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Equation of a plane that is parallel to yz-plane

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the vector equation of a plane that contains the point P(2,-3,0) and is parallel to the yz-plane

    2. Relevant equations

    Vector equation is in the form... Pi: r = point + t(u) + s(v) s,t element of real numbers

    3. The attempt at a solution

    We know that the direction vectors (u and v) for a plane parallel to the yz-plane don't have x = 0 (no x component).

    So the simplest answer to this would be...

    Pi: r = P(2,-3,0) + t(0,1,0) + s(0,0,1) s,t element of R

    However, another possible answer would be...

    Pi: r = P(2,-3,0) + t(0,1,1) + s(0,1,2) s,t element of R

    Although the second answer is a little more complex, it defines a plane that is parallel to the yz-plane.

    However, my teacher insists that the plane I defined (second answer) isn't parallel to the yz-plane. Is there some way for me to prove that it is? (I am trying to explain my visualization to her but she is very insistent that my answer is wrong).
     
    Last edited: Jan 9, 2013
  2. jcsd
  3. Jan 9, 2013 #2

    Dick

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    A plane is parallel to the yz plane if it's tangent vectors are normal to the x unit vector (1,0,0). Show that's true for dPi/ds and dPi/dt. And also show (0,1,1) and (0,1,2) are linearly independent, just to make sure you've got a plane and not a line.
     
  4. Jan 10, 2013 #3

    HallsofIvy

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    Note, by the way. that, in two dimensions, a line that is "parallel to the x-axis", that is, parallel to the line y= 0, is of the form y= c for some number c. Similarly a line that is "parallel to the y-axis", that is, parallel to the line x= 0, is of the for x= c for some number c.

    In three dimensions, the "xy-plane" is the plane z= 0. Any plane parallel to it has equation z= c for some number c.
     
    Last edited: Jan 10, 2013
  5. Jan 11, 2013 #4
    Thanks for everyone's help!

    The way I proved it to my teacher is by showing that the cross product of the direction vectors of each plane are a multiple of each other.

    e.g. n1 = dir1 x dir2
    n2 = dir2 x dir3

    if n1 = kn2 then Pi: r1 = t(dir1) + s(dir2) is parallel to Pi: r2 = a(dir2) + b(dir3)
     
  6. Jan 11, 2013 #5

    haruspex

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    To make it parallel to the yz plane you just need x constant. To pass through the given point the constant must be 2. So the generic form is (2, s, t).
     
  7. Jan 12, 2013 #6

    HallsofIvy

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    Actually, the simplest way to define the cross product is to define [itex]\vec{i}\times \vec{j}= \vec{k}[/itex], [itex]\vec{j}\times\vec{k}= \vec{i}[/itex], [itex]\vec{k}\times\vec{i}= \vec{j}[/itex] (i.e. cyclically) and extend to all vectors by defining it to be both "linear" and "anti-commutative".
     
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