Let $\pi:Y\rightarrow \mathbb{P}^3$ be the blow-up of two points $p,q\in\mathbb{P}^3$, and then of the strict transform of the line $L$ spanned by them. Now, Let $E\_p,E\_q, E\_{p,q}$ be respectively the exceptional divisors over $p,q$ and $L$. Therefore $E\_p,E\_q$ are isomorphic to the blow-up of a point in $\mathbb{P}^2$ and $E\_{p,q}\cong \mathbb{P}^1\times\mathbb{P}^1$. Let $\sigma\_{p,q}\subset E\_{p,q}$ be the class of the $\mathbb{P}^1$ which is not contracted by $\pi$. Therefore:


$$E\_p\cdot \sigma\_{p,q} = E\_q\cdot\sigma\_{p,q} = 1.$$


What is the intersection number $E\_{p,q}\cdot \sigma\_{p,q}$?



A question about an intersection number
If $E$ is the exceptional divisor of a smooth blowup, the restriction of $O(E)$ to $E$ is the relative $O(-1)$ for $E = P\_Z(N)$, where $Z$ is the center of the blowup and $N$ is the normal bundle. In your case $Z$ is the proper preimage of the line $L$ in the blowup of two points, consequently $N = O\_L(-1) \oplus O\_L(-1)$. This means that $O\_E(E)$ is the line bundle $O(-1,-1)$. Hence $E\cdot\sigma = -1$. 


