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Complex 3d vector intersection formula.. 
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#1
Feb812, 02:58 PM

P: 103

ok here goes...
In a three dimentional environment. i am standing at point (0,0,0) and there is someone else standing at (10,0,0) I start moving with a velocity of (1,2,3)/s and the other guy wants to meet me. I know that he is approaching the point of intersection at 4m/s (that is cumulative). We wind up meeting at the same spot at the same time. My question is how do you get that spot in 3d space and at what time did you meet? This seems that it can be done since we have the initial positions and the velocities. But since you only know the overall velocity of the other guy how do you know how to break it down into the velocities on the X, Y and Z without knowing how much time has passed? 


#2
Feb812, 04:15 PM

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Interesting problem. Are we to assume you and the other guy both move in a single direction? You say that you moved from (0, 0, 0) with velocity vector <1, 2, 3> m/s so at any time, t, you will be a position x= t, y= 2t, z= 3t. The other person moves with speed 4 m/s so at any time, t, he will be at distance 4t from (10, 0, 0). In other words, his position will be on the circle with center at (10, 0, 0) and radius 4t. That is given by [itex](x 10)^2+ y^2+ z^2= 16t^2[/itex]. If you put x= t, y= 2t, z= 3t into that equation, you will get a quadratic equation for t.



#3
Feb812, 04:19 PM

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Set up a family of lines start at (10,0,0) which intersect the line (1,2,3)t. The family can be define as a function of angle with the line between (0,0,0) and (10,0,0). For any point of intersection get the times that the two vectors hit the point of intersection. Solve the equations for the times being equal.



#4
Feb812, 08:25 PM

P: 103

Complex 3d vector intersection formula..



#5
Feb912, 12:56 PM

P: 103




#6
Feb912, 03:24 PM

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Start at (10,0,0) and go along (B) the third side (at angle a) and get the point of intersection with the line along side A. Compute the distances along A and B and get the times. These times will depend on angle a. Find the value of angle a needed to make these times equal. 


#7
Feb912, 05:12 PM

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#8
Feb1212, 08:23 PM

P: 103

I will let you know as soon as i find out! 


#9
Feb1212, 08:24 PM

P: 103

I will also check out HallsofIvy solve to see if i can get anywhere with it. 


#10
Feb1312, 03:35 PM

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#11
Feb1412, 08:12 AM

P: 103

I get this: [itex]13t^220t + 100 = 16t^2[/itex] guess im looking for the next step... I can get it down to: [itex]t^2  (20/3)t = (100/3)[/itex] (I kept it as fraction to keep it clean) ideas? I know it is something simple 


#12
Feb1412, 08:37 AM

P: 103

can someone double check me? I got
T = 6.25 which means Person 1 (from 0,0,0) traveled a total of 23.38125m and Person 2 traveled 25m.. Sounds close but does not feel right... 


#13
Feb1412, 12:15 PM

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#14
Feb1412, 04:12 PM

P: 103

initial (x,y,z) of person 1 initial (x,y,z) of person 2 (x,y,z) Velocity of person 1 Overall Velocity of person 2 I want to result in: The time of collision (assuming that the lines intersect) and The (x,y,z) of collision (assuming that the lines intersect) 


#15
Feb1512, 03:44 PM

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#16
Feb1512, 05:47 PM

P: 144

Your speed component away from xaxis is √13 , so in order to meet, the other person must have the same, meaning he is moving at a speed √3 parallell to xaxis. So you approach each other with a speed of 1+√3, and will meet when t = 10/(1+√3).



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