Answer
Using the identity:
$\left( {{\bf{A}} \times {\bf{B}}} \right) \times {\bf{C}} = {\bf{B}}\left( {{\bf{A}}\cdot{\bf{C}}} \right) - {\bf{A}}\left( {{\bf{B}}\cdot{\bf{C}}} \right)$
we show that
${\bf{r}}' = \frac{{\left( {{\bf{J}} \times {\bf{r}}} \right)}}{{||{\bf{r}}|{|^2}}}$
Work Step by Step
Recall the identity:
$\left( {{\bf{A}} \times {\bf{B}}} \right) \times {\bf{C}} = {\bf{B}}\left( {{\bf{A}}\cdot{\bf{C}}} \right) - {\bf{A}}\left( {{\bf{B}}\cdot{\bf{C}}} \right)$
We have ${\bf{J}} = {\bf{r}} \times {\bf{r}}'$. So,
${\bf{J}} \times {\bf{r}} = \left( {{\bf{r}} \times {\bf{r}}'} \right) \times {\bf{r}}$.
Using the identity above we obtain
${\bf{J}} \times {\bf{r}} = {\bf{r}}'\left( {{\bf{r}}\cdot{\bf{r}}} \right) - {\bf{r}}\left( {{\bf{r}}'\cdot{\bf{r}}} \right)$
Since ${\bf{r}}$ and ${\bf{r}}'$ are perpendicular, so ${\bf{J}} \times {\bf{r}} = {\bf{r}}'||{\bf{r}}|{|^2}$. Hence, ${\bf{r}}' = \frac{{\left( {{\bf{J}} \times {\bf{r}}} \right)}}{{||{\bf{r}}|{|^2}}}$
