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Homework Help: Equation of a plane

  1. Apr 29, 2005 #1
    Here is the problem I need help with:
    Find the page perpedicular to the plane x+3y-z-7=0 and containing the points (2,0,5) and (0,2,-1).

    The answer given in the book is 2x-y-z+1=0. But when I plotted x+3y-z-7=0 and 2x-y-z+1=0, by a PC runnig Maple, they don't look perpendicular to each other.

  2. jcsd
  3. Apr 29, 2005 #2


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    Have you found the vectors normal to the planes? They must be perpendicular. I am not familiar with your program, but 2D drawings of 3D objects can be deceiving. I don't think you want to go by looks.
  4. Apr 29, 2005 #3
    the answer is correct . You can check it by finding the dot product of the direction ratios, it gives zero which proves that the two normals of the plane are prependicular to each other.

    Method I To Find The Eqn of The Plane:

    Let the eqn be ax+by+cz+d=0

    then a+3b-c=0 ( using dot product of the direction ratios of the normals of the two planes)

    also, 2a+5c+d=0

    and 2b-c+d=0

    All the above conditions are being satisfied by the answer given in your book and is correct.
  5. Apr 29, 2005 #4
    So what about the graph? I used Maple to draw both the planes in 3D and, as I told you, the angle between the two planes is less than 90 degrees. Am I missing something?
  6. Apr 29, 2005 #5
    Check your scales in your program
  7. Apr 30, 2005 #6
    You're right. The problem was with the scales. I used to work only with 2D curves and I don't know how long it will take to get used to these 3D surfaces.
    Now, may I ask you how the problem can be solved?
    When I write the conditions of the required plane, I find a system with 4 unkowns and 3 equation. After solving them I get a plane wich doesn't satisfy one of the conditions :cry:

    Thank you very much
  8. May 1, 2005 #7


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    The procedure for finding the equation of the second plane is outlined as follows:

    1. Find a normal vector to the first plane. Since the planes are perpendicular, this vector will lie in the second plane.

    2. Find a vector that is perpendicular to the normal of the first plane and perpendicular to another vector that lies in the second plane. This vector will be normal to the second plane.

    3. Form the dot product of the normal to the second plane with a vector from an arbitrary point in that plane to one of the specified points in that plane.

    #1 can be accomplished by finding two vectors in the first plane using the coordinates of three points in that plane and taking the cross product. Three points are easily found by finding the intercepts of the plane with the coordinate axes. Find x when y = z = 0, etc. Any multiple of the resulting vector will be in the same direction. Reduce the vector to make all the components integers and you should see an interesting relationship between those components and the coefficients of the variables in the defining equation of the plane.

    #2 can be accomplished by forming the cross product of the vector obtained in #1 with a vector between the two points specified in the second plane. Reduce that vector to integer components and you will see a relationship between the components and the coefficients of the variables in the equation you know to be the answer to the problem. You are not done yet though. There are infinitely many planes that share that same normal, and you have to find the ONE plane that is asked for.

    #3 is accomplished by writing the vector from an arbitrary point {x,y,z} to either of the known points in the second plane and setting the dot product of that vector with the normal vector found in #2 to zero. This product must be zero for all vectors in that plane. This dot product will yield the equation of the second plane.
  9. May 1, 2005 #8


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    Here's an alternate approach similar to those previously given except based on the operator:

    Given: Plane W ≡ {x + 3*y - z - 7 = 0}
    Required: Plane T ⊥ W containing Points P1=(2, 0, 5) and P2=(0, 2, -1)

    Step #1:
    Given Plane W ≡ {f(x,y,z) = 0}, then (∇f) ⊥ W and parallel to required Plane T. For this case:
    f(x,y,z) = x + 3*y - z - 7
    ::: ⇒ (∇f) = <1, 3, -1> ::: <---- || Plane T

    Step #2:
    If required Plane T contains Points P1 and P2, then Vector V = (P1P2) lies in T. For this case:
    V = <2, 0, 5> ♠ <0, 2, -1> = <2, -2, 6> ::: <---- in Plane T

    Step #3:
    U = {V × (∇f)} ⊥ Plane T
    U = <1, 3, -1> × <2, -2, 6> = <16, -8, -8> ::: <---- ⊥ Plane T

    Step #4:
    Equation for required Plane T thus given by:
    U⋅(<x, y, z> ♠ P1) = 0
    ::: ⇒ <16, -8, -8>(<x, y, z> ♠ <2, 0, 5>) = 0
    ::: ⇒ 16*x - 8*y - 8*z + 8 = 0
    ::: ⇒ 2*x - y - z + 1 = 0

    Last edited: May 1, 2005
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