Calc III: sphere & plane intersection.

[hq1330]

Consider the intersection of the sphere (x-3)^2+(y+2)^2+(z-1)^2=13 with the plane x+y=0

a) This intersection should be a familiar curve. Describe the curve.

b) Substituing x=-y or y=-x into the equation of the sphere does not give this curve. Explain the difference.


I know that the intersectionis a circle but I am having trouble proving that. I rearranged the above equation and set it equal to x+y:

(x-3)^2+(y+2)^2+(z-1)^2-13=x+y

The next thing I did was put the like terms together to give the following equation:

x^2-6x+9-x+y^2+4y+4+y+z^2-2z+1-13=0

I then simplified this equation:

X^2-7x+9+y^2+3y+4+(z-1)^2-13=0

Now I am stuck. This just looks like a smaller sphere.
Also, I would plug in x=-y or vice versa but the second part of the question says that will not give me the result that I want.

 

[Mark44]

The intersection of a sphere and a plane will be one of the following.

1. The empty set (the two don't actually intersect).
2. A single point.
3. A circle.

 

[tiny-tim]

Hi hq1330!

(try using the X² tag just above the Reply box )

Consider the intersection of the sphere (x-3)^2+(y+2)^2+(z-1)^2=13 with the plane x+y=0

a) This intersection should be a familiar curve. Describe the curve.

b) Substituing x=-y or y=-x into the equation of the sphere does not give this curve. Explain the difference.
…
Also, I would plug in x=-y or vice versa but the second part of the question says that will not give me the result that I want.

I think this question is unnecessarily obscure.

A curve, in three dimensions, requires two equations to describe it, so of course substituing x=-y or y=-x into one equation, which will only give one equation, cannot give the curve.

(in fact, it will give the two-dimensional projection of the curve onto the xz or yz plane)

Sooo … plug away!

(also, the method you used gives a whole surface, including eg the intersection of the sphere (x-3)²+(y+2)²+(z-1)²=14 with the plane x+y=1, doesn't it? )