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Math Help - intersection of a plane and a sphere

  1. #1
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    intersection of a plane and a sphere

    Hello,

    I am having some trouble finding the curve, C, of intersection of the plane x+y+z=1 and the sphere x^2+y^2+z^2=1. I then have to find the center of the curve C and the radius of the circle as well as find two perpendicular unit vectors parallel to the plane of C? They even give me one of those vectors and ask for me to use that to find the other, v1= (i-j)/sqrt(2). Then I have to use these results to construct a parametrization of C.

    Any help with this is appreciated.
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  2. #2
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    Re: intersection of a plane and a sphere

    Quote Originally Posted by lytwynk View Post
    Hello,

    I am having some trouble finding the curve, C, of intersection of the plane p: x+y+z=1 and the sphere x^2+y^2+z^2=1. I then have to find the center of the curve C and the radius of the circle as well as find two perpendicular unit vectors parallel to the plane of C? They even give me one of those vectors and ask for me to use that to find the other, v1= (i-j)/sqrt(2). Then I have to use these results to construct a parametrization of C.

    Any help with this is appreciated.
    Only some hints:

    1. The curve of intersection between a sphere and a plane is a circle.

    2. The normal vector of the plane p is \vec n = \langle 1,1,1 \rangle

    3. The midpoint of the sphere is M(0, 0, 0) and the radius is r = 1.

    4. A straight line through M perpendicular to p intersects p in the center C of the circle. You should come out with C \left(\frac13, \frac13, \frac13 \right). So the distance of C to M is d = \frac13 \cdot \sqrt{3}

    5. Use r and d to determine the radius of the circle.

    6. A vector \vec v parallel to p is perpendicular to \vec n, that means \langle 1,1,1 \rangle \cdot \vec v = 0

    I'll show you how to get \vec v in general:

    \langle a,b,c \rangle \cdot \vec v = 0~\implies~ \vec v = \langle -b, a, 0 \rangle ~\vee~\vec v = \langle -c, 0, a \rangle ~\vee~\vec v = \langle 0, -c, b \rangle

    7. For the parametrization of the curve C you have to use the sine and cosine function.
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  3. #3
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    Re: intersection of a plane and a sphere

    Quote Originally Posted by earboth View Post
    Only some hints:

    4. A straight line through M perpendicular to p intersects p in the center C of the circle. You should come out with C \left(\frac13, \frac13, \frac13 \right). So the distance of C to M is d = \frac13 \cdot \sqrt{3}

    5. Use r and d to determine the radius of the circle.

    6. A vector \vec v parallel to p is perpendicular to \vec n, that means \langle 1,1,1 \rangle \cdot \vec v = 0

    I'll show you how to get \vec v in general:

    \langle a,b,c \rangle \cdot \vec v = 0~\implies~ \vec v = \langle -b, a, 0 \rangle ~\vee~\vec v = \langle -c, 0, a \rangle ~\vee~\vec v = \langle 0, -c, b \rangle

    7. For the parametrization of the curve C you have to use the sine and cosine function.

    I have trouble understanding mainly point 4.
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  4. #4
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    Re: intersection of a plane and a sphere

    Quote Originally Posted by lytwynk View Post
    I have trouble understanding mainly point 4.

    The plane x+y+z=1 is \frac{1}{\sqrt{3}} units from (0,0,0).

    The center of the circle will be on the line <t,t,t> at a distance of \frac{1}{\sqrt{3}} units from (0,0,0).

    So the center is \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)~.

    Thus the radius if r=\sqrt{1-1/3}.
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  5. #5
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    Re: intersection of a plane and a sphere

    OK, I got it now... I just was thinking a bit too hard. Now though how do you parametrize the curve, I get how to use sine and consine functions for a curve but I am just confused about how to set it up because it is on a plane other then for example the xy plane.
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