How do I find a parametrization of the intersection or part of the intersection of the sphere x2 + y2 + z2 = 4 and plane x + y + z = 1 with respect to a parameter t?

Question

How do I find a parametrization of the intersection  or part of the intersection of the sphere x2 + y2 + z2 = 4 and plane x + y + z = 1 with respect to a parameter t?

anonymous · Answer

first solve for one of the variables in the plane eq and substitue that into the sphere eq.  What remains will be a circle.  get it into std form and then let's talk.

anonymous · Answer

ok, so that's $$x^2+y^2+(1-y-x)^2=4$$

anonymous · Answer

yep now work it out and get it into std form (x-h)^2 + (y-k)^2 = r^2

anonymous · Answer

once you have it in that form, set (x-h)=rcos(t) and (y-k)=rsin(t) and solve for x and y to get your parameterization.

anonymous · Answer

does that make sense?

anonymous · Answer

Yes, that does make sense. Tank you, I just need figure out its standard form and from there I understand.

anonymous · Answer

coolio!

anonymous · Answer

I've been unable to find its standard form and when plotting the equation in wolfram alpha it looks geometrically like a ellipse http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%2B%281-y-x%29%5E2%3D4.

anonymous · Answer

how can a plane and sphere intersect to give an ellipse?  doesn't make sense...

anonymous · Answer

the sphere has it's center at (0, 0, 0) and a radius of 4.  the normal to the plane has to go through the center of the circle.  the shape will be a circle in 3d, however, you may have to rotate it.  that is why in the x, y plane it shows as an ellipse. the center of the circle is at (1/3, 1/3, 1/3).  this is because x+y+z-1=0 and the normal to this plane is the vector (1,1,1).  but this would involve z, which is ok.  you could do a change of basis to get it to be a circle in x' and y'.