I have the following vector valued function: $$\vec{r}(t)= \begin{bmatrix}\vec{i} & \vec{j} & \vec{k} \end{bmatrix} \begin{bmatrix}f(t) \\ g(t) \\ 0 \end{bmatrix}$$ I have to calculate the unit tangent vector which I presume would be as follows: $$\frac{d}{dt}\vec{r}(t)= \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \end{bmatrix} \begin{bmatrix} f'(t) \\ g'(t)\\ 0 \end{bmatrix}$$
The formula for unit tangent vector is given as follows: $$\vec{T}(t) = \frac{\vec{r'}(t)}{||\vec{r'}(t)||}$$
Substituting in what I have from my $\vec{r'}(t)$ I get the following:
$$\vec{T}(t) = \frac{\begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \end{bmatrix}\begin{bmatrix} f'(t) \\ g'(t)\\ 0 \end{bmatrix}}{\sqrt{(f'(t)^2) +(g'(t))^2+(0)^2}}$$ This would be my final answer right due in part with some simplifications. is this the right method to calculate the Unit Tangent Vector.