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Distance from point to plane

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\vec{PM_1}=d*\vec{n_0}[/tex]

    Here are the both pictures.

    Picture 1.


    Picture 2


    2. Relevant equations

    [tex]d = \frac{\left | a x_1 + b y_1 + c z_1+d \right |}{\sqrt{a^2+b^2+c^2}}.[/tex]


    3. The attempt at a solution

    Why in my book says that, if d > 0 then the coordinate M is of opposite site of the plane (relative to the (0,0,0)), and if it is d < 0 it is on same site of the plane (relative to (0,0,0)) when d is always positive?
     
  2. jcsd
  3. Apr 18, 2008 #2

    HallsofIvy

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    Bad editing? I suspect they meant
    [tex]d= \frac{ax_1+ by_1+ cz_1+ d}{\sqrt{a^2+ b^2+ c^3}}[/tex]
    without the absolute value! And then |d| is the distance.
     
  4. Apr 19, 2008 #3
    if d > 0 then the coordinate M is of opposite site of the plane (relative to the (0,0,0)), and if it is d < 0 it is on same site of the plane (relative to (0,0,0))? Where is the logic?
     
  5. Apr 19, 2008 #4
    and d up in the fraction should be D:

    like this: [tex]
    d= \frac{ax_1+ by_1+ cz_1+ D}{\sqrt{a^2+ b^2+ c^3}}
    [/tex]
     
  6. Apr 19, 2008 #5

    VietDao29

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    No, it's not like that.. It depends on whether D is negative or positive..

    Say, I have a 3-variable function:

    [tex]d(x, y, z) = \frac{ax + by + cz + D}{\sqrt{a ^ 2 + b ^ 2 + c ^ 2}}[/tex]

    where a, b, c, D are all coefficients of the plane function: ax + by + cz + D = 0, and [tex](x, y, z) \in \mathbb{R} ^ 3[/tex] is the co-ordinate of a point in 3D space.

    There should be a theorem in your books, which states:

    Two points A(x0, y0, z0), B(x1, y1, z1) are on the same side of the plane: ax + by + cz + D = 0, if and only if:

    d(x0, y0, z0) d(x1, y1, z1) > 0..

    They are not on the same side (i.e, one is on one side, and the other one is on another side), if and only if:

    d(x0, y0, z0) d(x1, y1, z1) < 0..

    -------------------------

    Back to your question:

    The origin O has co-ordinate (0, 0, 0), so [tex]d(0, 0, 0) = \frac{D}{\sqrt{a ^ 2 + b ^ 2 + c ^ 2}}[/tex], which will take the sign of D, since the square root in the denominator is always positive, hence, does not affect the sign of d(0, 0, 0).

    So, a random point with co-ordinate (x, y, z), and the origin are on the same side of the plane (whose function is: ax + by + cz + D = 0), if and only if:

    D * d(x, y, z) > 0..

    They lie on 2 different sides if and only if: D * d(x, y, z) < 0..

    Is this clear? :)
     
    Last edited: Apr 19, 2008
  7. Apr 19, 2008 #6
    d is vector? I don't think so, that d have co-ordinates, because
    [tex]
    \vec{PM_1}=d*\vec{n_0}
    [/tex]
    To be honest I don't understand what are you talking about.
     
  8. Apr 19, 2008 #7

    VietDao29

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    d is not a vector, it's a function.. Just like those of f(x), f(x, y), and f(x, y, z)..

    d(x, y, z) is just a 3-variable function, and it returns real numbers..

    Do you know the general method to check whether 2 points are on the same sides of a plane, or on 2 different sides?

    BTW, is that book in Russian? :eek:
     
  9. Apr 19, 2008 #8
    No, I don't know the general method to check whether 2 points are on same sides of plane.

    The book is in Macedonian, which have similar letter with Russian. :smile:
     
  10. Apr 19, 2008 #9

    HallsofIvy

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    That question is different, and simpler, than determining the distance from a point to a plane!

    The equation of a plane can be written [itex]A(x-x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex]
    Any point on the plane makes the left side of that 0, of course. Any point not on the plane make it either positive or negative. Two points are on the same side of the plane if they make the same sign: [itex](x_1,y_1,z_1)[/itex] and [itex](x_2,y_2,z_2)[/itex] are on the same side of the plane if and only if [itex]A(x_1-x_0)+ B(y_1- y_0)+ C(z_1- z_0)[/itex] and [itex]A(x_2-x_0)+ B(y_2- y_0)+ C(z_2- z_0)[/itex] have the same sign.

    That is, of course, essentially the same as the numerator of your "distance" formula without the absolute value. Since the denominator is positive, the sign of fraction is the same as the sign of the numerator.
     
  11. Apr 19, 2008 #10
    And what is numerator and denominator?
     
    Last edited: Apr 19, 2008
  12. Apr 19, 2008 #11
    If
    [tex]
    \vec{PM_1}=d*\vec{n_0}
    [/tex]
    then d is not vector, but it is module. How is possible than that d<0 ?
     
  13. Apr 20, 2008 #12

    VietDao29

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    In a fraction, a numerator is the number on top, the denominator is the one that lies at bottom..

    Say, [tex]\frac{2}{5}[/tex], then 2 is the numerator, and 5 is the denominator. Or, generally, in a fraction, we have: [tex]\frac{\mbox{numerator}}{\mbox{denominator}}[/tex]

    No, d is a scalar, in different words, a real number. It is positive when [tex]\vec{PM_1}[/tex], and [tex]\vec{n_0}[/tex] point in the same direction; and negative, otherwise..
     
  14. Apr 20, 2008 #13
    Ok, thanks. And where is the logic about d>0 and d<0 ?
     
  15. Jul 1, 2008 #14
    a distance is always absolute and unsigned and positive, |d| is the distance of whatever point from a plane, the signed distance is the negative or positive values of d, so -d will be on one side of the plane and it will be |d| units from that plane, if +d then it is on other side but also |d| units from plane,

    in other words -2 is exactly 2 units from 0, +2 is also exactly 2 units from 0, but is situated on the other side of zero 0.

    that is probably the simplest way it could be explained.
     
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