Intersection of Two Coplanar Lines in 3 Dimensions

In summary, to find the point of intersection of two lines in three-dimensional space, the parametric equations for the two lines can be set equal and then solved for the common parameter values. The resulting values can then be substituted back into one of the parametric equations to find the coordinates of the point of intersection.
  • #1
dimensionless
462
1

Homework Statement


Two lines in space are in the same plane. Line AB passes through points [tex]A(x,y,z)[/tex] and [tex]B(x,y,z)[/tex], and line CD passes through points [tex]C(x,y,z)[/tex] and [tex]D(x,y,z)[/tex]. Determine if these two lines are parallel. If they are not, determine the x,y,z coordinates where these two lines intersect.

Homework Equations


The parametric equations for lines passing through points [tex]A[/tex] and [tex]B[/tex] are:

[tex]x = A_{x} + (B_{x}-A_{x})t[/tex]
[tex]y = A_{y} + (B_{y}-A_{y})t[/tex]
[tex]x = A_{z} + (B_{z}-A_{z})t[/tex]

Symmetric equations for the same point:

[tex]\frac{x - A_{x}}{(B_{x}-A_{x})} = \frac{y - A_{y}}{(B_{y}-A_{y})} = \frac{z - A_{z}}{(B_{z}-A_{z})}[/tex]

The Attempt at a Solution



I tried setting
[tex]x = A_{x} + (B_{x}-A_{x})t[/tex]
and
[tex]x = C_{x} + (D_{x}-C_{x})t[/tex]
equal. This resulted in:

[tex]t = \frac{C_{x}-A_{x}}{(B_{x}-A_{x})-(D_{x}-C_{x})}[/tex]

When I worked this out, I got numbers that did not seem to fit. The symmetric equations did not seem to fit either. I also tried writing the equation down as:

[tex]z(x,y) = Ex + Fy + G[/tex]

but this is for a surface.

I have a feeling the answer is pretty simple, but for some reason I'm not finding it it. Any help is appreciated.
 
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  • #2
quick question do A and B have to have the same coordinates (x,y,z) or is that something you made up?
im thinking A and B should be distinct points like [itex] A = (x_{0},y_{0},z_{0}) [/itex] and [itex] B=(x_{1},y_{1},z_{1}) [/itex]

or is that A and B having identical coords part of the question?

ok on to the question ... try writing down the equations of both the lines
to be parallel the direction vectors must be equal
 
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  • #3
dimensionless said:
I tried setting
[tex]x = A_{x} + (B_{x}-A_{x})t[/tex]
and
[tex]x = C_{x} + (D_{x}-C_{x})t[/tex]
equal. This resulted in:

[tex]t = \frac{C_{x}-A_{x}}{(B_{x}-A_{x})-(D_{x}-C_{x})}[/tex]

This won't work. There is no reason the parameter for line A-B has the same value as the parameter for line C-D at the point of intersection. Case in point: A=(1,0,0), B=(2,0,0), C=(0,-2,0), D=(0,-1,0). The lines A-B and C-D obviously intersect at the origin with different parameters 't'.

You first have to determine if the lines intersect in three space. What vector expression tells you two vectors are parallel (or anti-parallel)?
 
  • #4
stunner5000pt said:
quick question do A and B have to have the same coordinates (x,y,z) or is that something you made up?
im thinking A and B should be distinct points like [itex] A = (x_{0},y_{0},z_{0}) [/itex] and [itex] B=(x_{1},y_{1},z_{1}) [/itex]

or is that A and B having identical coords part of the question?

A and B are distinct points.
 
  • #5
D H said:
This won't work. There is no reason the parameter for line A-B has the same value as the parameter for line C-D at the point of intersection. Case in point: A=(1,0,0), B=(2,0,0), C=(0,-2,0), D=(0,-1,0). The lines A-B and C-D obviously intersect at the origin with different parameters 't'.

You first have to determine if the lines intersect in three space. What vector expression tells you two vectors are parallel (or anti-parallel)?

To get a direction vector, I presume that I would use the equation
[tex]x = A_{x} + (B_{x}-A_{x})t[/tex]
without the [tex]A_{x}[/tex] and without the [tex]t[/tex]. So, I would use
[tex]v = (B_{x}-A_{x})\hat{i}+ ...[/tex]
This would tell me if they were parallel.

I'm still stuck, though, when it comes to finding a generalized expression for the point of intersection in three dimensions.
 
  • #6
I was able to derive the answer for this. I used the two parametric vector equations:
[tex]v_{1} = (m_{1x}t_{1}+b_{1x})\hat{i}+ (m_{1y}t_{1}+b_{1y})\hat{j}+ (m_{1z}t_{1}+b_{1z})\hat{k}[/tex]
[tex]v_{2} = (m_{2x}t_{2}+b_{2x})\hat{i}+ (m_{2y}t_{2}+b_{2y})\hat{j}+ (m_{2z}t_{2}+b_{2z})\hat{k}[/tex]

And then normalize them by dividing each slope by the magnitude of all three slopes
[tex]M = \frac{1}{\sqrt{m_{x}^2+m_{y}^2+m_{z}^2}}[/tex]
[tex]v_{1} = (\frac{m_{1x}}{M_{1}}t_{1}+b_{1x})\hat{i}+ (\frac{m_{1y}}{M_{1}}t_{1}+b_{1y})\hat{j}+ (\frac{m_{1z}}{M_{1}}t_{1}+b_{1z})\hat{k}[/tex]
[tex]v_{2} = (\frac{m_{2x}}{M_{2}}t_{2}+b_{2x})\hat{i}+ (\frac{m_{2y}}{M_{2}}t_{2}+b_{2y})\hat{j}+ (\frac{m_{2z}}{M_{2}}t_{2}+b_{2z})\hat{k}[/tex]

and then setting the various vector components equal

[tex]\frac{m_{1x}}{M_{1}}t_{1}+b_{1x} = \frac{m_{2x}}{M_{2}}t_{2}+b_{2x}[/tex]
[tex]\frac{m_{1y}}{M_{1}}t_{1}+b_{1y} = \frac{m_{2y}}{M_{2}}t_{2}+b_{2y}[/tex]
[tex]\frac{m_{1z}}{M_{1}}t_{1}+b_{1z} = \frac{m_{2z}}{M_{2}}t_{2}+b_{2z}[/tex]

These equations can then be solved to find the point of intersection.
 
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1. What is the equation for finding the intersection of two coplanar lines in 3 dimensions?

The equation for finding the intersection of two coplanar lines in 3 dimensions is known as the point-slope form. It is written as: (x,y,z) = (x1,y1,z1) + t(a,b,c) , where (x1,y1,z1) is a point on one of the lines and (a,b,c) is the direction vector of that line.

2. How do you determine if two coplanar lines are parallel or intersecting?

Two coplanar lines are parallel if their direction vectors are parallel, meaning they have the same slope. They are intersecting if their direction vectors are not parallel and they share a common point. This can be determined by solving the equations of the lines simultaneously.

3. Can two coplanar lines be skew (not parallel or intersecting) in 3 dimensions?

No, two coplanar lines cannot be skew in 3 dimensions. Skew lines are lines that do not lie in the same plane and therefore cannot be coplanar. In 3 dimensions, two lines are either parallel, intersecting, or they are the same line.

4. How many solutions are possible for the point of intersection of two coplanar lines in 3 dimensions?

There can be either one solution, no solution, or infinitely many solutions for the point of intersection of two coplanar lines in 3 dimensions. This depends on the relationship between the equations of the lines and whether they are parallel, intersecting, or the same line.

5. Can the intersection point of two coplanar lines be outside of the plane they lie in?

No, the intersection point of two coplanar lines must lie within the plane that they both lie in. This is because the equations of both lines are in terms of x, y, and z coordinates, which are all variables in the same 3-dimensional plane.

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