1.
Dựng \(\overrightarrow{DB'}=\overrightarrow{CB}\)
\(k\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{DB}\)
\(=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DA}+\overrightarrow{AB}\)
\(=2\overrightarrow{AB}+\overrightarrow{B'D}+\overrightarrow{DA}\)
\(=2\overrightarrow{AB}+\overrightarrow{B'A}\)
\(=2\overrightarrow{AB}+2\overrightarrow{AB}=4\overrightarrow{AB}\)
\(\Rightarrow k=4\)
Gọi M là trung điểm IB
\(\left|\overrightarrow{AB}+\overrightarrow{AI}\right|=\left|2\overrightarrow{AM}\right|=2AM\)
Ta có \(\overrightarrow{AM}^2=\left(\overrightarrow{MI}+\overrightarrow{IA}\right)^2=MI^2+IA^2-2MI.IA.cos90^o=\dfrac{1}{16}a^2+\dfrac{3}{4}a^2=\dfrac{13}{16}a^2\)
\(\Rightarrow AM=\dfrac{\sqrt{13}}{4}a\Rightarrow\left|\overrightarrow{AB}+\overrightarrow{AI}\right|=\dfrac{\sqrt{13}}{2}a\)