Finding the Equation of a 3D Line from Two Points

  • Thread starter Thread starter tandoorichicken
  • Start date Start date
  • Tags Tags
    3d Line Points
Click For Summary

Homework Help Overview

The discussion revolves around finding the equation of a line in three-dimensional space given two points. Participants are exploring the parametric representation of the line and the implications of different parameterizations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various forms of the line equation, questioning the correctness of specific parameterizations and their implications for the endpoints of the line. There is also a mention of the validity of different parameterizations that represent the same line.

Discussion Status

The discussion is active, with participants providing different perspectives on the parameterization of the line. Some guidance has been offered regarding the nature of vector equations and the relationship between the points and the parameter.

Contextual Notes

There is an emphasis on the need for multiple equations to describe a line in three-dimensional space, as well as the consideration of how different parameterizations can yield the same geometric line.

tandoorichicken
Messages
245
Reaction score
0
How do you find the equation of a line between two points in three dimensional space? I sort of forgot. =\
 
Physics news on Phys.org
c(t) = (P-Q)t+P
 
0rthodontist said:
c(t) = (P-Q)t+P

Shouldn't that be c(t) = (P-Q)t+Q

because at t = 0 you should get Q, and at t = 1 you should get P which isn't what happens in your equation.
 
It really doesn't matter how you parametrize it so long as it is the same line. c(t) = .3455(P-Q)t+(P+Q)/2 is equally valid.
 
d_leet said:
Shouldn't that be c(t) = (P-Q)t+Q

because at t = 0 you should get Q, and at t = 1 you should get P which isn't what happens in your equation.

Orthodontist's form, c= (P- Q)t+ P, gives P when t=0 and Q when t= -1. That perfectly valid.

By the way, Tandoorichicken, since a line is one-dimensional, in a three dimensional space you need more than one equation. Orthodontist gave a "vector" equation where P and Q are the position vectors of two points and t is a parameter. Writing that in component form gives three parametric equations. Given the single variable t, you can calculate x, y, and z.
 
Last edited by a moderator:

Similar threads

Replies
17
Views
3K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
8
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K