Let $l\subset\mathbb{P}^3$. How many planes contain $l$? What is the equation of a plane $\pi$ such that $l\subset\pi$?
I am studying Alegbraic Geometry and I am doing some exercises, but I still have some difficult to visualize this. I know that a line in $\mathbb{P}^3$ has the parametric equation
$ l: \begin{cases} x_0 =a_{0}s+b_{0}t \\ x_1 =a_{1}s+b_{1}t \\ x_2 =a_{2}s+b_{2}t \\ x_3 =a_{3}s+b_{3}t \end{cases} $
and this is the projectivization of the plane $\operatorname{span}(a,b)$. A point $p$ is a line in $\mathbb{K}^4$. Now, in $\mathbb{K}^4$ I now that there exists an unique plain that contains a line and a point -if the point is not in the line-, but I don't know how to proceed in $\mathbb{P}^{3}$. Can someone help me? Thanks before!
Instead of thinking about planes containing the line, you might think about planes containing the points $a$ and $b$. A point lies on a plane if the inner product of their coordinate vectors is zero. So you are looking for $\pi_i$ such that
$$\sum_{i=0}^4\pi_ia_i= \sum_{i=0}^4\pi_ib_i=0$$
So that's two linear equations in four unknown coordinates. Which leaves you with a two-dimensional space of solutions. You might find that as the kernel of a matrix, using Gaussian elimination or some similar technique.
Of course the coordinates of the plane are again homogeneous, so multiples of the same coordinate vector represent the same plane. Therefore you have one real degree of freedom without geometric relevance, and one real degree of freedom accounting for different planes.
Essentially you can pick any two linearly independent vectors as the basis of the solution space. Then any plane through your two points will be a linear combination of these two planes, similar to how every point on the line is a linear combination of the two points $a$ and $b$.