Xét tam giác vuông \(IAB:\)
\(IB^2=IA^2+AB^2=\left(\dfrac{3a}{2}\right)^2+16a^2=\dfrac{73a^2}{4}\)
\(\Rightarrow IB=\dfrac{a\sqrt{73}}{2}\)
\(tan\widehat{AIB}=\dfrac{AB}{IA}=\dfrac{4a}{\dfrac{3a}{2}}=\dfrac{8}{3}\)
\(\Rightarrow tan\widehat{DIB}=tan\left(180^o-\widehat{AIB}\right)=-tan\widehat{AIB}=-\dfrac{8}{3}\)
\(1+tan^2\widehat{DIB}=\dfrac{1}{cos^2\widehat{DIB}}\)
\(\Rightarrow cos^2\widehat{DIB}=\dfrac{1}{1+tan^2\widehat{DIB}}=\dfrac{1}{1+\dfrac{64}{9}}=\dfrac{9}{73}\)
\(\Rightarrow cos\widehat{DIB}=-\dfrac{3}{\sqrt{73}}\) (\(\widehat{DIB}\) thuộc góc phần tư số \(II\))
\(\Rightarrow cos\left(\widehat{\overrightarrow{IB};\overrightarrow{ID}}\right)=cos\widehat{DIB}=-\dfrac{3}{\sqrt{73}}\)
\(\left(\overrightarrow{IA}+\overrightarrow{IB}\right).\overrightarrow{ID}=\overrightarrow{IA}.\overrightarrow{ID}+\overrightarrow{IB}.\overrightarrow{ID}\)
\(=IA.ID.cos180^o+IB.ID.cos\left(\widehat{\overrightarrow{IB};\overrightarrow{ID}}\right)\)
\(=\dfrac{3a}{2}.\dfrac{3a}{2}.\left(-1\right)+\dfrac{a\sqrt{73}}{2}.\dfrac{3a}{2}.\left(-\dfrac{3}{\sqrt{73}}\right)\)
\(=-\dfrac{9a^2}{4}-\dfrac{9a^2}{4}=-\dfrac{9a^2}{2}\)
