I've been stumped on this all day, so hopefully someone can help me. I have two vector functions r1(t) and r2(t). r1(t)=< t,t^2,t^3 > r2(t)=< 1+2t,1+6t, 1+14t > How do i tell if the paths intersect - not collide. For the record, they don't collide, but I need proof that they don't intersect (which I'm pretty sure they don't?).
What's the difference between intersection and collision? Anyway, if there is some t for which: r<sub>1</sub>(t)=r<sub>2</sub>(t) then the two intersect. So, set the components of each vector equal to each other and solve for t. If you find a t for which the two vectors are equal (in each component), then you have proved that they intersect.
Re: Re: vector function question if the two objects meet at the same time, then they collide. your solution, tom, will find these collisions. i think what tempus means by intersect, is the intersection of the trajectories. your solution will not find these.
i found it. and yes, i wanted path intersections - not collisions. the answer is 2 points. r1(2)=r2(1/2), which is (2,4,8), and r1(1)=r2(0), which is (1,1,1).
To find where (if) the trajectories intersect, treat the two parameters as distinct variables. Solve <t, t<sup>2</sup>, t<sup>3</sup>>= <1+ 2s, 1+6s, 1+ 14s> which is the same as the three equations: t= 1+ 2s t<sup>2</sup>= 1+ 6s t<sup>3</sup>= 1+ 14s That is three equations in 2 unknowns. In general, two arbitrary one-dimensional trajectories in three dimensions do NOT intersect. From the first equation, 2s= t-1 so s= (t-1)/2. Put that into the second equation to get t<sup>2</sup>- 6(t-1)/2- 1= 0 or t<sup>2</sup>- 3t+ 2= (t-1)(t-2)= 0. There are two solutions for t: t= 1, in which case s= (1-1)/2= 0, and t= 2, in which case s= (2-1)/2= 1/2. When t= 1, the object in the first trajectory is at <1, 1, 1> and when s= 0, the object in the second trajectory <1, 1, 1> so <1,1,1> is on both trajectories. When t= 2, the object in the first trajectory is at <2, 4, 8> and when s= 1/2, the object in the second trajectory is at <2, 4, 8> so <2, 4, 8> is on both trajectories. Yes, the trajectories intersect!
that's exactly how i solved it Please Register or Log in to view the hidden image! but thanks Please Register or Log in to view the hidden image!.
