Shortest Distance between 2 skew lines (vectors)

In summary: Also, the minimum of the squared distance is the same as the minimum of the distance itself, so you don't lose any information by using the squared distance.
  • #1
thomas49th
655
0

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
 
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  • #2
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.
 
  • #3
thomas49th said:

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
 
  • #4
HallsofIvy said:
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
 
  • #5
Ray Vickson said:
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.
 
Last edited:
  • #6
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).
 
  • #7
thomas49th said:
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
 
  • #8
why would you want to minimize d(s,t)^2 and not d(s,t). am I missing something stupidly ovbious?
 
  • #9
thomas49th said:
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.
 

What is the concept of "shortest distance" between two skew lines?

The shortest distance between two skew lines is the shortest possible distance between any two points on the two lines. It is the perpendicular distance between the two lines, measured from a point on one line to the other line.

How do you determine the shortest distance between two skew lines?

To determine the shortest distance between two skew lines, we first find the closest points on each line to the other line. Then, we calculate the distance between these two points using the distance formula. This distance is the shortest distance between the two skew lines.

What is the formula for calculating the shortest distance between two skew lines?

The formula for calculating the shortest distance between two skew lines involves finding the closest points on each line to the other line and using the distance formula to calculate the distance between these points. The formula is:
d = |(a1 - a2) * n| / |n|, where a1 and a2 are the closest points on each line and n is the direction vector of either line.

Can the shortest distance between two skew lines ever be zero?

No, the shortest distance between two skew lines will never be zero. This is because skew lines, by definition, do not intersect and therefore do not share any common points. The distance between two lines is only zero when they intersect.

Are there any real-life applications of finding the shortest distance between two skew lines?

Yes, there are several real-life applications of finding the shortest distance between two skew lines. One example is in computer graphics, where it is used to calculate the distance between two objects in a 3D space. It is also used in physics and engineering for calculating the distance between two non-intersecting objects, such as wires or pipes.

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