I have this two equations:
$$x^2+y^2+z^2=1$$ $$x+y+z=0$$
And I have to do the intersection and get the parametric equations for $x,y,z$. So I did a first step $$z=-x-y$$ and I put it in my first equation. Then my final equation was $$ x^2+xy+y^2=1/2$$ but I don't know how I should proceed know. I saw in wolfram alpha that it is a ellipse with a rotation but I don't know how I should work with $xy$.
Thanks in advance for the help,I appreciate it!
In polar coordinates,
$$\rho^2+\rho^2\cos\theta\sin\theta=\frac12$$ or
$$\rho=\frac1{\sqrt{2(1+\cos\theta\sin\theta)}}.$$
Then
$$\begin{cases}x=\dfrac{\cos\theta}{\sqrt{2(1+\cos\theta\sin\theta)}}, \\y=\dfrac{\sin\theta}{\sqrt{2(1+\cos\theta\sin\theta)}}, \\z=-\dfrac{\cos\theta+\sin\theta}{\sqrt{2(1+\cos\theta\sin\theta)}}.\end{cases}$$
The solution is more elegant in spherical coordinates: $\rho=1$, and
$$(\cos\theta+\sin\theta)\sin\phi+\cos\phi=0.$$
$$\begin{cases}x=-\cos\theta\sin(\text{arccot}(\cos\theta+\sin\theta)) \\y=-\sin\theta\sin(\text{arccot}(\cos\theta+\sin\theta)), \\z=\cos(\text{arccot}(\cos\theta+\sin\theta)).\end{cases}$$
But a yet more elegant solution is obtained by means of a 3D rotation that brings the normal to the plane in the vertical direction. The change of variable reads
$$\begin{cases}w=\dfrac{x+y+z}{\sqrt3},\\ v=\dfrac{x-y}{\sqrt2},\\ u=\dfrac{x-2y+z}{\sqrt6}.\end{cases}$$
The sphere has the equation
$$x^2+y^2+z^2=u^2+v^2+w^2=1$$ and the plane $$w=0.$$
Then
$$u^2+v^2=1$$ is a circle, which can be generated by $$\begin{cases}u=\cos\theta,\\v=\sin\theta,\\ w=0.\end{cases}$$
The final equations are obtained by multiplying by the inverse rotation matrix, which is also the transpose.