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I am trying to understand the Shortest dostance between two skew lines. I know the process. But I can not understand the process. My doubts are following.

First: A line segment , which is perpendicular to both lines , is drawn . (My doubt --- How you are sure to have a segment like that? There may not exist such kind of segment.)

Second: Two points P & Q, one from each line, are taken and we determine the length of the projection of PQ on the segment. The length of the projection on the segment , which is perpendicular to both the lines, is nothing but the length of the segment .This length is shortest distance. ( My question : PQ and the segment, which is perpendicular to both the lines, may not be coplanar. Then how we take the projection of PQ on the segment. )

Can someone please help me to clear my doubts?

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Consider two parallel planes. The distance between these planes is the length of the common perpendicular segment connecting the two planes. Now if we draw two lines one on each plane we can define the distance between the lines to be the distance between the two parallel planes.

If you start with two sqew lines you can construct parallel planes on which these two lines locate.

Thus the distance between sqew lines is well defined and the process works as you have explained.

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  • $\begingroup$ That's how I can construct infinitely many pair of parallel planes which will contain those two lines. Can you assure me that for every case the perpendicular distance between two parallel planes will remain same?@Mohammad Riazi-Kermani $\endgroup$ – sani Mar 11 at 8:59
  • $\begingroup$ I do not agree with infinitely many parallel planes. I think there are only two parallel planes containing the skew lines. $\endgroup$ – Mohammad Riazi-Kermani Mar 11 at 11:55
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Let $r$, $s$ be two skew lines.

First: existence of a common perpendicular. Take point $A$ on $r$ and construct line $s'$, passing through $A$ and parallel to $s$. The plane $\alpha$ containing $r$ and $s'$ is parallel to $s$. Take then points $B$ and $C$ on $s$ and let $B'$, $C'$ be their perpendicular projections on $\alpha$: line $B'C'$ is then parallel to $s$ and $s'$ and must therefore meet line $r$ at some point $E$. The line through $E$ parallel to $BB'$ lies on plane $BCB'C'$ and meets line $s$ at some point $F$. Line $EF$ is the requested common perpendicular: it is parallel to $BB'$, hence perpendicular to both $s$ and $\alpha$.

Second: minimum distance. Take now any pair of points $P\in r$ and $Q\in s$, different from $E$ and $F$ defined above: let's show that $PQ>EF$. If either $P$ or $Q$ (but not both) is the same as $E$ and $F$, then $PQ$ is the hypotenuse and $EF$ the leg of a right triangle, hence the thesis follows. Otherwise, draw line $PP'$ parallel to $s$, such that $PP'FQ$ is a parallelogram with $FP'=QP$: you can then prove as before that $FP'>EF$, i.e. $PQ>EF$.

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Here is a picture that might help. enter image description here

Skew lines with a red and blue plane drawn where each line is completely within its plane and the red and blue planes are parallel. The perpendicular plane also completely includes the L2 line and it intersects the bottom plane. It forms a line of intersection with the bottom plane and that line intersects L1 at point C.

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