6. In $\triangle A B C$, $\angle B A C=120^{\circ}, A B=2, A C=1, D$ is a point on side $B C$ (including endpoints), then the range of $\overrightarrow{A D} \cdot \overrightarrow{B C}$ is
A. $[2,5]$
B. $[-5,3]$
C. $[0,5]$
D. $[-5,2]$
Given $\overrightarrow{B D}=x \overrightarrow{B C} \Rightarrow \overrightarrow{A D}=x \overrightarrow{A C}+(1-x) \overrightarrow{A B}, x \in[0,1]$,
thus $\overrightarrow{A D} \cdot \overrightarrow{B C}=[x \overrightarrow{A C}+(1-x) \overrightarrow{A B}](\overrightarrow{A C}-\overrightarrow{A B})=x-4(1-x)-(1-2 x)=7 x-5 \in[-5,2]$, so the answer is $D$.