Answer
The solution is ${\bf{r}}\left( t \right) = \left( {1 - \cos t} \right){\bf{u}}$.
Work Step by Step
We have ${\bf{r}}'\left( t \right) = \left( {\sin t} \right){\bf{u}}$ with initial condition ${\bf{r}}\left( 0 \right) = {\bf{0}}$, where ${\bf{u}}$ is a constant vector in ${\mathbb{R}^3}$.
The solution is obtained by integrating ${\bf{r}}'\left( t \right)$:
${\bf{r}}\left( t \right) = \smallint \left( {\sin t} \right){\bf{u}}{\rm{d}}t = - {\bf{u}}\cos t + {\bf{c}}$,
where ${\bf{c}}$ is a constant vector.
Since ${\bf{r}}\left( 0 \right) = {\bf{0}}$, we have ${\bf{r}}\left( 0 \right) = - {\bf{u}} + {\bf{c}} = {\bf{0}}$. So, ${\bf{c}} = {\bf{u}}$.
${\bf{r}}\left( t \right) = - {\bf{u}}\cos t + {\bf{u}}$,
Hence, the solution is
${\bf{r}}\left( t \right) = \left( {1 - \cos t} \right){\bf{u}}$
