I'm currently taking multivariable calculus (the only math course after AP Calc that is offered by our school) as a senior in high school, and I've run into the following issue. We were discussing the idea of the distance between two skew lines in threedimensional space, and we defined the distance as the length of the mutually perpendicular line segment, and said that this is equivalent to the distance between the two planes, each containing one of the lines, that are parallel to each other. The issue I have is with the idea that the mutually perpendicular segment is unique. This seems to be true intuitively, and we've made many arguments that seem to show it, but nobody, including the teacher, has so far been able to prove it. Most textbooks I've looked at seem to simply assume that it's true and then move on to calculating the distance, but I want to actually prove that it works.
Does anyone know of a way to prove that there is always exactly one mutually perpendicular line segment (or line) joining two skew lines?
Unique Distance Between 2 Skew Lines
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 jestingrabbit
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Re: Unique Distance Between 2 Skew Lines
You'll want to parametrise the lines then look at lines, L, going from one line to the other, that are perpendicular to one of the lines but not the other. As you vary the starting point of the perp, you'll vary the dot product between the direction of L and the direction of the other line. Have a look at how it varies, and you should see there's only one place where the dot product is 0.
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Re: Unique Distance Between 2 Skew Lines
There's a few different ways to prove it. Jestingrabbit gave you one good one.
If you happen to know that the angle sum of a quadrilateral is [imath]2\pi[/imath] if and only if it is planar, then this follows immediately. (If it were false, consider two such mutual perpendiculars, and derive a contradiction). However, you're relatively unlikely to have seen that theorem in high school geometry, so you're probably not familiar with it.
Here's a more (for me) intuitive proof:
First note that a mutual perpendicular has the property that it is a local minimizer of the length of a segment between the two lines; moving either endpoint slightly will increase the length of the segment.
Let one of the lines be parametrized by a variable t. (I.e., we have a function p(t) that maps each real number t to a point on the line). For each t, there is a unique point on the second line that is closest to p(t); specifically, it is the point where the perpendicular from p(t) to the second line intersects the second line. Let's call that point q(t). Now, a few observations: first, p(t) is obviously a linear function of t. It's a littler harder to see, but q(t) is a linear function of t as well (you can actually get the expression with some linear algebra if you want). If we write down the distance squared between p(t) and q(t), we get a function that is quadratic in t (because it's a sum of squares of linear terms). Quadratic functions have a single unique global minimum, so there's a unique minimizer, so there's a unique perpendicular.
It's not too painful to go through all this and put in formulas, but there's really no reason to.
If you happen to know that the angle sum of a quadrilateral is [imath]2\pi[/imath] if and only if it is planar, then this follows immediately. (If it were false, consider two such mutual perpendiculars, and derive a contradiction). However, you're relatively unlikely to have seen that theorem in high school geometry, so you're probably not familiar with it.
Here's a more (for me) intuitive proof:
First note that a mutual perpendicular has the property that it is a local minimizer of the length of a segment between the two lines; moving either endpoint slightly will increase the length of the segment.
Let one of the lines be parametrized by a variable t. (I.e., we have a function p(t) that maps each real number t to a point on the line). For each t, there is a unique point on the second line that is closest to p(t); specifically, it is the point where the perpendicular from p(t) to the second line intersects the second line. Let's call that point q(t). Now, a few observations: first, p(t) is obviously a linear function of t. It's a littler harder to see, but q(t) is a linear function of t as well (you can actually get the expression with some linear algebra if you want). If we write down the distance squared between p(t) and q(t), we get a function that is quadratic in t (because it's a sum of squares of linear terms). Quadratic functions have a single unique global minimum, so there's a unique minimizer, so there's a unique perpendicular.
It's not too painful to go through all this and put in formulas, but there's really no reason to.
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Re: Unique Distance Between 2 Skew Lines
Here's a more visual argument:
Imagine parameterizing the lines as a+tu and b+sv. (I'll use boldface letters for vectors.) Then the distance between the points a+tu and b+sv on the lines is the same as the distance between the point a and a point b+svtu on the plane parameterized by b+svtu. Similarly, a mutually perpendicular line segment between the points a+tu and b+sv corresponds to a line segment between the point a and a point b+svtu which is perpendicular to both the directions u and v in the plane, i.e. which is perpendicular to the plane.
So the fact that there is a unique minimum distance, and that distance is the length of the unique mutually perpendicular segment, follows directly from the fact that there is a unique minimum distance from a point to a plane, and that this distance is the length of a line segment from the point to the plane which is perpendicular to the plane.
Did that make sense? There is also a proof using partial derivatives, but I'm guessing you haven't seen those yet.
Imagine parameterizing the lines as a+tu and b+sv. (I'll use boldface letters for vectors.) Then the distance between the points a+tu and b+sv on the lines is the same as the distance between the point a and a point b+svtu on the plane parameterized by b+svtu. Similarly, a mutually perpendicular line segment between the points a+tu and b+sv corresponds to a line segment between the point a and a point b+svtu which is perpendicular to both the directions u and v in the plane, i.e. which is perpendicular to the plane.
So the fact that there is a unique minimum distance, and that distance is the length of the unique mutually perpendicular segment, follows directly from the fact that there is a unique minimum distance from a point to a plane, and that this distance is the length of a line segment from the point to the plane which is perpendicular to the plane.
Did that make sense? There is also a proof using partial derivatives, but I'm guessing you haven't seen those yet.
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Re: Unique Distance Between 2 Skew Lines
The line segment must be some scalar multiple of the cross product of the direction vectors of the lines (this is the nontrivial step; I hope you believe that orthogonality is a necessary condition). Now one can draw a plane containing one of the lines L1 consisting of all points such that the vector from that point to L1 is a scalar multiple of this cross product. And under the proper nondegeneracy conditions this plane intersects the second line L2 once.
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