Answer
\begin{array}{l}
T(0)=\frac{1}{\sqrt{5}} j-\frac{2}{\sqrt{5}} k \\
N(0) \frac{-2}{3} i-\frac{2}{3} j-\frac{1}{3} k \\
B(0)=\left\langle-\frac{\sqrt{5}}{3}, \frac{4 \sqrt{5}}{15}, \frac{2 \sqrt{5}}{15}\right\rangle
\end{array}
Work Step by Step
We find:
\[
\mathbf{r}(t)=\left\langle 2 \cos t, 2 \cos t+\frac{3}{\sqrt{5}} \sin t, \cos t-\frac{6}{\sqrt{5}} \sin t\right\rangle
\]
\[
\begin{array}{l}
\left\langle-2 \sin t,-2 \sin t+\frac{3}{\sqrt{5}} \cos t,-\sin t-\frac{6}{\sqrt{5}} \cos t\right\rangle=\mathrm{r}^{\prime}(t) \\
\left\langle-2 \cos t,-2 \cos t-\frac{3}{\sqrt{5}} \sin t,-\cos t+\frac{6}{\sqrt{5}} \sin t\right\rangle =\mathrm{r}^{\prime \prime}(t)\\
\qquad \begin{aligned}
\mathrm{T}(0)=& \frac{\mathrm{r}^{\prime}(0)}{\left\|\mathrm{r}^{\prime}(0)\right\|} \\
=& \frac{(0) \mathrm{i}+(3 / \sqrt{5}) \mathrm{j}-(6 / \sqrt{5}) \mathrm{k}}{\sqrt{(9 / 5)+(36) / 5}} \\
=& \frac{1}{\sqrt{5}} \mathrm{j}-\frac{2}{\sqrt{5}} \mathrm{k} \\
=& \frac{\mathrm{r}^{\prime \prime}(0)}{\left\|\mathrm{r}^{\prime \prime}(0)\right\|} \\
=& \frac{-2 \mathrm{i}-2 \mathrm{j}-\mathrm{k}}{\sqrt{4+4+1}} \\
=& \frac{-2}{3} \mathrm{i}-\frac{2}{3} \mathrm{j}-\frac{1}{3} \mathrm{k}
\end{aligned} \\
\begin{aligned}
\mathrm{B}(0)=\mathrm{T}(0) \times \mathrm{N}(0) & \\
=\frac{1}{3 \sqrt{5}}\left|\begin{array}{rrr}
\mathrm{N} & 1 & \mathrm{j} \\
1 & -2 & 1 \\
-2 & -2 & -1
\end{array}\right| \\
=&\left\langle-\frac{\sqrt{5}}{3}, \frac{4 \sqrt{5}}{15}, \frac{2 \sqrt{5}}{15}\right\rangle
\end{aligned}
\end{array}
\]
