What is the (parametric) intersection of a plane and a sphere? Can someone please show me how to prove that the intersection of the plane
$$x+y+z=0$$  and the sphere  
$$x^2+y^2+z^2=1$$
can be expressed as  
$$x(t)=\frac{\cos t-\sqrt3 \cdot\sin t}{\sqrt6}$$ 
$$y(t)=\frac{\cos t+\sqrt3 \cdot\sin t}{\sqrt6}$$   
$$z(t)=\frac{-2\cos t}{\sqrt6}$$ 
Ps: Also, why am I not getting the correct notation? I am using a macbook pro (safari) if that is a concern?
 A: You can solve for $z=-x-y$ and plug in $x^2+y^2+(-x-y)^2=1$, which rearranges to $2x^2+2xy+2y^2=1$.  Hence for any fixed $z$ you have a rotated ellipse.
Let $u=(x+y/2)$.  Then $u^2=x^2+xy+y^2/4$, so the equation becomes $2u^2+\frac{3}{2}y^2=1$.  In standard form this is:  $$\frac{u^2}{(\frac{1}{\sqrt{2}})^2}+\frac{y^2}{(\frac{\sqrt{2}}{\sqrt{3}})^2}=1$$
We can solve this parametrically as $u=\frac{1}{\sqrt{2}}\sin t$, $$y=\sqrt{\frac{2}{3}}\cos t$$  We can now find $$x=u-y/2=\frac{1}{\sqrt{2}}\sin t-\frac{1}{\sqrt{6}}\cos t$$ and $$z=-x-y=\frac{-1}{\sqrt{2}}\sin t +(-\frac{1}{\sqrt{6}}-\sqrt{\frac{2}{3}})\cos t$$  
Alas, you posted your revisions after I'd made my choices so my solution will not agree with yours.  There are three choices at the first step (solve for $z,y,x$), and two choices at the second step (replace $x$ or $y$).  One of those six choices might give the result you have.  :-)
Followup: A cross term ($xy$) is a rotation and can always be eliminated by a change of variables, pointing in the directions of the major and minor axes.  There is a systematic way to do this; see here or here, or you can do it by the seat of your pants, like I did.  I didn't directly rotate, instead I sheared, which made the computations a bit simpler.
Edit: fix sign error, thanks @Mhenni.
Edit 2: add general method
A: (Originally I had $u,v=x+y,x-y$, but after going through half the problem, I considered that $2u,2v=x+y,x-y$ would be a little more convenient.) 
amanda, after your substitution, we consider $x^2+xy+y^2={1\over2}$. 
Here are is a nice choice for the next substitution, 
\begin{align}
2u&=x+y,\\
2v&=x-y.
\end{align}
We next consider 
\begin{align}
4u^2&=x^2+y^2+2xy,\\
4v^2&=x^2+y^2-2xy.
\end{align}
To get $x^2+y^2+xy$, we need $2u^2+2v^2+(u^2-v^2)$. Thus our problem is transformed to 
$$6u^2+2v^2=1.$$  
Then we have parametric solutions 
$$u=\sqrt{1\over6}\cos t,\ v=\sqrt{1\over2}\sin t.$$
From there, we find for instance 
$$x=u+v={\cos t+\sqrt3\sin t\over\sqrt6},$$
and the rest should be straightforward. 
To get the sign to match what is shown in your equations, I can think of several ways to go. 


*

*You could simply note the symmetry in $x$ and $y$ and switch them (which suggests that if we had made the substitution $2v=y-x$ instead of $x-y$, that would have worked also). 

*From $6u^2+2v^2=1$, we could have chosen $v=-\sqrt{1\over2}\sin t$. 

*If we make the substitution $t=-s$, then we have $x={\cos (-s)+\sqrt3\sin (-s)\over\sqrt6}={\cos s-\sqrt3\sin s\over\sqrt6}$. Now, call it $t$ again. 

