Let $x^2+y^2+z^2=1$ be the unit sphere over $\mathbb{C}$. Prove that the sphere has 2 rulings by straight lines.
I have learnt the Segre embedding $\mathbb{P}^1(\mathbb{C})\times \mathbb{P}^1(\mathbb{C}) \rightarrow \mathbb{P}^3(\mathbb{C})$ given by $((x_0:x_1),(y_0:y_1)) \mapsto (x_0y_0:x_0y_1:x_1y_0:x_1y_1):=(X:Y:Z:W)$ which defines the Segre variety $XW=YZ$ and it has exactly two one-parameter families of straight lines, namely by fixing $(x_0:x_1)$ and $(y_0:y_1)$ respectively.
However, I am confused as to how the above problem can be phrased and solved by Segre embedding: homogenizing $x^2+y^2+z^2=1$ gives $x^2+y^2+z^2=w^2$ and we can factorize the equation as $(x+iy)(x-iy)=(w-z)(w+z)$. This motivates us to set $X=x+iy, Y=w-z, Z=w+z, W=x-iy$ and hence $((x_0:x_1),(y_0:y_1)) \mapsto (x_0y_0:x_0y_1:x_1y_0:x_1y_1) = (x+iy, w-z, w+z, x-iy)$. But this looks very messy and doesn't seem easy to solve for the two families of lines.
Just by inspection I can find three families of straight lines:
$l_1: x=1, y=iz,$
$l_2: y=1, x=iz,$
$l_3: z=1,x=iy.$
However, by the above discussion, there should be only two families of straight lines. Can anyone please tell me what did I do wrong? Thanks in advance.