Answer
$\dfrac{\pi}{2}$
Work Step by Step
Here, $r_1'(t)=-\sin t i+\cos t j+k; r_1'(0)=0 i+ j+k$
and $r_2'(t)=i+2t j+3t^2 k; r_1'(0)=i+0 j+0 k$
Apply the dot product formula to find the angle.
$\theta=\cos^{-1} [\dfrac{p \cdot q}{|p||q|}]$
Now, we have
$\theta=\cos^{-1} [\dfrac{(0)(1)+(1)(0)+(1)(0)}{|1||1|}]$
or, $\theta=\cos^{-1} [0]$
Hence, $\theta=\dfrac{\pi}{2}$
