Vector of shortest distance between two skew lines

In summary: This is an easy problem to solve if you just have two points. Take the dot product of the vectors between the points, and that will give you the distance between the points.
  • #1
Krushnaraj Pandya
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Homework Statement


how to write the vector equation of the line of shortest distance between two skew lines in the shortest and most efficient way?
(The exact lines given in a particular problem in my book can be referenced- L1=(3i+8j+3k)+λ(3i-j+k) and L2=(-3i-7j+6k)+μ(-3i+2j+4k) )
2. Relevant methods
The vector we want would be perpendicular to both the lines, so we can use dot product=0 for both lines which gives us 2 equations in l,m and n where vector=li+mj+nk. We can find distance between the lines from a routine formula then equate the modulus of the vector sqrt(l^2+m^2+n^2) to that distance

The Attempt at a Solution


combined above
 
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  • #2
I don't see any attempted solution, so where do you need guidance? Here's an idea, though: the Cross Product of two vectors gives a vector perpendicular to both vectors. Which part of the line equation represents the direction vector?
 
  • #3
scottdave said:
I don't see any attempted solution, so where do you need guidance? Here's an idea, though: the Cross Product of two vectors gives a vector perpendicular to both vectors. Which part of the line equation represents the direction vector?
Under relevant methods I've outlined a complete solution. The main question is in the first line "solution in the shortest and most efficient way"- meaning I already know a tedious way to solve it that I outlined, I want to know about other ways which would be shorter and less tedious.
 
  • #4
scottdave said:
I don't see any attempted solution, so where do you need guidance? Here's an idea, though: the Cross Product of two vectors gives a vector perpendicular to both vectors. Which part of the line equation represents the direction vector?
That was what I was looking for in part, taking the cross product reduces the work a lot by directly giving us the direction ratios of the required vector. But we still don't know the point it passes through on one of those lines
 
  • #5
Krushnaraj Pandya said:
That was what I was looking for in part, taking the cross product reduces the work a lot by directly giving us the direction ratios of the required vector. But we still don't know the point it passes through on one of those lines
The same technique I outlined in your thread about the distance between a line and a point can be used here. Find a point P on one of the lines, and a point Q on the other line. Then find the vector projection of PQ in the direction of one of the lines. Now find the magnitude of PQ (the hypotenuse) and the magnitude of the vector projection (one leg of the right triangle). The distance between the two skew lines is the length of the other leg.
 
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  • #6
Krushnaraj Pandya said:
That was what I was looking for in part, taking the cross product reduces the work a lot by directly giving us the direction ratios of the required vector. But we still don't know the point it passes through on one of those lines
As @Mark44 replied elsewhere: https://www.physicsforums.com/goto/post?id=6097457#post-6097457

Also: Check wikipedia and or google 'skew lines' and distance.
 
Last edited:
  • #7
Krushnaraj Pandya said:

Homework Statement


how to write the vector equation of the line of shortest distance between two skew lines in the shortest and most efficient way?
(The exact lines given in a particular problem in my book can be referenced- L1=(3i+8j+3k)+λ(3i-j+k) and L2=(-3i-7j+6k)+μ(-3i+2j+4k) )
So, I was playing around with this example problem, to see just how involved the solution might be.

For the cross product:
## (3\hat i - \hat j+ \hat k ) \times (-3\hat i+2\hat j+4\hat k) = -6\hat i -15\hat j+3\hat k##​

But notice that the vector from the given point on Line 1 to the given point on Line 2 is the same as that cross product.
## (-3\hat i -7 \hat j + 6\hat k ) - (3\hat i+ 8\hat j +3 \hat k) = -6\hat i -15\hat j+3\hat k ##​

So we know each of those points lies on the line which is mutually perpendicular to Line 1 and Line 2 and a vector from one of those points to the other is the vector of shortest distance between these two skew lines.

Not a very general example.
 
  • #8
Krushnaraj Pandya said:

Homework Statement


how to write the vector equation of the line of shortest distance between two skew lines in the shortest and most efficient way?
(The exact lines given in a particular problem in my book can be referenced- L1=(3i+8j+3k)+λ(3i-j+k) and L2=(-3i-7j+6k)+μ(-3i+2j+4k) )
2. Relevant methods
The vector we want would be perpendicular to both the lines, so we can use dot product=0 for both lines which gives us 2 equations in l,m and n where vector=li+mj+nk. We can find distance between the lines from a routine formula then equate the modulus of the vector sqrt(l^2+m^2+n^2) to that distance

The Attempt at a Solution


combined above

Instead of attempting fancy and sophisticated geometric constructions, the easiest way is just to compute the distance between two points (one on each line) and then minimize it. It is, in fact, easier to minimize the square of the distance; that is an equivalent problem. Do you see why?

A point on L1 has the form ##\vec{x}_1 = (3+3s, 8-s,3+s)## and a point on L2 has the form ##\vec{x}_2 = (-3-3t, -7+2t, 6+4t)##, where I write ##s## and ##t## instead of your ##\lambda## and ##\mu.## The squared distance between the points is
$$S = \sum_{i=1}^3 [x_1(i)-x_2(i)]^2 = (6+3s+3t)^2 +(15-s-2t)^2 + (-3 +s + 4t)^2.$$
We can expand this out to get
$$S = 11 s^2 + 14 s t + 29 t^2 + 270 \hspace{4ex}(1)$$ The contour curves (the ##s-t## curves of constant values for ##S##) are ellipses with center at ##(s,t) = (0,0)##, and the values of ##S## get larger as we go away from this center. Thus, the least value of ##S## is ##S = 270,## obtained at ##s = t = 0##. The two points are ##\vec{x}_{1,best} = (3,8,3)## on L1 and ##\vec{x}_{2,best} = (-3,-7,6)## on L2.

Note: normally, minimization problems like this would be handled using calculus, but because of the special structure of this type of problem it can be handled using plain algebra alone, essentially by rewriting a quadratic of the form
$$S = a s^2 + b st + c t^2 + ps + qt + r$$ to first get rid of the linear terms ##ps## and ##qt##, thus getting a simple "pure" quadratic similar to (1) above (but with different numbers). Then find the center of the ellipse of constant ##S##. No calculus is needed at all.
 
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  • #9
Ray Vickson said:
Instead of attempting fancy and sophisticated geometric constructions, the easiest way is just to compute the distance between two points (one on each line) and then minimize it. It is, in fact, easier to minimize the square of the distance; that is an equivalent problem. Do you see why?
Yes I do, I studied similar problems in "application of derivatives", I didn't know it would be so useful everywhere.
Ray Vickson said:
The contour curves (the s−ts−ts-t curves of constant values for SSS) are ellipses with center at (s,t)=(0,0)(s,t)=(0,0)(s,t) = (0,0),
Suppose I didn't realize that the equation I have is an ellipse, how would we proceed using calculus then?
 
  • #10
Ray Vickson said:
S=as2+bst+ct2+ps+qt+rS=as2+bst+ct2+ps+qt+r​
S = a s^2 + b st + c t^2 + ps + qt + r to first get rid of the linear terms pspsps and qtqtqt,
How do we remove these linear terms? I don't recall.
 
  • #12
Mark44 said:
The same technique I outlined in your thread about the distance between a line and a point can be used here. Find a point P on one of the lines, and a point Q on the other line. Then find the vector projection of PQ in the direction of one of the lines. Now find the magnitude of PQ (the hypotenuse) and the magnitude of the vector projection (one leg of the right triangle). The distance between the two skew lines is the length of the other leg.
I got the correct answer from this as well, thank you very much :D
 

Related to Vector of shortest distance between two skew lines

What is a vector of shortest distance between two skew lines?

A vector of shortest distance between two skew lines is a vector that connects two points on the two lines that are closest to each other. It represents the shortest distance between these two lines and is perpendicular to both lines.

How is the vector of shortest distance between two skew lines calculated?

The vector of shortest distance between two skew lines can be calculated by finding the cross product of the direction vectors of the two lines. This cross product will result in a vector that is perpendicular to both lines and represents the shortest distance between them.

Can the vector of shortest distance between two skew lines be zero?

No, the vector of shortest distance between two skew lines cannot be zero. Skew lines, by definition, do not intersect and therefore, there will always be a non-zero distance between them.

Is the vector of shortest distance unique?

Yes, the vector of shortest distance between two skew lines is unique. This is because the cross product of two non-parallel vectors always results in a vector that is perpendicular to both of them, and therefore, there can only be one vector that represents the shortest distance between two skew lines.

What is the significance of the vector of shortest distance between two skew lines?

The vector of shortest distance between two skew lines is significant because it can be used to find the distance between the two lines, as well as the point of closest approach between them. It is also an important concept in vector algebra and has various applications in mathematics and physics.

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