What Determines the Locus of Midpoints Between Two Lines in Space?

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Discussion Overview

The discussion revolves around determining the locus of midpoints between two lines in three-dimensional space, specifically focusing on the implications of non-coplanarity and the geometric nature of midpoints as points or planes depending on the configuration of the lines. The scope includes mathematical reasoning and conceptual clarification.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Participants discuss the parametric equations of two lines and the conditions under which they are non-coplanar.
  • One participant shows that the locus of midpoints between a fixed point on one line and a moving point on another line is a line.
  • Another participant notes that the locus of midpoints between two moving points on the respective lines results in a plane, raising questions about the geometric interpretation.
  • There is a discussion about the independence of vectors associated with the midpoints, which is necessary to confirm the locus as a plane.
  • Participants explore the implications if the two lines intersect, suggesting that the locus of midpoints would then be a plane that contains the intersection point.
  • One participant expresses confusion about the concept of a "support vector" being the null vector in the context of intersecting lines.
  • Clarifications are made regarding the nature of midpoints when lines are non-intersecting versus intersecting.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical reasoning behind the locus of midpoints, but there are unresolved questions regarding the geometric implications of intersecting lines and the role of vector independence. The discussion remains open regarding the exact nature of the locus when lines intersect.

Contextual Notes

There are limitations in understanding the geometric interpretations of midpoints, particularly in distinguishing between cases of intersecting and non-intersecting lines. The discussion also reflects varying degrees of familiarity with vector independence and its implications for the dimensionality of the locus.

Perlita
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Hello everyone,
Here's an exercise I have to solve:
Given an orthonormal basis (O;i,j,k) and two lines m and n whose respective parametric equations are:
x = 2+t x = 4-3t'
y = 3-2t (t belongs to R) and y = 5-8t' (t' belongs to R)
z = 5-t z = 7-t'

1)Show that Lines m and n are non-coplanar.(easy)
2) Show that point A(2,3,5) belongs to line m.

Let B be a point on line n.
Find the locus of point I, midpoint of segment AB, while B moves along line n.

3) Let M be a point of line m, and B be a point of line n.
Find the locus of the midpoint of segment MB while M and B move along lines m and n respectively.
The answers I gave:
2)By substitution in the equation of line m, we see that A belongs to m.
point A = (2,3,5) corresponds to t=0
point B has the form (4,5,7) + (-3,-8,-1)t'

I = (3,4,6) + (-3/2,-4,-1/2)t'
The result shows that the locus of points halfway between a given point and a line is a line.

3)point M = (2,3,5) + (+1,-2,-1)t
point B = (4,5,7) + (-3,-8,-1)t'

Midpoint of MB: (3,4,6) + (1/2,-1,-1/2)t + (-3/2,-4,-1/2)t'
The result shows that the locus of the midpoint of MB is a plane (but how to explain clearly to the teacher that it's a plane??)Well, I hope that I answered correctly on the two parts.And my second question is:
why this difference between parts 2 and 3? why the locus in part 3 is not a line as well (as in part 2)? I mean that in the two parts we're finding the locus of a midpoint...
-
Thanks for helping.
 
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Re: locus of a point in Space

Perlita said:
Hello everyone,
Here's an exercise I have to solve:
Given an orthonormal basis (O;i,j,k) and two lines m and n whose respective parametric equations are:
x = 2+t x = 4-3t'
y = 3-2t (t belongs to R) and y = 5-8t' (t' belongs to R)
z = 5-t z = 7-t'

1)Show that Lines m and n are non-coplanar.(easy)
2) Show that point A(2,3,5) belongs to line m.

Let B be a point on line n.
Find the locus of point I, midpoint of segment AB, while B moves along line n.

3) Let M be a point of line m, and B be a point of line n.
Find the locus of the midpoint of segment MB while M and B move along lines m and n respectively.
The answers I gave:
2)By substitution in the equation of line m, we see that A belongs to m.
point A = (2,3,5) corresponds to t=0
point B has the form (4,5,7) + (-3,-8,-1)t'

I = (3,4,6) + (-3/2,-4,-1/2)t'
The result shows that the locus of points halfway between a given point and a line is a line.

3)point M = (2,3,5) + (+1,-2,-1)t
point B = (4,5,7) + (-3,-8,-1)t'

Midpoint of MB: (3,4,6) + (1/2,-1,-1/2)t + (-3/2,-4,-1/2)t'
The result shows that the locus of the midpoint of MB is a plane (but how to explain clearly to the teacher that it's a plane??)Well, I hope that I answered correctly on the two parts.And my second question is:
why this difference between parts 2 and 3? why the locus in part 3 is not a line as well (as in part 2)? I mean that in the two parts we're finding the locus of a midpoint...
-
Thanks for helping.

Welcome to MHB, Perlita! :)

Your answers to parts 2 and 3 look fine.
To complete part 3, you need that the vectors (1/2,-1,-1/2) and (-3/2,-4,-1/2) are independent. I presume you already did something similar for part 1?

As for the difference, well, algebraically the mid-point in part 3 has 2 degrees of freedom, since there are 2 variables t and t'.

To understand geometrically, consider for instance the lines that are the x-axis and a line parallel to the y-axis, say at z=2.
If we fix one point on the one line, and drag the other point along the other line, the mid-point describes indeed a line.
If we then move the point on the first line, all those points are translated in another direction.
Can you "see" it?
 
Re: locus of a point in Space

I like Serena said:
Welcome to MHB, Perlita! :)

Your answers to parts 2 and 3 look fine.
To complete part 3, you need that the vectors (1/2,-1,-1/2) and (-3/2,-4,-1/2) are independent. I presume you already did something similar for part 1?

As for the difference, well, algebraically the mid-point in part 3 has 2 degrees of freedom, since there are 2 variables t and t'.

To understand geometrically, consider for instance the lines that are the x-axis and a line parallel to the y-axis, say at z=2.
If we fix one point on the one line, and drag the other point along the other line, the mid-point describes indeed a line.
If we then move the point on the first line, all those points are translated in another direction.
Can you "see" it?
Thanks a lot! yes I can see it :)
but why should I show that the vectors (1/2,-1,-1/2) and (-3/2,-4,-1/2) are independent in part 3?
And what if the two lines m and n are intersecting? what will be the locus of the midpoint of segment MB?
 
Re: locus of a point in Space

Perlita said:
Thanks a lot! yes I can see it :)
but why should I show that the vectors (1/2,-1,-1/2) and (-3/2,-4,-1/2) are independent in part 3?

Suppose one is a multiple of the other, then you do not have a plane but a line.
And what if the two lines m and n are intersecting? what will be the locus of the midpoint of segment MB?

Then the support vector in your equation will be the null vector.
The result is a plane that contains the origin.
 
Re: locus of a point in Space

I like Serena said:
Suppose one is a multiple of the other, then you do not have a plane but a line.

Then the support vector in your equation will be the null vector.
The result is a plane that contains the origin.
why the support vector is the null vector?
if we take for example the two lines
(d): x=3t+2 (d'): x=t'+1
y=-t-1 t belongs to R and y=2t'-3 t' belongs to R
z= t+1 z=-t'+2

these two lines intersect at I(2;-1;1) for t'=1 and t=0
And points M and N moving along d and d' respectively. We need the locus of the midpoint of MN.
why the support vector is null? I can't see it!
 
Re: locus of a point in Space

Perlita said:
why the support vector is the null vector?
if we take for example the two lines
(d): x=3t+2 (d'): x=t'+1
y=-t-1 t belongs to R and y=2t'-3 t' belongs to R
z= t+1 z=-t'+2

these two lines intersect at I(2;-1;1) for t'=1 and t=0
And points M and N moving along d and d' respectively. We need the locus of the midpoint of MN.
why the support vector is null? I can't see it!

My mistake.
What you do have, is that the intersection point of the 2 lines is in the plane.
And when you have 2 non-intersecting lines, the midpoint where they are closest, is in the plane.
 
Re: locus of a point in Space

So what's the final answer, if the lines are intersecting?
Thanks
 
Re: locus of a point in Space

Perlita said:
So what's the final answer, if the lines are intersecting?
Thanks

If the lines are intersecting, the space of midpoints is the plane that the 2 lines span.
 

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