# Closest approach of two skew lines in R3

by E'lir Kramer
Tags: closest, lines, skew
 P: 66 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 $f : \Re \to \Re^{n}$ and $g : \Re \to \Re^{n}$ be two differentiable curves with f'(t) ≠ 0 and g'(t) ≠ 0 for all all $t \in \Re$. Suppose the two points $p = f(s_{0})$ and $q = g(t_{0})$ 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 $f'(s_{0}), g'(t_{0})$. Hint: the point $(s_{0}, t_{0})$ must be a critical point for the function $\rho : \Re^{2} -> \Re$ defined by $\rho(s, t) = \left \| f(s) - g(t) \right\|^{2}$ From an earlier problem I have proved three lemmas to my satisfaction: Lemma 1: $- =$ Lemma 2: If $n(t) =$ for some constant c, then $n'(t) =$ Lemma 3: if $n(t) = = \left \| f(t) \right \|^{2}$, then $n'(t) = 2$. Now the problem statement formally is that we must show that $= 0$, $= 0$. This is just a matter of algebra and the observation that $\rho`(s_{0}, t_{0}) = 0$ with respect to either partial derivative. We get this from the hint, but the hint was unnecessary. $\rho = + - 2$ Applying lemmas 2 and 3, and differentiating first wrt s: $\frac{d\rho}{ds}(s, t) = 2 - 2 \\ = 2 \\ \frac{d\rho}{dt}(s, t) = 2 - 2 \\ = 2$ When we plug in $(s_{0}, t_{0})$ to these equations, we know that they are equal to 0 by observation that $(s_{0}, t_{0})$ is a critical point. At the same time, the expression for $f(s) - g(t)$, which appears in both equations, turns into $f(s_{0}) - g(t_{0}) = p - q$ and the equations become $0 = < p - q, g'(t_{0}) >$ and $0 = < p - q, f'(s_{0})>$, 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 $\Re^{3}$ defined by $f(s) = (s, 2s, -s)$ and $g(t) = (t+1, t-2, 2t+3)$. 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: $f'(s) = (1, 2, -1) \\ g'(t) = (1, 1, 2)$ for all s and t. Somehow I've got to use these equations in reverse and solve for $s_{0}$ and $t_{0}$. With that observation I have two equations of two unknowns: $= 0 =$. 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.
 Homework Sci Advisor HW Helper Thanks P: 9,849 Can't you just substitute for f(s0), g(t0) in etc. and expand the inner products? That will give you two equations in two unknowns.
 P: 66 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 $x \bullet y = x_{1}y_{1} + x_{2}y_{2} + ... + x_{n}y_{n}$. 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?
 Sci Advisor HW Helper Thanks P: 25,228 Closest approach of two skew lines in R3 Yes, it will depend on the inner product. Whether two particular vectors are orthogonal depends on the inner product.
Homework
 P: 66 I'm stuck again. If expand the dot product, I get these equations: $f_{1}(s_{1}) - g_{1}(t_{1}) + 2( f_{2}(s_{2})-g_{2}(t_{2}) ) - f_{3}(s_{3}) - g_{3}(t_{3}) = 0 \\ f_{1}(s_{1}) - g_{1}(t_{1}) + f_{2}(s_{2}) - g_{2}(t_{2}) + 2( f_{3}(s_{3}) - g_{3}(t_{3})) = 0 \\$ 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?