To prove:the identity given above where $ a, b $ are vectors, $ B $ is the unit bivector in the $ a\wedge b $ plane and $\theta $ is the angle between $ a$ and $ b$. (From "Geometric Algebra for Physicists" by Doran and Lasenby).
Expanding the L.H.S i get $$ \frac{1+b.a}{|a+b|} - \frac{|a\wedge b|}{|a+b|}B $$ The R.H.S gives me by definition, $$ \cos(\theta/2) - \sin(\theta/2)B $$ Using grade projection, we should have $$ \frac{1+b.a}{|a+b|} = \cos(\theta/2) $$ and $$ \frac{|b\wedge a|}{|a+b|} = \sin(\theta/2) $$ But i can't think of an easy way to prove either. I am trying to prove it using geometry and the rules of GA, rather than trigonometry.

  • $\begingroup$ Please check theta=0 and a=b. can this be true for any a, b pair? $\endgroup$ – ahala Jul 15 '13 at 4:02
  • $\begingroup$ For theta = 0 the plane of rotation is not defined anyway $\endgroup$ – user997712 Jul 18 '13 at 7:35

Remember that a rotation can be performed through a composition of two reflections. Let $c$ be the vector in the $a \wedge b$ plane that has an angle $\theta/2$ with both $a$ and $b$. Then a rotation that would rotate $a$ to $b$ can be seen as

$$\underline R(s) = \hat c \hat a s \hat a \hat c$$

for any vector $s$. (Here, we're choosing trivially to reflect over $a$ and then reflect over the angle bisector.)

The question then becomes how we can compute the vector $c$ that bisects that angle. In fact, $c = \hat a + \hat b$ does this nicely (not well defined if $\hat b = - \hat a$, but then the plane is not well-defined anyway). Then $\hat c = (\hat a + \hat b)/|\hat a + \hat b|$ and we get the resulting rotor to be

$$R = \hat c \hat a = \frac{\hat a \hat a + \hat b \hat a}{|\hat a + \hat b|}$$

which takes the form you want. It is, however, not immediately clear to me how to generalize the problem to using non-unit vectors. This argument that $c$ is the right vector to reflect over doesn't work if you don't use unit vectors.

  • $\begingroup$ Upon a careful reading of the text, i think $ a $ and $ b $ are meant as unit vectors, although it is not explicitly stated in the question. Thanks for laying out the geometric picture. $\endgroup$ – user997712 Jul 18 '13 at 7:33

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