How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$ and find it arc length? How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$?
Should I make a new equation $x^{2} + y^{2} + z^{2} = x + y + z$?
Or do I need to use spherical coordinates to parametrize $x^{2} + y^{2} + z^{2} = 1$ ?
Also, I want to find the arc length of this parametric equation.
 A: The intersection will be an ellipse in 3-space.  Therefore, it is parameterisable by a single variable $t$.
First, solve the equation of the plane for $z$.
$$z = 1-x-y.$$
Then substitute into the equation of the sphere:
$$x^2+y^2+(1-x-y)^2 = 1.$$
Write this equation in the form:
$$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2} = 1.$$
Then, the parameterisation is given by
$$x = x_0 + a\cos(t),\quad y = y_0 + b\sin(t),\quad z = 1-x_0-y_0-a\cos(t)-b\sin(t), \quad 0 \leq t \leq 2\pi$$
A: That curve is a circle, centered at $C=\left(\frac13,\frac13,\frac13\right)$. Consider the vector $w=\left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right)$, which is an unit vector orthogonal to the line defined by the origin and $C$. Then$$\|C+tw\|=1\iff\frac{\sqrt{3+9t^2}}3=1\iff t=\pm\sqrt{\frac23}.$$So, the radius of your circle is $\sqrt{\frac23}$. Now, let $w'=\left(\frac1{\sqrt6},\frac1{\sqrt6},-\frac2{\sqrt6}\right)$ which is also an unit vector and which is orthogonal to $w$ and also to the line defined by the origin and $C$. Then the parameterization that you're after is\begin{multline}\theta\mapsto C+\cos(\theta)w+\sin(\theta)w'=\\=\frac13\left(1+\sin (\theta )+\sqrt{3} \cos (\theta),1+\sin (\theta )-\sqrt{3} \cos(\theta ),1-2 \sin (\theta )\right).\end{multline}
A: For symmetry, that's a circle with center on the diagonal plane $x+y+z=1$, so at ${\bf c} =(1/3,1/3,1/3)$.
Its radius will be the distance between the above center and a point on the plane and the sphere like $(1,0,0)$, i.e. $r=\sqrt{4/9+1/9+1/9} = \sqrt{6}/3$.
Then take the normal vector to the plane $\bf n =(1,1,1)$, a vector parallel to the plane (orthogonal to $\bf n$), like $\bf u = (-1,1,0)$ and a vector orthogonal to $\bf n, \bf u$ , e.g. $\bf v = \bf u \times \bf n = (1,1, -2) $.
Then you can write
$$
\left( {x,y,z} \right) = {\bf c} + r{{\bf u} \over {\left| {\bf u} \right|}}\cos \alpha  + r{{\bf v} \over {\left| {\bf v} \right|}}\sin \alpha 
$$
A: Yet another answer.
The unit sphere in $\mathbb{R}^n$ can be described by $\|x\|^2 = 1$, and the equivalent plane by $e^T x = 1$ where $e=(1,...,1)$. Let $Q$ be an orthogonal rotation such that $Qe = \sqrt{n}e_1$, where $e_1 = (1,0,...,0)$.
Using the transformation $y=Qx$ we get the equivalent equations $\|y\|^2 = 1$ and $e_1^Ty = {1 \over \sqrt{n}}$. Equivalently
$y_2^2+\cdots+ y_n^2 = 1-{1 \over n}$, so we see that this is an $n-1$ dimensional $\sqrt{1-{1 \over n}}$ sphere centered at $({1 \over \sqrt{n}},0,...,0)$.
Mapping back we see that the intersection is parameterised by
${1 \over n} e + \sum_{k=2}^n y_kQ^T e_k$, where $y_2^2+\cdots+ y_n^2 = 1-{1 \over n}$.
For $n=3$ we can parameterise this as
${1 \over 3}(1,1,1) + {\sqrt{2 \over 3}}\cos \theta \ u +{\sqrt{2 \over 3}}\sin \theta \ v$ where ${1 \over \sqrt{3}} (1,1,1),u,v$ are orthonormal.
