At what point do the curves r1 and r2 intersect

At what point do the curves r1 = < t, 3 - t, 15 + t^2 >
and r2 = < 5 - s, s - 2, s^2 > intersect?


( ____, ____, ____)

Find their angle of intersection, θ correct to the nearest degree.
θ = ____°

When x, y, and z coordinates are all equal at the same time

x-coordinates: t = 5 - s
y-coordinates: 3 - t = s - 2 -----> -t = s - 5 ----> t = 5 - s
z-coordinates: 15 + t² = s²

Notice that equations we get from x- and y-coordinates both give us t = 5 - s
So we will replace t with this value in equation we derived from z-coordinate:
15 + (5-s)² = s²
15 + 25 - 10s + s² = s²
40 - 10s = 0
10s = 40
s = 4

t = 5 - s = 1

So we will get point of intersection when t = 1 and s = 4
(t, 3-t, 15+t²) = (1, 3-1, 15+1) = (1, 2, 16)
(5-s, s-2, s²) = (5-4, 4-2, 4²) = (1, 2, 16)

Point of intersection: (1, 2, 16)

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To find angle of intersection, we find gradient vectors at point (1, 2, 16)
Angle between curves at intersection = angle between gradient vectors.

r₁ = < t, 3-t, 15+t² >
dr₁/dt = < 1, -1, 2t >
u = dr₁/dt at t = 1
u = < 1, -1, 2 >

r₂ = < 5-s, s-2, s² >
dr₂/dx = < -1, 1, 2s >
v = dr₂/dx at s = 4
v = < -1, 1, 8 >

We use dot product to find θ
cos θ = u.v / (||u|| ||v||)

u.v = < 1, -1, 2 > . < -1, 1, 8 > = -1 - 1 + 16 = 14
||u|| = √(1+1+4) = √6
||v|| = √(1+1+64) = √66

cos θ = 14 / (√6√66)
cos θ = 14 / (6 √11)
cos θ = 7 / (3 √11)

θ = arccos(7/(3√11)) = 45.289377545 = 45° (to the nearest degree)

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