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Scalar Equation of a Plane

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the Scalar Equation of a Plane containing the points

    P(1,1,-1)
    Q(0,1,1)

    2. Relevant equations

    ax + by+ cz = d

    3. The attempt at a solution

    PQ = [-1,0,2]T

    [x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[a,b,c]T

    ^ This is the vector equation.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 30, 2012 #2

    SammyS

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    Hello KingKai. Welcome to PF !

    First of all, you should know that this plane is not unique --- there are an infinity of planes passing through these two points.

    The vector [a,b,c]T is normal (perpendicular) to the plane, it's not in the plane.
     
  4. May 30, 2012 #3
    Correct me if I am wrong, but the normal to a plane is determined by taking the cross product of the two given direction vectors in the vector equation of the plane. The normal to a plane is not included in the vector equation at all.

    Since only 1 direction vector is given in the problem statement, then it is correct that there would be infinitely many planes that satisfy these conditions. That being said, I would like to obtain the scalar equation of this plane and have difficulty doing so.
     
  5. May 31, 2012 #4

    SammyS

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    You gave the scalar equation of the plane as ax + by + cz = d .

    Then the vector [a,b,c] is normal to the plane. Therefore, it should not be in the vector equation the way you have it.
     
  6. May 31, 2012 #5
    Okay, then if I modify the vector equation to

    [x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[x0,y0,z0]T

    and the scalar equation of the plane is still in form

    ax + by + cz = d

    Then how, with the above posted conditions, would I attain a scalar equation for this plane?

    The main reason that I am confused is because only 1 direction vector is given to me, the other has infinite possibilities, so how do I derive a scalar equation that satisfies these conditions?
     
  7. May 31, 2012 #6
    The Big T represents transpose.
     
  8. May 31, 2012 #7

    SammyS

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    If you mean that the arbitrary point (x0,y0,z0) is to be in the plane, then the vector equation would be

    [x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[x0-1,y[SUB ]0[/SUB]-1,z0+1]T .

    However, (besides the fact that this is not a scalar equation) this doesn't ensure that the point (x0,y0,z0) is not on the line determined by points P & Q.

    Remember, vector [a,b,c]T must be perpendicular to vector [-1,0,2]T . That implies that the scalar product of these two vectors must be zero. That will give you a relationship among the quantities a, b, and c.

    You will get the same relationship is you plug the coordinates of P and then Q into the scalar equation, ax + by+ cz = d .
     
  9. May 31, 2012 #8

    HallsofIvy

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    If you mean that you want a scalar version of this then you have that [-1, 0, 2] and [a, b, c] are vectors in the plane. Their cross product:
    [tex]\left|\begin{array}{ccc}i & j & k \\ -1 & 0 & 2\\a & b & c\end{array}\right|= -2bi+ (2a- c)j- bk[/tex]
    is a normal vector. That, together with the point (1, 1, -1) in the plane gives you the equation.

     
  10. Jun 1, 2012 #9

    So, following this, I take the normal vector

    n = [-2b, 2a-c, -b]T

    and the point P(1,1,-1)

    and compile the scalar equation

    -2b(x-1) + (2a-c)(y-1) -b(z+1) = d

    -2bx + 2b +2ay -2a -cy +c -bz -b = d


    Would this be correct?
     
  11. Jun 1, 2012 #10

    SammyS

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    If your scalar equation for the plane is ax + by + cz = d, then the vector [a, b, c]T is not in the plane. That vector is normal to the plane.

    Like I suggested, plug (1,1,-1) into ax + by + cz = d. Then plug (0,1,1) into ax + by + cz = d.

    That gives a + b - c = d and b + c = d .

    That gives a relation between a & c, so you can eliminate one of them.
     
  12. Jun 1, 2012 #11


    a + b -c = d b + c = d
    c = d - b

    Sub c = d - b into

    a + b - c = d


    a + b - (d - b) = d

    a + b -d + b = d

    a + 2b = 2d From This the scalar equation would be:

    ax +2by = 2d True?
     
  13. Jun 1, 2012 #12

    SammyS

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    Where do z go ?


    If a + b - c = d and b + c = d, then a + b - c = b + c → a = 2c.

    Using that a = 2c and b + c = d gives:

    2cx + by + cz = b+c .

    You can play around with this to make it look nicer.
     
  14. Jun 1, 2012 #13
    Thank You SammyS for all your help!

    I'm having trouble typing out matrices on this forum, is there a thread you can refer me that will teach me the syntax of how I can type out proper matrices the way HoI did above?

    I have another question from my linear algebra textbook that I need help with but I can't type the matrix.
     
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