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Drawing a plane

  1. Jan 9, 2009 #1
    1. What are the steps that I should take in drawing a plane

    for example if the equation is given of a plane

    x=y

    in 3 space

    what will the plane look like...

    what should I do in terms of steps?

    if i say x = 0 so does y

    and z can be anything

    with this information i cant think of how i would go about drawing the plane
     
  2. jcsd
  3. Jan 9, 2009 #2

    HallsofIvy

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    Yes, if x= 0 then so is y while z can be anything: (0, 0, z) is a point on the plane for all z. If x= 1, the so is y while z can be anything: (1, 1, z) is a point on the plane for any z. (0, 0, 0), (0, 0, 1) and (1, 1, 0) are points in the plane, not on a single line. Because any plane can is determined by 3 such points, you can plot those three point and imagine a plane passing through them to see what it looks like.

    In this simple case, since z can be anything, the simplest way to see what the graph looks like is to draw the line y= x in the xy-plane and imagine it extended straight up.
     
  4. Jan 9, 2009 #3
    ok thanks.........one of the problems i was doing was

    sketch the curve with the given vector equation...

    r(t) = (sint)i + (sint)j + squareroot(2)(cost)k



    i read the section before this but im really struggling on what they are doing. There is only one example like this in the section and what they do is they just find relations between the parametric equations which I dont understand how they can assume. For example the solution for the above question is as follows:

    the parametric equations are

    x = sint
    y = sint
    z = sqrt(2)cos t

    Now they start looking for relationships between the parametric equations

    1. x = y
    NOW FOR SOME REASON THEY SAY: this means the curve is on the x=y plane ..

    ??? how can you assume that???

    2. x^2 + y^2 + z^2 = 2sin^2t + 2cos^2t = 2

    the curve lies on a sphere with radius 2 (since the above equation represents the sphere)

    again how can you assume the curve lies on the sphere ?


    3. this means draw that plane and the sphere and the intersection of that plane and sphere is how the curve can be sketched.


    Can anyone give me a rough idea of why these steps are taken.. im so confused even after reading the section
     
  5. Jan 10, 2009 #4
    Hmm...

    r(t) = | sin(t) |
    | sint(t) |
    | sqrt(2)cos(t) |

    Yes, ok.

    The above equation represents the possition vector of a point on the curve with respect to a varible t. In this case, the i and j componants will always be equal in magnitude;

    So if we draw our axis;

    j | /
    | /
    | /
    | /
    |/_______________ i

    Then our resultant diagonal is the result. Now if you imagine that diagonal line as a sheet of paper perpendicular to your computer screen this represents the k axis.

    We will no concern outselves with what that looks like;

    i or j |
    |
    |
    |
    |_____________ k

    Because both the i and j componants are the same, it doesn't matter which we are using to describe the resultant curve. We can actually call that axis the i-j plane and treat is as another set of axis (note: can't do this if the i-j plane was a curve)

    Hence with respect to this 2D axis we have;

    r(t) = (sint)j + sqrt(2)(cost)k

    Now if t is an angular quantatity as in normal polar co-ordinates (messured anti-clockwise from the positive k axis).

    Then as;

    t = 0, k = sqrt(2) as cos(0) = 1, as
    t = 1/2pi, sin(t)i = 1 and sin(t)j = 1 => The magnitude of this vector point in the i-j plane is sqrt( 12 + 12) which is the same as sqrt(2).

    Hence can you see that it describes a circular motion in three dimentions?

    The circle is being described in the i-j plane, by the k componant of the vector. Sorry that's not a very clear mathamatical way of showing it, but hopefully it will help you visulise what r(t) is. Once you can do that, it shouldn't be too hard to understand that in three diemensions a circle is just a 'hoop' that fits around a sphere in all three directions of rotation. And hence it 'lies on the [imaginary] sphere.

    As you should know, that x2i + y2j + z2k = r2 describes a sphere, then it doesn't take a lot to compare your two equations and their co-effs.

    I hope that helps,
    Haths
     
  6. Jan 10, 2009 #5

    HallsofIvy

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    They haven't "assumed" any thing. It is true because x= sin t= y are exactly the same for every t.

    [/quote]2. x^2 + y^2 + z^2 = 2sin^2t + 2cos^2t = 2

    the curve lies on a sphere with radius 2 (since the above equation represents the sphere)

    again how can you assume the curve lies on the sphere ?[/quote]
    [itex]x^2= sin^2 t[/itex]
    [itex]y^2= sin^2 t[/itex]
    so [itex]x^2+ y^2= sin^2 t+ sin^2 t= 2 sin^2 t[/itex]

    [itex]z= \sqrt{2} cos t[/itex] so [itex]z^2= 2 cos^2 t[/itex]
    [itex]x^2+ y^2+ z^2= 2 sin^2t + 2 cos^2 t= 2(sin^2 t+ cos^2 t)= 2[/itex]
    Since every point (x, y, z) on the curve satisfies [itex]x^2+ y^2+ z^2= 2[/itex] and so is on the sphere that equation defines.


     
  7. Jan 10, 2009 #6

    HallsofIvy

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    typo: you should not have the "i", "j", "k" here. These are not vectors.

     
  8. Jan 10, 2009 #7
    ok thanks for your help!
     
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