Cho biết hình lập phương \(ABCD.ABCD\) có tâm \(O.

Gọi hình lập phương có cạnh là \(a.\)

Xét hệ trục tọa độ như hình vẽ ta có:

\(\begin{array}{l}
B'\left( {0;\;0;\;0} \right),\;\;A'\left( {a;\;0;\;0} \right),\;C'\left( {0;\;a;\;0} \right),\;D'\left( {a;\;a;\;0} \right),\\
A\left( {a;\;0;\;a} \right),\;I\left( {\frac{a}{2};\;\frac{a}{2};\;0} \right),\;B\left( {0;\;0;\;a} \right),\;O\left( {\frac{a}{2};\;\frac{a}{2};\;\frac{a}{2}} \right).\\
\Rightarrow \overrightarrow {OI} = \left( {0;\;0;\;\frac{a}{2}} \right) \Rightarrow \overrightarrow {OM} = \frac{2}{3}\overrightarrow {OI} = \left( {0;\;0;\;\frac{a}{6}} \right).\\
\Rightarrow \left\{ \begin{array}{l}
{x_M} - {x_O} = 0\\
{y_M} - {y_O} = 0\\
{z_M} - {z_O} = \frac{a}{6}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_M} = \frac{a}{2}\\
{y_M} = \frac{a}{2}\\
{z_M} = \frac{{2a}}{3}
\end{array} \right. \Rightarrow M\left( {\frac{a}{2};\;\frac{a}{2};\;\frac{{2a}}{3}} \right).
\end{array}\)

\(\begin{align}  & \Rightarrow \overrightarrow{MA}=\left( \frac{a}{2};\ -\frac{a}{2};\ \frac{a}{3} \right),\ \overrightarrow{MB}=\left( -\frac{a}{2};\ -\frac{a}{2};\ \frac{a}{3} \right),\ \overrightarrow{MC'}=\left( -\frac{a}{2};\ \frac{a}{2};-\frac{2a}{3} \right),\ \overrightarrow{MD'}=\left( \frac{a}{2};\ \frac{a}{2};-\frac{2a}{3} \right). \\ & \Rightarrow {{\overrightarrow{n}}_{\left( MAB \right)}}=\left[ \overrightarrow{MA},\ \overrightarrow{MB} \right]=\left( 0;-\frac{{{a}^{2}}}{3};-\frac{{{a}^{2}}}{2} \right)=-{{a}^{2}}\left( 0;\ \frac{1}{3};\ \frac{1}{2} \right). \\ & {{\overrightarrow{n}}_{\left( MC'D' \right)}}=\left[ \overrightarrow{MC'},\ \overrightarrow{MD'} \right]=\left( 0;-\frac{2{{a}^{2}}}{3};-\frac{{{a}^{2}}}{2} \right)=-{{a}^{2}}\left( 0;\ \frac{2}{3};\ \frac{1}{2} \right). \\\end{align}\)

Gọi \(\alpha \) là góc giữa hai mặt phẳng \(\left( MAB \right)\) và \(\left( M'C'D' \right).\)

\(\begin{align}  & \Rightarrow \cos \alpha =\frac{\left| {{\overrightarrow{n}}_{\left( MAB \right)}}.{{\overrightarrow{n}}_{\left( MC'D' \right)}} \right|}{\left| {{\overrightarrow{n}}_{\left( MAB \right)}} \right|.\left| {{\overrightarrow{n}}_{\left( MC'D' \right)}} \right|}=\frac{\left| \frac{1}{3}.\frac{2}{3}+\frac{1}{2}.\frac{1}{2} \right|}{\sqrt{\frac{1}{9}+\frac{1}{4}}.\sqrt{\frac{4}{9}+\frac{1}{4}}}=\frac{17\sqrt{13}}{65}. \\ & \Rightarrow sin\alpha =\sqrt{1-{{\cos }^{2}}\alpha }=\frac{6\sqrt{13}}{65}. \\\end{align}\)

Chọn D.