Woke up again at 11, and unable to fall asleep. My sleep cycle is following Mr Murphy, most active and alert when am supposed to be sleeping. Was reading something online, and came across wikipedia's post on Rodrigues' rotation formula, and saw that they derived in a juvenile manner, using 11th grade math (http://en.wikipedia.org/wiki/Rodrigues'_rotation_formula#Derivation). So just felt like derived it in slightly more elegant manner, and am here. Will add figures that go with this later. Dont have scanner, nor pencil, paper or patience to draw lines now. Until then, use your imagination. :)
Let v be the vector to be rotated by angle $\theta$ about axis k.
Component of v along k,
$v_k = (v \cdot k) k$
Component perpendicular to k
$v_{\perp} = v - (v \cdot k) k$
After rotation by $\theta$,
Component perpendicular to k, along $v_{\perp}$
$v_{\perp} = |v_{\perp}| cos(\theta) \hat{v}_{\perp} = \frac{ \left( v - (v \cdot k) k \right)}{\left| v - (v \cdot k) k \right|} cos(\theta) $
$v_{\perp} = |v_{\perp}| cos(\theta) \hat{v}_{\perp} = \left| v - (v \cdot k) k \right| cos(\theta) \frac{ \left( v - (v \cdot k) k \right)}{\left| v - (v \cdot k) k \right|} $
$v_{\perp} = \left( v - (v \cdot k) k \right) cos(\theta) $Component perpendicular to k and $v_{\perp}$
$v_{\perp,2} = |v_{\perp,2}| sin(\theta) \hat{v}_{\perp,2} = \left| v - (v \cdot k) k \right| sin(\theta) \frac{ k \times \left( v - (v \cdot k) \right) } {\left| k \times \left( v - (v \cdot k) k \right|} $
as k is unit vector,
$ \left| k \times \left( v - (v \cdot k) k \right| = \left| \left( v - (v \cdot k) k \right| $
So
$v_{\perp,2} = \left( v - (v \cdot k) k \right) sin(\theta) $
also,
$k \times \left( v - (v \cdot k) \right) = k \times v$.
Now rotated vector is,
$v_{rot} = v_k + v_{\perp} + v_{\perp,2}$$ v_{rot} = (v_k \cdot k) k + \left( v - (v \cdot k) k \right) cos(\theta) + ( k \times v) sin(\theta) $
Simplifying
$ v_{rot} = v cos(\theta) + (v \cdot k) k ( 1 - cos(\theta)) + ( k \times v) sin(\theta) $
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