Point-Point Distance--3-D

The Line Element is

$$ds=\sqrt{dx^2+dy^2+dz^2},$$ (1)

so the Arc Length between the points $x_1$ and $x_2$ is

$$L = \int ds = \int_{x_1}^{x_2} \sqrt{1+{y'}^2+z'^2}\,dx$$ (2)

and the quantity we are minimizing is

$$f=\sqrt{1+{y'}^2+z'^2}.$$ (3)

Finding the derivatives gives

$${\partial f\over \partial y} = 0$$ (4)

$${\partial f\over \partial z} = 0$$ (5)

and

$${\partial f\over\partial y'} = {y'\over\sqrt{1+y'^2+z'^2}}$$ (6)

$${\partial f\over\partial z'} = {z'\over\sqrt{1+y'^2+z'^2}},$$ (7)

so the Euler-Lagrange Differential Equations become

$${d\over dx} \left({y'\over\sqrt{1+{y'}^2+z'^2}}\right) = 0$$ (8)

$${d\over dx} \left({z'\over\sqrt{1+{y'}^2+z'^2}}\right) = 0.$$ (9)

These give

$${y'\over\sqrt{1+{y'}^2+z'^2}}=c_1$$ (10)

$${z'\over\sqrt{1+{y'}^2+z'^2}}=c_2.$$ (11)

Taking the ratio,

$$z'={c_2\over c_1}y'$$ (12)

$${y'\over\sqrt{1+y'^2+\left({c_2\over c_1}\right)^2y'^2}} = c_1$$ (13)

$$y'^2={c_1}^2\left[{1+y'^2+\left({{c_2\over c_1}}\right)^2 y'^2}\right]= {c_1}^2+y'^2({c_1}^2+{c_2}^2),$$ (14)

which gives

$$y'^2 = {{c_1}^2\over 1-{c_1}^2-{c_2}^2} \equiv {a_1}^2$$ (15)

$$z'^2 = \left({c_2\over c_1}\right)^2 y'^2 = {{c_2}^2\over 1-{c_1}^2-{c_2}^2} \equiv {b_1}^2.$$ (16)

Therefore, $y'=a_1$ and $z'=b_1$, so the solution is

$$\left[{\matrix{x\cr y\cr z\cr}}\right]=\left[{\matrix{x\cr a_1x+a_0\cr b_1x+b_0\cr}}\right],$$ (17)

which is the parametric representation of a straight line with parameter $x \in [x_1,x_2]$. Verifying the Arc Length gives

$$L=\sqrt{1+{a_1}^2+{b_1}^2}\,(x_2-x_1)$$ (18)

where

$$\left[{\matrix{y_1\cr y_2\cr}}\right] = \left[{\matrix{x_1 & 1\cr x_2 & 1\cr}}\right]\left[{\matrix{a_1\cr a_0\cr}}\right]$$ (19)

$$\left[{\matrix{z_1\cr z_2\cr}}\right] = \left[{\matrix{x_1 & 1\cr x_2 & 1\cr}}\right]\left[{\matrix{b_1\cr b_0\cr}}\right].$$ (20)

See also Point-Point Distance--1-D, Point-Point Distance--2-D, Point-Quadratic Distance