Answer
Point of intersection is: $ \lt 1,0,4 \gt$ and the angle of intersection is $55^\circ$
Work Step by Step
From the given curves we have $r_1(t)=\lt t,1-t,3+t^2 \gt $ and $r_2(s)= \lt 3-s,s-2,s^2 \gt $
The given curves will intersect when their coordinates are to be equal.
Thus, $t=3-s,1-t=s-2,3+t^2=s^2$
Point of intersection is: $ \lt 1,0,4 \gt$
$r_1'(t)=\lt 1,-1,2t \gt $ and $r_2(s)= \lt -1,1,2s \gt $
and $r_1'(1)=\lt 1,-1,2 \gt $ and $r_2(s)= \lt -1,1,4 \gt $
Let $\theta$ be the angle between the two curves, then:
$\cos \theta=\dfrac{\lt 1,0,0 \gt \cdot\lt 1,2,1 \gt}{\sqrt {1+0+0}\sqrt {1+4+1}}$
$\implies \theta=\cos ^{-1}[\dfrac{\lt 1,-1,2 \gt \cdot\lt -1,1,4 \gt}{\sqrt {1+1+4}\sqrt {1+1+16}}]=\cos ^{-1}\dfrac{1}{\sqrt 3}\approx 55^\circ$
Hence, point of intersection is : $ \lt 1,0,4 \gt$ and the angle of intersection is $55^\circ$
