Answer
$${\bf{N}}\left( 1 \right) = - \frac{{\sqrt {14} }}{{14}}{\bf{i}} + \frac{{2\sqrt {14} }}{{14}}{\bf{j}} - \frac{{3\sqrt {14} }}{{14}}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = t{\bf{i}} + {t^2}{\bf{j}} + \ln t{\bf{k}},{\text{ }}t = 1 \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = {\bf{i}} + 2t{\bf{j}} + \frac{1}{t}{\bf{k}} \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{{{\bf{i}} + 2t{\bf{j}} + \frac{1}{t}{\bf{k}}}}{{\left\| {{\bf{i}} + 2t{\bf{j}} + \frac{1}{t}{\bf{k}}} \right\|}} = \frac{{{\bf{i}} + 2t{\bf{j}} + \frac{1}{t}{\bf{k}}}}{{\sqrt {1 + 4{t^2} + \frac{1}{{{t^2}}}} }} \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{i}} + 2t{\bf{j}} + \frac{1}{t}{\bf{k}}}}{{\frac{{\sqrt {{t^2} + 4{t^4} + 1} }}{t}}} \cr
& {\bf{T}}\left( t \right) = \frac{{t{\bf{i}} + 2{t^2}{\bf{j}} + {\bf{k}}}}{{\sqrt {4{t^4} + {t^2} + 1} }} \cr
& {\bf{T}}\left( t \right) = \frac{t}{{\sqrt {4{t^4} + {t^2} + 1} }}{\bf{i}} + \frac{{2{t^2}}}{{\sqrt {4{t^4} + {t^2} + 1} }}{\bf{j}} + \frac{1}{{\sqrt {4{t^4} + {t^2} + 1} }}{\bf{k}} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{t}{{\sqrt {4{t^4} + {t^2} + 1} }}{\bf{i}} + \frac{{2{t^2}}}{{\sqrt {4{t^4} + {t^2} + 1} }}{\bf{j}} + \frac{1}{{\sqrt {4{t^4} + {t^2} + 1} }}{\bf{k}}} \right] \cr
& {\text{Making the operations by hand and replacing }}\left( {{\text{or use a CAS}}} \right) \cr
& {\bf{T}}'\left( t \right) = \frac{{1 - 4{t^4}}}{{{{\left( {4{t^4} + {t^2} + 1} \right)}^{3/2}}}}{\bf{i}} + \frac{{2{t^3} + 4t}}{{{{\left( {4{t^4} + {t^2} + 1} \right)}^{3/2}}}}{\bf{j}} - \frac{{8{t^3} + t}}{{{{\left( {4{t^4} + {t^2} + 1} \right)}^{3/2}}}}{\bf{k}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 1 \cr
& {\bf{T}}'\left( 1 \right) = \frac{{1 - 4}}{{{{\left( {4 + 1 + 1} \right)}^{3/2}}}}{\bf{i}} + \frac{{2 + 4}}{{{{\left( {4 + 1 + 1} \right)}^{3/2}}}}{\bf{j}} - \frac{{8 + 1}}{{{{\left( {4 + 1 + 1} \right)}^{3/2}}}}{\bf{k}} \cr
& {\bf{T}}'\left( 1 \right) = - \frac{3}{{{{\left( 6 \right)}^{3/2}}}}{\bf{i}} + \frac{6}{{{{\left( 6 \right)}^{3/2}}}}{\bf{j}} - \frac{9}{{{{\left( 6 \right)}^{3/2}}}}{\bf{k}} \cr
& \left\| {{\bf{T}}'\left( 1 \right)} \right\| = \sqrt {{{\left( { - \frac{3}{{{{\left( 6 \right)}^{3/2}}}}} \right)}^2} + {{\left( {\frac{6}{{{{\left( 6 \right)}^{3/2}}}}} \right)}^2} + \left( { - \frac{9}{{{{\left( 6 \right)}^{3/2}}}}} \right)} = \sqrt {\frac{7}{{12}}} \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 = 1 \cr
& {\bf{N}}\left( 1 \right) = \frac{{{\bf{T}}'\left( 1 \right)}}{{\left\| {{\bf{T}}'\left( 1 \right)} \right\|}} = \frac{{ - \frac{3}{{{{\left( 6 \right)}^{3/2}}}}{\bf{i}} + \frac{6}{{{{\left( 6 \right)}^{3/2}}}}{\bf{j}} - \frac{9}{{{{\left( 6 \right)}^{3/2}}}}{\bf{k}}}}{{\sqrt {\frac{7}{{12}}} }} \cr
& {\bf{N}}\left( 1 \right) = \frac{{2\sqrt {21} }}{7}\left( { - \frac{3}{{{{\left( 6 \right)}^{3/2}}}}{\bf{i}} + \frac{6}{{{{\left( 6 \right)}^{3/2}}}}{\bf{j}} - \frac{9}{{{{\left( 6 \right)}^{3/2}}}}{\bf{k}}} \right) \cr
& {\text{Multiply and simplify by using a calculator}} \cr
& {\bf{N}}\left( 1 \right) = - \frac{{\sqrt {14} }}{{14}}{\bf{i}} + \frac{{2\sqrt {14} }}{{14}}{\bf{j}} - \frac{{3\sqrt {14} }}{{14}}{\bf{k}} \cr} $$
