How does using cross product to find shortest distance work?

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Discussion Overview

The discussion revolves around the method of using the cross product to find the shortest distance between two skew, non-intersecting lines. Participants explore the intuition and proof behind this method, as well as the geometric implications of the cross product in relation to non-coplanar vectors.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant proposes that the shortest distance can be found by first determining the common normal using the formula $ \vec{n} = \frac{\vec{v_1} \times \vec{v_2}}{|\vec{v_1} \times \vec{v_2}|} $ and seeks proof or intuition for this approach.
  • The same participant questions whether the angles between any position vector and the skew lines minimize the projection of the distance vector onto the normal.
  • Another participant expresses uncertainty about the cross product of two non-coplanar vectors and its relationship to the perpendicularity of the resulting vector.
  • A participant asserts that two distinct vectors form the basis for a plane, suggesting that they are always coplanar unless collinear.
  • Several participants struggle to visualize the scenario involving skew lines and discuss the implications of moving vectors to meet end to end, indicating a conceptual challenge in understanding the geometry involved.
  • One participant concludes that the discussion is about the directions of the vectors, suggesting that a vector parallel to one of the skew lines intersects the other, making the cross product's perpendicularity clearer.

Areas of Agreement / Disagreement

Participants express various uncertainties and challenges in visualizing the concepts, indicating that there is no consensus on the geometric interpretation of the cross product in this context. Multiple competing views and interpretations remain present.

Contextual Notes

Limitations include the participants' struggles with visualizing the geometry of skew lines and the implications of the cross product in three-dimensional space. There are unresolved questions about the nature of coplanarity and the conditions under which the cross product applies.

ognik
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A method for finding the shortest distance between 2 skew, non intersecting lines is to 1st find the common normal, using $ \vec{n} = \frac{\vec{v_1} \times \vec{v_2}}{|\vec{v_1} \times \vec{v_2}|} $ I'm looking for a proof or intuition as to why this is true please?

Then apparently we get the shortest distance (d) by projecting any distance vector onto this normal? A distance vector here is any point on one vector minus any point on the other vector. I can obviously choose points to get distance vectors of any magnitude. So my 2nd question is, I assume the angles between any position vector and the skew lines effectively minimise the projection?

But I did a sketch - View attachment 4918 - which doesn't convince me at all about the above ..., here $|\vec{P_2P_1}| \gt |d|$, multiplying that by $Cos \theta $ produces the line from $P_2 \perp \vec{v_1}$ - still > d?
 

Attachments

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What I'm not quite sure of is - I know the cross product of 2 coplaner vectors is perpendicular to the plane those vectors are in, but why is that the same for 2 non-coplaner vectors?
 
ognik said:
What I'm not quite sure of is - I know the cross product of 2 coplaner vectors is perpendicular to the plane those vectors are in, but why is that the same for 2 non-coplaner vectors?
The two vectors (unless they are colinear) form the basis vectors for a plane. Two distinct vectors are always coplanar.

-Dan
 
Thanks - I just can't visualise it in this case - where the vectors are skew and don't intersect ... the only surface I can visualise in this case is not flat?
 
ognik said:
Thanks - I just can't visualise it in this case - where the vectors are skew and don't intersect ... the only surface I can visualise in this case is not flat?
We can always "move" the vectors so that they meet end to end.

-Dan
 
topsquark said:
We can always "move" the vectors so that they meet end to end.

-Dan
Hi - after much thought I decided we must be talking about the directions of the vectors? IE there is a vector parallel to one of the vectors, which does intersect the other vector. Then the cross product makes sense as being perpendicular to both directions?
 
ognik said:
Hi - after much thought I decided we must be talking about the directions of the vectors? IE there is a vector parallel to one of the vectors, which does intersect the other vector. Then the cross product makes sense as being perpendicular to both directions?
In a word: Yup! (Sun)

-Dan
 

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