Answer
$\theta=\dfrac{\pi}{2}$
Work Step by Step
Here, we have $r_1'(t)=-\sin (t) i+\cos (t) j+k$ and
$ r_1'(0)=0 i+ j+k$
Now, $r_2'(t)=i+2t j+3t^2 k$ and
$ r_1'(0)=i+0 j+0 k$
Need to use dot product formula to find the angle.
$\theta=\cos^{-1} [\dfrac{O \cdot P}{|||P|}]$
$\theta=\cos^{-1} [\dfrac{(0)(1)+(1)(0)+(1)(0)}{|1||1|}]$
This gives: $\theta=\cos^{-1} [0] \implies \theta=\dfrac{\pi}{2}$
