I need help finding the angle of the curves $\displaystyle r_1(t) = (t, 1-t, 3+t^2), \ \ r_2(s) = (3-s, s-2, s^2)$ I found that they intersect at $\displaystyle x=1, \ \ y = 0, \ \ z= 4$ , and would the angle between them be $\displaystyle cos^{-1} \frac{(3)(0)+(1)(-2)+(3)(1)}{(\sqrt{1+9})(\sqrt{9+4+1})} = cos^{-1} \frac{1}{\sqrt{140}}=1.48618$ ?