1. In an isosceles trapezoid $A B C D$, given $A D / / B C, B C=2 A D=4, \angle A B C=60^{\circ}, \overrightarrow{C E}=\frac{1}{3} \overrightarrow{C D}$, then $\overrightarrow{C A} \cdot \overrightarrow{B E}=$
A. $-\frac{21}{2}$
B. -10
C. $-\frac{19}{2}$
D. -9
Given that the midpoint of $B C$ is $F$, then $A D \cong F C$
$\Rightarrow A D C F$ is a parallelogram.
Thus, $\overrightarrow{B E}=\overrightarrow{B C}+\frac{1}{3} \overrightarrow{C D}=\overrightarrow{B C}+\frac{1}{3} \overrightarrow{F A}$
$=\overrightarrow{B C}+\frac{1}{3}(\overrightarrow{B A}-\overrightarrow{B F})=\frac{1}{3} \overrightarrow{B A}+\frac{5}{3} \overrightarrow{B F}$,
$\overrightarrow{C A}=\overrightarrow{B A}-\overrightarrow{B C}=\overrightarrow{B A}-2 \overrightarrow{B F}$, hence $\overrightarrow{C A} \cdot \overrightarrow{B E}=\frac{1}{3}(\overrightarrow{B A}-2 \overrightarrow{B F})(\overrightarrow{B A}+5 \overrightarrow{B F})$
$=\frac{1}{3}\left(\overrightarrow{B A}^{2}-10 \overrightarrow{B F}^{2}+3 \overrightarrow{B A} \cdot \overrightarrow{B F}\right)=-10$, so the answer is $B$.