# Closest approach of two skew lines in R3

1. Feb 7, 2013

### E'lir Kramer

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: $<a,a> - <b,b> = <a+b, a-b>$

Lemma 2: If $n(t) = <c, f(t)>$ for some constant c, then $n'(t) = <c, f'(t)>$

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

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

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.

$\rho = <f(s), f(s)> + <g(t), g(t)> - 2<f(s), g(t)>$

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

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:

$<f(s_{0}) - g(t_{0}), (1, 2, -1)> = 0 = <f(s_{0}) - g(t_{0}), (1, 1, 2)>$. 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. Feb 7, 2013

### haruspex

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.

3. Feb 7, 2013

### E'lir Kramer

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?

4. Feb 7, 2013

### Dick

Yes, it will depend on the inner product. Whether two particular vectors are orthogonal depends on the inner product.

5. Feb 7, 2013

### haruspex

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).

6. Feb 7, 2013

### E'lir Kramer

Alright, thanks guys. I'll take it from here and see if I can finish it off. It should be easy now...

7. Feb 7, 2013

### E'lir Kramer

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?

8. Feb 7, 2013

### haruspex

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.