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Finding line of intersection of two planes

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Find parametric equations for the line of intersection of the
    planes x + y + z = 1 and
    r = (1, 0, 0) + [itex]\lambda[/itex](2, 1, 0) + [itex]\mu[/itex](0, 1, 1) where [itex]\lambda[/itex], [itex]\mu[/itex] [itex]\in[/itex] R

    2. Relevant equations



    3. The attempt at a solution

    I attempted to convert the 2nd plane equation to scalar form by finding the normal: (2, -1, 0) cross product with (0, 1, -1). That ended up being i + 2j + 2k, meaning that the plane had equation x + 2y + 2z = d, subbing in the point in the vector equation (1, 0, 0) gives d =1 therefore x + 2y + 2z = 1. I'm not sure what to do from here or if what i've done is either correct or necessary.

    Help please?
     
  2. jcsd
  3. Aug 25, 2012 #2
    the normal vector would be (-1,2,-2).
    And Cartesian equation would be x-2y+2z=1
     
  4. Aug 25, 2012 #3

    HallsofIvy

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    The parametric equations for the second plane are saying that [itex]x= 1+ 2\lambda+ 0\mu[/itex], [itex]y= 0+ 1\lambda+ 1\mu[/itex], and [itex]z= 0+ 0\lambda+ 1\mu[/itex]. Putting those into the equation of the first plane, x+ y+ z= 1 gives [itex](1+ 2\lambda)+ (\lambda+ \mu)+ (\mu)= 1+ 3\lambda+ 2\mu= 1[/itex] or [itex]3\lambda+ 2\mu= 0[/itex]. Solve that for either [itex]\lambda[/itex] or [itex]\mu[/itex] in terms of the other and replace in the parametric equation, getting linar equations in a single parameter- a line.

    I don't know if it was just a typo or copying mistake but the normal to the second plane is NOT the cross product of (2, -1, 0) and (0, 1, -1). I'm not sure where you got those vectors. Two vectors in the plane are (1, 0, 0)+ (2, 1, 0)= (3, 1, 0) (taking [itex]\lambda= 1[/itex] and [itex]\mu= 0[/itex] and another is (1, 0, 0)+ (0, 1, 1)= (1, 1, 1) (taking [itex]\lambda= 0[/itex] and [itex]\mu= 1[/itex]. The cross product of those vectors is normal to the plane.
     
  5. Aug 25, 2012 #4
    HallsofIvy
    I dont think cross product of 3,1,0 and 1,1,1 would be normal to the 2nd plane
    as the cross product of above is 1,-3,2 , but the normal vector should be 1,-2,2
     
    Last edited: Aug 25, 2012
  6. Aug 25, 2012 #5

    ehild

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    (2,1,0) and (0,1,1) are vectors which lay in the plane. You mixed the signs a bit.

    Kushan is right, (-1, 2,-2) is a normal vector of the plane.

    ehild
     
  7. Aug 25, 2012 #6
    ehild would it be -1,2,-2 or 1,-2,2 ?
     
  8. Aug 25, 2012 #7

    ehild

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    The normal of a plane is the cross product of two vectors laying in the plane, your vectors are position vectors of two points in the plane.

    ehild
     
    Last edited: Aug 25, 2012
  9. Aug 25, 2012 #8

    ehild

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    They are parallel vectors, both normal to the plane, one is -1 times the other.

    ehild
     
  10. Aug 25, 2012 #9
    Direction doesn't matter?
     
  11. Aug 25, 2012 #10

    Bacle2

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    Not for normality: if a vector v is normal to the surface, then so is the vector -v.
     
  12. Aug 25, 2012 #11
    Ok I used to imagine like the one originating from center and piercing though the plane perpendicularly
    So it is both way .
    Thank Ehild and Bacle2
     
  13. Aug 25, 2012 #12

    Bacle2

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    Actually,

    maybe thinking of it this way would help: imagine you just have , say the vector

    1,-2,2 on the plane. Now, give the plane a 180 turn,and you get the vector

    -1,2,-2 --which is the same as 1,-2,2, but with opposite orientation.

    Give 99% of the tanks to Ehild, he did most of the work.
     
  14. Aug 25, 2012 #13
    1/100 thanks Bacle
     
  15. Aug 25, 2012 #14
    Wow yep I can't even read the question properly, the normal is indeed (-1, 2,-2). So if I cross product (-1, 2, -2) with (1, 1, 1) I get 4i + 3j - 3k, which is the direction vector of the line. So that's half of what I need for the parametric equations, the other half is a point on the line, how do I get that?
     
  16. Aug 25, 2012 #15
    find the equation of line Spacec , using the equation of planes .
     
  17. Aug 25, 2012 #16

    ehild

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    You do not need that cross product really. You need to give the parametric equation of the line. Follow HallsofIvy's hint.

    ehild
     
  18. Aug 25, 2012 #17
    Ok so using HallsofIvy's method I get:

    λ = -2/3*μ

    x = 1 - 4/3*μ
    y = 5/3*μ
    z = -μ

    Is this right? Thanks for helping me out and that method was pretty easy but I would still like to know how to do it using the direction vector of the line and finding a point on it, just so I know another way to do it.
     
  19. Aug 25, 2012 #18
    You mean other method for solving this?
     
  20. Aug 25, 2012 #19
    Yea, by finding the vector equation of the line then putting it into parametric form. We got the direction vector of the line before from the cross product of the 2 plane normals. Now we need a point on the line but I'm not sure how to get that.
     
  21. Aug 25, 2012 #20
    put some x=0 in both plane then solve the two equation .
     
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