# 3D vector collsions- do u terat them differently?

1. Sep 23, 2006

### ku1005

3D vector collsions- do u terat them differently??

hi upon attemptin the following questions, in which i just answered the Questions in the same manner as i would with a 2D vector, i keep getting them wrong and was hoping sum1 could tell me where my thinking is flawed:

Proove that L1 and L2 intersect, finding th position vecor of the point:

L1 = 13i+j+8k + T(2i-j+3k)

L2= -5i+2j-3k + B(2i+j-k)

i simply said for them to intersect let them have the same x y z coordinates at some time/point T (ie let T and B = T)

so that (13+2T)i+(1-T)j+(8+3T)k = (-5+2T)i + (2+T)j + (-3-T)k

such that set each coordinate equal to each other ie

13+2T = -5+2T
1-T = 2+T
8+3T = -3-T and if T value is equal for each point therfore intersect for that value of T.....however since i am on here this obviously doesnlt work,
the answ is 3i+6j-7k

another question involves finding the least distance between 2 particles, where again i applied same principles for 2D vecotrs in 3D vectors (ie minimise a length between the 2 bodies using calculus) however it dosnlt work either , so i am hoping somethin in this Q will unlock the key to my mistakes in the others.....thanks

2. Sep 23, 2006

### HallsofIvy

Here's your problem. T and B do not have to be equal. You have, in fact, three equations for two unknown numbers, T and B, not one. Thats true even in 2D!

Try solving 13+ 2T= -5+ 2B, 1- T= 2+ B, 8+ 3T= -3- B for T and B. Of course, in general, you cannot solve 3 equations for 2 "unknowns" since, in general, two lines in 3 dimensions do not intersect. Solve two of the equations for T and B and then check to see if those values satisfy the third.

Determining whether 2 lines intersect is not the same as a "vector collision". If you think of the equations for the lines as giving the positions of particles, moving on those lines, as a function of time, then in order for the particles to collide they would have to reach the same point at the same time. If the two lines intersect, it is possible for the particles to pass through the same point at different times.

Show what you have done on this one. Certainly minimizing the distance (or distance square- its easier) will give the point of closest approach whether in 2D or 3D.

3. Sep 23, 2006

### ku1005

this was the questions, "an aircraft is maintaining a constant velocity of (80i+50j+5k) m/s, the units vecotrs being indictaed as on the right (ie a image showing that upwards = k axis, east = i axis and north = j axis. At 1 pm the aircraft is situated 5km south and 10km west of a trackin station situated at O (ie origin- 0,0,0)The aircrafts altitiude is 500m.

I could anwer all Q's (ie 6 part) except this one

e)Find the least distance between the aircraft and tracking station.

I did the follwing:

I let intial positon of the aircraft = A and the point at which the aircrft is closest = P.

Therefore expressed

Vector OA= (-10i-5j+0.5k)km (yes i used km this time since i used metres the other time an end up with the same incorrect answer)

Vector AP= (-10+0.08t)i + (-5+0.05t)j + (0.5 +0.005t)k

Given Vector OP = OA + AP
=(-20+0.08t)i + (-5+0.05t) j + (0.5 + 0.005t)k

Therefore need to minimise Ab value of OP = distance or (OP)^2 = D^2(distance squared)

(-20+0.08t)^2+(-10+0.05t)^2+(1 + 0.005t)^2 = D^2
where t= 234.73

which isnlt correct since the answer is ~1.52 km (ie once u sub answer t bak in of course)

4. Sep 23, 2006

### ku1005

and by the way, thanks heaps for that, i dunno where i got that T and B should be equal......since they are mutilplying diff velocity vecotrs (direction vecotrs of the line) adn therfore in no way should be equal, it is ikely that i this Q I am doin the same thing.....but i put all my working regardless...