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Lines and plane

  1. Dec 15, 2005 #1
    How can i proove if this line is perpendicular to this one?
    Line
    x=-2-4y, y=3-2t, z=1+2t
    Plane
    2x+y-z=5
    I don't care that much about the answer, I want the procedure.
     
  2. jcsd
  3. Dec 15, 2005 #2
    Not sure exactly what you mean, but a line is perpendicular to a plane if it is parallel to the normal of the plane.....
     
  4. Dec 15, 2005 #3

    mathwonk

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    i think he means this: the direction vector of your line is the vector of coefficients of the letter t, so (-4,-2,2).

    The "normal" vector (perpendicular) to your plane is the vector of coefficients of the letters, x,y,z, so (2,1,-1). these two vectors are proportional, by a scale factor of -2, so the answer is yes.
     
  5. Dec 15, 2005 #4


    Thanks, it is appriciated.
     
  6. Dec 15, 2005 #5
    How do we find the parametric equations for the line of intersection of the given planes
    7x-2y+3z = -2 and -3x+y+2z+5=0

    :approve:
     
  7. Dec 15, 2005 #6
    add 2 to both sides of the first equation.

    so we have 7x -2y + 3z + 2 = 0

    now we can set the equations equal to each other and that should give you the line of intersection. I'm not entirely sure of this though so I'm sorry if I'm wrong.
     
  8. Dec 15, 2005 #7

    Sorry that was wrong, I just remembered how to do this. Find the normal vectors to each plane, and take their cross product, and you know that this vector will be parallel to the line in which they intersect so you can just find one point of intersection and write the parametric equations for the line knowing the parallel vector.
     
  9. Dec 16, 2005 #8

    HallsofIvy

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    No, that would work perfectly well. From 7x-2y+3z = -2 you get
    7x- 2y+ 3z+ 2= 0. Since we want -3x+y+2z+5=0 on the line of intersection also: 7x- 2y+ 3z+ 2= -3x+ y+ 2z+ 5, one equation in three unknowns. You can use that together with either of the orignal equations to have 2 equations in 3 unknowns. Solve for two of x,y,z in terms of the other and use the third as "parameter".
    Or course, you could just as easily use the two given equations to solve for two of the unknowns in terms of the other. I don't see any reason to take a cross-product of two vectors.
    Here's how I would do the problem:
    Since -3x+y+2z+5=0 and 7x-2y+3z = -2, multiply the first equation by 2 to get -6x+ 2y+ 4z+ 10= 0 and write the second equation as 7x- 2y+ 3z+ 2= 0. Now add those two equations: x+ 7z+ 12= 0. x= -7z- 12.
    Put that back into the first equation: -3(-7z-12)+ y+ 2z-5= 0 which gives y= -2z+ 5-21z- 36= -23z-31.

    Assuming my arithmetic is correct (which I wouldn't guarentee), parametric equations for the line of intersection are:
    x= -7t- 12
    y= -23t- 31
    z= t.
     
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