Intersection of a sphere and a plane How can I find the intersection between the sphere $x^2+y^2+z^2=1$ and the plane $x+y+z=1?$
Context
This is related to a computation of surface integral using Stokes' theorem, Calculate the surface integral $\iint_S (\nabla \times F)\cdot dS$ over a part of a sphere
 A: If we visualize it, it's pretty easy to believe that the intersection should be a circle living in space.

The real question is, what type of description do you want?  Here are a few possibilities.
Clearly, the circle lies in the plane, which has normal vector $\langle 1,1,1 \rangle$.
Also, symmetry dictates that the center must lie on the line where $x=y=z$.  Since the center is on the plane $x+y+z=1$, we have the center is $(1/3,1/3,1/3)$.  The radius is just the distance between center and any known point on the circle, like $(1,0,0)$.  Thus, the radius is $r=\sqrt{2/3}$.  I would say that this center, radius, and normal provide a complete description of the circle.
Another alternative is to provide a parametrization.  This can be accomplished via
$$p(t) = c + r\cos(t)u + r\sin(t)v,$$
where $c$ is the center, $r$ is the radius, and $u$ and $v$ are perpendicular unit vectors that are both perpendicular to the normal vector for the plane.  Specifically,
$$p(t) = \langle 1/3,1/3,1/3 \rangle + \cos(t)\langle 1,-1,0 \rangle/\sqrt{3} + 
\sin(t)\langle 1,1,-2 \rangle/3.$$
That's the parametrization I used to create the image.

A: If you set $(x,y,z)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)+(u,v,w)$ you are left with the constraints $u+v+w=0$ and $u^2+v^2+w^2=\frac{2}{3}$, from which it follows that $u^2+uv+v^2=\frac{1}{3}$ and $w=-(u+v)$.
A: Consider a change of coordinates. Let $Q = \begin{bmatrix} 
{1 \over \sqrt{3}} & {1 \over \sqrt{2}} & {1 \over \sqrt{6}} \\
{1 \over \sqrt{3}} & -{1 \over \sqrt{2}} & {1 \over \sqrt{6}} \\
{1 \over \sqrt{3}} & 0 & -{2 \over \sqrt{6}} \\
\end{bmatrix}$. Note that $Q$ is orthogonal, $Q^T Q = I$, and
$Q (1,0,0)^T = {1 \over \sqrt{3}}(1,1,1)^T$.
The purpose is to rotate the plane so that it is perpendicular to the $x$ axis. Any suitable rotation will do, this was the easiest for me to compute.
The surface of the sphere is invariant under rotations, so this works nicely.
Let $D = \{x \in \mathbb{R}^3 | \|x\|=1 \}$, $P = \{x \in \mathbb{R}^3 | \langle (1,1,1)^T , x \rangle =1 \}$. Since $Q$ is orthogonal, we have
$Q D = D$, and we have $P = Q P'$, where $P'=\{ x \in \mathbb{R}^3 | x_1 = {1 \over \sqrt{3}} \}$.
Hence $D \cap P = Q (D \cap P') $. We see that
$D \cap P' = \{ ( {1 \over \sqrt{3}}, x_2,x_3)^T | x_2^2+x_3^2 = 1-{1 \over 3} \} $.
Consequently we have $D \cap P = Q \{ ({1 \over \sqrt{3}}, \sqrt{2 \over 3} \cos \theta, \sqrt{2 \over 3} \sin \theta ) \}_{\theta \in [0, 2 \pi)}$
A: compute the intersection point of the line $\vec{x}=(0;0;0)+t(1;1;1)$ with the given plane. This is the midpoint of a circle (if he exists).
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Lets
$\ds{\quad\vec{n}_{1}
     = {1 \over \root{3}}\,\pars{1,1,1}\,,\quad
     \vec{n}_{2}={1 \over \root{6}}\,\pars{2,-1,-1}\quad}$
and $\ds{\quad\vec{n}_{3} = \vec{n}_{1}\times\vec{n_{2}}\ .\quad}$

It is clear that
  $\ds{\vec{n}_{i}\cdot\vec{n}_{j}=\delta_{ij}\,.\quad}$ Lets
  $\ds{\quad\vec{r}\equiv\pars{x,y,z}=
    {1 \over \root{3}}\,\vec{n}_{1} + \mu_{2}\,\vec{n}_{2} + \mu_{3}\,\vec{n}_{3}
    \quad}$
  which satisfies $\ds{x + y + z = 1}$. In addition, $\ds{r^{2} = 1}$ yields

$$
{1 \over 3} + \mu_{2}^{2} + \mu_{3}^{2}= 1\qquad\imp\qquad
{2 \over 3}=\mu_{2}^{2} + \mu_{3}^{2}
=\verts{\vec{r} - {\root{3} \over 3}\,\vec{n}_{1}}^{2}
$$

It's equivalent to
$$
\pars{x - {\root{3} \over 3}}^{2} + \pars{y - {\root{3} \over 3}}^{2}
+\pars{z - {\root{3} \over 3}}^{2}=\pars{\root{6} \over 3}^{2}
$$


It describes a circle with center at
  $\ds{\quad\pars{{\root{3} \over 3},{\root{3} \over 3},{\root{3} \over 3}}\quad}$
  and radius $\ds{\quad{\root{6} \over 3}}$.

A: Because someone has to: An algebraic geometry approach. Let,
$$
\begin{align*}
p &:= x^2 + y^2 + z^2 - 1,\\
q &:= x + y + z -1.
\end{align*}
$$
Form the ideal,
$$
I = \left< p, q\right> \subset \mathbb{R}\left[x,y,z\right].
$$
A Groebner basis for this ideal, with respect to a degree reverse lexicographic order, is then,
$$
G = \left< y^2 + yz + z^2 - y - z, x + y + z - 1 \right>.
$$
The first polynomial in $G$ has two variables and the second has three. Fix an $x$; you are left with two polynomials in two unknowns.
