1. ## Curve parametrization

Hi, I would like to parametrize the intersection between the unit sphere ( $x^2 + y^2 + z^2 =1$ ) and the plane $x+y+z =0$*
I came with
$*x(t) = \frac{cos(t)}{\sqrt{2}}$
$y(t) = \frac{sin(t)}{\sqrt{2}}$
$z(t) = -\frac{(cos(t)+sin(t)}{\sqrt{2}}$
It almost works, but it seem there are scale problems and I can't solve it.

2. Hi

u $(\frac{1}{\sqrt{2}} , 0 , -\frac{1}{\sqrt{2}})$ and v $(0 , \frac{1}{\sqrt{2}} , -\frac{1}{\sqrt{2}})$ are 2 direction vectors of the plane $x + y + z = 1$

A parametrization of the plane is

$x(t) = \frac{t}{\sqrt{2}}$

$y(t) = \frac{t'}{\sqrt{2}}$

$z(t) = -\frac{t}{\sqrt{2}} - \frac{t'}{\sqrt{2}}$

Substituting in the equation of the sphere gives

$t'^2 + t t' + t^2 - 1 = 0$

Discriminant is $-3 t^2 +4$

It is positive for $-\frac{2}{\sqrt{3}} \leq t \leq \frac{2}{\sqrt{3}}$

Let $t = \frac{2}{\sqrt{3}} \cos(\theta)$

Then $t' = -\frac{\cos(\theta)}{\sqrt{3}} \pm |\sin(\theta)|$

You just have to substitute t and t' in the expression of x(t), y(t) and z(t)