Shortest Distance between 2 skew lines (vectors)

[thomas49th]

Homework Statement

x1 = (0,0,4) + s(2,0,-1)
x2 = (-4,2,2) + t(-5,1,1)

Homework Equations

The Attempt at a Solution

First of all I find the common perpendicular (vector cross product)
Make it unit
Find a arbitrary line joining the two 2 lines (set s = t= 0)

Then scalar produce the unit normal and the line between the vectors to get the minimum distance. I get 2/sqrt(14), while the answers say just sqrt(14). How so?

Also I am just doing the method, but what does it mean geometrically. Why find a line perpendicular, or use a unit vector?

Thanks
Thomas

 

[HallsofIvy]

The shortest distance from a point to a line is always along the line perpendicular to the given line. That is because the hypotenuse of a right triangle is longer than either leg.

As a result of that, the shortest distance between two skew lines is along the line perpendicular to both given lines.

 

[Ray Vickson]

Homework Statement

x1 = (0,0,4) + s(2,0,-1)
x2 = (-4,2,2) + t(-5,1,1)

Homework Equations

The Attempt at a Solution

First of all I find the common perpendicular (vector cross product)
Make it unit
Find a arbitrary line joining the two 2 lines (set s = t= 0)

Then scalar produce the unit normal and the line between the vectors to get the minimum distance. I get 2/sqrt(14), while the answers say just sqrt(14). How so?

Also I am just doing the method, but what does it mean geometrically. Why find a line perpendicular, or use a unit vector?

Thanks
Thomas

I did it another way: I expressed the distance^2 between x1 and x2 as a function of s and t, then minimized it. My distance answer agrees with yours.

RGV

 

[thomas49th]

The shortest distance from a point to a line is always along the line perpendicular to the given line. That is because the hypotenuse of a right triangle is longer than either leg.

As a result of that, the shortest distance between two skew lines is along the line perpendicular to both given lines.

I know that the shortest distance between 2 skew lines is perpendicular, but why does finding the "common perpendicular" help. Let's go to a 2D graph. Let there be 2 parallel lines (2D equiv of skew lines). Taking the vector product produces a vector into or out of the graph (from our view on a 2D graph you cannot see the vector because it goes "into" or "out of" the graph).
How does finding this vector help. For that matter, how does finding an arbitray line between the 2 vectors help

Thanks
Thomas

 

[thomas49th]

I did it another way: I expressed the distance^2 between x1 and x2 as a function of s and t, then minimized it. My distance answer agrees with yours.

RGV

That's how the problem sheet answer is done (I don't like doing it that way... I DON'T understand why you square it) but they get sqrt(14). I get 2/sqrt(7) or sqrt(7/2) . Hmm.

 

[Quinzio]

Your answer is correct.
Geometrically meaning: imagine the 2 parallel planes that each contains one of the lines.
Easier: imagine 2 parallel planes and on each plane draw one line.
Connect the 2 lines by a 3rd line AB. In A draw the perpendicular of the 2 planes.
The length of the perpendicular is the projection of AB onto the perpendicular of the planes in A (or B, the same).

 

[Ray Vickson]

That's how the problem sheet answer is done (I don't like doing it that way... I DON'T understand why you square it) but they get sqrt(14). I get 2/sqrt(7) or sqrt(7/2) . Hmm.

Minimizing d(s,t)^2 gives the same solution as minimizing d(s,t). What could be simpler?

RGV

 

[thomas49th]

why would you want to minimize d(s,t)^2 and not d(s,t). am I missing something stupidly ovbious?

 

[SammyS]

why would you want to minimize d(s,t)^2 and not d(s,t). am I missing something stupidly obvious?

The square of the distance does not have square roots, so it's much easier to work with.