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

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SUMMARY

This discussion focuses on finding the coordinates of a point #3 (x, y, z) on a line in 3D space, given two known points and a specified distance D from point #1. The line can be represented using parametric equations derived from the vector between the two points. The distance formula in 3D space is applied to derive the necessary equations, leading to two potential solutions for point #3, one on each side of point #1.

PREREQUISITES
  • Understanding of 3D coordinate systems
  • Familiarity with parametric equations
  • Knowledge of distance formulas in Euclidean space
  • Basic algebra for solving equations
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  • Study the derivation of parametric equations for lines in 3D space
  • Learn about vector operations in three dimensions
  • Explore applications of the distance formula in 3D geometry
  • Investigate the concept of direction vectors and their role in line equations
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Mathematicians, physics students, computer graphics developers, and anyone working with 3D modeling or simulations will benefit from this discussion.

ronatu
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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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