1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Distance Formula in 3 Dimensions

  1. Oct 25, 2009 #1
    Is it possible to change:
    _______
    distance = \/X2 + Y2

    To:
    _______________
    distance = \/X^2 + Y^2 + Z^2

    And get the distance between a point in 3 dimensional space and a the point of origin, just as the first equation does in 2 dimensional space?

    Thanks!
     
  2. jcsd
  3. Oct 25, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that's correct. On way to see that is to "drop" a perpendicular from (x, y, z) to (x, y, 0). The x-axis, from (0,0,0) to (x,0,0), the line from (x,0,0) to (x,y,0), and the direct line from (0,0,0) to (x, y, z) form a right triangle with legs of length x and y and hypotenuse of length [itex]\sqrt{x^2+ y^2}[/itex]. Now that line from (0,0,0) to (x,y,0), the line from (x,y,0) to (x,y,z), and the line from (0,0,0) to (x,y,z) form a right triangle with legs of length [itex]\sqrt{x^2+ y^2}[/itex] and z.

    So you can use the Pythagorean theorem again to get that the distance from (0,0,0) to (x,y,z), the hypotenuse is [itex]\sqrt{(\sqrt{x^2+y^2}^2+ z^2}= \sqrt{x^2+ y^2+ z^2}[/itex].
     
  4. Oct 25, 2009 #3
    Ok cool!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Distance Formula in 3 Dimensions
  1. Distance formula (Replies: 3)

Loading...