Intersection of Two Coplanar Lines in 3 Dimensions

[dimensionless]

Homework Statement

Two lines in space are in the same plane. Line AB passes through points $$A(x,y,z)$$ and $$B(x,y,z)$$, and line CD passes through points $$C(x,y,z)$$ and $$D(x,y,z)$$. 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 $$A$$ and $$B$$ are:

$$x = A_{x} + (B_{x}-A_{x})t$$
$$y = A_{y} + (B_{y}-A_{y})t$$
$$x = A_{z} + (B_{z}-A_{z})t$$

Symmetric equations for the same point:

$$\frac{x - A_{x}}{(B_{x}-A_{x})} = \frac{y - A_{y}}{(B_{y}-A_{y})} = \frac{z - A_{z}}{(B_{z}-A_{z})}$$

The Attempt at a Solution

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

$$t = \frac{C_{x}-A_{x}}{(B_{x}-A_{x})-(D_{x}-C_{x})}$$

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:

$$z(x,y) = Ex + Fy + G$$

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.

 

[stunner5000pt]

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 $A = (x_{0},y_{0},z_{0})$ and $B=(x_{1},y_{1},z_{1})$

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

 

[D H]

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

$$t = \frac{C_{x}-A_{x}}{(B_{x}-A_{x})-(D_{x}-C_{x})}$$

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)?

 

[dimensionless]

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 $A = (x_{0},y_{0},z_{0})$ and $B=(x_{1},y_{1},z_{1})$

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

A and B are distinct points.

 

[dimensionless]

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
$$x = A_{x} + (B_{x}-A_{x})t$$
without the $$A_{x}$$ and without the $$t$$. So, I would use
$$v = (B_{x}-A_{x})\hat{i}+ ...$$
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.

 

[dimensionless]

I was able to derive the answer for this. I used the two parametric vector equations:
$$v_{1} = (m_{1x}t_{1}+b_{1x})\hat{i}+ (m_{1y}t_{1}+b_{1y})\hat{j}+ (m_{1z}t_{1}+b_{1z})\hat{k}$$
$$v_{2} = (m_{2x}t_{2}+b_{2x})\hat{i}+ (m_{2y}t_{2}+b_{2y})\hat{j}+ (m_{2z}t_{2}+b_{2z})\hat{k}$$

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

and then setting the various vector components equal

$$\frac{m_{1x}}{M_{1}}t_{1}+b_{1x} = \frac{m_{2x}}{M_{2}}t_{2}+b_{2x}$$
$$\frac{m_{1y}}{M_{1}}t_{1}+b_{1y} = \frac{m_{2y}}{M_{2}}t_{2}+b_{2y}$$
$$\frac{m_{1z}}{M_{1}}t_{1}+b_{1z} = \frac{m_{2z}}{M_{2}}t_{2}+b_{2z}$$

These equations can then be solved to find the point of intersection.