Intersection of vector functions???

Question

At what points do the curves given by vector functions r(t) = and r(s)=<3-s, s-2, s^2> intersect? Find their angle of intersection correct to the nearest degree.

Answer 1

Uh... point of notation here, r is the name of the function, not t. r(t) denotes the function r evaluated at t, so if r(t)=, then r(s) must equal . Presumably you meant r(t)= and s(t)=<3-t, t-2, t²>.

Anyway, if r and s intersect, then that means that ∃t₁, t₂ such that r(t₁)=s(t₂). Equating the x-components, this means that t₁=3-t₂. So this means that we are looking for points where r(3-t)=s(t). This means that the z-components must be equal, so r_z(3-t)=s_z(t). Thus:

3+(3-t)²=t²
3+9-6t+t²=t²
12-6t=0
t=2

So the only possible intersection is where t₂=2, and t₁=3-2=1. Evaluating the vectors:

r(1)=<1, 0, 4>
s(2)=<1, 0, 4>

So the curves do in fact intersect at the point <1, 0, 4>. This is the only point of intersection.

To find the angle of intersection, we first find the velocity vectors:

r'(t)=<1, -1, 2t>
s'(t)=<-1, 1, 2t>

Evaluating them at the point of intersection (which as you recall, occurred at t₁=1 and t₂=2):

r'(1)=<1, -1, 2>
s'(2)=<-1, 1, 4>

So now we find the angle between these vectors, which is given by:

cos θ = r'(1)·s'(2)/(||r'(1)||*||s'(2)||)

We have that:

r'(1)·s'(2) = -1-1+8 = 6
||r'(1)||=√(1+1+4)=√6
||s'(2)||=√(1+1+16)=√18=3√2

Thus:

cos θ = 6/(√6*3√2) = √3/3 = 1/√3
θ = arccos (1/√3) ≈ 55°

And we are done.