Distance between two skew lines given by parametric equations (and more)

In summary, the conversation is about a calculus homework problem involving finding a normal vector and projecting lines. The equations for finding the projection are discussed, along with the attempt at a solution. The focus is on finding the correct normal vector, as it is essential for solving the problem correctly. A diagram is also provided to help visualize the problem.
  • #1
Interesting
14
0

Homework Statement


L1 L2
x= 1+t x= 1+2s
y= 1+6t y= 5+15s
z= 2t z= -2+6s

My professor said to find a normal vector and project the lines, but I'm new to this calculus and all these words are just a fuzzy cloud over my head.

Homework Equations



Projection of c onto L1 -> (C dot L1 / magnitude C) (L1 / magnitude L1)
Or (C dot L1)C/(magnitude C^2)

The Attempt at a Solution



L1xL2= a vector of <6,2,3> = C
The direction vector for L1 = <1,6,2>
L2 = <2,15,6>

So I think I need to project one onto the other...
But when I do that I get another vector, that's <144/7,48/7,72/7>. Do I find the magnitude of this, and that should be the distance? I'm lost. The magnitude of that vector is 24, which is a clean answer so it could be right...
 
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  • #2
Interesting said:

Homework Statement


L1 L2
x= 1+t x= 1+2s
y= 1+6t y= 5+15s
z= 2t z= -2+6s

My professor said to find a normal vector and project the lines, but I'm new to this calculus and all these words are just a fuzzy cloud over my head.

Homework Equations



Projection of c onto L1 -> (C dot L1 / magnitude C) (L1 / magnitude L1)
Or (C dot L1)C/(magnitude C^2)

The Attempt at a Solution



L1xL2= a vector of <6,2,3> = C
The direction vector for L1 = <1,6,2>
L2 = <2,15,6>

So I think I need to project one onto the other...
But when I do that I get another vector, that's <144/7,48/7,72/7>. Do I find the magnitude of this, and that should be the distance? I'm lost. The magnitude of that vector is 24, which is a clean answer so it could be right...
You normal vector is wrong. If c is normal (perpendicular) to L1 or L2 then
c.L1 = 0
c.L2 = 0
 
  • #3
But, I think here's what you are doing
http://img230.imageshack.us/img230/9503/64421739uy1.gif
 
Last edited by a moderator:

FAQ: Distance between two skew lines given by parametric equations (and more)

1. What are the parametric equations for two skew lines?

The parametric equations for two skew lines can be written as:
Line 1: x = x1 + at, y = y1 + bt, z = z1 + ct
Line 2: x = x2 + su, y = y2 + sv, z = z2 + sw

2. How do you find the distance between two skew lines using parametric equations?

To find the distance between two skew lines, you can use the formula:
d = |(x1-x2)a + (y1-y2)b + (z1-z2)c| / √(a²+b²+c²)
Where a, b, and c are the direction ratios of Line 1 and x1, y1, z1 and x2, y2, z2 are any two points on Line 2.

3. Can the distance between two skew lines be negative?

No, the distance between two skew lines cannot be negative. It is always a positive value because it represents the shortest distance between the two lines.

4. What is the significance of the direction ratios in the parametric equations of skew lines?

The direction ratios in the parametric equations of skew lines represent the slope of the lines in the x, y, and z directions. They determine the direction and orientation of the lines in space.

5. Are there any other methods for finding the distance between two skew lines besides using parametric equations?

Yes, there are other methods for finding the distance between two skew lines, such as using vector projection or the shortest distance formula. However, the use of parametric equations is a commonly used method and can be easily applied in various situations.

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