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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

    Mark44

    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} ;]


    http://img222.imageshack.us/my.php?image=13210615.gif"
     
    Last edited by a moderator: Apr 24, 2017
  5. Mar 2, 2009 #4

    Mark44

    Staff: Mentor

    Of the two answers shown: 16/sqrt(14) and sqrt(14), I agree with the latter.
     
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