Vector of shortest distance between two skew lines

[Krushnaraj Pandya]

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

 

[scottdave]

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?

 

[Krushnaraj Pandya]

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.

 

[Krushnaraj Pandya]

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

 

[Mark44]

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.

 

[SammyS]

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.

 

[SammyS]

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.

 

[Ray Vickson]

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.

 

[Krushnaraj Pandya]

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.

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?

 

[Krushnaraj Pandya]

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.

 

[Krushnaraj Pandya]

As @Mark44 replied elsewhere: https://www.physicsforums.com/goto/post?id=6097457#post-6097457

Also: Check wikipedia and or google 'skew lines' and distance.

It was quite helpful, thank you very much :D

 

[Krushnaraj Pandya]

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