Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$ 
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$

My attempt:


*

*$x^2 + y^2 + z^2 = 1$

*$x+y+z=0$


$$(2) \implies z = -(x+y)$$
$$(1) \implies x^2+y^2+(x+y)^2 = 1$$
$$2x^2 + 2y^2 + 2xy = 1$$
This is the curve in the xy-plane. Now if I could get y as a function of x, I could easily parametrize the curve but I am not able to do that. 
Is there an easier way to solve the problem?
 A: $2x^2+2xy+2y^2=1$ can be rewritten as $(2x+y)^2+3y^2=2$ and  this  can be parameterised 
\begin{eqnarray*}
\frac{2x+y}{\sqrt{2}} &=& \cos \theta  \\
\sqrt{\frac{3}{2}} y &=& \sin \theta . \\
\end{eqnarray*}
now substitute these into $x+y+z=0$ and we have 
\begin{eqnarray*}
x&=&\frac{1}{\sqrt{2}} \cos \theta- \frac{1}{\sqrt{6} } \sin \theta \\
y &=&\sqrt{ \frac{2}{3}} \sin \theta \\
z&=& - \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{6}} \sin \theta.
\end{eqnarray*}
Alternatively 
\begin{eqnarray*}
\left(\begin{array}{c}
x  \\ y \\ z
\end{array}\right)=\left(\begin{array}{c}
\frac{1}{\sqrt{2}}  \\ 0 \\ -\frac{1}{\sqrt{2}}
\end{array}\right) \cos \theta+\left(\begin{array}{c}
-\frac{1}{\sqrt{6}}  \\ \sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{6}}
\end{array}\right) \sin \theta
\end{eqnarray*}
now observe that these are orthogonal unit vectors and so this is a circle.
A: HINT: it is $$y_{1,2}=-\frac{x}{2}\pm\sqrt{\frac{1}{2}-\frac{3}{4}x^2}$$ and with that $y$ we can compue $z$
A: I would go about this geometrically.
First of all, the intersection is a circle. This is true in general for (non-tangent) intersection of plane and sphere. Here it is especially easy to see because the center of the sphere is the origin and the origin happens to lie in the plane.
Knowing this, we can parametrize this circle using rotation. If we were inside $\mathbb R^2$, this would just be $(\cos\varphi,\sin\varphi) = \cos\varphi (1,0) + \sin\varphi(0,1)$. But in our situation, we need two orthogonal unit vectors $e_1$ and $e_2$ that lie in $x+y+z = 0$ plane. Then the circle will be parametrized by $$\cos\varphi\ e_1 + \sin\varphi\ e_2.$$
We are free to choose $e_1$ as we like, so let us set $z = 0$. From the equation of the plane we get $y = -x$, i.e. $e_1 = \lambda(1,-1,0)$. From the condition $\|e_1\| = 1$ we get $\lambda = \frac{\sqrt 2} 2$. 
To get $e_2$, we notice that $e_2$ should be orthogonal to both $e_1$ and $n = (1,1,1)$ (this second vector comes from the equation $x+y+z = 0$ since it is normal to the plane). So, you can calculate $e_2$ using cross product: $e_2 = \mu (e_1\times n)$ 
(Or by guessing it: since $z$-component of $e_1$ is $0$, it makes no difference what $z$-component of $e_2$ will be for orthogonality to $e_1$, so we choose $(1,1,z)$ since $(1,1)$ is orthogonal to $(1,-1)$ and then let $z = -2$ so that $e_2$ will lie in the plane). 
Taking into account that $\|e_2\| = 1$, we finally get $e_2= \frac{\sqrt 6}6(1,1,-2)$.
Thus, the intersection of the sphere and the plane is given by
$$\frac{\sqrt 2}{2}(\cos\varphi+\frac{\sqrt 3}3\sin\varphi, -\cos\varphi+\frac{\sqrt 3}3\sin\varphi, -\frac{2\sqrt 3}3\sin\varphi),\ \varphi\in[0,2\pi].$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Find the curve of intersection between $\ds{x^{2} + y^{2} + z^{2} = 1}$ and $\ds{x + y + z = 0}$.

\begin{align}
1 & = x^{2} + y^{2} + \pars{-x - y}^{2} =
2x^{2} + 2y^{2} + 2xy
\\[5mm] & =
2\bracks{{x + y \over 2} + {x - y \over 2}}^{2} +
2\bracks{{x + y \over 2} - {x - y \over 2}}^{2} + 
2\bracks{{x + y \over 2} + {x - y \over 2}}\bracks{{x + y \over 2} - {x - y \over 2}}
\\[5mm] & =
{3 \over 2}\pars{x + y}^{2} + {1 \over 2}\pars{x - y}^{2}.
\\ & \text{which is}\ identically\ satisfied\ \text{by}\qquad
\left\{\begin{array}{rcrcr}
\ds{x} & \ds{+} & \ds{y} & \ds{=} & \ds{{\root{6} \over 3}\cos\pars{\theta}}
\\
\ds{x} & \ds{-} & \ds{y} & \ds{=} & \ds{\root{2}\sin\pars{\theta}}
\end{array}\right.
\end{align}

$$
\bbx{x =
{\root{6} \over 6}\,\cos\pars{\theta} +
{\root{2} \over 2}\,\sin\pars{\theta}\,,\qquad
y = {\root{6} \over 6}\,\cos\pars{\theta} -
{\root{2} \over 2}\,\sin\pars{\theta}}
$$
