# Homework Help: Lin algebra question

1. May 15, 2005

### EvaBugs

To find the point of intersection of two lines, do I use the same method as in finding the intersection of a line and a plane?

2. May 15, 2005

### p53ud0 dr34m5

what method do you use to find the intersection of a line and a plane?

3. May 15, 2005

### EvaBugs

The line intersects the xy-plane when z=0

4. May 15, 2005

### p53ud0 dr34m5

well, heres what i do to find the point of intersection of two lines. i just set them equal to each other.

5. May 15, 2005

### neurocomp2003

ooh a graphics question...sorry i can't remember the solution off the top of my head its in my graphics book but assuming you no that the 2 lines intersect then the solution
evolves using the parametrics

aight i go get the book...

This ist he distance between 2 lines: obviously if there is an intersection
then d=0 but the solution points still hold true to what your looking for

|t1| = |V1.V1 ,-V1.V2|^-1 * |(P2-P1).V1|
|t2|.... |V1.V2 ,-v2.v2|..........|(P2-P1).V2|
hope this doesn't look ugly

t1,t2 will give you the parametric solution plug into one and you get your point.

6. May 15, 2005

### EvaBugs

I'm sorry, but I don't really understand the forumla. Is that the distance formula?

7. May 15, 2005

### EvaBugs

Should I find the line that is the cross-product of the normal vectors?

8. May 15, 2005

### neurocomp2003

its a matrix form(2 equation Linear system)formula for distance...and if the distance is zero than the parameters t1,t2 find the same point.....sorry but i don't know how to make the spaces stick.

The formula(from a math game book but found in any text)
takes the parametric form of lines and then solves the dist...using minimization and PDEs it results in the 2 equation linear system above.

Last edited: May 15, 2005
9. May 15, 2005

### OlderDan

What form are you equations in? Lines in three dimensions are usually written by specifying a point on the line as a vector, plus a multiple of a vector parallel to the line. The multiplying constant is called a parameter, often designated by t

$$\overrightarrow r = \overrightarrow r _0 + t\overrightarrow r _\parallel$$

Another way of specifying a line is to eliminate the parameter and write the multiple equalities

$$\frac{{x - x_0 }}{a} = \frac{{y - y_0 }}{b} = \frac{{z - z_0 }}{c}$$

Whichever way your lines are written, to find their intersection you need to have the lines passing through the same point in space. For the first form of the equation, that means

$$\overrightarrow r_1 = \overrightarrow r _{0,1} + t_1\overrightarrow r _{\parallel,1} = \overrightarrow r_2 = \overrightarrow r _{0,2} + t_2\overrightarrow r _{\parallel,2}$$

I think this is probably easier to deal with than the second form. It is a vector equation, which is really three equations. The only way two vectors can be equal is if their individual components are each equal.

Last edited: May 15, 2005
10. May 15, 2005

### OlderDan

Another way to look at this situation:

If your equations are in vector form, consider the vector

$$\overrightarrow r_2 - \overrightarrow r_1$$

The two lines go through the points that are defined by these vectors. For the lines to intersect, they must be coplaner with this difference vector. Do you know how to find out if three vectors lie in the same plane?