Find Equations for Point #3 on a Line in 3D Space

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To find the coordinates of point #3 on a line in 3D space at a distance D from point #1, start with the known coordinates of points #1 (x1, y1, z1) and #2 (x2, y2, z2). The line can be expressed using parametric equations derived from the vector connecting the two points. The distance formula in 3D space is used to set the distance from point #1 to point #3 equal to D, leading to an equation in terms of a parameter t. Solving this equation yields two possible solutions for point #3, located on either side of point #1 along the line. This method effectively determines the coordinates of point #3 based on the specified distance.
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I have a simple question.

Assume we have 2 points in 3D space with known coordinates:
#1 (x1,y1,z1)
and
#2 (x2,y2,z2)
How to find coordinates of point #3 (x,y,z) belongs to the same line with known distance D from point#1?
Equation for transformation will be helpful...

Thanks in advance.
 
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A line in 3d space can be written as two functions that goes through the origin has two equations:

y = mx

and

z/(y^2 + x^2)^1/2 = +- C

Where C is a constant.

Finally the distance between two points in 3d space is given by

(x^2 + y^2 + z^2)^1/2

Work around that...
 
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For example, find the points on the line through (1, 0, 1) and (2, 2, 2) that are a distance d from (1, 0, 1). The vector from (1, 0, 1) to (2, 2, 2) is (2-1)i+ (2-0)j+ (2-1)k= i+ 2j+ k so parametric equations of the line are x= t+ 1, y= 2t, z= t+ 1. The distance from any point (x, y, z) to (1, 0, 1) is \sqrt{(x-1)^2+ y^2+ (z-1)^2}. Replace x, y, z with their expressions in terms of the parameter t, set equal to d and solve for t:
\sqrt{(t+1-1)^2+ (2t)^2+ (t+1-1)^2}= \sqrt{4t^2}= 2|t|= d[/itex]<br /> There are two solutions because there are two such points, one on either side of (1, 0, 1).
 
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