Use the following scheme to prove that in E3 a straight line is the shortest distance between two points. Let X : [a,b] → E3 be a curve, and set x(a) = p, x(b) = q.

Hi everyone - I wanted to make sure I was on the right track. I believe I need to show that the distance between p and q is less than or equal to the arclength of some L - thereby showing that the shortest distance is always the straight line between them. Here's what I have so far:

d(p,q) = q - p

X(t) = p + t(q - p) = p(1 - t) + tq (for t:[0,1])

X'(t) = q - p

L = q - p

Thereby proving that L = d(p,q).

But I feel like that was too easy. Did I calculate the arclength right? Thanks very much for any help!


There are several things wrong.

First, $d(p,q) = |p-q|$; $p-q$ is a vector, $|p-q|$ is its length.

Second, you did not show how you got the arc length computation from the formula for $X'(t)$; and since you got a vector $p-q$, something is wrong.

Third, and most important, once you've correctly shown that the arc length of $X(t)$ equals $|p-q|$, you then have to show that every other path from $p$ to $q$ has arc length $\ge |p-q|$.

  • $\begingroup$ Okay, I understand the piece about the distance. It sounds like X(t) was defined correctly, and I found X'(t) okay. So L is the square root of the dot product of X'(t), so I should have L=(q^2+p^2)^1/2. I am not seeing the relation to distance yet. $\endgroup$ – Carolyn Apr 18 '16 at 0:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.