#### Answer

$\langle \frac{1}{\sqrt 2}, 0,\frac{-1}{\sqrt 2}\rangle$

#### Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
$\textbf{r}(t) = \langle sin(t), cos(t), e^{-t}\rangle$
$\textbf{r}'(t) = \langle cos(t), -sin(t), -e^{-t}\rangle$
$|\textbf{r}'(t)| = \sqrt {(cos(t))^2 + (-sin(t))^2 + (-e^{-t})^2} = \sqrt {1 + e^{-2t}}$
$\textbf{T}(t) = \frac{\textbf{r}'(t)}{|\textbf{r}'(t)|} = \frac{\langle cos(t), -sin(t), -e^{-t}\rangle}{ \sqrt {1 + e^{-2t}}}$
$\textbf{T}(0) = \frac{\langle cos(0), -sin(0), -e^{-0}\rangle}{ \sqrt {1 + e^{-2(0)}}} = \langle \frac{1}{\sqrt 2}, 0,\frac{-1}{\sqrt 2}\rangle$