(1) 右図から A を原点として方向と大きさを考えると AB→=2(1,0)AC→=2(1,1)\overrightarrow{AB} = 2(1,0) \qquad \overrightarrow{AC} = 2(1,1)AB=2(1,0)AC=2(1,1) これから AB→⋅AC→=2(1,0)⋅2(1,1)=4\overrightarrow{AB} \cdot \overrightarrow{AC} = 2(1,0) \cdot 2(1,1) = 4AB⋅AC=2(1,0)⋅2(1,1)=4 (2) AB→=2(1,0)AD→=2(0,1)\overrightarrow{AB} = 2(1,0) \qquad \overrightarrow{AD} = 2(0,1)AB=2(1,0)AD=2(0,1) これから AB→⋅AD→=2(1,0)⋅2(0,1)=0\overrightarrow{AB} \cdot \overrightarrow{AD} = 2(1,0) \cdot 2(0,1) = 0AB⋅AD=2(1,0)⋅2(0,1)=0 (3) AC→=2(1,1)CB→=2(0,−1)\overrightarrow{AC} = 2(1,1) \qquad \overrightarrow{CB} = 2(0,-1)AC=2(1,1)CB=2(0,−1) これから AC→⋅CB→=2(1,1)⋅2(0,−1)=−4\overrightarrow{AC} \cdot \overrightarrow{CB} = 2(1,1) \cdot 2(0,-1) = -4AC⋅CB=2(1,1)⋅2(0,−1)=−4