Cho hình hộp ABCD.A’B’C’D’ có AB=3, AD=4, AA’=5. Biết góc BAD = 90°, góc BAA’ = góc DAA’ = 60°. Số đo góc giữa vectơ AC’ và vectơ D’C bằng bao nhiêu độ?

Đáp án: 130

\[\begin{array}{l}\overrightarrow {AC'} {\,^2} = \,{\left( {\overrightarrow {AB}  + \overrightarrow {AD}  + \overrightarrow {AA'} } \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\, = \,A{B^2} + A{D^2} + A{{A'}^2} + 2AB.AD.\cos 90^\circ  + 2AB.AA'.\cos 60^\circ  + 2AD.AA'.\cos 60^\circ \end{array}\]

         \[ = 9 + 16 + 25 + 15 + 20 = \,85\, \Rightarrow \,AC' = \sqrt {85} \]

\[{\overrightarrow {D'C} ^2} = {\overrightarrow {A'B} ^2} = A'{B^2} = A{A'^2} + A{B^2} - 2.AA'.AB.\cos 60^\circ  = \,25 + 9 - 15 = 19\,\]\[ \Rightarrow \,D'C = \sqrt {19} \]

\[\overrightarrow {AC'} .\,\overrightarrow {D'C} \,\, = \,\,\left( {\overrightarrow {AC}  + \overrightarrow {AA'} } \right)\left( {\overrightarrow {AC}  - \overrightarrow {AD'} } \right) = \left( {\overrightarrow {AC}  + \overrightarrow {AA'} } \right)\left( {\overrightarrow {AC}  - \overrightarrow {AA'}  - \overrightarrow {AD} } \right)\]

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\left( {\overrightarrow {AC}  + \overrightarrow {AA'} } \right)\left( {\overrightarrow {AC}  - \overrightarrow {AA'} } \right) - \,\left( {\overrightarrow {AC}  + \overrightarrow {AA'} } \right)\overrightarrow {AD} \]

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,A{C^2} - A{A'^2} - \overrightarrow {AC} .\overrightarrow {AD}  - \overrightarrow {AA'} .\overrightarrow {AD} \]

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,25\, - 25\, - \,AC\,.\,AD\,.\,\frac{{AD}}{{AC}} - AA'.AD.\cos 60^\circ \]

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \, - 16 - 10 =  - 26\]

\[\cos \left( {\overrightarrow {AC'} ,\overrightarrow {D'C} } \right) = \frac{{\overrightarrow {AC'} .\,\overrightarrow {D'C} }}{{AC'.\,D'C}} =  - \frac{{26}}{{\sqrt {85} .\sqrt {19} }}\,\].

\[ \Rightarrow \,\left( {\overrightarrow {AC'} ,\overrightarrow {D'C} } \right) \approx 130^\circ \,\]