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Distance from plane to origin

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the distance D from the origin of any point (x , y , z) lying on the plane x + y + z= 1 satisfies
    D^2 + x^2 + y^2 +(x + y -1)^2 .
    By considering partial derivatives, find the point that is closest to the origin. Prove that this distance is genuinely minimal.

    2. Relevant equations

    3. The attempt at a solution
    I did the first part. Knowing Distance is D^2 = x^2 + y^2 + z^2 then by re-arranging x + y + z =1 to -z = x + y - 1.
    z^2 = (-z)^2 = (x + y - 1)^2
    D^2 = x^2 + y^2 + (x + y - 1)^2
    Not sure how to do the next part. How does considering partial derivatives help me find a point closest to the origin?
    Any help would be great.
  2. jcsd
  3. May 25, 2009 #2


    Staff: Mentor

    Shouldn't this be an equation? Namely, D^2 = x^2 + y^2 +(x + y -1)^2.

    You have D2 as a function of x and y. You're looking for the smallest value of D2 from the origin to the plane. You can use partial derivatives to find the minimum point in a way analogous to finding the minimum value of a function of one variable (i.e., finding the derivative, and setting it to zero to get the critical points, and then checking with the second derivative to see if you have a local maximum or local minimum).
  4. May 25, 2009 #3


    User Avatar
    Science Advisor

    The distance from a plane P to the origin is equal to a.x + b.y + c.z = d where <a,b,c> is
    a normal vector and d is the distance to the origin.

    We do this through the notion of a Euclidean Inner Product. Basically what happens here is
    that in three dimensional geometry we express the inner product in the form of
    <a,b> = |a||b|cos(a,b) where |a| is the length vector of a and |b| is the lebgth vector of
    b. cos(a,b) is the cosine of the angle between a and b is denoted in that way.
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