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Find the equation of plane

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data

    << Moderator Note -- This poster has received an infraction for deleting their OP after they received help. Their OP is restored below >>
     
    Last edited by a moderator: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2

    Simon Bridge

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    You need three, not co-linear, points to define a plane.
    You have a line of them - so pick two from the line.
    The first plane also includes the origin - so there's your three points for that plane.
    The second plane is perpendicular to the first ...

    http://jtaylor1142001.net/calcjat/Solutions/VPlanes/VP3Pts.htm
     
  4. Nov 6, 2011 #3
    pooo
     
    Last edited: Nov 6, 2011
  5. Nov 6, 2011 #4

    Simon Bridge

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    The equation of the plane perpendicular to vector (a,b,c) through point (x0,y0,z0) is a(x-x0)+b(y-y0)+c(z-z0)=0

    So your strategy is to find a vector perpendicular to the plane. The cross-product of any two vectors in the plane (so long as they make a V shape) will give you that. You can make two vectors in the plane from three points in the plane.

    So - step-by-step:

    1. pick three points in the plane.
    eg. for the first one you could choose:
    P=(0,0,0), Q=(1,1,1), R=(2,3,4)

    2. make two vectors from these:
    This means that u=PQ=(1,1,1) and v=PR=(2,3,4) which is the advantage of picking the origin as one of your points.

    3. form the cross product:

    [itex]{\bf w} = {\bf u} \times {\bf v}[/itex]
    (the link shows you how to do this from a determinant.)

    This will give you the vector perpendicular to the plane in form w=ai+bj+ck

    4. substitute and simplify:
    You want an equation of form:
    Ax+By+Cz=D

    5. Check your work:
    Substitute all three points into the expression to show they make it true.

    hint: for the second plane, use the (a,b,c) from the first one to find a third point.
     
    Last edited: Nov 6, 2011
  6. Nov 6, 2011 #5
    trollll
     
    Last edited: Nov 6, 2011
  7. Nov 6, 2011 #6

    Simon Bridge

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    What were the three points you picked to define the second plane?
    (... and how did you choose them?)
    I get (4,1,-2) as the normal to the second plane.
     
    Last edited: Nov 6, 2011
  8. Nov 6, 2011 #7
    trollllll
     
    Last edited: Nov 6, 2011
  9. Nov 6, 2011 #8

    Simon Bridge

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    How can you possibly calculate the difference between a point you know and one you don't know?

    hint: plane2 is perpendicular to plane1 - (1,-2,1) is a vector perpendicular to plane1 ...
     
  10. Nov 6, 2011 #9
    trollll
     
    Last edited: Nov 6, 2011
  11. Nov 6, 2011 #10

    Simon Bridge

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    That formula is for the magnitude of the cross product - the cross product itself is a vector so you need the direction as well.

    Stop thinking about formulas and look at the geometry.
    What is the vector (1,-2,1) to do with plane2?

    OR

    consider the line parameterized by:

    (2,3,4)+(1,-2,1)t

    where does this line go in relation to plane2?
    can you use this to find a third point in plane2?

    answer:
    because the two plane are perpendicular to each other, any normal vector to plane1 will point along the surface of plane2. So we can use it to find the third point ...

    Q=(1,1,1)
    R=(2,3,4)
    P=(2,3,4)+(1,-2,1)=(3,1,5)

    then follow the same method as before.
    u=QR=(2,3,4)-(1,1,1)=(1,2,3) [final minus initial]
    v=QP=(3,1,5)-(1,1,1)=(2,0,4)

    w = u x v

    w is the determinant of a 3x3 matrix - the first row is the unit vectors (i,j,k), the second is u and the third is v.

    (I recall having trouble with this - I used to use three pencils and sheets of stiff paper.)
     
    Last edited: Nov 6, 2011
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