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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!

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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 $\ge |p-q|$.

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  • $\begingroup$ 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

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