Given a vector function in terms of arc length, determine the two planes which intersects to create the curve (Describe a curve)

Question

I don't understand how the prof concluded that the curve is the intersection of $x + z = 1, x + 2y = 2$. I know how to find a line of intersection of two planes. However, I don't seem to be able to figure out the reverse.

Answer 1

The point $(x,y,z)$ on the curve is given by $\left(\frac{2s}{3}, 1-\frac{s}{3}, 1+\frac{-2s}{3}\right)$. Thus \begin{align*} x&=\frac{2s}{3}\\ y&=1-\frac{s}{3}\\ z&=1+\frac{-2s}{3} \end{align*} So we have one free parameter say $z$ (and that is why the curve is a line), and we have $s=\frac{3-3z}{2}$. With this we get $$x=\frac{2}{3}\left(\frac{3-3z}{2}\right)=1-z \implies \color{blue}{x+z=1}$$ and $$x+2y=\frac{2s}{3}+2-\frac{2s}{3}=2$$