How can I find the intersection between the sphere $x^2+y^2+z^2=1$ and the plane $x+y+z=1?$
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
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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 $w=-(u+v)$.
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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.
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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 $D \cap P = Q \{ ({1 \over \sqrt{3}}, \sqrt{2 \over 3} \cos \theta, \sqrt{2 \over 3} \sin \theta ) \}_{\theta \in [0, 2 \pi)}$
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 describes a circle with center at $\ds{\quad\pars{{\root{3} \over 3},{\root{3} \over 3},{\root{3} \over 3}}\quad}$ and radius $\ds{\quad{\root{6} \over 3}}$.
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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.
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).