Equations plane in projective space

Question

suppose I have a point $(0:0:y:z)$ and a line described by $x_2=x_3=0$ in $\mathbb{P}^3(\mathbb{C})$. How do I write the equation of the plane that contains both? Then I want the intersection with the line described by $x_0=x_2, \space x_1=x_3$. I get the plane $x_2 t = x_3 s$ is it correct? What is the intersection?

Answer 1

The first line is described by as the intersection (meet) of the planes $x_2=0$ and $x_3=0$. Every plane that contains this line is a linear combination of these two planes, i.e., has an equation of the form $\lambda x_2+\mu x_3=0$, with $\lambda$ and $\mu$ not both zero. Substituting the coordinates of the point into this equation yields $\lambda y+\mu z=0$. We can take $\lambda=z$ and $\mu=-y$ to yield the equation $zx_2-yx_3=0$. Now you have a system of three linear equations for the line-plane intersection that I assume you know how to solve.

Using the method that I mentioned in my comment, if $\mathbf a$ and $\mathbf b$ are two planes, then their line of intersection has the dual Plücker matrix $L^* = \mathbf a\mathbf b^T-\mathbf b\mathbf a^T$. Further, if $\mathbf p$ is a point not on this line, then the plane spanned by the point and line is $L^*\mathbf p$. Here we have $\mathbf a = (0:0:1:0)$ and $\mathbf b = (0:0:0:1)$, so $L^*(0:0:y:z)=(0:0:z:-y)$. (It’s much less work to compute this directly from the above expression for $L^*$ instead of constructing the matrix $L^*$ explicitly first.) The intersection of the three planes is then a null vector of $$\begin{bmatrix}1&0&-1&0\\0&1&0&-1\\0&0&z&-y\end{bmatrix},$$ which can be pretty much read directly from this matrix: $(y:z:y:z)$.