Finding shortest distance between skew lines, checking work.

In summary, the conversation is about finding the shortest distance between two skew lines using different methods. The original poster shares their method of using vector products and asks for feedback on their answer. Other users suggest alternative methods and discuss the calculations, ultimately concluding that the original answer is correct and the book's solution is wrong. They also discuss the ease of using different methods and the importance of having a calculator.
  • #1
RoboNerd
410
11
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!
 
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  • #2
RoboNerd said:
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
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
Math_QED said:
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
mfb said:
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
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
Math_QED said:
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
mfb said:
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
Math_QED said:
Then calculate the distance between that plane and a random point on RS.
Which method do you use for that?
 
  • #10
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
@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
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
mfb said:
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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1. How do you find the shortest distance between two skew lines?

To find the shortest distance between two skew lines, you can use the formula d = |(a1 - a2) · (b1 x b2)| / ||b1 x b2||, where a1 and a2 are points on each line, and b1 and b2 are the respective direction vectors of the lines. This formula takes into account the perpendicular distance between the lines and the direction of the lines.

2. Can you explain the concept of skew lines?

Skew lines are two lines in three-dimensional space that do not intersect and are not parallel. They have different slopes and do not lie in the same plane. This means they are not coplanar and will never intersect or be parallel, creating a unique challenge in finding the shortest distance between them.

3. What is the significance of finding the shortest distance between skew lines?

The shortest distance between skew lines is important in various applications, such as engineering and physics, where the exact distance between two non-intersecting lines needs to be determined. This can also help in finding the closest point of intersection between the lines, which can be useful in solving problems involving motion or velocity.

4. How can you check your work when finding the shortest distance between skew lines?

After using the formula to calculate the shortest distance between skew lines, you can check your work by plugging in the points and direction vectors of the lines into the distance formula and comparing the result to your calculated distance. You can also use a graphing calculator or software to visually confirm the distance and intersection point between the lines.

5. Are there any special cases to consider when finding the shortest distance between skew lines?

Yes, there are a few special cases to consider. If the lines are parallel, the shortest distance will be the perpendicular distance between them. If the lines are skew but have a common perpendicular, the shortest distance will be the distance between the common perpendicular and the intersection point of the lines. Lastly, if the lines are skew but intersect at a point, the shortest distance will be 0, as the lines are no longer skew and are now coplanar.

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