# Closest approach of two skew lines in R3

• E'lir Kramer
In summary, the conversation is about a problem in Advanced Calculus of Several Variables, C. H. Edwards Jr. The problem involves proving that the vector p-q is orthogonal to both velocity vectors f'(s0) and g'(t0), where p and q are two points closer than any other pair of points on the two differentiable curves f and g. The conversation also includes the solution to the problem, which involves three lemmas and using algebra to show that < p - q, g'(t0) > and < p - q, f'(s0) > both equal to 0. The second part of the problem involves applying the result from the first part to find the closest pair of points on two skew straight
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.

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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.

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?

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

E'lir Kramer said:
Well, the thing is, he hasn't defined the inner product in the problem statement.
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).

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

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?

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.

## 1. What is the closest distance between two skew lines in R3?

The closest distance between two skew lines in R3 is equal to the length of the shortest segment connecting the two lines. This distance can be calculated using the formula d = |(P2-P1) * n| / |n|, where P1 and P2 are two points on the respective lines and n is a normal vector to both lines.

## 2. Can the closest distance between two skew lines be negative?

No, the closest distance between two skew lines cannot be negative. This is because the distance is defined as a positive value and represents the shortest distance between the two lines. If the lines do not intersect, the closest distance will always be positive.

## 3. How do you determine if two lines in R3 are skew?

Two lines in R3 are skew if they do not intersect and are not parallel. This means that they do not lie on the same plane and do not have any points in common. To determine if two lines are skew, you can check if their direction vectors are not parallel and if they have a common point. If these conditions are not met, the lines are skew.

## 4. Is there a formula for finding the closest point between two skew lines in R3?

Yes, there is a formula for finding the closest point between two skew lines in R3. This point can be found by setting the two lines equal to each other and solving for the parameters of both lines. The resulting point will be the closest point between the two lines.

## 5. Can two skew lines in R3 ever intersect?

No, two skew lines in R3 can never intersect. Skew lines are defined as lines that do not lie on the same plane and do not have any points in common. Therefore, they can never intersect. If two lines in R3 do intersect, they are either parallel or the same line.

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