Parametrization - circle defined by plane intersection sphere

[forty]

Show that the circle that is in the intersection of the plane x+y+z=0 and the sphere x²+y²+z²=1 can be expressed as:

x($$\vartheta$$) = (cos($$\vartheta$$)-(3)^1/2sin($$\vartheta$$)) / (6^1/2)y($$\vartheta$$) = (cos($$\vartheta$$)+(3)^1/2sin($$\vartheta$$)) / (6^1/2)z($$\vartheta$$) = -(2cos($$\vartheta$$)) / (6^1/2)

I'm really stuck on how to do this. I have already looked over a couple of threads on this forum as well as others however they all do the intersection between a sphere and a plane.

I know this is the unit sphere centered at (0,0,0) and I know the plane is 45 degrees to all axis with its normal vector defined as (1,1,1). I've tried changing co-ordinate systems as well as using some trig substitutions. I've also tried projecting the intersection onto a plane but can't really work out how to do that properly.

Any help on this would be greatly appreciated, thanks!

 

[fantispug]

Well have you checked it satisfies the equations, and that each parameter ranges over the complete circle? Alternatively you could just rotate your coordinate system (and hence parameters) so the plane has a nicer equation, and is easier to check.

 

[forty]

How would I change my parameter so I could rotate it to make it nicer?