Equation of a line segment in 3D?

[question99]

Is there an equation for say a line segment in 3D Cartesian Coordinates that is between these two points?

Point One: x=0, y=0, z=0
Point Two: x=5, y=5, z=5

Does it involve inequalities?

 

[robphy]

For a segment, you need to somehow restrict the allowed values of x,y,z.
So, you could require 0<x<5, 0<y<5, 0<z<5... for the line that passes through (0,0,0) and (5,5,5).
You could also do it parametrically... but that too has a restriction (which could be interpreted as an inequality.. e.g.
0<t<1, x=5t,y=5t,z=5t).

Maybe, you could so some crazy parametrization where t runs from -infinity to infinity
and asymptotically meets the end points... something akin to x=Aarctan(t),y=Barctan(t),z=Carctan(t), for all t.

 

[HallsofIvy]

A one dimensional figure, such as a line or line segment cannot be written in a single equation in three dimensions. You can write three parametric equation in a single parameter or write it in "symmetric" form.

The line through (x_1, y_1, z_1) and (x_2, y_2, z_2) can be written as
x= (x_2- x_1)t+ x_1
y= (y_2- y_1)t+ y_1
z= (z_2- z_1)t+ z_1

To write it in "symmetric" form, solve each of those for t,
t= \frac{x- x_1}{x_1- x_1}
t = \frac{y- y_1}{y_2- y_1}
t= \frac{z- z_1}{z_2- z_1}
and set them equal:
\frac{x- x_1}{x_1- x_1}= \frac{y- y_1}{y_2- y_1}= \frac{z- z_1}{z_2- z_1}

 

[da_nang]

\vec x (t) = \vec{x}_1 + (\vec{x}_2 - \vec{x}_1)f(t) where 0 \leq f(t) \leq 1.

If say f(t) = \frac{1}{\pi}\arctan(t) + \frac{1}{2}, then you've covered the entire real line.

As for an equation, \vec{x}_1 + (\vec{x}_2 - \vec{x}_1)(\frac{1}{\pi}\arctan(t) + \frac{1}{2}) - \vec x = 0 works, yes?