Answer
(a) the arc length function is
$s\left( t \right) = t||{\bf{v}}||$
(b) the arc length parametrization is ${{\bf{r}}_1}\left( s \right) = \left( {1,2,3} \right) + \frac{s}{{5\sqrt 2 }}\left( {3,4,5} \right)$
Work Step by Step
(a) We have the parametrization of a line ${\bf{r}}\left( t \right) = {\bf{w}} + t{\bf{v}}$.
The derivative is ${\bf{r}}'\left( t \right) = {\bf{v}}$. So, the arc length function is
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = t||{\bf{v}}||$
If ${\bf{v}}$ is a unit vector, then $s=t$. This implies that ${\bf{r}}\left( t \right)$ is parametrized by arc length. Conversely, if ${\bf{r}}\left( t \right)$ is parametrized by arc length, then we must have $||{\bf{v}}|| = 1$. Hence, ${\bf{r}}\left( t \right)$ is an arc length parametrization if and only if ${\bf{v}}$ is a unit vector.
(b) We have the parametrization of a line ${\bf{r}}\left( t \right) = {\bf{w}} + t{\bf{v}}$, where ${\bf{w}} = \left( {1,2,3} \right)$ and ${\bf{v}} = \left( {3,4,5} \right)$.
${\bf{r}}\left( t \right) = \left( {1,2,3} \right) + t\left( {3,4,5} \right)$
The derivative is ${\bf{r}}'\left( t \right) = \left( {3,4,5} \right)$.
Evaluate the arc length function:
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_0^t \sqrt {\left( {3,4,5} \right)\cdot\left( {3,4,5} \right)} {\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_0^t \sqrt {50} {\rm{d}}u = 5\sqrt 2 t$
The inverse of $s\left( t \right)$ is $t = \frac{s}{{5\sqrt 2 }}$. Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {1,2,3} \right) + t\left( {3,4,5} \right)$ gives
${{\bf{r}}_1}\left( s \right) = \left( {1,2,3} \right) + \frac{s}{{5\sqrt 2 }}\left( {3,4,5} \right)$
Thus, the arc length parametrization is ${{\bf{r}}_1}\left( s \right) = \left( {1,2,3} \right) + \frac{s}{{5\sqrt 2 }}\left( {3,4,5} \right)$
