Answer
$$ r'(t) = \lt 4, − 5, 9 \gt, \quad T(t)= \frac{ \lt 4, − 5, 9\gt}{\sqrt{122}}, \quad T(1)=\frac{ \lt 4, − 5, 9\gt}{\sqrt{122}}.$$
Work Step by Step
Since $ r(t) = \lt 3 + 4t, 3 − 5t, 9t\gt$, then $ r'(t) = \lt 4, − 5, 9 \gt$ and $\|r'(t)\|=\sqrt{ 122} $
Hence, we have
$$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt 4, − 5, 9\gt}{\sqrt{122}}, \quad T(1)=\frac{ \lt 4, − 5, 9\gt}{\sqrt{122}}.$$
