Formula for a straight line?

[TiPeRa]

What is the formula for a straight line, in 3D? In 2D it is y=mx+c right.

TIA

-TiPeRa

[thinktank2]

In 3-D

If you the coordinates of two points through which the line is passing.

let's say (x1,y1,z1) and (x2,y2,z2)

Then Equation of the line is

[x - x1]/[x2 - x1] = [y - y1]/[y2 - y1] = [z - z1]/[z2 - z1]

[thinktank2]

In 3-D geometry we have what is called as Directional Cosines and Directional Ratios.

The Cosine of angles which these lines make with the x,y and z axis are called directional cosines. Any set of numbers that are propotional to the Directional Cosines are called Directional Ratios.

If we assume a as the angle the line makes with x- axis, b as the angle it makes with y - axis and g as the angle with z - axis. Then

Cos a , Cos b and Cos g are called the directional Cosines of the lines.

It is common to denote Cos a as l , Cos b as m and Cos g as n.

Any set of numbers propotional to these are called directional ratios. For example 2l,2m,2n are directional ratios of the line.

let's say a,b,c are the Directional Ratios of this line. and it passes through (x1,y1,z1) then it's equation is..

[x-x1]/a = [y-y1]/b = [z-z1]/c

[Guv]

It seems to me that I remember equations like the following.

X/A + Y/B + Z/C = Constant.
A*X + B*X + C*X = Constant. which is really equivalent to the above.

Anything equivent to one of the above has to be correct.

[CornedBee]

Or use a 3D vector form:
Points = Starting Point + t * Directional Vector
code for monospacing
in 3D:

    /5\       / 1\
X = |3| + t x |-3|
    \7/       \ 6/

top is x, mid is y, bottom is z axis