Average distance point to line 3d maybe

[Guest]

We have a 3d line with both end points known and another point with its coordinates known in 3d, how to find average distance to all points of said line from said other point? Can you connect with me, maybejosiah@aol.com if you have an answer?

[Guest]

let the coordinates of all three points be [tex](x_i,y_i,z_i)[/tex] where [tex]i=1,2,3[/tex] such that the third point is external to the line. Now, the parametric representation of any general point on the line is [tex]P(u)=\begin{cases} x=x_1+u(x_2-x_1) \\ y = y_1+u(y_2-y_1) \\ z=z_1+u(z_2-z_1) \end{cases}[/tex], where [tex]u\in[0,1][/tex]. Now, the distance from the third point to this general point is [tex]\sqrt{\left\{x_3-x_1-u(x_2-x_1)\right\}^2+\left\{y_3-y_1-u(y_2-y_1)\right\}^2+\left\{z_3-z_1-u(z_2-z_1)\right\}^2}[/tex].

So, taking the average over the entire interval, we get
[tex]\frac{1}{1-0}\int\limits_{0}^{1}\sqrt{\left\{x_3-x_1-u(x_2-x_1)\right\}^2+\left\{y_3-y_1-u(y_2-y_1)\right\}^2+\left\{z_3-z_1-u(z_2-z_1)\right\}^2}\,du[/tex]
[tex]=\boxed{\int\limits_{0}^{1}\sqrt{\left\{x_3-x_1-u(x_2-x_1)\right\}^2+\left\{y_3-y_1-u(y_2-y_1)\right\}^2+\left\{z_3-z_1-u(z_2-z_1)\right\}^2}\,du}[/tex]

[Guest]

Alright thanks maybe or maybe not, maybe.