Đáp án C

Chọn C

Giả sử tứ diện đều \(ABCD\) cạnh \(a.\)
\( \Rightarrow AM = DM = \sqrt {M{C^2} + C{D^2} - 2MC.CD.\cos \widehat {MCD}} = \dfrac{{a\sqrt {13} }}{4}\)
\(\cos \widehat {BAM} = \dfrac{{A{B^2} + A{M^2} - M{B^2}}}{{2AB.AM}} = \dfrac{{5\sqrt {13} }}{{26}}\)
Ta có: \(\cos \left( {\overrightarrow {AB},\overrightarrow {DM} } \right) = \dfrac{{\overrightarrow {AB}.\overrightarrow {DM} }}{{AB.DM}} = \dfrac{{\overrightarrow {AB}.\overrightarrow {DM} }}{{a.\dfrac{{a\sqrt {13} }}{4}}} = \dfrac{{4\overrightarrow {AB}.\overrightarrow {DM} }}{{\sqrt {13} {a^2}}}\)
Mà \(\overrightarrow {AB}.\overrightarrow {DM} = \overrightarrow {AB}.\overrightarrow {AM} - \overrightarrow {AB}.\overrightarrow {AD} = AB.AM.\cos \left( {\overrightarrow {AB},\overrightarrow {AM} } \right) - AB.AD.\cos \left( {\overrightarrow {AB},\overrightarrow {AD} } \right) = \dfrac{1}{8}{a^2}\)
Vậy \(\cos \left( {\overrightarrow {AB},\overrightarrow {DM} } \right) = \dfrac{{\sqrt {13} }}{{26}} > 0\) nên \(\cos \left( {AB,DM} \right) = \dfrac{{\sqrt {13} }}{{26}}.\)