Intersection/Collision of two lines in R^3

[pearss]

Homework Statement

Determine whether r1 and r2 collide or intersect:

r1(t) = <t^2 + 3 , t + 1 , 6t^-1 >

r2(t) = <4t , 2t -2 , t^2 - 7>

I am completely lost in this problem and was hoping for a just a hint at where to begin. I'm unsure what it even means if two lines collide or intersect.

I've done a similar problem that read:

Determine if

r1(t) = < 1 , 0 , 1 > + t<3, 3, 5 >

and

r2(t) = < 3, 6, 1 > +t<4, -2, 7>

intersect.

I did it by multiplying the scalars out and adding the two vectors. Then setting the x components of the two lines equal to each other...same with y and z. This gives me three equations with which i use to solve for t1 and t2. Finally, plugging the t values into the third equation will prove whether or not the lines intersect if the equation is satisfied with the two t values.

I'm unsure what 'collision' is. Do i approach this problem the same way?

Thanks all

 

[Dick]

Yes, do it the same way for intersect. 'Collide' I think means that they intersect with the same value of t in each equation. I.e. they are at the same place at the same time.

 

[pearss]

so if i find the value for t1 to be 3 and the value for t2 to be 3 and they satisfy all equaitons for x, y and z then these lines collide because both "t" values are the same and all intersections are the same?

 

[Dick]

so if i find the value for t1 to be 3 and the value for t2 to be 3 and they satisfy all equaitons for x, y and z then these lines collide because both "t" values are the same and all intersections are the same?

Yes, t=3 is a collision. There MIGHT be more intersections that aren't collisions. But in this case I don't think there are.

 

[Dick]

so if i find the value for t1 to be 3 and the value for t2 to be 3 and they satisfy all equaitons for x, y and z then these lines collide because both "t" values are the same and all intersections are the same?

Yes, t=3 is a collision. There MIGHT be more intersections that aren't collisions. But in this case I don't think there are.