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 from a point to a plane

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the distance from (3,0,10) to the plane 2x+3y+z=2

    2. Relevant equations

    D= absv(ax+by+cd)/ sqrt(a^2+b^2+c^2)

    3. The attempt at a solution

    I found vector n= 2i + 3j + k

    Thus, the distance is 16/sqrt(14). Can you guys check the result for me ? I am not sure if I am right.
  2. jcsd
  3. Mar 2, 2009 #2


    Staff: Mentor

    Your vector n is the normal to the plane. Can you show us how you got the distance you show?
  4. Mar 2, 2009 #3
    That's how I have solved this question :
    [; ax+by+cz+d=0 \rightarrow 2x+3y+z-2=0 \\
    (x_0,y_0,z_0) \rightarrow (3,0,10) \\
    D=\frac{a\cdot x_0+b\cdot y_0+c\cdot z_0-d}{\sqrt{a^2+b^2+c^2}}
    = \frac{2\cdot3+3\cdot0+1\cdot10-2}{\sqrt{2^2+3^2+1^2}}=\frac{14}{\sqrt{14}}=\sqrt{14} ;]

    Last edited by a moderator: Apr 24, 2017
  5. Mar 2, 2009 #4


    Staff: Mentor

    Of the two answers shown: 16/sqrt(14) and sqrt(14), I agree with the latter.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook