How do you find the slope of a line in 3D coordinate system given two points?

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Discussion Overview

The discussion revolves around the concept of finding the slope of a line in a 3D coordinate system given two points. Participants explore the definition of slope in relation to direction and angles, particularly in the context of geometry and vector representation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants suggest that the slope is direction-dependent and cannot be defined without a reference direction.
  • One participant proposes finding the direction of the line by forming a vector from two points and projecting it onto a unit direction vector.
  • Another participant emphasizes that a "slope" is connected to an angle and requires two lines to establish a reference for comparison.
  • A further contribution explains the concept of unit vectors in three dimensions and notes that the direction cosines depend on multiple angles, complicating the definition of slope.

Areas of Agreement / Disagreement

Participants express differing views on the definition of slope in 3D, with some agreeing that a reference line is necessary while others focus on vector representation. The discussion remains unresolved regarding a consensus on how to define slope in this context.

Contextual Notes

Limitations include the dependence on the choice of reference direction and the ambiguity in defining slope without additional context or reference lines.

Lizwi
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Given two points in 3D, is there a way to find the slope of that line.
 
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Lizwi said:
Given two points in 3D, is there a way to find the slope of that line.

Hey Lizwi.

Your question is incomplete because the slope is usually with respect to some direction.

What you can do is find the direction of the line (take two points on the line with a difference of unit parameter: i.e. take t = 1 and t = 0 and form a vector from those two points), and then project this vector onto some direction vector with length 1.

Alternatively you can simply take your vector calculated with t = 0 and t = 1 and find the length of this vector: this is equivalent to the concept of the total differential that is used in multivariable calculus.

In other geometries, you need to consider the basis, determinant and the stuff in tensor theory but for this example for a normal cartesian geometry, you take the difference of t = 1 and t = 0 and use the fact that the rate of change is constant across the line to give you this 'tangent' vector.
 
Lizwi said:
Given two points in 3D, is there a way to find the slope of that line.

Adding to chiro's post:

A "slope" is always connected to some ANGLE.

And, in order to talk about angles, you need TWO lines.

Thus, you cannot talk about "slope", either, without having some reference line in addition to the line specified by the two points.

Agreed?
 
Consider a unit vector in the direction of a given line. In two dimensions, it is [itex]<cos(\theta), sin(\theta)>[/itex] where [itex]\theta[/itex] is the angle the line makes with the x-axis. (And the slope is [itex]tan(\theta)[/itex].)

A unit vector in three dimensions is given by [itex]<cos(\alpha), cos(\beta), cos(\gamma)>[/itex] (the "direction cosines" of the direction) where [itex]\alpha[/itex] is the angle the line makes with the x axis, [itex]\beta[/itex] is the angle the line makes with the y axis, and [itex]\gamma[/itex] is the angle the line makes with the z axis. Because this is a unit vector, these must satisfy [itex]cos^2(\alpha)+ cos^2(\beta)+ cos^2(\gamma)= 1[/itex] but this still depends upon two angles so we cannot give a single number, slope or other, that determines the direction of the line.

(Note that in two dimensions we can take [itex]\gamma= 0[/itex] so [itex]cos^2(\alpha)+ cos^2(\beta)= 1[/itex] which means that [itex]cos(\beta)= sin(\alpha)[/itex].)
 

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