Information given: $$x+y+z = 0$$ $$x^2 + y^2 + z^2 = 1$$ We have the portion of the intersection of these two equation going from $(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$ to $(0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
What I need to find: The parameterization of this intersection.
Where I'm at: I know that I need to take the first vector $\vec{a} = (0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$ and find a vector ($\vec{b}$) that is perpendicular to that vector and on the curve of the intersection.
The problem is that I don't seem to quite understand how to do it
After that I would only need to take this equation $\vec{r}(t) = cos(t)\vec{a} + sin(t)\vec{b}$ to find parameterization, because it's a circle.
Solving $$x+y+z=0$$ and $$y-z=0$$
We find that $(-2,1,1)$ is a solution.
Normalize the vector.
$$\left(\frac{-2}{\sqrt6}, \frac{1}{\sqrt6},\frac{1}{\sqrt6}\right)$$
is perpendicualr to $a$ and lies on the curve of intersection.