Answer
$\langle 0,0,-1\rangle$
Work Step by Step
$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
$\textbf{r}(t) = \langle cos(2t), 4, 3sin(2t)\rangle$
$\textbf{r}'(t) = \langle -2sin(2t), 0, 6cos(2t)\rangle$
$|\textbf{r}'(t)| = \sqrt {(-2sin(2t))^2 + (0)^2 + (6cos(2t))^2} = \sqrt {4sin^2(2t) + 36cos^2(2t)}$
$\textbf{T}(t) = \frac{\textbf{r}'(t)}{|\textbf{r}'(t)|} = \frac{\langle -2sin(2t), 0, 6cos(2t)\rangle}{\sqrt {4sin^2(2t) + 36cos^2(2t)}}$
$\textbf{T}(\pi/2) =\frac{\langle -2sin(2(\pi/2)), 0, 6cos(2(\pi/2))\rangle}{\sqrt {4sin^2(2(\pi/2)) + 36cos^2(2(\pi/2))}} = \langle 0,0,-1\rangle$
