Answer
The decomposition of ${\bf{a}}$:
${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}} = \frac{5}{3}{\bf{T}} + \frac{{10\sqrt 2 }}{3}{\bf{N}}$,
where ${\bf{T}} = \left( {\frac{2}{3},\frac{2}{3}, - \frac{1}{3}} \right)$ and ${\bf{N}} = \left( { - \frac{1}{{3\sqrt 2 }},\frac{{13}}{{15\sqrt 2 }},\frac{{16}}{{15\sqrt 2 }}} \right)$.
Work Step by Step
We have at some time $t$, ${\bf{v}} = \left( {2,2, - 1} \right)$. So, the unit tangent vector is
${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {2,2, - 1} \right)}}{{\sqrt {\left( {2,2, - 1} \right)\cdot\left( {2,2, - 1} \right)} }} = \frac{1}{3}\left( {2,2, - 1} \right)$
${\bf{T}} = \left( {\frac{2}{3},\frac{2}{3}, - \frac{1}{3}} \right)$
By Eq. (2) of Theorem 1 we have
${a_{\bf{T}}} = \frac{{{\bf{a}}\cdot{\bf{v}}}}{{||{\bf{v}}||}} = \frac{{\left( {0,4,3} \right)\cdot\left( {2,2, - 1} \right)}}{{\sqrt {\left( {2,2, - 1} \right)\cdot\left( {2,2, - 1} \right)} }} = \frac{5}{3}$
${a_{\bf{N}}} = \sqrt {||{\bf{a}}|{|^2} - |{a_{\bf{T}}}{|^2}} $
${a_{\bf{N}}} = \sqrt {\left( {0,4,3} \right)\cdot\left( {0,4,3} \right) - \frac{{25}}{9}} = \sqrt {25 - \frac{{25}}{9}} = \frac{{10\sqrt 2 }}{3}$
By Eq. (3) of Theorem 1, we have
${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - \left( {\frac{{{\bf{a}}\cdot{\bf{v}}}}{{{\bf{v}}\cdot{\bf{v}}}}} \right){\bf{v}}$
${a_{\bf{N}}}{\bf{N}} = \left( {0,4,3} \right) - \left( {\frac{{\left( {0,4,3} \right)\cdot\left( {2,2, - 1} \right)}}{{\left( {2,2, - 1} \right)\cdot\left( {2,2, - 1} \right)}}} \right)\left( {2,2, - 1} \right)$
${a_{\bf{N}}}{\bf{N}} = \left( {0,4,3} \right) - \frac{5}{9}\left( {2,2, - 1} \right) = \left( { - \frac{{10}}{9},\frac{{26}}{9},\frac{{32}}{9}} \right)$
Since ${a_{\bf{N}}} = \frac{{10\sqrt 2 }}{3}$, so
${\bf{N}} = \frac{3}{{10\sqrt 2 }}\left( { - \frac{{10}}{9},\frac{{26}}{9},\frac{{32}}{9}} \right) = \left( { - \frac{1}{{3\sqrt 2 }},\frac{{13}}{{15\sqrt 2 }},\frac{{16}}{{15\sqrt 2 }}} \right)$
Thus, the decomposition of ${\bf{a}}$:
${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}} = \frac{5}{3}{\bf{T}} + \frac{{10\sqrt 2 }}{3}{\bf{N}}$,
where ${\bf{T}} = \left( {\frac{2}{3},\frac{2}{3}, - \frac{1}{3}} \right)$ and ${\bf{N}} = \left( { - \frac{1}{{3\sqrt 2 }},\frac{{13}}{{15\sqrt 2 }},\frac{{16}}{{15\sqrt 2 }}} \right)$.
