Complex 3d vector intersection formula

In summary: I am trying to find the angle and i was thinking of using the tangent and arctangent functions to figure it out. In summary, the conversation revolves around two people in a three-dimensional environment, one at (0,0,0) and the other at (10,0,0), and their movements with given velocities. The question is how to find the point of intersection and the time at which they meet, using the initial positions and velocities. Solutions include setting up a family of lines and using trigonometry to find the angle needed to make the times equal.
  • #1
Physicist1231
103
0
ok here goes...

In a three dimensional 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?
 
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  • #2
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
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
mathman said:
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.

interesting concept. I am unfamiliar with setting up a family of lines. Can you go into more detail?
 
  • #5
HallsofIvy said:
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.

Yes

HallsofIvy said:
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.

I think you have that right. His position would land on the permimiter of the circle with a radius of 4t.

HallsofIvy said:
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.

... i got a pretty big equation for that but since there are two (or more) unknowns it seems unsolvable but i KNOW it can be done. I actually did this once before but someone (a friend of mine...) deleted my work...
 
  • #6
Physicist1231 said:
interesting concept. I am unfamiliar with setting up a family of lines. Can you go into more detail?

Construct the triangle where one side (C) is the line from (0,0,0) to (10,0,0). One angle is given by the angle at (0,0,0) between the given side and a line along (0,0,0) to (1,2,3) - this will be along side A. Then let a be the angle at (10,0,0). You now have defined a complete triangle.

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
Physicist1231 said:
Yes



I think you have that right. His position would land on the permimiter of the circle with a radius of 4t.



... i got a pretty big equation for that but since there are two (or more) unknowns it seems unsolvable but i KNOW it can be done. I actually did this once before but someone (a friend of mine...) deleted my work...
On the contrary, there is only one variable, t- putting x= t, y= 2t, z= 3t into [itex](x- 10)^2+ y^2+ z^2= 16t[/itex], [itex]t^2- 20t+ 100+ 4t^2+ 9t^2= 16t^2[/itex] which reduces to the quadratic equation [itex]t^2+ 10t- 50= 0[/itex]. That's not hard to solve!
 
  • #8
HallsofIvy said:
On the contrary, there is only one variable, t- putting x= t, y= 2t, z= 3t into [itex](x- 10)^2+ y^2+ z^2= 16t[/itex], [itex]t^2- 20t+ 100+ 4t^2+ 9t^2= 16t^2[/itex] which reduces to the quadratic equation [itex]t^2+ 10t- 50= 0[/itex]. That's not hard to solve!

very nice! I will try this out in my project (no not for school or a grade:))

I will let you know as soon as i find out!
 
  • #9
mathman said:
Construct the triangle where one side (C) is the line from (0,0,0) to (10,0,0). One angle is given by the angle at (0,0,0) between the given side and a line along (0,0,0) to (1,2,3) - this will be along side A. Then let a be the angle at (10,0,0). You now have defined a complete triangle.

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.

Yeah that is what i have been trying to do but i seem to be missing something...

I will also check out HallsofIvy solve to see if i can get anywhere with it.
 
  • #10
Physicist1231 said:
Yeah that is what i have been trying to do but i seem to be missing something...

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

Could you show your work?
 
  • #11
HallsofIvy said:
On the contrary, there is only one variable, t- putting x= t, y= 2t, z= 3t into [itex](x- 10)^2+ y^2+ z^2= 16t[/itex], [itex]t^2- 20t+ 100+ 4t^2+ 9t^2= 16t^2[/itex]

Ok i got that far...

HallsofIvy said:
which reduces to the quadratic equation [itex]t^2+ 10t- 50= 0[/itex]. That's not hard to solve!

Can you show this because I can't seem to get there...

I get this:

[itex]13t^2-20t + 100 = 16t^2[/itex]

guess I am 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
 
Last edited:
  • #12
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
Physicist1231 said:
I get this:

[itex]13t^2-20t + 100 = 16t^2[/itex]

That should be ##14t^2## on the left, not ##13t^2##.
 
  • #14
LCKurtz said:
That should be ##14t^2## on the left, not ##13t^2##.

Thank you for that! I still do not seem to be getting a clear cut answer. What I am striving for a formula that i can use for this and just plug in the following:

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
Physicist1231 said:
Thank you for that! I still do not seem to be getting a clear cut answer. What I am striving for a formula that i can use for this and just plug in the following:

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)

You need to get the velocity vector for person 2 as a function of angle that he makes with the line from him to person 1. Once you have that you can calculate the time it takes to hit the velocity vector line from person 1 and also the time time person 1 takes to reach that point. Adjust the angle to make the times equal.
 
  • #16
Your speed component away from x-axis is √13 , so in order to meet, the other person must have the same, meaning he is moving at a speed √3 parallell to x-axis. So you approach each other with a speed of 1+√3, and will meet when t = 10/(1+√3).
 

1. What is a complex 3d vector intersection formula?

A complex 3d vector intersection formula is a mathematical equation that is used to find the point where two or more 3-dimensional vectors intersect in space. It takes into account the direction and magnitude of each vector and calculates the coordinates of the intersection point.

2. How is the complex 3d vector intersection formula derived?

The complex 3d vector intersection formula is derived from the principles of vector algebra and geometry. It involves solving equations for the coordinates of the intersection point using dot and cross products of the given vectors.

3. What is the importance of the complex 3d vector intersection formula?

The complex 3d vector intersection formula is important in various fields such as computer graphics, robotics, and physics. It allows for the accurate calculation of intersection points between objects in 3-dimensional space, which is essential for tasks such as collision detection and object tracking.

4. Are there any limitations to the complex 3d vector intersection formula?

Yes, the complex 3d vector intersection formula has limitations. It may not work for vectors that are parallel or have a zero magnitude. In addition, it assumes that the vectors are infinitely thin and do not have any width or thickness. Therefore, it may not accurately represent the intersection point in some real-world scenarios.

5. How can the complex 3d vector intersection formula be applied in real life?

The complex 3d vector intersection formula can be applied in various real-life situations, such as in computer graphics software for creating 3-dimensional models, in robotics for motion planning and navigation, and in physics for calculating the path of moving objects in 3-dimensional space. It is also used in video games for simulating realistic collisions between objects.

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