv\;\ni |m|=|v|=1}_{R^{\dagger}= vm = R^{-1}}{\Large\color{red}{\dagger}}^{RR^{\dagger}=RR^{-1}=1}_{(R_1R_2)^{\dagger} = {R_2}^{\dagger}{R_1}^{\dagger}}$ --->  $ B = ae_1e_2 + be_2e3 + ce_3e_1 = (be_1 + ce_2 + ae_3)I $>  $ \hat{B} = \frac{B}{-B^2}$>  $ B = vI , B^2 = -1 = -|B|^2$>  $ v = BI^{-1}$>  $ v = |v|\big(\cos(\phi_v)\sin(\theta_v)\,e_1 + \cos(\phi_v)\sin(\theta_v)\,e_2 + \sin(\phi_v)\,e_3 \big)$>  $ w = |w|\big(\cos(\phi_w)\sin(\theta_w)\,e_1 + \cos(\phi_w)\sin(\theta_w)\,e_2 + \sin(\phi_w)\,e_3 \big)$<jupyter_code>import clifford as cl 
import clifford.g3 as g3 
import numpy as np
import math as mt
import matplotlib.pyplot as plt
import ipympl
%matplotlib widget
u = g3.e3
v = np.sin(1.3)*np.cos(np.pi/4)*g3.e1 + np.sin(1.3)*np.sin(np.pi/4)*g3.e2 + np.cos(1.3)*g3.e3
w = np.sin(1.3)*np.cos(np.pi/2)*g3.e1 + np.sin(1.3)*np.sin(np.pi/2)*g3.e2 + np.cos(1.3)*g3.e3
# rotor ~
~(w*v)

ax = plt.figure(figsize=(8,8)).add_subplot(projection='3d')
ax.set_xlim3d(xmin=-1.5,xmax=1.5)
ax.set_ylim3d(ymin=-1.5,ymax=1.5)
ax.set_zlim3d(zmin=-1.5,zmax=1.5)
# v
ax.quiver3D(0,0,0,0.68134,0.68134,0.2675,arrow_length_ratio=0.1)
ax.text(0.68134,0.68134,0.2675,'v')
# w
ax.quiver3D(0,0,0,0,0.96356,0.2675,arrow_length_ratio=0.1)
ax.text(0,0.96356,0.2675,'w')
# u
ax.quiver3D(0,0,0,0.0,0.0,1.,arrow_length_ratio=0.1,color='k')
ax.text(0,0,1,'u')

# wvuvw
(v*w).inv()*u*(v*w)
ax.quiver3D(0,0,0,-0.36451,-0.12938,0.92217,arrow_length_ratio=0.1,color='b')
ax.text(-0.36451,-0.12938,0.92217,'wvuvw')

# nomal vector vw plane
ax.quiver3D(0,0,0,-0.07549,-0.18226,0.65651,arrow_length_ratio=0.1,color='r')
ax.text(-0.07549,-0.18226,0.65651,'$\hat{n}_{\perp_{vw}}$')

# B = vI \iff  B^{-1} = v  |** scalar = 0
# B*I^{-1} \to  (v ^ w)*I^{-1} = (v \times w) ** scalar = 0
x = np.linspace(-1,1,100)
x,y = np.meshgrid(x,x)
z = (x*0.07549 + y*0.18226)/0.65651
ax.plot_surface(x,y,z,alpha=0.3)
# -----------------
# RuR^{\dagger} = e^(-\theta \hat{B}) u e^{\theta \hat{B}}
# B = \hat{v^w}
ax.quiver3D(0,0,0,-0.37134,-0.1445,0.91718,arrow_length_ratio=0.1,color='magenta')
B = (v^w).normal()
Rd = np.e**((np.pi/4)*B)
R = np.e**(-(np.pi/4)*B)
R*u*Rd
#np.arccos([1,(v|w).value[0]])/np.pi<jupyter_output><empty_output><jupyter_text>Wedge product vs Cross product>  $ a \wedge b =  \left | \begin{array}{}e_{23} & e_{31} & e_{12} \\a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \\ \end{array}\right | \iff a \times b =  \left | \begin{array}{}i & j & k \\a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \\ \end{array}\right |$ >  $\therefore v \times w = (u \wedge w)I^{-1}$>  $\therefore B = vI \quad v = BI^{-1}$<jupyter_code>u= g3.e1 + g3.e3
v = g3.e1
w = np.cos(np.pi/4)*g3.e1 + np.sin(np.pi/4)*g3.e2
(v*w).inv()*u*v*w
~(v*w)*u*v*w<jupyter_output><empty_output>
StarcoderdataJupyterStructuredCleanDedup