3. Given $\overrightarrow{O P}=(2,1), \overrightarrow{O A}=(1,7), \overrightarrow{O B}=(5,1)$, let $X$ be a point on the line $O P$ ($O$ being the origin), then the value of $\angle A X B$ when $\overrightarrow{X A} \cdot \overrightarrow{X B}$ is minimized is
A. $90^{\circ}$
B. $\arccos \frac{4 \sqrt{17}}{17}$
C. $\arccos \left(-\frac{4 \sqrt{17}}{17}\right)$
D. $\pi+\arccos \left(-\frac{4 \sqrt{17}}{17}\right)$
3. $\mathrm{C}$ hint, Let $\overrightarrow{O X}=\left(x_{0}, y_{0}\right), \overrightarrow{O P}=(2,1)$, then $\frac{x_{0}}{2}=\frac{y_{0}}{1}$, so $x_{0}=2 y_{0}$, then $\overrightarrow{X A}=$ $\left(1-2 y_{0}, 7-y_{0}\right), \overrightarrow{X B}=\left(5-2 y_{0}, 1-y_{0}\right) \quad$ so $\quad \overrightarrow{X A} \cdot \overrightarrow{X B}=\left(5-2 y_{0}\right)\left(1-2 y_{0}\right)+\left(1-y_{0}\right)(7-$ $\left.y_{0}\right)=5 y_{0}^{2}-20 y_{0}+12$. Therefore, when $y_{0}=2$, $\overrightarrow{X A} \cdot \overrightarrow{X B}$ is minimized, at this time, $\overrightarrow{O X}=(4,2), \overrightarrow{X A}=(-3,5), \overrightarrow{X B}=(1,-$ 1), $|\overrightarrow{X A}|=\sqrt{34},|\overrightarrow{X B}|=\sqrt{2}$, thus, $\cos \angle A X B=\frac{\overrightarrow{X A} \cdot \overrightarrow{X B}}{|\overrightarrow{X A}| \cdot|\overrightarrow{X B}|}=-\frac{4 \sqrt{17}}{17}$.