I'd like to solve the following equation for $c$, where $a$, $c$, and $d$ are position vectors represented by quaternions with $w$ (the real component) set to $0$ and $b$ is a unit quaternion: $$a+(b*c*b^{-1})=d$$
The $(b*c*b^{-1})$ part of the equation is used to rotate $c$ by $b$, as described here: http://www.euclideanspace.com/maths/algebra/realNormedAlgebra/quaternions/transforms/
In other words: $$originalPosition+transformPosition=newPosition$$
Basic algebra leads me to believe that the following is correct, but I'm not getting the correct answer, probably because quaternion multiplication isn't commutative: $$c=\frac{d-a}{b*b^{-1}}$$
Plus, $b*b^{-1}$ is a quaternion with $x$, $y$, and $z$ equal to $0$ and $w$ equal to $1$, effectively dividing by $1$. So I'm basically computing: $$c=d-a$$
What is the correct equation for $c$?
$$a+(b*c*b^{-1})=d$$ $$b*c*b^{-1}=d-a$$ $$b^{-1}b*c*b^{-1}=b^{-1}(d-a)$$ $$c*b^{-1}=b^{-1}(d-a)$$ $$c*b^{-1}b=b^{-1}(d-a)b$$ $$c=b^{-1}(d-a)b$$