in this example we're going to look at how intersection and collision questions can be extended to 3d with the the important notation that simply graphing paths is not going to be as helpful if we imagine drawing some path in 3d with three axes here versus some other trajectory we can see very quickly that our human eyeballs are not good at determining whether this is a crossing a near miss or anything else we really do need to go down to the algebra down to the numbers or down to the symbols to be able to recognize whether we have an intersection a collision or nothing at all so let's look at that through the lens of an example we have two trajectories again here r of t and u of t and we're asking whether the trajectories cross so cross is the same as intersection and that we saw earlier was the same point and possibly different times so we have to allow for that so we're going to take the same approach we used earlier and we're going to imagine that we're tracing out the r path and that r path making it look like a vector over top here we're going to use t for that and we're going to use a different time value for u we're going to use a different variable we're going to call it s and we're interested in having the same x y and now z value at r of t [Music] and some other point u at some other time value s so we're going to set up our equations there and see what we can find to start or look at the x coordinates and again r of t on the left hand side each time and u of s on the right hand side the x coordinate we all have to have t equals let's gather these 1 plus 2 not ts but s's and then repeat we're going to have the y coordinates equal to each other t for the r coordinates 1 plus 6 not t but s's for the u path and last but not least we're going to have our z's and they are going to involve t cubed and 1 plus 14 somethings but it's not t's it's going to be s's again as well and there we are these are the conditions which if we can find a solution represent an intersection between the two paths let's see how we can work through this uh it looks like equation one is going to be the easiest to work with simply because it's all linear again so if we look at equation one and it's not super critical actually we have t equals a t squared here we might as well make our lives a little bit easier and simply sub this t into this value and see what happens we could also solve for s and do that into equation 2 as well there's really no right way to do this there's pros and cons or how the equations are going to look either way we'll try this easy one though we're going to sub t into here so we're going to have t squared but t squared is the same as 1 plus 2s it has to be to satisfy equation 1 and then the other side of this is 1 plus 6s if we do that then we want to be able to solve this it's just a quadratic but we have to expand it out a bit so we're gonna have one plus four s we've got a perfect square here four s squared one plus six s and if we bring everything over to the side we'll have four s squared we have a plus one plus one oh that's got to cancel and we have 4s minus 6s that'll be minus 2s by the time we're done and that is nice because there's an s factor 4s minus 2 and we can even take that one step further if we want we can divide by 2 and we'll have s and 2s minus 1. and that gives us either s equals 0 or 2s minus 1 equals 0 which is equivalent to 2s equals 1 which is the same as s equals 1 half all right so to satisfy one and two so let's put that in here to satisfy one and two here we have to have one of these two s values and we can get a little more precise because to satisfy one we have to satisfy equation one specifically so t would have to be equal one plus two times s let's put the one in here and one in here one plus two times s but s 0 so t would have to equal 1 and here t would have to equal 1 plus 2 s 1 plus 2 times a half that would be 2. so summarizing we have s equals zero t equals one or s equals a half t equals two those are two possibilities that satisfy the x and y coordinates what we need to do next is still check that the x or the z's rather intersect so i still need to check the z's are equal we don't have to look at any other values because these are the only pairs that are going to satisfy one and two we're actually going to grab this and copy it over to the next page and then we're gonna look at our cases so check three with s equals zero t equals one the left-hand side is t cubed that's one cubed that's one and the right hand side therefore it's going to be similar it's going to be 1 plus 14 times s but s is 0. no that's 1 as well excellent s equals 0 t equals 1 gives the same z value check and then we check 3 with the other value other pair values that are time options with s equals a half and t equals two left hand side of this equation is t cubed but t cubed is the same as two cubed for this point and that's eight while the right hand side is one plus fourteen s but s is a half for this test point fourteen times a half that's much better known as seven plus one is eight these are equal i should emphasize that up here as well so s equals a half t equals 2 also gives the same z so what we found are two sets of time values which produced the same x y and z coordinates and the question was well where do these actual intersections occur for let's just do a double check here it's always reassuring to see our numbers work out all the way through it so 4 s equals 0 t equals 1 we have r of 1 is equal to so the t remember went into the r function and it tells us that at time one for that trajectory you're going to be at one one squared one cubed well that's conveniently one one one and u at s equals zero so a little bit earlier we think u pass through that same coordinate and one plus two t but s is zero so it's one plus zero one one plus zero hey you see the pattern coming up here one one one so both of those trajectories go through the same point at different times just to compare and contrast for s equals one-half and t equals one again try not to get these mixed up it's awkward with the labeling list that i did there but r which we use the time value t when it's evaluated at one sorry at two my apologies the second solution was one half and two r of two is going to be two two squared two cubed or two four and eight and u of one half is going to be one plus two halves that's one plus one hey that's two excellent one plus three sorry one plus three because one is six halves not surprisingly because we saw these earlier we get the same coordinates for all three so we get our intersections at two four eight and one one one just summarize that here so we don't have to use a new page intersections at one one and one and two four and eight exactly two intersections between these two trajectories here which is a nice segue into what about collisions do these particles collide if they do at what time and at what xy location where the collisions occur well the nice thing about doing this in this process in this order where we allowed different time parameters for the two trajectories s and t is that if there was a collision we would have gotten a solution just with the same s and t values we saw that earlier in another example so there are no solutions for the same x y zed at the same time we would have found them earlier with the s equal to t as a condition and that tells us right away that there's only the two intersections that we found earlier and specifically answering this question there are no collisions at all that's it so again knowing how to set these up knowing how to set the conditions for collisions versus intersections is what we're looking for you to retain out of this the logic behind it and also where possible if you have the graphical understanding as well where you can imagine the trajectories going through space and time what does it mean to have those paths those trajectories intersect versus having the particles collide in the in a visual sense as well