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A plane and a point

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data

    I have a plane: 20x+15y+10z=4
    Find the point on the plane where it´s closest to the point (5,6,7)

    2. Relevant equations

    any ideas?

    3. The attempt at a solution
    No idea!
     
  2. jcsd
  3. Mar 24, 2009 #2

    jambaugh

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    Is this a calculus problem or a linear algebra problem?

    If a calculus problem express the distance between the point and an arbitrary point on the plane as a function of x,y,z. Then impose the conditions on the plane to reduce to two variables. Then you optimize the distance (or more easily the square of the distance) which will occur when the partial derivatives are zero.

    If a linear algebra or analytic geometry problem then note that the point on the plane where this is the case will be such that the line through it and the point (5,6,7) will be orthogonal to the plane and thus parallel to the vector <20,15,10>.
     
  4. Mar 24, 2009 #3
    It´s a linear algebra problem and I don´t know how to do it. I have no formulas at all.
     
  5. Mar 24, 2009 #4

    jambaugh

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    The coefficients of the linear equation defining the plane are coordinates of a vector normal (perpendicular) to the plane.

    Call this vector N.

    Now express a vector pointing from the given point to an arbitrary point (x,y,z) on the plane. That'd be V= <5-x,6-y,7-z>.

    Now solve the system of equations V = cN along with the equation for the plane. The solution will give x,y, and z for the point you're looking for and the magnitude of V will give you the distance.
     
  6. Mar 25, 2009 #5
    cN=V
    then
    c(20,15,10)=(5-x, 6-y, 7-z)

    Then what?

    20c=5-x
    15c=6-y
    10c=7-z
    ?????????
     
  7. Mar 25, 2009 #6

    jambaugh

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    Right, but also include the equation requiring (x,y,z) lie on the plane:
    20x + 15y + 10z = 4.
    That's 4 equations and 4 unknowns c,x,y, and z.
    I suggest you solve the three for x,y, and z in terms of c then substitute those into the plane equation to find c.
     
  8. Mar 25, 2009 #7

    HallsofIvy

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    What are parametric equations of the line through (5, 6, 7) with direction vector <10, 15, 10>? Where does that line intersect 20x+ 15y+ 10z= 4?

    (Since the length of the direction vector is irrelevant to the line, the "c" in cN= V can be anything. I am suggesting you take it to be equal to 1.)
     
  9. Mar 25, 2009 #8
    I´m sorry, I don´t understand you :/
     
  10. Mar 26, 2009 #9

    jambaugh

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    You've got three equations which you can rewrite:
    x = 5 - 20c
    y = 6 - 15c
    z = 7 - 10c

    you've got the equation for the plane:
    20x + 15y + 10z = 4.

    Substitute (replace x in the plane equation with (5-20c) and so on) and you have an equation you can solve for the value c.
    when you know c you know x, y, and z, which are the coordinates of the point you are looking for.
     
  11. Mar 26, 2009 #10
    What I need is the way to solve this problem. Step by step.
    I don´t need to understand it. I can´t see why you don´t want to help me.
     
  12. Mar 26, 2009 #11
    It is a forum rule, not to simply give out answers. So, you better try to understand it for people here won't do the hw for you!
     
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