I am trying to understand an article by Gibbons, Rychenkova and Goto, called "Hyperkähler quotient construction of BPS Monopole Moduli Spaces". I will paraphrase the relevant notions and formulas in order to get to my question.

Let $M = \mathbb{H}$ and $q$ be a quaternionic coordinate on $M$. Then the flat metric is $$ g = dq \odot d\bar{q}.$$ Moreover, if we decompose $$-\frac{1}{2} dq \wedge d\bar{q} = \omega_I i + \omega_J j + \omega_K k,$$ the tuple $(g,\omega_I,\omega_J,\omega_K)$ is a flat hyperkähler structure on $M$.

With respect to the action of $U(1)$ that maps $q$ to $qe^{it}$, the moment map is $$ \mu = \frac{1}{2} q i \bar{q}.$$

It is natural to introduce other coordinates, which are adapted to the moment map $\mu$. One may write $$q = ae^{i \psi/2}$$ where $a$ is a pure imaginary quaternion and $\psi \in (0,4\pi]$. Let $$\mathbf{r} = qi\bar{q} = ai\bar{a} = -aia.$$

Then the authors write that a "short calculation" shows that, in the new coordinates $(\mathbf{r},\psi)$, the metric $g$ can be written as $$ g = \frac{1}{4}\left(\frac{1}{r} d\mathbf{r}^2 + r(d\psi + \omega.d\mathbf{r})^2\right),$$ where $r = |\mathbf{r}|$ is the Euclidean norm of $\mathbf{r}$ and $$\tag{1}\label{Bog}\nabla \times \omega = \nabla\left(\frac{1}{r}\right).$$

I attempted the calculation. After completing the square, I got that $$ g = \frac{1}{4}\left(\frac{1}{r} d\mathbf{r}^2 + r(d\psi + \frac{1}{r}(aid\bar{a}-dai\bar{a}))^2\right),$$ which would agree with what the authors got, assuming that $$ \omega.d\mathbf{r} = \frac{1}{r}(aid\bar{a}-dai\bar{a})$$ satisfies the Bogomolny equations \eqref{Bog}.

I tried to prove that, but somehow I got stuck. I obtained for instance that $$*d\left(\frac{1}{r}\right) = - \frac{d\mathbf{r}\,\mathbf{r} \wedge d\mathbf{r}}{2r^3}$$ and that $$d(\omega.d\mathbf{r}) = -\frac{dr}{r^2} \wedge (aid\bar{a} - dai\bar{a}) + \frac{2}{r} daid\bar{a},$$ but I could not yet show that $$*d(1/r) = d(\omega.d\mathbf{r}),$$ despite a few algebraic manipulations attempts.

Edit: I was able to prove, after some time, that $$d(\omega.d\mathbf{r}) = rd\left( \frac{1}{r} \right) \wedge (\omega.d\mathbf{r}) + *d\left( \frac{1}{r} \right).$$ Am I missing something?