So I have these two planes:
$$\pi_1: X = (1,0,0) + \lambda(0,1,1) + \gamma(1,2,1)$$ $$\pi_2: X = (0,0,0) + \lambda(0,3,0) + \gamma(-2,-1,-1)$$
I need to find two points $A$ and $B$ of the intersection of the two planes and then determinate the line that passes between these two points. I already know one way to solve this:
I can find the normal vector of the two planes by doing the cross product of its direction vectors (for $\pi_1$ we have $\vec v_1 = (0,1,1), \vec v_2 = (1,2,1)$ and for $\pi_2$ we have $\vec v_3 = (0,3,0), \vec v_4 = (-2,-1-1)$. If I take the cross product of each pair of vectors, I'm gonna have the normal vectors to the planes $\vec n_1, \vec n_2$. Then, I suppose that exists the direction vector of the line that passes by $A$ and $B$, let's call this vector $\vec v$. I then, can take the dot product of this vector with $\vec n_1$ and $\vec n_2$, so I'll find the vector in question. But then, for the line, I still need to find a point of it, so I can create the vector equation. How could I do it?
Also, is there another way of doing this, without determining the line equation from the vectors? (I really need another way too, so I can understand it better)