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Symmetric point of plane

  1. May 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the coordinates of the symmetric point of the point M(3,4,7) from the plane 2x-y+z+9=0

    2. Relevant equations


    3. The attempt at a solution

    I found the equation of the plane which the symmetric point is staying at:

    2x-y-z+27=0

    Also I found the distance between M(3,4,7) and the symmetric point:

    [tex]\sqrt{(x-3)^2+(y-4)^2+(z-7)^2}=2*3*\sqrt{6}[/tex]

    I need one more condition to solve the system.

    The coordinate from my text book results which I need to find is: [tex]M_s(-9,10,1)[/tex]
     
  2. jcsd
  3. May 17, 2008 #2

    tiny-tim

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    Hi Physicsissuef! :smile:

    Hint: if the symmetric point is N, and the origin is O, find P such that OP is parallel to MN! :smile:
     
  4. May 17, 2008 #3
    I don't understand you. :smile: What is N? Is N the point that we are looking for? What is M?
     
  5. May 17, 2008 #4

    HallsofIvy

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    M is the point (3, 4, 7). You said that! N is the point you are looking for- the point on the opposite side of the given plane at the same distance from the plane as M- the "symmetric point". Of course, the line connecting those two points will be perpendicular to the plane so:
    1) Find the equation of the line through (3, 4, 7) normal to the plane 2x-y+z+9=0.

    2) Find the point, O, at which that line intersects the plane.

    3) Find the point, N, on that line at the same distance as (3, 4, 7) from O.
     
  6. May 17, 2008 #5
    How will I find the equation of plane when I need [itex]a(a_1,a_2,a_3)[/itex] or 2 points. Even if I find the equation of the line, How will I find O, what formula? Thanks.
     
  7. May 17, 2008 #6

    HallsofIvy

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    You don't need to find the equation of a plane- you are given the only plane you need. After you have found the equation of the line through the given point, perpendicular to the given plane, you solve the "simultaneous equations". For example, it your line is given by "Ax+ By+ Cz+ D= 0 and Px+ Qy+ Rz+ S= 0" (the line is the intersection of those two planes) then, including the equation for the given plane, you have 3 linear equations for the three unknown values x, y, and z. If your line is given by "x= at+ x0, y= bt+ y0, z= ct+ z0" (parametric equations) replacing the x, y, z in the equation of the given plane gives a single equation for t. Solve that, then calculate the corresponding x, y, and z. If you line is given by "(Ax+B)/C= (Dy+ E)/F= (Gz+ H)/I" ('symmetric' equations) set each of the three fractions equal to t and solve for the x, y, z to get parametric equations.
     
  8. May 17, 2008 #7
    ohhh... Instead of line I wrote plane... The equation of the line will be:

    [tex]\frac{x-3}{a_1}=\frac{y-4}{a_2}=\frac{z-7}{a_3}[/tex]

    Is a(2,-1,1)? If a(2,-1,1) the system got no solution...
     
  9. May 18, 2008 #8
    Even if I find the point [itex]N ( \frac{22}{4} , \frac{11}{4} , \frac{33}{4})[/itex] and make
    [tex]\sqrt{(x-\frac{22}{4})^2+(y- \frac{11}{4})^2+(z-\frac{33}{4})^2}=\sqrt{(3-\frac{22}{4})^2+(4-\frac{11}{4})^2+(7-\frac{33}{4})^}[/tex]
    I don't know why but the system got no solution...
     
  10. May 18, 2008 #9

    Defennder

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    I don't know what you meant by a(2,-1,1), but that appears to be the vector which is normal to the plane. You should write the equation of the line in vector form like this:

    [tex]\left(\begin{array}{c}x\\y\\z\end{array} \right) = \left( \begin{array}{c}3\\4\\7\end{array} \right) \ + \ k \left( \begin{array}{c}2\\-1\\1\end{array} \right) , \ k \in \Re[/tex].

    You should now find the point on the plane which is closest to M(3,4,7). That point satisfies the above vector equation of a line, as well as the equation of the plane 2x-y+z+9=0. Substituting the former into the latter, solve for k. Plug in the value of k into the vector equation of the line, that would give you the coordinate of the point on the plane closest to M. Let this point be B.

    Now note that [tex]\vec{MB} \ = \ \vec{BN}[/tex] where N is the required symmetric point. You have the value of k which you found earlier. Note that [tex]\vec{MB} = k \left( \begin{array}{c}2\\-1\\1\end{array} \right) = \vec{BN}[/tex] Now solve for N and you'll find the answer.
     
  11. May 18, 2008 #10
    a is vector parallel to the line... Why I can't find the equation of line like that?
     
  12. May 18, 2008 #11

    Defennder

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    Expressing the equation of line in vector form allows you to solve for one variable k. This variable then allows you to find the point on the plane, and it can later also be used to find the 'mirrored' point.
     
  13. May 18, 2008 #12
    The point B is [tex](0, \frac{11}{2}, \frac{11}{2})[/tex] and again the system got no solution...
     
  14. May 18, 2008 #13

    Defennder

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    No, that is not B. B is a point on the plane and must hence satisfy the equation of the plane 2x-y+z+9=0 which (0, 11/2, 11/2) does not. How did you get that value anyway?
     
  15. May 18, 2008 #14
    [tex]
    x=x_1+ta_1
    [/tex]
    [tex]
    y=y_1 + ta_2
    [/tex]
    [tex]
    z=z_1+ta_3
    [/tex]
    [tex]
    Ax+By+Cz+D=0
    [/tex]
    [tex]
    A(x_1+ta_1)+B(y_1 + ta_2)+C(z_1+ta_3)+D=0
    [/tex]
    [tex]
    (Aa_1+Ba_2+Ca_3)t + Ax_1+By_1+Cz_1+D=0
    [/tex]
    [tex]
    a=Aa_1+Ba_2+Ca_3
    [/tex]
    [tex]
    b=Ax_1+By_1+Cz_1+D
    [/tex]

    Ohh... I see now I got miscalculation

    the equation of line

    [tex]\frac{x-3}{2}=\frac{y-4}{-1}=\frac{z-7}{1}[/tex]

    [tex]
    x=3+2t ; y=4-t ;
    z=7+t ;
    a=4+1+1=6 ;
    b=6-4+7+9=18 ;
    t=\frac{-18}{6}=-3 ;

    x=3-6=-3 ;

    y=4+3=7 ;

    z=7-3=4 ;
    [/tex]

    B(-3,7,4)

    But still I got no solution to the system. I don't know what is the problem...

    Edit: Finally I got the solution. Thanks for the help...
     
    Last edited: May 18, 2008
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