# Locus of a point in Space

#### Perlita

##### New member
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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: locus of a point in Space

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?

#### Perlita

##### New member
Re: locus of a point in Space

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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: locus of a point in Space

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.

#### Perlita

##### New member
Re: locus of a point in Space

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!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: locus of a point in Space

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.

#### Perlita

##### New member
Re: locus of a point in Space

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: locus of a point in Space

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.