Cho hình lập phương \(ABCD \cdot A'B'C'D'\). Xác định góc \(\left( {\overrightarrow {AB} ,\overrightarrow {A'D'} } \right),\left( {\overrightarrow {AB} ,\overrightarrow {A'C'} } \right)\).

Đặt \(\overrightarrow {OA}  = \vec a,\overrightarrow {OB}  = \vec b,\overrightarrow {OC}  = \vec c\).

Khi đó, \(\left| {\vec a\left|  =  \right|\vec b\left|  =  \right|\vec c} \right| = 1\) và \(\vec a \cdot \vec b = \vec a \cdot \vec c = \vec b \cdot \vec c = 0\).

Ta có: \({\rm{cos}}\left( {\overrightarrow {OM} ,\overrightarrow {AC} } \right) = \frac{{\overrightarrow {OM}  \cdot \overrightarrow {AC} }}{{\left| {\overrightarrow {OM} \left|  \cdot  \right|\overrightarrow {AC} } \right|}}\).

Mặt khác, do \(\overrightarrow {OM}  = \frac{1}{2}\left( {\overrightarrow {OA}  + \overrightarrow {OB} } \right) = \frac{1}{2}\left( {\vec a + \vec b} \right)\) và \(\overrightarrow {AC}  = \overrightarrow {OC}  - \overrightarrow {OA}  = \vec c - \vec a\) nên \(\overrightarrow {OM}  \cdot \overrightarrow {AC}  = \frac{1}{2}\left( {\vec a + \vec b} \right) \cdot \left( {\vec c - \vec a} \right)\) \( = \frac{1}{2}\left( {\vec a \cdot \vec c - {{\vec a}^2} + \vec b \cdot \vec c - \vec b \cdot \vec a} \right) =  - \frac{1}{2}.\)

Ta lại có: \(\left| {\overrightarrow {OM} \left| { = OM = \frac{{\sqrt 2 }}{2};} \right|\overrightarrow {AC} } \right| = AC = \sqrt 2 \).

Do đó, \({\rm{cos}}\left( {\overrightarrow {OM} ,\overrightarrow {AC} } \right) = \frac{{\frac{{\overrightarrow {OM} }}{{\overrightarrow {AC} }} \cdot \overrightarrow {AC} }}{{\left| {\overrightarrow {OM} \left|  \cdot  \right|\overrightarrow {AC} } \right|}} = \frac{{\frac{{ - 1}}{2}}}{{\frac{{\sqrt 2 }}{2} \cdot \sqrt 2 }} = \frac{{ - 1}}{2}\).

Vậy \(\left( {\overrightarrow {OM} ,\overrightarrow {AC} } \right) = {120^ \circ }\).

Trả lời: \(n =  - 0,5\)

Vì \(MN//{A^\prime }{C^\prime }\) nên $\left(\overrightarrow{MN},\overrightarrow{C^{\text{'}}B}\right)=\left(\overrightarrow{A^{\text{'}}{C}^{\text{'}}},\overrightarrow{C^{\text{'}}B}\right)={180}^{°}-\widehat{A^{\text{'}}{C}^{\text{'}}B}={120}^{°}$.

Ta có: \(MN = \frac{{a\sqrt 2 }}{2},{C^\prime }B = a\sqrt 2 \). Suy ra $\overrightarrow{MN}\cdot \overrightarrow{C^{\text{'}}B}=\left|\overrightarrow{MN}\right|\cdot \left|\overrightarrow{C^{\text{'}}B}\right|\cdot \cos \left(\overrightarrow{MN},\overrightarrow{C^{\text{'}}B}\right)=\frac{a\sqrt{2}}{2}\cdot a\sqrt{2}\cdot \cos {120}^{°}=-\mathrm{0,5}{a}^2$

Vậy \(n =  - 0,5\).