Intuition on the skew Line distances So I am trying to get a better understanding of how to find the shortest distance between two lines. I know now that if I find two parallel planes containing these two lines, then find the distance between the two planes I have the answer. My question is really why is the distance of the two planes the shortest distance between two skew lines? Please don't show me a formula which is a result of this logic, but instead derive this logic. 
Thank you in advance
 A: First, consider two distinct parallel lines. They determine a unique plane that contains both of them. From a point on one line draw a perpendicular line in the plane and that line intersects the other parallel line at another point. The distance between the two points is thus the distance between the two parallel lines. Second, consider two lines that are not parallel. If they intersect, then the two lines determine a unique plane that contains both of them and the point of intersection shows the shortest distance between the two lines is zero.
Second, comes a key idea. Given any line and any point not on that line, the shortest distance from the point to the line is along the perpendicular dropped from the point to the line because in a right triangle the hypotenuse is longer than any of the two other sides. Motivated by this, and reversing roles, given any point on any given line, consider the locus of all points such that the line determined by it and the given point is perpendicular to the given line. The locus is a plane perpendicular to the given line and passes through the given
point. Notice that give any line, all planes perpendicular to it are parallel
to each other. Also given any plane, all lines perpendicular to it are parallel
to each other.
Third, consider two given lines that are not parallel and don't intersect. They
are called skew lines. Now consider the two families of perpendicular plane
of the two lines. The two families are disjoint since all lines perpendicular to
a plane must be parallel to each other but the two given lines are not parallel.
Thus picking a plane from each family leads to a pair of intersecting planes. The intersection of such a pair is a line. All these lines in one of the planes are
parallel to each other and thus all of the lines are parallel to each other.
This family of parallel lines determines a family of parallel planes. Each of the
two skew lines is contained in a unique member of the family of parallel planes. Now, pick one of the two skew lines and project perpendicularly all the points of the other line onto the plane containing the first skew line. These points form a line which intersects the first skew line in a single point. This point and the point it was projected from on the other skew line are the points of closest approach for the two skew lines. Also, since the projection was perpendicular, the distance between the two planes is the distance between the two lines.
Notice that the key idea applies to two parallel planes also. What is the  distance between two such planes? Pick any point on the first plane and the shortest distance between it and a point on the second plane is achieved using the base point of the perpendicular line dropped to the second plane.They form a minimum pair of points. This shortest distance is the same for any point we pick because the situation is translation invariant. We can use a rigid motion to move any minimum pair to any other minimum pair because the they form a rectangle and opposite sides of any rectangle have equal lengths.
One minor detail. Given a plane and a point not on the plane, how to determine the closest point in the plane to the first point? One way is to pick a line through the first point not parallel to the plane so it intersects the plane in a point. Draw a line in the plane through this point and it determines a second plane with the first point. Draw a circle in the second plane with center at the first point and passing through the second point. This circle intersects the line at a third point with the line being a chord of the circle. Now determine from the first point the closest point to the perpendicular bisector of the two points in the plane using the key idea. This closest point is now also the closest point in the plane.
A: The shortest distance from a point to a line is measured along a perpendicular to that line, so given a pair of skew lines $\mathscr l$ and $\mathscr m$, the closest point $M$ of $\mathscr m$ to $\mathscr l$ will lie on some line that’s perpendicular to $\mathscr l$, and similarly for the closest point $L$ of $\mathscr l$ to $\mathscr m$. The line $\overline{LM}$ is thus perpendicular to both $\mathscr l$ and $\mathscr m$. The perpendicular plane $P$ to $\overline{LM}$ through $L$ contains $\mathscr l$ and the perpendicular plane $Q$ through $M$ contains $\mathscr m$. These two planes are clearly parallel, and the distance between them is $LM$, which is also equal to the distance between the lines.
