A particle of mass $m$ and charge $q$ has position vector $r(t)$ and moves in a constant, uniform magnetic field B so that its equation of motion is $\mathbf{\ddot r}=q\mathbf{\dot r\times B}$.
Let $\mathbf L = m\mathbf{r\times \dot r}$ be the particle’s angular momentum.
Show that $\mathbf{L\cdot B}+\frac12q|\mathbf{r\times B}|^2$ is a constant of the motion. Explain why the kinetic energy $T$ is also constant, and
show that it may be written in the form $T=\frac12m\mathbf{u\cdot((u\cdot v)v−}{r^2\mathbf{\ddot u}})$,
where $\mathbf{v=\dot r}, r=|\mathbf r| $ and $\mathbf u = \frac{\mathbf r}r$.
[Hint: Consider $\mathbf{u\cdot\dot u}$]
This may be slightly too physicsy to post on here but my issue is mathematical.
I have no issue with the first two parts, it is the "show that it may be written in the form $T=\frac12m\mathbf{u\cdot((u\cdot v)v−}{r^2\mathbf{\ddot u}})$" that I am having issues with.
I am looking at a solution and in the last part it uses that $\mathbf{\dot r\cdot r}=\dot rr$- where has this come from?
Any help is appreciated, thanks
It is not entirely clear from your question what the final solution was but $$ \mathbf{\dot{r}}\cdot \mathbf{r} = \dot{\left(\matrix{x \\ y \\z}\right)}\cdot (\matrix{x & y & z}) = \dot{x}x + \dot{y}y + \dot{z}z = \frac{1}{2}\frac{d}{dt}(x^2+y^2+z^2) = \frac{1}{2}\frac{d}{dt}r^2 = r\dot{r} $$