Answer
The decomposition of ${\bf{a}}$ at $t = \frac{\pi }{2}$:
${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$
$\left( {0,0, - \frac{\pi }{2}} \right) = - \frac{\pi }{{2\sqrt 3 }}{\bf{T}} + \frac{\pi }{{\sqrt 6 }}{\bf{N}}$,
where ${\bf{T}} = \left( {\frac{1}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$ and ${\bf{N}} = \left( {\frac{1}{{\sqrt 6 }}, - \frac{1}{{\sqrt 6 }}, - \frac{1}{3}\sqrt 6 } \right)$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,\cos t,t\sin t} \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {1, - \sin t,\sin t + t\cos t} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {0, - \cos t,2\cos t - t\sin t} \right)$, respectively.
At $t = \frac{\pi }{2}$, we get ${\bf{v}}\left( {\frac{\pi }{2}} \right) = \left( {1, - 1,1} \right)$ and ${\bf{a}}\left( {\frac{\pi }{2}} \right) = \left( {0,0, - \frac{\pi }{2}} \right)$. Thus, the unit tangent vector is
${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {1, - 1,1} \right)}}{{\sqrt {\left( {1, - 1,1} \right)\cdot\left( {1, - 1,1} \right)} }}$
${\bf{T}} = \left( {\frac{1}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$
By Eq. (2) of Theorem 1 we have
${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = \left( {0,0, - \frac{\pi }{2}} \right)\cdot\left( {\frac{1}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$
${a_{\bf{T}}} = - \frac{\pi }{{2\sqrt 3 }}$
Next, we use Eq. (3) of Theorem 1 to find
${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$
${a_{\bf{N}}}{\bf{N}} = \left( {0,0, - \frac{\pi }{2}} \right) + \frac{\pi }{{2\sqrt 3 }}\left( {\frac{1}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$
${a_{\bf{N}}}{\bf{N}} = \left( {0,0, - \frac{\pi }{2}} \right) + \left( {\frac{\pi }{6}, - \frac{\pi }{6},\frac{\pi }{6}} \right) = \left( {\frac{\pi }{6}, - \frac{\pi }{6}, - \frac{\pi }{3}} \right)$
Since ${\bf{N}}$ is an unit vector, so
${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}|| = \sqrt {{{\left( {\frac{\pi }{6}} \right)}^2} + {{\left( { - \frac{\pi }{6}} \right)}^2} + {{\left( { - \frac{\pi }{3}} \right)}^2}} $
${a_{\bf{N}}} = \frac{\pi }{{\sqrt 6 }}$
To find ${\bf{N}}$ we use the equation
${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \frac{{\sqrt 6 }}{\pi }\left( {\frac{\pi }{6}, - \frac{\pi }{6}, - \frac{\pi }{3}} \right)$
${\bf{N}} = \left( {\frac{1}{{\sqrt 6 }}, - \frac{1}{{\sqrt 6 }}, - \frac{1}{3}\sqrt 6 } \right)$
Thus, we obtain the decomposition of ${\bf{a}}$ at $t = \frac{\pi }{2}$:
${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$
$\left( {0,0, - \frac{\pi }{2}} \right) = - \frac{\pi }{{2\sqrt 3 }}{\bf{T}} + \frac{\pi }{{\sqrt 6 }}{\bf{N}}$,
where ${\bf{T}} = \left( {\frac{1}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$ and ${\bf{N}} = \left( {\frac{1}{{\sqrt 6 }}, - \frac{1}{{\sqrt 6 }}, - \frac{1}{3}\sqrt 6 } \right)$.
