Consider the interseccion of the surfaces:
$x+y+z=1$ and $x^2+y^2=1$.
I need to parametrize the curve of intersection, and know it is an ellipse on the plane,so I tried with
$\gamma(t)=(\cos t,\sin t,1-\cos t -\sin t),0 \leq t \leq 2\pi$
But what I'm not sure about is the eccentricity of the ellipse, I mean, should it be of the following form?
$\gamma(t)=( k\cos t,l\sin t,1-k\cos t -l\sin t)$ with $k,l\in \mathbb{R}$ not necessarily 1?
$\mathbb A$ is one of rotational transforms that convert $(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3})$ into $(0,0,1)$.
$\mathbb A=$ \begin{bmatrix} \frac1{\sqrt2} & -\frac1{\sqrt2} & 0\\ \frac1{\sqrt6} & \frac1{\sqrt6} & -\frac2{\sqrt6}\\ \frac1{\sqrt3} & \frac1{\sqrt3} & \frac1{\sqrt3}\\ \end{bmatrix}
$\mathbb A$ would transform $\left(\cos t, \sin t, -\cos t-\sin t,0\right)$ [note that I shifted the z coordinate by $-1$ to make things more visible] into $$\left(\frac{1}{\sqrt2}\cos t-\frac{1}{\sqrt2}\sin t,\frac{\sqrt3}{\sqrt2}\cos t+\frac{\sqrt3}{\sqrt2}\ \sin t,0\right)$$
Now let's eliminate the parameter $t$, we substutue
$$x=\frac1{\sqrt2}(\cos t - \sin t)$$ $$y=\frac{\sqrt3}{\sqrt2}(\cos t + \sin t)$$
And we get this equation
$$\left(\frac{y}{\sqrt3}\right)^2+\left(x\right)^2=\left(\frac{\cos t - \sin t}{\sqrt2}\right)^2+\left(\frac{\cos t + \sin t}{\sqrt2}\right)^2=1$$