Do tam giác SAB đều \(\Rightarrow SI\perp AB\Rightarrow SI\perp\left(ABC\right)\)
\(SI=\frac{a\sqrt{3}}{2}\) ; \(BC=AB.tan60^0=a\sqrt{3}\)
\(V_{S.ABC}=\frac{1}{3}SI.\frac{1}{2}AB.BC=\frac{a^3}{4}\)
Qua S kẻ đường thẳng vuông góc SB cắt AB kéo dài tại P
\(\Rightarrow IH//SP\Rightarrow IH//\left(SCP\right)\Rightarrow d\left(IH;SC\right)=d\left(IH;\left(SCP\right)\right)=d\left(I;\left(SCP\right)\right)\)
\(BP=\frac{SB}{cos60^0}=2a\)
Từ I kẻ \(IQ\perp CP\) , từ \(I\) kẻ \(IK\perp SQ\Rightarrow IK\perp\left(SCP\right)\)
\(\Rightarrow IK=d\left(I;\left(SCP\right)\right)\)
\(IP=BP-IB=\frac{3a}{2}\) ; \(IQ=IP.sin\widehat{BPC}=IP.\frac{BC}{CP}=\frac{IP.BC}{\sqrt{BP^2+BC^2}}=\frac{3a\sqrt{21}}{14}\)
\(\frac{1}{IK^2}=\frac{1}{SI^2}+\frac{1}{IQ^2}\Rightarrow IK=\frac{SI.IQ}{\sqrt{SI^2+IQ^2}}=\frac{3a\sqrt{3}}{8}\)
