已知向量(oldsymbol{a},oldsymbol{b},oldsymbol{c})满足(left | oldsymbol{a}
ight |=2),(oldsymbol{b}=oldsymbol{a}cdotoldsymbol{b}=3),(oldsymbol{c}=xoldsymbol{a}+yoldsymbol{b})((x>0,y>0)),若向量(oldsymbol{c}-2oldsymbol{a})与向量(oldsymbol{c}-dfrac{2}{3}oldsymbol{b})的夹角为(dfrac{pi}{3}),则(left | oldsymbol{c}-dfrac{1}{2}oldsymbol{a}
ight|)的取值范围是((qquad))
(mathrm{A}.left[sqrt{7}-2,sqrt{7}+2
ight]) (qquadmathrm{B}.left( 3,sqrt{7}+2
ight]) (qquadmathrm{C}.[1,3)) (qquadmathrm{D}.left(sqrt{3},sqrt{7}+2
ight])}
解析:
由题易知(oldsymbol{b}=3),且(oldsymbol{a},oldsymbol{b})的夹角为(dfrac{pi}{3}),若记$$ overrightarrow{OA}=oldsymbol{a}, overrightarrow{OB}=oldsymbol{b}, overrightarrow{OC}=oldsymbol{c},$$如图所示.
图中$$ overrightarrow{OA}=dfrac{1}{2} overrightarrow{OA_1}=2 overrightarrow{OA_2}, overrightarrow{OB}=dfrac{3}{2} overrightarrow{OB_1}=dfrac{3}{4} overrightarrow{OB_2}.$$ 则由题可知 $$left< overrightarrow{CA_1}, overrightarrow{CB_1} ight>=dfrac{pi}{3}.$$又因$C$点在角$angle AOB$内部,因此点$C$的轨迹是以$A_1B_2$为直径的一段圆弧$($不含端点$)$.所以$$ 3=| overrightarrow{A_2A_1}|<| overrightarrow{CA_2}|=left | oldsymbol{c}-dfrac{1}{2}oldsymbol{a} ight|leqslant |A_2E|+2=sqrt{7}+2.$$ 所以$left | oldsymbol{c}-dfrac{1}{2}oldsymbol{a} ight|$的取值范围为$left( 3,sqrt{7}+2 ight]$.