Find the point which lie in a 3d line.

  • Context: Undergrad 
  • Thread starter Thread starter parch
  • Start date Start date
  • Tags Tags
    3d Line Point
Click For Summary

Discussion Overview

The discussion revolves around determining whether a given point in 3D space lies on the line segment defined by two other points in the same space. Participants explore mathematical representations and algorithms for this problem.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant seeks a simple algorithm to determine if a point lies on the line segment between two known points in 3D space.
  • Another participant suggests that any point on the line can be expressed in the form t(a,b,c) + (1-t)(d,e,f) for some t, indicating that the line is defined by the two points.
  • A participant attempts to express the condition for the third point (x,y,z) to lie on the line segment, proposing that t can be represented as t(u,v,w) where u, v, and w should be between 0 and 1.
  • There is a clarification that if t exists such that t(a,b,c) + (1-t)(d,e,f) = (x,y,z), then the point (x,y,z) is on the line, and if t is between 0 and 1, it lies between the two endpoints.
  • Another participant questions the interpretation of t as a vector, asserting that t is a scalar value, which leads to confusion about the representation of t(u,v,w).

Areas of Agreement / Disagreement

Participants express differing views on the representation of t, with some asserting it as a scalar and others attempting to define it as a vector. The discussion remains unresolved regarding the correct interpretation of t.

Contextual Notes

There is ambiguity in the definitions and representations of t, as well as the conditions under which the point (x,y,z) is considered to lie on the line segment.

Who May Find This Useful

Individuals interested in geometric representations in 3D space, mathematical modeling of lines, and those working on related problems in physics or computer graphics may find this discussion relevant.

parch
Messages
17
Reaction score
0
hi

I know two points in 3d space.I know another point in the space and need to know whether that point lie in between the two points(i.e same line). can anyone give me simple algorithm other than logic.


Thanks in advance.
 
Physics news on Phys.org
If the coordinates of your first two points are (a,b,c) and (d,e,f) then any point lying on the line going through them will be of the form

t(a,b,c) + (1-t)(d,e,f)

for some t. You can easily see that this is a line, and that (a,b,c) and (d,e,f) are on the line

So you simply have to find whether such a t exists for your third point
 
hi shredder,

I got ur answer, please check whether i am correct.

(x,y,z) is new point where need to find it is in that path.

t(a,b,c) + (1-t)(d,e,f)=(x,y,z)

from this i can get the t, assume my t as t(u,v,w). so these u , v and w should be inbetween 0 and 1,so that t lies in the path, ie in the line.
 
parch said:
hi shredder,

I got ur answer, please check whether i am correct.

(x,y,z) is new point where need to find it is in that path.

t(a,b,c) + (1-t)(d,e,f)=(x,y,z)

from this i can get the t, assume my t as t(u,v,w). so these u , v and w should be inbetween 0 and 1,so that t lies in the path, ie in the line.
What are "u, v, and w"?

If there exist any t at all satisfying t(a,b,c) + (1-t)(d,e,f)=(x,y,z), then (x,y,z) is on the line through (a,bc) and (d,e,f). If t lies between 0 and 1, then (x,y,z) is not only on that line, it lies between (a,b,c) and (d,e,f)- i.e. it is on the line segment with endpoints (a,b,c) and (d,e,f).
 
from this t(a,b,c) + (1-t)(d,e,f)=(x,y,z), the unknown t will be a vector.which i gave the as (u,v,w). could you come some more detail please..
 
hey thanks... got it...
 
parch said:
from this t(a,b,c) + (1-t)(d,e,f)=(x,y,z), the unknown t will be a vector.which i gave the as (u,v,w). could you come some more detail please..
No, t is not a vector- t is a number. If t were a vector "1- t" would make no sense.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
17
Views
3K
  • · Replies 7 ·
Replies
7
Views
8K