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Equation of a plane

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Find an equation of the plane for the plane that passes through the point (1, -1, 1) and contains the line with symmetric equations x=2y=3z.


    2. Relevant equations
    ax+by+cz=d
    n=<a, b, c>
    n [itex]\bullet[/itex] (r -r0) = 0


    3. The attempt at a solution
    I just have no idea where to start. How do I work through these kinds of problems?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 13, 2011 #2

    ehild

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    There are two many parameters in your equation of the plane. It would be equivalent to write 1 instead of d.
    Plug in (1,-1,1) for x,y,z into the equation of the plane. You get a relation among a,b,c.
    The points of the line are also points of the plane. Replace y=x/2, z=x/3 in the equation. It has to be true for all x values. You can find the parameters from this condition.

    ehild
     
  4. Sep 13, 2011 #3
    I got the equation of the plane right out of the book, im confused... and why would you write 1 instead of d ?

    plugging the numbers in, youd get a-b+y=d but you said to put 1 for d, so a-b+y=1 ?

    and replace y=x/2, z=x/3 in what equation
     
  5. Sep 13, 2011 #4

    ehild

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    The normal vector can have any length, it does not influence its direction, that it is normal to the plane.
    In the equation ax+by+cz=d, a,b,c are the components of a normal vector. In vector form, it is rn=r0n where r0 is a vector of the plane. If you divide the equation by any number, the normal vector changes length, but the direction is the same. But you can not divide by d when it is zero. So you are right, it is better to keep d as you can not exclude that it is zero.

    So you have an equation for a, b, c, d: a-b+c=d. You can express one of the parameters with the others: For example, c=d-a+b.

    Replace the points of the line into the equation for the plane ax+by+cz=d. It will be

    a x+b/2 x +c/3 x=d. Plug in the expression of c. You get an equation that must hold for all values of x. What you get when x =0? What you get when x=1?

    ehild
     
  6. Sep 13, 2011 #5
    When x=0 don't you just get d=0 ? And when x=1 you have the same equation jut with all the x's as 1 I don't see how/why this helps
     
  7. Sep 13, 2011 #6
    Could i do this (if not why?) :

    Use x=2y=3z and switch to parametric eqns and get x=t , y=(1/2)t , z=(1/3)t and so the direction vector is equal to <1, 1/2, 1/3> which is your a, b, and c. So use that for the eqn:
    a(x-x0)+b(y-y0)+z(z-z0)=0
    so then you'd get x+(1/2)y+(1/3)z=(5/6)
     
  8. Sep 13, 2011 #7

    HallsofIvy

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    That doesn't work because the plane [itex]a(x- x_0)+ b(y- y_0)+ c(z- z_0)= 0[/itex] or, equivalently, [itex]ax+ by+ cz= const[/itex] has <a, b, c> as a normal vector. The plane you calculated has the given line perpendicular to the plane, not in the plane.

    What I would do is this: taking t= 0 in the parametric equations x= 0, y= 0, z= 0 so (0, 0, 0) is a point on that line and so in the desired plane. Since we are given that (1,-1, 1) is also in the plane, the vector <1-0, -1- 0, 1- 0>= <1, -1, 1>, is in the plane. We now have two vectors, <1, -1, 1> and the direction vector of the given line, <1, 1/2, 1/3> in the plane. Their cross product is perpendicular to both and so perpendicular to the plane. You now have a point in the plane and a vector perpendicular to the plane.
     
  9. Sep 13, 2011 #8

    ehild

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    You have found d : d=0, so
    a-b+c=0 and *
    a x+b/2 x +c/3 x = 0.
    The last equation holds for all x values. If you plug in x=1 you get
    a+b/2+c/3=0**

    Solve the system of equations * and **. One parameter is arbitrary, find the others in terms of that parameter.

    ehild
     
  10. Sep 15, 2011 #9
    i used HallsofIvy's method:

    so the answer i got is (5/6)(x-1) - (2/3)(y+1) + (3/2)(z-1) = 0

    is that right ?
     
  11. Sep 15, 2011 #10

    ehild

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    Does it contain the line x=2y=3z?

    ehild
     
  12. Sep 15, 2011 #11
    how do i check that?
     
  13. Sep 15, 2011 #12

    ehild

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    That line goes through the origin. So x=0, y=0, z=0 has to be a point of the plane. Is it?

    ehild
     
  14. Sep 15, 2011 #13
    ...no
     
  15. Sep 15, 2011 #14

    ehild

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    Check that vector product. You have only a sign error.


    ehild
     
  16. Sep 15, 2011 #15
    oh, ok i see what i did. i did the plane's vector thing cross the dir vector of the line but its the other way around. how would i know that on a test?
     
  17. Sep 15, 2011 #16

    HallsofIvy

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    No, that was not your error. Doing cross product "the other way around" gives a vector pointing in the opposite direction but still perpendicular to the plane.

    That would be the same as multiplying the entire equation by -1: -a(x- 1)- b(y-1)- c(z-1)= 0 rather than a(x-1)+ b(x-1)+ c(x-1)= 0 but, in fact, they are the same equation.

    Go ahead and do the cross product again. You have an error in the last component alone.
     
  18. Sep 15, 2011 #17

    ehild

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    The order does not matter when you get the normal vector. The negative of a normal vector is also a normal vector. There is one sign wrong in your equation: It should be
    (5/6)(x-1) -(2/3)(y+1) - (3/2)(z-1) = 0.
    And it would be nicer to write the equation in the standard form: ax+by+cz=d.

    ehild
     
  19. Sep 15, 2011 #18
    Ohhhh I'm just dumb and did the cross product wrong. I got it now thanks!
     
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