At what point do the curves r1(t) = r 1 t

t, 2 − t, 15 + t2 t t t 2

and r2(s) = r 2 s

5 − s, s − 3, s2 s s s 2

intersect? (x, y, z) = x y z

Find their angle of intersection, θ, correct to the

nearest degree.θ = °

θ θ

Consider., (t)-(t, 2-t,15+t2), rs(s) (5-s,5-3,s) These are intersect means, ri (t)-(s) i coefficients j coefficients k coefficients – s2. 15+t2 s Substitute t-5-s in 15+ 15+25+52-10s-32 40 = 10s From first equation, Thus, t-1,s -4. Therefore n(t)-(t, 2-t,15+t2-1, 1, 16) S, S The angle between these interesting vectors is,

S. Here S, is a slope vector at t-1. s(t) dt 15+t dt =(1,-1, 2t S2 is a slope vector at s-1. (n(s)) = S-4 -(-1,1. 2s), Therefore, the angle is, cos e 1-1+16 14 396 14 396