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Distance Between 2 Lines: Vectors

Date: 8/19/96 at 23:29:28
From: Anonymous
Subject: Shortest Distance...

What is the shortest distance between 2 lines?

Date: 8/20/96 at 8:16:26
From: Doctor Anthony
Subject: Re: Shortest Distance...

I am not sure how much vector work you have done, but I will assume a 
knowledge of scalar products of vectors, and the vector equation of 
straight lines.  

In 3D space the shortest distance between two skew lines is in the 
direction of the common perpendicular. (There is one and only one such 
direction, as can be seen if you move one line parallel to itself 
until it intersects the other line. These two lines would now define a 
plane, and the perpendicular to this plane is the direction of the 
common perpendicular).  

You now take any point on one line, and any point on the other line, 
and write down the vector joining these two points.  Finally you find 
the component of this vector in the direction of the common 
perpendicular.  This is done by finding the scalar product of the 
vector with the UNIT vector in the direction of the common 
perpendicular.  The result of the scalar product is the shortest 
distance you require.  

I will illustrate the method by means of an example. 

Find the shortest distance between the lines:

x/1 = (y-3)/1 = z/(-1)

(x-5)/3 = (y-8)/7 = (z-2)/(-1)

First we require the vector perpendicular to both (1,1,-1) and 

Let the common perpendicular be (p,q,r). The scalar product of this 
with both (1,1.-1) and (3,7,-1) will be zero, so:

   p+q-r = 0 and 3p+7q-r = 0

Note that although there are apparently 3 unknowns and only two 
equations, these are homogeneous equations (having 0 on the right hand 
side), so we could find values of p/r and q/r and hence the ratios 
p:q:r which is all that we require.  Using the determinant method for 
solving, we have:

    p       -q          r
 ------ = -------  =  ------
|1  -1|   |1  -1|     |1   1|
|7  -1|   |3  -1|     |3   7|

   p/6  =  -q/2    =   r/4

   p/3  =  q/-1    =   r/2   and so  p:q:r = 3:-1:2

So the common perpendicular is the vector (3,-1,2)

As a UNIT vector this is (1/sqrt(14)){3,-1,2}

Next we have point (0,3,0) on line (1) and (5,8,2) on line (2).  
The vector joining these points is (5,5,2)  and now scalar product 
this with the unit vector of the common perpendicular.

Scalar product = (1/(sqrt(14)){5*3 + 5*(-1) + 2*2}
               = (1/sqrt(14)){15 - 5 + 4}
               = 14/sqrt(14)
               = sqrt(14)    

   and this is the shortest distance required.

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Linear Algebra

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