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^{2}+y

^{2}+z

^{2}=1 can be expressed as:

x([tex]\vartheta[/tex]) = (cos([tex]\vartheta[/tex])-(3)

^{1/2}sin([tex]\vartheta[/tex])) / (6

^{1/2})

y([tex]\vartheta[/tex]) = (cos([tex]\vartheta[/tex])+(3)

^{1/2}sin([tex]\vartheta[/tex])) / (6

^{1/2})

z([tex]\vartheta[/tex]) = -(2cos([tex]\vartheta[/tex])) / (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!