giúp tui với

Câu 2: a) Ta có: \[ \overrightarrow{AC} + \overrightarrow{BD} = (\overrightarrow{AD} + \overrightarrow{DC}) + (\overrightarrow{BA} + \overrightarrow{AD}) = \overrightarrow{AD} + \overrightarrow{DC} + \overrightarrow{BA} + \overrightarrow{AD} = \overrightarrow{AD} + \overrightarrow{BA} + \overrightarrow{DC} + \overrightarrow{AD} = \overrightarrow{AD} + \overrightarrow{BC} \] b) Ta có: \[ \overrightarrow{AB} - \overrightarrow{CD} = \overrightarrow{AB} + (-\overrightarrow{CD}) = \overrightarrow{AB} + \overrightarrow{DC} = \overrightarrow{AC} + \overrightarrow{DB} \] Câu 3: Gọi G là trọng tâm tam giác AB'D', ta có: \[ \overrightarrow{A'G} = \frac{1}{3} (\overrightarrow{A'A} + \overrightarrow{A'B'} + \overrightarrow{A'D'}) \] Mặt khác, ta có: \[ \overrightarrow{A'C} = \overrightarrow{A'A} + \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CC'} \] Do đó: \[ \overrightarrow{A'C} = \overrightarrow{A'A} + \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CC'} = \overrightarrow{A'A} + \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CC'} = 3 \overrightarrow{A'G} \] Câu 1: a) Ta có: \[ \overrightarrow{A'B} = \overrightarrow{A'A} + \overrightarrow{AB} \] \[ \overrightarrow{D'C'} = \overrightarrow{D'D} + \overrightarrow{DC} \] Do đó: \[ \overrightarrow{A'B} \cdot \overrightarrow{D'C'} = (\overrightarrow{A'A} + \overrightarrow{AB}) \cdot (\overrightarrow{D'D} + \overrightarrow{DC}) = \overrightarrow{A'A} \cdot \overrightarrow{D'D} + \overrightarrow{A'A} \cdot \overrightarrow{DC} + \overrightarrow{AB} \cdot \overrightarrow{D'D} + \overrightarrow{AB} \cdot \overrightarrow{DC} = 0 + 0 + 0 + a^2 = a^2 \] \[ \overrightarrow{D'A} = \overrightarrow{D'D} + \overrightarrow{DA} \] \[ \overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} \] Do đó: \[ \overrightarrow{D'A} \cdot \overrightarrow{BC} = (\overrightarrow{D'D} + \overrightarrow{DA}) \cdot (\overrightarrow{BA} + \overrightarrow{AC}) = \overrightarrow{D'D} \cdot \overrightarrow{BA} + \overrightarrow{D'D} \cdot \overrightarrow{AC} + \overrightarrow{DA} \cdot \overrightarrow{BA} + \overrightarrow{DA} \cdot \overrightarrow{AC} = 0 + 0 + 0 + a^2 = a^2 \] b) Ta có: \[ (\overrightarrow{A'D}, \overrightarrow{B'C'}) = 90^\circ \] \[ (\overrightarrow{AD'}, \overrightarrow{BD}) = 90^\circ \]