Cho tứ diện đều S.ABC là trung...

A

- Hướng dẫn giải

Đáp án A

$\begin{array}{l}\overrightarrow{DM}=\frac{1}{2}\left(\overrightarrow{DB}+\overrightarrow{DC}\right)\\ =\frac{1}{2}\left(\overrightarrow{AB}-\overrightarrow{AD}+\overrightarrow{AC}-\overrightarrow{AD}\right)\\ =\frac{1}{2}\overrightarrow{AB}-\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AC}\\ \overrightarrow{AB}.\overrightarrow{DM}=\frac{1}{2}{\overrightarrow{AB}}^2-\overrightarrow{AB}.\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}.\overrightarrow{AC}\\ =\frac{1}{2}{a}^2-a.a.\cos {60}^0+\frac{1}{2}a.a.\cos {60}^0\\ =\frac{1}{4}{a}^2\\ \Rightarrow a.\frac{a\sqrt{3}}{2}\cos \left(\overrightarrow{AB};\overrightarrow{DM}\right)=\frac{1}{4}{a}^2\\ \iff \cos \left(\overrightarrow{AB};\overrightarrow{DM}\right)=\frac{\sqrt{3}}{6}\\ \Rightarrow \cos \left(AB;DM\right)=\frac{\sqrt{3}}{6}\end{array}$