Answer
The unit tangent vector at $t = \pi $ is ${\bf{T}}\left( \pi \right) = \left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},0} \right)$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\sin t,t,\cos t} \right)$. The tangent vector is
${\bf{r}}'\left( t \right) = \left( {\cos t,1, - \sin t} \right)$
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {\cos t,1, - \sin t} \right)}}{{\sqrt {\left( {\cos t,1, - \sin t} \right)\cdot\left( {\cos t,1, - \sin t} \right)} }}$
${\bf{T}}\left( t \right) = \frac{{\left( {\cos t,1, - \sin t} \right)}}{{\sqrt {{{\cos }^2}t + 1 + {{\sin }^2}t} }} = \left( {\frac{{\cos t}}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}, - \frac{{\sin t}}{{\sqrt 2 }}} \right)$
The unit tangent vector at $t = \pi $ is ${\bf{T}}\left( \pi \right) = \left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},0} \right)$.
