The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Length of a Line Segment in Three Dimensions

Date: 04/16/2004 at 13:58:26
From: Catherine
Subject: Length of a line segment in three dimensions

What is the length of a line segment in three dimensions with 
endpoints (1, 0, 2) and (1, 4, 5)?

Date: 04/16/2004 at 16:00:33
From: Doctor Douglas
Subject: Re: Length of a line segment in three dimensions

Hi, Catherine.

Thanks for writing to the Math Forum.

You can apply the Pythagorean Theorem twice in succession.  Consider
the box below:

    B-------C    We want to know the distance between A and G.       
   /|      /|    We know the length (AD), height (DH) and width (HG).
  A-------D |    
  | F-----|-G    The vertices AHG form a right triangle, because
  |/      |/     the segment GH is perpendicular to the plane ADHE.

We apply the Pythagorean theorem first to triangle AHG, and in doing
so, find we need to apply it again to get the distance AH:

  (AG)^2 = (GH)^2 + (AH)^2                     Pythagoras (AHG)
         = (GH)^2 + [(AD)^2 + (DH)^2]          Pythagoras (ADH)
         = width^2 + length^2 + height^2

The body diagonal distance (AG) is simply the square root of this, or

  distance = sqrt[width^2 + length^2 + height^2].

Let's think of the length of the box in terms of the x-coordinates of
the points, the height in terms of y, and the width in terms of z.  In
other words, think of the box as being on this axis system:

       y  z
       | /
  -----+----- x
     / |

Since each of those three distances is one-dimensional, we can think
of the length of the box as the difference in the two x-coordinates,
the width as the difference in the two z-coordinates, and the height
as the difference in the two y-coordinates.  Thus, modifying the above
distance formula:

  distance = sqrt[width^2 + length^2 + height^2]

  distance = sqrt[(z2 - z1)^2 + (x2 - x1)^2 + (y2 - y1)^2]

In other words, we've discovered that by using the Pythagorean theorem
twice, we can find the distance between any two points (x1,y1,z1) and
(x2,y2,z2) in three dimensional space by calculating:

  distance = sqrt[(z2 - z1)^2 + (x2 - x1)^2 + (y2 - y1)^2]

Are you familiar with the distance formula in two dimensions?  Given
two points (x1,y1) and (x2,y2), the distance between them is:

  distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

Can you see how the three dimensional formula is a logical extension
of this formula?

Now, can you use the three dimensional distance formula to find the
distance between your two points?
- Doctor Douglas, The Math Forum 
Associated Topics:
High School Polyhedra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.