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Homework Help: Closest approach of two skew lines in R3

  1. Feb 7, 2013 #1
    Hello all, and thanks again to all the help I've been getting with this book. This is a two part problem in Advanced Calculus of Several Variables, C. H. Edwards Jr. I have the first part and the second part should be easy, but I find I'm stumped.

    Since the second part builds on the solution of the first part, I'll give both problem statements and the proof of my first part.

    II.1.2a: Let [itex]f : \Re \to \Re^{n}[/itex] and [itex]g : \Re \to \Re^{n}[/itex] be two differentiable curves with f'(t) ≠ 0 and g'(t) ≠ 0 for all all [itex] t \in \Re [/itex]. Suppose the two points [itex] p = f(s_{0})[/itex] and [itex] q = g(t_{0}) [/itex] are closer than any other pair of points on the two curves. Then prove that the vector p - q is orthogonal to both velocity vectors [itex]f'(s_{0}), g'(t_{0})[/itex]. Hint: the point [itex] (s_{0}, t_{0}) [/itex] must be a critical point for the function [itex]\rho : \Re^{2} -> \Re [/itex] defined by [itex] \rho(s, t) = \left \|
    f(s) - g(t)

    From an earlier problem I have proved three lemmas to my satisfaction:

    Lemma 1: [itex]<a,a> - <b,b> = <a+b, a-b>[/itex]

    Lemma 2: If [itex] n(t) = <c, f(t)> [/itex] for some constant c, then [itex] n'(t) = <c, f'(t)>[/itex]

    Lemma 3: if [itex] n(t) = <f(t), f(t) > = \left \| f(t) \right \|^{2}[/itex], then [itex] n'(t) = 2<f'(t), f(t)> [/itex].

    Now the problem statement formally is that we must show that [itex] <p-q, f'(t)> = 0[/itex], [itex] <p-q, g'(t)> = 0[/itex].

    This is just a matter of algebra and the observation that [itex]\rho`(s_{0}, t_{0}) = 0 [/itex] with respect to either partial derivative. We get this from the hint, but the hint was unnecessary.

    [itex] \rho = <f(s), f(s)> + <g(t), g(t)> - 2<f(s), g(t)>[/itex]

    Applying lemmas 2 and 3, and differentiating first wrt s:
    [itex] \frac{d\rho}{ds}(s, t) = 2<f'(s), f(s)> - 2<f'(s), g'(t)> \\
    = 2<f(s) - g(t), f'(s)> \\
    \frac{d\rho}{dt}(s, t) = 2<g'(t), g(t)> - 2<f'(s), g'(t)> \\
    = 2<f(s) - g(t), g'(t)>[/itex]

    When we plug in [itex](s_{0}, t_{0})[/itex] to these equations, we know that they are equal to 0 by observation that [itex](s_{0}, t_{0})[/itex] is a critical point. At the same time, the expression for [itex]f(s) - g(t)[/itex], which appears in both equations, turns into [itex]f(s_{0}) - g(t_{0}) = p - q[/itex] and the equations become [itex] 0 = < p - q, g'(t_{0}) >[/itex] and [itex] 0 = < p - q, f'(s_{0})> [/itex], QED.

    Now the part I can't get is II.1.2b: Apply the result of (a) to find the closest pair of points on the "skew" straight lines in [itex] \Re^{3} [/itex] defined by [itex]f(s) = (s, 2s, -s)[/itex] and [itex]g(t) = (t+1, t-2, 2t+3)[/itex].

    So far I just have some aimless restating of facts that I know. All of them are available as results of part A above, except for two:

    [itex] f'(s) = (1, 2, -1) \\
    g'(t) = (1, 1, 2)[/itex] for all s and t.

    Somehow I've got to use these equations in reverse and solve for [itex]s_{0}[/itex] and [itex]t_{0}[/itex]. With that observation I have two equations of two unknowns:

    [itex] <f(s_{0}) - g(t_{0}), (1, 2, -1)> = 0 = <f(s_{0}) - g(t_{0}), (1, 1, 2)>[/itex]. But now I am truly stuck. I need to get these variables out of the functions, but g(t) is not linear, so I can't easily construct an inverse function for it.
    Last edited: Feb 7, 2013
  2. jcsd
  3. Feb 7, 2013 #2


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    Can't you just substitute for f(s0), g(t0) in <f(s0) - g(t0), (1,2,−1)> etc. and expand the inner products? That will give you two equations in two unknowns.
  4. Feb 7, 2013 #3
    Well, the thing is, he hasn't defined the inner product in the problem statement. In this book so far, <> has been a generalization of what he calls the "usual inner product", which I understand is what most people just call the inner product, defined as [itex]x \bullet y = x_{1}y_{1} + x_{2}y_{2} + ... + x_{n}y_{n} [/itex]. In fact he's mentioned nothing of an inner product in the problem statement at all, and the only reason I felt justified in assuming the existence of one is that the definition of orthogonality requires that an inner product be defined, and he's talking about orthogonality.

    Would any inner product I chose to define give the same results in this problem?
  5. Feb 7, 2013 #4


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    Yes, it will depend on the inner product. Whether two particular vectors are orthogonal depends on the inner product.
  6. Feb 7, 2013 #5


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    Not explicitly, perhaps, but the second part of the question is clearly to be interpreted as using the Euclidean metric (else it is not solvable).
  7. Feb 7, 2013 #6
    Alright, thanks guys. I'll take it from here and see if I can finish it off. It should be easy now...
  8. Feb 7, 2013 #7
    I'm stuck again. If expand the dot product, I get these equations:

    [itex] f_{1}(s_{1}) - g_{1}(t_{1}) + 2( f_{2}(s_{2})-g_{2}(t_{2}) ) - f_{3}(s_{3}) - g_{3}(t_{3}) = 0 \\
    f_{1}(s_{1}) - g_{1}(t_{1}) + f_{2}(s_{2}) - g_{2}(t_{2}) + 2( f_{3}(s_{3}) - g_{3}(t_{3})) = 0 \\[/itex]

    Which is six variables. I can undo f and g componentwise and have done so. But I simply don't have enough equations to solve this, do I?
  9. Feb 7, 2013 #8


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    Why six? It's just s0 and t0 isn't it?
    And you can't write out the inner product expansion until after you have substituted for f and g. If f(s)=(s,2s,−s) and v=(1,2,−1) then <f(s), v> = s+4s+s = 6s.
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