I am learning Geometric Algebra by reading New Foundations for Classical Mechanics, by David Hestenes. Chapter 3-2 studies the motion of a particle with constant gravitational force and on page 129 the maximum range of the particle is derived. There are two parts I am stumped on for deriving equation (2.12) which is as follows:
$$ \begin{equation} r_{max} =\frac{2v_{0}^{2}}{g}\frac{1}{| \hat{\mathrm{r}} -\hat{\mathrm{g}}| ^{2}} =\frac{v_{0}^{2}}{g}\frac{1}{1-\hat{\mathrm{g}} \cdotp \hat{\mathrm{r}}} \end{equation} $$
Equation (2.8) is the equation for range, $r$, which is:
$$ \begin{equation} r=\frac{2(\hat{\mathrm{g}} \land \mathrm{v}_0) \cdot (\mathrm{v}_0 \land \hat{\mathrm{r}})}{|\mathrm{g} \land \hat{\mathrm{r}} |^2} \end{equation} $$
The numerator of (2.8) is "maximized" to get equation (2.10). To maximize, uses a vector identity to resolve that $2(\hat{\mathrm{g}} \land \mathrm{v}_0) \cdot (\mathrm{v}_0 \land \hat{\mathrm{r}})=\mathrm{v}_0^{2}[\hat{\mathrm{g}} \cdot \hat{\mathrm{r}} - \hat{\mathrm{g}} \cdot (\mathrm{v}_0 \hat{\mathrm{r}} \mathrm{v}_0)]$. This I understand.
By maximizing, he finds that the numerator has a maximum value when $\hat{\mathrm{g}} \cdot (\hat{\mathrm{v}_0} \hat{\mathrm{r}} \hat{\mathrm{v}_0}) = 1$, then concluded that $-\hat{\mathrm{g}} \hat{\mathrm{v}_0} = \hat{\mathrm{v}_0} \hat{\mathrm{r}}$, which says that $\hat{\mathrm{v}_0}$ can be expressed as a function of $\hat{\mathrm{v}_0}$ and $\hat{\mathrm{g}}$ and writes it as (2.11): $$ \hat{\mathrm{v}}_0=\frac{\hat{\mathrm{r}}-\hat{\mathrm{g}}}{|\hat{\mathrm{r}}-\hat{\mathrm{g}}|} $$
My Confusion
Then, (2.11) is substituted into (2.8) and gets equation (2.12). I have attempted this multiple times to get the same result but do not get an equation that looks anything like (2.12). I feel as though I am missing some key geometric algebra identities that'll lead the solution. I would appreciate assistance, thank you.
EDIT
Here is the pages from the book!
