# Quaternion Division

If $q$ and $r$ are quaternions, and $p$ is a point, applying $q$ then $r$ to $p$ is:

$$(qr)p\dfrac{1}{qr}$$

What if I want to go the other way? Instead of concatenating rotations, I want to remove them.

So, I have $q$ and $m$, and I know that $qx =m$, but I don't know what $x$ is. How can I discover $x$? (How can I do quaternion divison?)

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In algebra, multiplication is denoted by juxtaposition, not by an asterisk. The inverse of the operation $p\mapsto qpq^{-1}$ is $p\mapsto q^{-1}pq$. The solution of $qx=m$ is $x=q^{-1}m$. Note that $$(a+bi+cj+dk)^{-1}=(a^2+b^2+c^2+d^2)^{-1}(a-bi-cj-dk).$$

All this can be found in Wikipedia.

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I would note that the situation is similar to matrix multiplication: since multiplication here is noncommutative, you have to distinguish between "premultiplying" with the inverse $p^{-1}q$ and "postmultiplying" with the inverse $qp^{-1}$.

Note also that

$$(q\cdot r)^{-1}=r^{-1}\cdot q^{-1}$$

The only other thing I can say is that what you're essentially doing is the quaternion analog of a "similarity transformation" of a matrix.

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This is described in Wikipedia under the section Conjugation, the norm, and reciprocal. You just need the reciprocal of q.

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The definition of quaternion division also applies to complex numbers and real numbers. If I write a quaternion as a scalar and a three vector like so: $q=(s,\vec{V})$, then the inverse is $q^{-1}=(s,-\vec{V})/(s^2 + V \cdot V)$. For the reals, the inverse is 1/s. I agree with other posters that writing 1/q is in error. The 1/x notation is fine so long as one's algebra commutes, which is true for the real and complex numbers which are subgroups of quaternions.

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