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Homework Help: Intersection of a sphere and plane problem

  1. Nov 5, 2006 #1
    Hello,

    I was wondering if anyone could offer some advice on this one. I have a multi-part problem, in which I can't get the second part. It starts like this:

    (a)
    Find the equation of the sphere passes through the point (6,-2,3) and has a center of (-1,2,1).

    So I did this find and came up with:

    (x+1)²+(y-2)²+(z-1)² = 69

    (b)
    Find the curve in which this sphere intersects the yz-plane.

    So this is where I am stuck...
     
  2. jcsd
  3. Nov 5, 2006 #2

    quasar987

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    The zy plane is the set [itex]\Pi_{zy}[/itex] of all points (x,y,z ) of R^3 for which x=0.

    and your sphere (I havent checked if you found the correct radius) is the set [itex]\mathbb{S}_{\sqrt{69}}[/itex] of all points (x,y,z) of R^3 that obey the equation (x+1)²+(y-2)²+(z-1)² = 69.

    So, if you define the set [itex]\mbox{Intersection}(\Pi_{zy},\mathbb{S}_{\sqrt{69}})[/itex] of as the points (x,y,z) of R^3 that obey both to the x=0 and the (x+1)²+(y-2)²+(z-1)² = 69 condition, then points of this set are both the plane and the sphere. Logical?
     
    Last edited: Nov 5, 2006
  4. Nov 5, 2006 #3
    Hmm.. This is a little confusing for me. Sorry about that. So... I am not sure what the notation [itex]\Pi_{zy}[/itex] means, but I am guessing that it means the set of points on a sphere in the zy plane? But what you are basically telling me to do, or to think about, is to let x = 0, then I can solve a system of 2 equations and two unknowns to solve for y and z when x = 0. How does that sound?
     
  5. Nov 6, 2006 #4

    HallsofIvy

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    First, do you understand that the intersection is a curve- one dimensional- and so can't be written as a single equation in 3 dimensions?

    Yes, you can let x= 0. But then you can't "solve a system of two equations and two unknowns to solve for y and z when x= 0" because, first, you dont have two equations, and second, the intersection is not a single point!

    letting x= 0 you get 1+ (y- 2)2+ (z- 1)2= 69 or
    (y- 2)2+ (z- 1)2= 68. That's the equation for a circle. A standard parametrization for a circle is to let [itex]\theta[/itex] be the angle a radius makes with an axis. In particular, what must y and z be, in terms of [itex]\theta[/itex] so that your equation becomes
    [itex]68cos^2(\theta)+ 68sin^2(\theta)= 68[/itex]?
     
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