4. Given the parabola $C: y^{2}=4 x, F$ is the focus of $C$, a line $l$ passing through $F$ intersects the parabola $C$ at points $A$ and $B$, with a slope of 1, then the angle between vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ is $\qquad$ .
4. $\pi-\arctan \frac{4 \sqrt{2}}{3} \cdot l$ The equation is $y=x-1$, which is $1=x-y$, so $y^{2}=4 x \cdot 1=4 x(x-y)$. Dividing both sides by $x^{2}$ gives $\left(\frac{y}{x}\right)^{2}+4\left(\frac{y}{x}\right)-4=0$. By Vieta's formulas, $k_{C A}+k_{C B}=-4, k_{C A} \cdot k_{C B}=4$, thus $\tan \angle A O B=\frac{k_{O A}-k_{C B}}{1+k_{O A} \cdot k_{C B}}=$ $\frac{\sqrt{\left(k_{O A}+k_{C B}\right)^{2}-4 k_{O A} k_{Q B}}}{1+k_{O A} \cdot k_{C B}}=-\frac{4 \sqrt{2}}{3}$, so $\angle A O B=\pi-\arctan \frac{4 \sqrt{2}}{3}$.