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Câu 2. a) Ta có: - $\overrightarrow{AB} \cdot \overrightarrow{BC} = |\overrightarrow{AB}| \cdot |\overrightarrow{BC}| \cdot \cos(90^\circ) = a \cdot a \cdot 0 = 0$ - $\overrightarrow{AB} \cdot \overrightarrow{BD} = |\overrightarrow{AB}| \cdot |\overrightarrow{BD}| \cdot \cos(45^\circ) = a \cdot a\sqrt{2} \cdot \frac{\sqrt{2}}{2} = a^2$ - $(\overrightarrow{AB} + \overrightarrow{AD}) \cdot (\overrightarrow{BD} + \overrightarrow{BC}) = \overrightarrow{AB} \cdot \overrightarrow{BD} + \overrightarrow{AB} \cdot \overrightarrow{BC} + \overrightarrow{AD} \cdot \overrightarrow{BD} + \overrightarrow{AD} \cdot \overrightarrow{BC} = a^2 + 0 + 0 + a^2 = 2a^2$ - $(\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD}) \cdot (\overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC}) = \overrightarrow{AB} \cdot \overrightarrow{DA} + \overrightarrow{AB} \cdot \overrightarrow{DB} + \overrightarrow{AB} \cdot \overrightarrow{DC} + \overrightarrow{AC} \cdot \overrightarrow{DA} + \overrightarrow{AC} \cdot \overrightarrow{DB} + \overrightarrow{AC} \cdot \overrightarrow{DC} + \overrightarrow{AD} \cdot \overrightarrow{DA} + \overrightarrow{AD} \cdot \overrightarrow{DB} + \overrightarrow{AD} \cdot \overrightarrow{DC} = 0 + a^2 + 0 + 0 + 0 + a^2 + 0 + 0 + 0 = 2a^2$ b) Ta có: - $\overrightarrow{ON} \cdot \overrightarrow{AB} = |\overrightarrow{ON}| \cdot |\overrightarrow{AB}| \cdot \cos(45^\circ) = \frac{a}{2} \cdot a \cdot \frac{\sqrt{2}}{2} = \frac{a^2\sqrt{2}}{4}$ - $\overrightarrow{NA} \cdot \overrightarrow{AB} = |\overrightarrow{NA}| \cdot |\overrightarrow{AB}| \cdot \cos(135^\circ) = (a - BN) \cdot a \cdot (-\frac{\sqrt{2}}{2}) = -\frac{a(a - BN)\sqrt{2}}{2}$ c) Ta có: - $\overrightarrow{MA} \cdot \overrightarrow{MB} + \overrightarrow{MC} \cdot \overrightarrow{MD} = MA \cdot MB \cdot \cos(\angle AMB) + MC \cdot MD \cdot \cos(\angle CMD)$ - Vì M nằm trên đường tròn nội tiếp hình vuông nên $\angle AMB = \angle CMD = 90^\circ$ và $MA = MB = MC = MD = \frac{a\sqrt{2}}{2}$ - Do đó, $\overrightarrow{MA} \cdot \overrightarrow{MB} + \overrightarrow{MC} \cdot \overrightarrow{MD} = \frac{a\sqrt{2}}{2} \cdot \frac{a\sqrt{2}}{2} \cdot 0 + \frac{a\sqrt{2}}{2} \cdot \frac{a\sqrt{2}}{2} \cdot 0 = 0$ Đáp số: a) $\overrightarrow{AB} \cdot \overrightarrow{BC} = 0$, $\overrightarrow{AB} \cdot \overrightarrow{BD} = a^2$, $(\overrightarrow{AB} + \overrightarrow{AD}) \cdot (\overrightarrow{BD} + \overrightarrow{BC}) = 2a^2$, $(\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD}) \cdot (\overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC}) = 2a^2$ b) $\overrightarrow{ON} \cdot \overrightarrow{AB} = \frac{a^2\sqrt{2}}{4}$, $\overrightarrow{NA} \cdot \overrightarrow{AB} = -\frac{a(a - BN)\sqrt{2}}{2}$ c) $\overrightarrow{MA} \cdot \overrightarrow{MB} + \overrightarrow{MC} \cdot \overrightarrow{MD} = 0$