The unit normal vector is then,

\[\begin{array}{c}\vec T'\left( t \right) = \left\langle { - 2\cos \left( {2t} \right), - 2\sin \left( {2t} \right),0} \right\rangle \hspace{0.5in}\left\| {\vec T'\left( t \right)} \right\| = \sqrt {4{{\cos }^2}\left( {2t} \right) + 4{{\sin }^2}\left( {2t} \right)} = 2\\ \require{bbox} \bbox[2pt,border:1px solid black]{{\vec N\left( t \right) = \frac{1}{2}\left\langle { - 2\cos \left( {2t} \right), - 2\sin \left( {2t} \right),0} \right\rangle = \left\langle { - \cos \left( {2t} \right), - \sin \left( {2t} \right),0} \right\rangle }}\end{array}\] Show Step 3

Finally, the binormal vector is,

\[\begin{align*}\vec B\left( t \right) & = \vec T\left( t \right) \times \vec N\left( t \right) = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - \sin \left( {2t} \right)}&{\cos \left( {2t} \right)}&0\\{ - \cos \left( {2t} \right)}&{ - \sin \left( {2t} \right)}&0\end{array}} \right|\\ & = {\sin ^2}\left( {2t} \right)\vec k - \left( { - {{\cos }^2}\left( {2t} \right)\vec k} \right) = \left( {{{\sin }^2}\left( {2t} \right) + {{\cos }^2}\left( {2t} \right)} \right)\vec k = \require{bbox} \bbox[2pt,border:1px solid black]{{\vec k = \left\langle {0,0,1} \right\rangle = \vec B\left( t \right)}}\end{align*}\]