Finding shortest distance between skew lines, checking work.

  • #1
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Hi everyone.

I was working on a problem for days.

The problem statement is: "Consider points P(2,1,3), Q(1,2,1), R(-1,-1,-2), S(1,-4,0). Find the shortest distance between lines PQ and RS."

Now, I did the following formula: PS dot (PQ x RS) / magnitude of (PQ x RS). (For skew lines)

Now, I have that
PQ = Q-P = (1,2,1) - (2,1,3) = (-1,1,-2) and
RS = S - R = (1,-4,0)-(-1,-1,-2) = (2,-3,2)
PS = S - P = (1,-4,0) - (2,1,3) = (-1,-5,-3).

PQ x RS = < -1, -5, -3>
Thus, I have PS * (PQ x RS) / | PQ x RS | = <-1,-5,-3> dot <-4,-2,1> (the result of my crossing operation) / sqrt( 4^2 + 2^2 + 1). This results in 11 / sqrt(21).

This is a strange result to me as the book's solution is sqrt(3). Could anyone please give me some feedback as to what I have done (right or wrong)?

Thank you in advance!
 

Answers and Replies

  • #2
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PQ x RS = < -1, -5, -3>
I get completely different values here.
Edit: Ah, just a typo, the line afterwards has the right result.

I get the same result as you.
 
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  • #3
Math_QED
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I didn't check your work, but I think there is an easier method.

Find the equation of the plane that contains the line PQ and the vector that gives the direction of RS. Try to imagine that. Then, take a random point on the line RS and calculate the distance between the plane and this point (there is this nice formula that you can use)
 
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  • #4
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Find the equation of the plane that contains the line PQ and the vector that gives the direction of RS. Try to imagine that. Then, take a random point on the line RS and calculate the distance between the plane and this point (there is this nice formula that you can use)
That leads to exactly the calculations RoboNerd did.
 
  • #5
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That leads to exactly the calculations RoboNerd did.
There is no need to use vector products using this method, which is, in my opinion, easier.
 
  • #6
mathman
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I checked - your result looks correct (noting the typo). I also tried the other three possible vectors (PR,QR,QS) and got the same result.
 
  • #7
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There is no need to use vector products using this method, which is, in my opinion, easier.
The vector product is the easiest method to get the normal vector I think.
 
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  • #8
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The vector product is the easiest method to get the normal vector I think.
You don't need to find the normal vector in my method. Just find the equation of the plane that contains PQ and the direction of RS. Then calculate the distance between that plane and a random point on RS. I got 11/√21 too.
 
  • #9
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Then calculate the distance between that plane and a random point on RS.
Which method do you use for that?
 
  • #10
wrobel
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Assume we have two lines. One of them passes through the points ##A,B## and the second one passes through the points ##C,D##. Then the distance between the lines is calculated by the following formula

##h=\frac{|(\boldsymbol{AB}\times \boldsymbol{CD},\boldsymbol{AC})|}{|\boldsymbol{AB}\times \boldsymbol{CD}|}##
 
  • #11
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@wrobel: That approach has been posted in post#1 already, and calculated by three users in this thread. How does repeating the formula help?
 
  • #14
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OK then. Thank you very much to everyone who took the time out of their busy lives to help me out.

I am glad that I did not mess up and that the book's answer is wrong.

Thank you again!
 
  • #16
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One method uses a normal vector, the other first requires to convert the plane to the form ax+by+cz=d which is more effort than finding the normal vector (because you can directly read off the normal vector from that equation but not the equation from the normal vector).
Depends whether you have a calculator ;)
 
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