Chọn B
Cách 1. Gọi $N$ là trung điểm $AD$ ta có: $MN\text{//}AC$ $\Rightarrow \left( \,\widehat{AC;BM} \right)=\,\left( \,\widehat{MN;BM} \right)$. Ta tính góc$\,\widehat{BMN}$. Ta có: $BM=BN=\dfrac{a\sqrt{3}}{2}$ (trung tuyến tam giác đều).
$MN=\dfrac{AC}{2}=\dfrac{a}{2}$.
Áp dụng định lý cosin cho $\Delta BMN$, ta được:
$\cos \,\widehat{BMN}=\dfrac{B{{M}^{2}}+M{{N}^{2}}-B{{N}^{2}}}{2BM.MN}=\dfrac{MN}{2BM}=\dfrac{\sqrt{3}}{6}>0$.
Vậy $\cos \left( \,\widehat{AC;BM} \right)=\dfrac{\sqrt{3}}{6}.$
Cách 2. $\cos \varphi =\left| \cos \left( \overrightarrow{AC},\overrightarrow{BM} \right) \right|=\dfrac{\left| \overrightarrow{AC}.\overrightarrow{BM} \right|}{\left| \overrightarrow{AC} \right|.\left| \overrightarrow{BM} \right|}=\dfrac{\left| \overrightarrow{AC}.\left( \overrightarrow{CM}-\overrightarrow{CB} \right) \right|}{a.\dfrac{a\sqrt{3}}{2}}$
$=\dfrac{\left| \overrightarrow{AC}.\overrightarrow{CM}-\overrightarrow{AC}.\overrightarrow{CB} \right|}{\dfrac{{{a}^{2}}\sqrt{3}}{2}}=\dfrac{\left| a.\dfrac{a}{2}\cos {{120}^{0}}-a.a.\cos {{120}^{0}} \right|}{\dfrac{{{a}^{2}}\sqrt{3}}{2}}=\dfrac{\left| -\dfrac{{{a}^{2}}}{4}+\dfrac{{{a}^{2}}}{2} \right|}{\dfrac{{{a}^{2}}\sqrt{3}}{2}}=\dfrac{\dfrac{{{a}^{2}}}{4}}{\dfrac{{{a}^{2}}\sqrt{3}}{2}}=\dfrac{\sqrt{3}}{6}$.
