Answer
$${\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} = \frac{{\left\langle {t, - 3,0} \right\rangle }}{{\sqrt {{t^2} + 9} }}{\text{ and }}{\bf{N}}\left( t \right) = \frac{{\left\langle {3,t,0} \right\rangle }}{{\sqrt {{t^2} + 9} }}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {\frac{{{t^2}}}{2},4 - 3t,1} \right\rangle \cr
& {\text{Calculate }}{\bf{v}}\left( t \right){\text{ and }}\left| {{\bf{v}}\left( t \right)} \right| \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left\langle {\frac{{{t^2}}}{2},4 - 3t,1} \right\rangle \cr
& {\bf{v}}\left( t \right) = \left\langle {t, - 3,0} \right\rangle \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{t^2} + 9} \cr
& \cr
& {\text{Find the unit tangent vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} = \frac{{\left\langle {t, - 3,0} \right\rangle }}{{\sqrt {{t^2} + 9} }} \cr
& \cr
& {\text{Find the principal unit normal vector }}{\bf{N}}\left( t \right) \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left\langle {\frac{t}{{\sqrt {{t^2} + 9} }}, - \frac{3}{{\sqrt {{t^2} + 9} }},0} \right\rangle \cr
& {\bf{T}}'\left( t \right) = \left\langle {\frac{9}{{{{\left( {{t^2} + 9} \right)}^{3/2}}}},\frac{{3t}}{{{{\left( {{t^2} + 9} \right)}^{3/2}}}},0} \right\rangle \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \sqrt {\frac{{81}}{{{{\left( {{t^2} + 9} \right)}^3}}} + \frac{{9{t^2}}}{{{{\left( {{t^2} + 9} \right)}^3}}}} \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \frac{{3\sqrt {9 + {t^2}} }}{{{{\left( {{t^2} + 9} \right)}^{3/2}}}} \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \frac{3}{{{t^2} + 9}} \cr
& {\text{Then}} \cr
& {\bf{N}}\left( t \right) = \frac{{{t^2} + 9}}{3}\left\langle {\frac{9}{{{{\left( {{t^2} + 9} \right)}^{3/2}}}},\frac{{3t}}{{{{\left( {{t^2} + 9} \right)}^{3/2}}}},0} \right\rangle \cr
& {\bf{N}}\left( t \right) = \frac{1}{{{{\left( {{t^2} + 9} \right)}^{1/2}}}}\left\langle {3,t,0} \right\rangle \cr
& {\bf{N}}\left( t \right) = \frac{{\left\langle {3,t,0} \right\rangle }}{{\sqrt {{t^2} + 9} }} \cr} $$
