Answer
The normal vector is ${\bf{N}}\left( t \right) = \left( {0, - \sin 2t, - \cos 2t} \right)$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {4,\sin 2t,\cos 2t} \right)$.
The tangent vector is ${\bf{r}}'\left( t \right) = \left( {0,2\cos 2t, - 2\sin 2t} \right)$.
We find the unit tangent vector ${\bf{T}}\left( t \right)$:
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {0,2\cos 2t, - 2\sin 2t} \right)}}{{\sqrt {\left( {0,2\cos 2t, - 2\sin 2t} \right)\cdot\left( {0,2\cos 2t, - 2\sin 2t} \right)} }}$
${\bf{T}}\left( t \right) = \frac{{\left( {0,2\cos 2t, - 2\sin 2t} \right)}}{{\sqrt {4{{\cos }^2}2t + 4{{\sin }^2}2t} }} = \left( {0,\cos 2t, - \sin 2t} \right)$
So, ${\bf{T}}'\left( t \right) = \left( {0, - 2\sin 2t, - 2\cos 2t} \right)$
By Eq. (6), the normal vector ${\bf{N}}\left( t \right)$ is given by
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \frac{{\left( {0, - 2\sin 2t, - 2\cos 2t} \right)}}{{\sqrt {\left( {0, - 2\sin 2t, - 2\cos 2t} \right)\cdot\left( {0, - 2\sin 2t, - 2\cos 2t} \right)} }}$
${\bf{N}}\left( t \right) = \frac{{\left( {0, - 2\sin 2t, - 2\cos 2t} \right)}}{{\sqrt {4{{\sin }^2}2t + 4{{\cos }^2}2t} }} = \left( {0, - \sin 2t, - \cos 2t} \right)$
