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Equation of plane

  1. May 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane which passes through M(2,3,1) and the line

    2x-y+3z-1=0 & \\
    5x+4y-z-7=0 &

    2. Relevant equations

    B_1 & C_1\\
    C_1 & C_2
    C_1 & A_1\\
    C_2 & A_2
    A_1 & B_1\\
    A_2 & B_2

    3. The attempt at a solution

    first I find the equation of the line:


    What should I do next?
  2. jcsd
  3. May 12, 2008 #2
    Are you familiar with cross products? (You must be, to some extent, as you have posted determinants. A solution of this form I have never seen before.)
    If yes, then [tex] N1 \\ X \\ N2 [/tex] will be the normal to your plane, where N1 and N2 are the normals to the lines given.
    Last edited: May 12, 2008
  4. May 12, 2008 #3
    [itex]n_1 x n_2[/itex] will give me vector perpendicular to this ones, but how it will help me?
  5. May 12, 2008 #4
    from that, you will have a, b, and c of the following equation for a plane.

    [tex] a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=d [/tex]

    In other words, you will have the normal to the plane you're trying to solve for. Substitution of the point they give you will solve it.
  6. May 12, 2008 #5
    [tex]n_1 x n_2=(-1,-5,-3)[/itex]


    Hmm... And what about x,y,z, should I substitute for M?
  7. May 12, 2008 #6


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    Unfortunately, any line, R3 has an normals in an infinite number of directions- it doesn't make sense to talk about "the" normal to a line.
  8. May 12, 2008 #7


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    Choose any two points on the line. For example, if you choose x= 0, you can solve the two equations 2(0)-y+3z-1=0 and 5(0)+4y-z-7=0 for y= 2 and z= 1: (0, 2, 1) is on the line. If you take x= 1, you can solve 2(1)- y+ 3z- 1= 0 and 5(1)+ 4y- z- 7= 0 for y= -23/11 and z= -2/11. The cross product of the vectors from (2, 3, 1) to those points will give you the normal to the plane.
  9. May 12, 2008 #8
    I retract having said 'normals' to the line. I got 'normal-happy' , what with the terminology for planes.
  10. May 12, 2008 #9
    But the plane don't necessarily need to intersect the line in 2 points. It can intersect it in one point...
  11. May 12, 2008 #10


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    The plane intersects the line at an infinite number of points, i.e. the line is contained in the plane.
  12. May 12, 2008 #11


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    Otherwise, you would need more information than you are given to determine the plane uniquely.
  13. May 12, 2008 #12


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    What plane are you talking about? You were told to find the plane that contains that line. Every point on the line is in that plane.
  14. May 13, 2008 #13
    Yes, sorry. My fault.

    Will it be correct if I show it like this:

    [tex]a \circ n=0[/tex]

    [tex](a_1,a_2,a_3) \circ (A,B,C)=0[/tex]

    So one of the conditions is:


    the second condition is:


    and the third condition:


    -A-5B-3C=0 & \\
    2A+3B+C+D=0 & \\
    -7B-3C+D=0 &

    The result is:


  15. May 13, 2008 #14
    Just to mention that in the first post the line is:

    x+y-2z+1=0 & \\
    2x-y+z-4=0 &

    Sorry, for mistake, I type the wrong line...
  16. May 14, 2008 #15
    And can I ask you something?

    Two planes to be parallel, must this condition be satisfied?



    [tex]\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}[/tex], since in my book doesn't look like that...
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