\(V_{OABC}=\dfrac{1}{6}OA.OB.OC=\dfrac{2a^3}{3}\)
\(AC=\sqrt{OA^2+AC^2}=a\sqrt{5}\)
Trong tam giác vuông OAC: \(OA^2=AH.AC\Rightarrow AH=\dfrac{OA^2}{AC}=\dfrac{OA^2}{\sqrt{OA^2+OC^2}}=\dfrac{a\sqrt{5}}{5}\)
\(\Rightarrow\dfrac{AH}{AC}=\dfrac{1}{5}\)
Do I là trung điểm AB \(\Rightarrow\dfrac{AI}{AB}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{V_{AOIH}}{V_{OABC}}=\dfrac{\dfrac{1}{3}d\left(O;\left(ABC\right)\right).\dfrac{1}{2}AH.AI.sin\widehat{BAC}}{\dfrac{1}{3}d\left(O;\left(ABC\right)\right).\dfrac{1}{2}AC.AB.sin\widehat{BAC}}=\dfrac{AH}{AC}.\dfrac{AI}{AB}=\dfrac{1}{10}\)
\(\Rightarrow V_{AOIH}=\dfrac{1}{10}.\dfrac{2a^3}{3}=\dfrac{a^3}{15}\)