How Do You Calculate the Shortest Distance Between Two Skew Lines in Space?

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SUMMARY

The shortest distance between two skew lines in space, represented by the lines A={a+s*u|s ∈ R} and B={b+t*v|t ∈ R}, is calculated using the formula \(\frac{|(u \times v) \bullet (a-b)|}{||u \times v||}\). In this context, vectors u and v are linearly independent, and the cross product \(u \times v\) yields a vector perpendicular to both lines. The discussion highlights the importance of understanding vector operations and the geometric interpretation of the distance between skew lines.

PREREQUISITES
  • Understanding of vector operations, specifically cross products and dot products.
  • Familiarity with the concept of skew lines in three-dimensional space.
  • Knowledge of linear independence in vector spaces.
  • Basic principles of geometry related to distance calculations in three dimensions.
NEXT STEPS
  • Study the properties of vector cross products and their geometric interpretations.
  • Learn about linear independence and its implications in vector spaces.
  • Explore the derivation of the distance formula between skew lines in greater detail.
  • Investigate applications of skew line distance calculations in physics and engineering.
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Mathematicians, physics students, and engineers who require a solid understanding of vector geometry and its applications in calculating distances in three-dimensional space.

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1. The problem statement,
all variables and given/known data

There are 4 vectors a,b,u,v in space and {u,v} are linearly independen
The groups A={a+s*u|s E R} B = {b+t * v|t E R}are two lines in space

The Two A , B lines does not meet and the "part/line/segment (I don't know how it called ) " PQ is the smallest part that is ends are [tex]Q \in B ,P \in A[/tex] show that the length of PQ is"
[tex]\frac{|(u \times v)\bullet(a-b)|}{||u \times v||}[/tex]



Homework Equations



all calc/..

The Attempt at a Solution


I mange to get to the point of
[tex]u \times v =(s)[/tex]
and then
[tex]\frac{L(s)*L(a-b)*cos(w)}{L(s)}=L(a-b)*cos(w)[/tex]
bUT I don't know what's next
 
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ThankYou said:
I mange to get to the point of
[tex]u \times v =(s)[/tex]
and then
[tex]\frac{L(s)*L(a-b)*cos(w)}{L(s)}=L(a-b)*cos(w)[/tex]
bUT I don't know what's next

Huh?:confused: How are you managing to get to this point? What do [itex]w[/itex] and [itex]L[/itex] represent?

What is the general formula for the distance between two points [itex]\textbf{r}_1[/itex] and [itex]\textbf{r}_2[/itex]? What if the points lie on the lines given in the problem (i.e. [itex]\textbf{r}_1=\textbf{a}+s\textbf{u}[/itex] and [itex]\textbf{r}_2=\textbf{b}+t\textbf{v}[/itex])? How would you go about minimizing that distance?
 
Yes I've managed to do it...
I've done:
[tex]u \times v[/tex] and got the vector that is vertical to both u , v
Then I've made a plane that use this vector and use point a, then I've used the forumala of plane distance from point in the first line.
Thank you/
 

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