Answer
$${\bf{N}}\left( {\frac{{3\pi }}{4}} \right) = \frac{{\sqrt 2 }}{2}{\bf{i}} - \frac{{\sqrt 2 }}{2}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 6\cos t{\bf{i}} + 6\sin t{\bf{j}} + {\bf{k}},{\text{ }}t = \frac{{3\pi }}{4} \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = - 6\sin t{\bf{i}} + 6\cos t{\bf{j}} \cr
& {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}'\left( t \right)} \right\|}},{\text{ }}{\bf{r}}'\left( t \right) \ne 0 \cr
& {\bf{T}}\left( t \right) = \frac{{ - 6\sin t{\bf{i}} + 6\cos t{\bf{j}}}}{{\left\| { - 6\sin t{\bf{i}} + 6\cos t{\bf{j}}} \right\|}} = \frac{{ - 6\sin t{\bf{i}} + 6\cos t{\bf{j}}}}{{\sqrt {36{{\sin }^2}t + 36{{\cos }^2}t} }} \cr
& {\bf{T}}\left( t \right) = \frac{{ - 6\sin t{\bf{i}} + 6\cos t{\bf{j}}}}{6} \cr
& {\bf{T}}\left( t \right) = - \sin t{\bf{i}} + \cos t{\bf{j}} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ { - \sin t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\cos t} \right]{\bf{j}} \cr
& {\bf{T}}'\left( t \right) = - \cos t{\bf{i}} - \sin t{\bf{j}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = \frac{{3\pi }}{4} \cr
& {\bf{T}}'\left( {\frac{{3\pi }}{4}} \right) = - \cos \left( {\frac{{3\pi }}{4}} \right){\bf{i}} - \sin \left( {\frac{{3\pi }}{4}} \right){\bf{j}} \cr
& {\bf{T}}'\left( {\frac{{3\pi }}{4}} \right) = \frac{{\sqrt 2 }}{2}{\bf{i}} - \frac{{\sqrt 2 }}{2}{\bf{j}} \cr
& \left\| {{\bf{T}}'\left( {\frac{{3\pi }}{4}} \right)} \right\| = \sqrt {{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( { - \frac{{\sqrt 2 }}{2}} \right)}^2}} = 1 \cr
& {\text{Finding the Principal Unit Normal Vector}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr
& {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = \frac{{3\pi }}{4} \cr
& {\bf{N}}\left( {\frac{{3\pi }}{4}} \right) = \frac{{{\bf{T}}'\left( {3\pi /4} \right)}}{{\left\| {{\bf{T}}'\left( {3\pi /4} \right)} \right\|}} = \frac{{\frac{{\sqrt 2 }}{2}{\bf{i}} - \frac{{\sqrt 2 }}{2}{\bf{j}}}}{1} \cr
& {\bf{N}}\left( {\frac{{3\pi }}{4}} \right) = \frac{{\sqrt 2 }}{2}{\bf{i}} - \frac{{\sqrt 2 }}{2}{\bf{j}} \cr} $$
