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Homework Help: Converting parametric to cartesian equation

  1. Oct 27, 2008 #1
    1. I found the parametric equation of a plane;

    [tex]\left(\begin{array}{ccc}x\\y\\z\end{ar ray}\right) =[/tex] [tex]\left(\begin{array}{ccc}1\\2\\3\end{ar ray}\right)[/tex] +s[tex]\left(\begin{array}{ccc}1\\1\\0\end{ar ray}\right)[/tex] +t [tex]\left(\begin{array}{ccc}2\\1\\-1\end{ar ray}\right)[/tex]

    s,t ∈ R.

    I was asked to find a Cartesian equation. So I write down the three equations;

    x=1+s+2t
    y=2+s+t
    z=3−t

    I don't understand how this set can be solved. What's the aim? Do I need to eliminate s,t from the equations?


    3. The attempt at a solution

    1) x=1+s+2t
    2) y=2+s+t
    3) z=3−t

    If I subtract (1) and (2) => x–y=t–1
    Subtracting the third from the second => y-z = s-1+2t
    And I can also find an expression for t, from the (3) => t=3-z

    Could you please show me how we can use this info to find the cartesian equation of the plane.

     
  2. jcsd
  3. Oct 27, 2008 #2

    Office_Shredder

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    You have t=3-z. This is an excellent start, as you can go back to say x-y=t-1 and exchange t with 3-z. Then you have an equation in x,y and z
     
  4. Oct 27, 2008 #3
    So as soon as we have an equation in x,y and z it will be the cartesian equation?
    So, therefore x–y+z=2 is the cartesian equation of this plane.
     
  5. Oct 27, 2008 #4

    Mark44

    Staff: Mentor

    That will work. A plane can also be defined by a point in the plane and a vector that is normal to the plane. In your case, the point (1, 2, 3) is in the plane. To get a normal, cross the two other vectors in your parametric equation, (1, 1, 0) and (2, 1, -1). The cross product will give you a vector (A, B, C). The equation of the plane will be A(x - 1) + B(y - 2) + C(z - 3) = 0.

    What I've described is probably a bit more complicated, but has the advantage of helping you comprehend the plane geometrically.
     
  6. Oct 27, 2008 #5
    Oh, thanks very much!
    Yes, I know how to do it by finding the normal using the cross product, I just wasn't sure how to solve those set of equations.
     
  7. Oct 28, 2008 #6
    Just out of curiosity, I tried the method you suggested:

    direction vectors of the plane (vectors parallel to the plane) are:

    [tex]u = \left(\begin{array}{ccc}1\\1\\0\end{ar ray}\right)[/tex] and [tex]v = \left(\begin{array}{ccc}2\\1\\-1\end{ar ray}\right)[/tex]

    so that a normal vector is u ×v (cross product) = [tex]\left(\begin{array}{ccc}-1\\-1\\-1\end{ar ray}\right)[/tex]

    So the cartesian equation is -x - 13y −z = d, satisfying any point on the plane. For example if I substitute a point on the plane (1,2,3), [which is the position vector];

    -1-2-3 = d
    d = -6

    => -x - 13y −z = -6

    But this is different from what we got previously, x–y+z=2. Why is it different?
    Thanks again!
     
  8. Oct 28, 2008 #7

    HallsofIvy

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    ??Certainly not! Where in the world would you get the "13" from? The cross product Of [itex]\vec{i}+ \vect{j}[/itex] and [itex]2\vec{i}+ \vec{j}- \vec{k}[/itex] is [itex]-\vec{i}+ \vec{j}- \vec{k}[/itex] or [itex]\vec{i}- \vec{j}+ \vec{k}[/itex]} depending on the order of multiplication.

    So the cartesian equation is -x - 13y −z = d, satisfying any point on the plane. For example if I substitute a point on the plane (1,2,3), [which is the position vector];

    -1-2-3 = d
    d = -6

    => -x - 13y −z = -6

    But this is different from what we got previously, x–y+z=2. Why is it different?
    Thanks again![/QUOTE]
     
  9. Oct 28, 2008 #8
    oops there was a typo in my previous post...so I'll start over.

    I want the general cartesian equation of the plane, Ax+By+Cz = D

    I found the cross product of the two direction vectors;

    [tex]\left(\begin{array}{ccc}i&j&k\\1&1&0\\2&1&-1\end{ar ray}\right)[/tex]

    i [tex]\left(\begin{array}{ccc}1&0\\1&-1\end{ar ray}\right)[/tex], -j [tex]\left(\begin{array}{ccc}1&0\\2&-1\end{ar ray}\right)[/tex], k [tex]\left(\begin{array}{ccc}1&1\\2&1\end{ar ray}\right)[/tex]

    Cross Product = (-1,-1,-1)

    The cartesian equation is -x-y-z = D

    To find a D I just substitute point (1,2,3)
    -1-2-3 = -6

    Therefore => -x-y-z = -6

    But the problem is that this is different from the equation we obtained previously, x–y+z=2. Why is that? I don't understand it.
     
  10. Oct 28, 2008 #9

    HallsofIvy

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    No, the "j" determinant is not 1, so the "j" component is not -1.

     
  11. Oct 28, 2008 #10
    I understand. It's (-1,1,-1)

    => -x+y-z = -2

    Why is it different from the other equation we obtained (y+z=2)? It is still a multiple of that equation. It looks like it has been multiplied through by -1.
     
  12. Oct 28, 2008 #11

    Mark44

    Staff: Mentor

    -x + y -z = -2 is different from x + y + z = 6. Geometrically they represent different planes in R^3.
     
  13. Oct 28, 2008 #12
    I know! I'm talking about -x+y-z = -2 VS x–y+z=2 ?
     
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