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Find some vector function whose image is the intersection of two surfaces

  1. Oct 20, 2009 #1
    Hi all, I'm quite new here, but it's been a while since I've been browsing through these forums for past answered questions for calculus and physics, but now comes the time where I'm the one needing help that's not been questioned yet.


    1. The problem statement, all variables and given/known data

    Find some* vector funcion r with domain (-∞,∞) whose image is the intersection of the two surfaces:
    x²+y²+z²=9,
    x+y+z=3.
    *: vector fuction must not contain inverse trig functions or +/ square routes.

    2. Gameplan

    Well, what I started to do was draw a picture of the plane cutting through the sphere, luckily the radius of it was 3 and the x,y,z intercepts of the plane landed on the sphere too. So I thought I could set the equations equal to eachother and from there find a parametric function that I can put into a vector function form.

    3. The attempt at a solution

    What I had found for setting both equations equal to eachother is
    x²+y²+z²=3x+3y+3z
    =
    (x-3/2)²+(y-3/2)²+(z-3/2)²=27/4
    from this, this gives me an equation on a sphere which confuses me becuase I thought that the intersection of two surfaces would be a line (which I hoped to have happen when setting both equations equal to eachother)
    Also, I've found the centroid of triangle formed by the x,y,z intercepts of the plane which is (1,1,1). The closest point from the centroid to the surface of the sphere is (sqrt(3),sqrt(3),sqrt(3)).
    Much help would be appreciated

    Thanks
     
  2. jcsd
  3. Oct 20, 2009 #2

    lanedance

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    Homework Helper

    hi ramb, hope this helps..

    ok so first off thinking about this geometrically as you've pointed out its a plane intersecting a sphere of radius 3, which will give a circle

    so you're trying to parametrise a vector function which traces out that circle

    as a check you know both the sphere & the plane cross the x,y,z axis at the points
    (3,0,0)
    (0,3,0)
    (0,0,3)
    and these points will be on the circle, in fact these points should actually be enough to work out the circle, though not sure if thats the easiest way

    i think the point you found on the plane (1,1,1) is actually the centre of the circle

    this give the radius as
    sqrt(|(3,0,0)-(1,1,1)|^2) = sqrt(2^2+1+1)= sqrt(6)...

    from here we can parameterise

    so why can't you use square roots or trig functions? and any ideas what the preferred method is?
     
    Last edited: Oct 20, 2009
  4. Oct 20, 2009 #3

    lanedance

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    sorry that said inverse trig functions, now worries then...

    you know the centre of your circle, radius & its orientation, now find 2 perpindicular vectors to represent the radial direction & parameterise using sinusoids
     
    Last edited: Oct 20, 2009
  5. Oct 20, 2009 #4

    HallsofIvy

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    Staff Emeritus
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    Setting the two equations equal does not guarentee that either equation is satisfied separately.

    Try this- since x+ y+ z= 3 is linear, solve for one of the variables, say z. Replace z in the equation of the sphere by that. Now you have a single equation in x and y. Can you write x and y in terms of a single parameter, t, say, so that equation is satisfied?

    You may have to "rotate" the coordinate system to get something simple.
     
  6. Oct 20, 2009 #5
    Hi, thanks for your responses. I understand that I can use Halls of Ivy's method once I parameterize x and y, but I am still confused about an equation of a circle in 3D space, the fact that it's tilted in 3D will make projections on x y and z all ellipses. Must I put the equation in cartesian first, then go to the parametric?
     
  7. Oct 20, 2009 #6

    lanedance

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    if the vector to the centre of the circle is c, and if you find 2 perindicular vectors, in the direction from the centre to the circle edge, r1 and r2then you can parametrise as follows

    [tex] \textbf{p}(t) = \textbf{c} + \textbf{r}_1 cos(t) + \textbf{r}_2 sin(t) [/tex]

    not the following is also true
    [tex] \textbf{r}_2 \bullet \textbf{r}_1 = 0 [/tex]
    and as the centre point vector is perindicular to teh plane of teh circle then (in this case)
    [tex] \textbf{c} \bullet \textbf{r}_1 = \textbf{c} \bullet \textbf{r}_2 = 0 [/tex]
     
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