# Basic Differential geometry: Shortest path between two points in R^3 is straight.

Given two points P and Q in $\mathbb{R^3}$, we want to show that the shortest distance between them is through a straight line.

let $c(a) = P$ and $c(b) = Q$ and $c(t)\neq P$ for $t>a$(One question I have is: Why is this condition important?). Define $f(t) = |c(t)-c(a)|$. Compute $f'(t)$ and prove that $|f'|\leq |c'|$. (Can someone give a hint on how to go about doing this after computing f'? Now I think we can use the formula $f(b)-f(a) = \int_{a}^b f'(t)dt$ and use the previous inequality to show that $|Q-P|\leq \int_{a}^{b}|c'|dt$ but I'm not entirely sure.

Any help would be appreciated.

• Have you tried computing $f'(t)$? What have you got so far? – Rahul Jan 22 '15 at 9:25
• I did compute f'. the result is pretty messy so I will type it out later. but basically I considered the R^3 norm using the three component functions of c and took the derivative. – Bobby Jones Jan 22 '15 at 19:32