Góc giữa SA và (SCA) là góc \(\widehat{SAH}=60^{\circ}\)
\(\frac{1}{AH^{2}}=\frac{1}{AB^{2}}+\frac{1}{AC^{2}}=\frac{1}{a^{2}}+\frac{1}{(a\sqrt{3})^{2}}=\frac{1}{2a^{2}}\Rightarrow AH=\frac{a\sqrt{3}}{2}\)
\(\Rightarrow SH=AH.\tan 60^{\circ}=\frac{3a}{2}\)
Từ đó \(V_{S.ABC}=\frac{1}{3}.SH.S_{\triangle ABC}=\frac{1}{3}.\frac{3a}{2}.\frac{1}{2}.a.a\sqrt{3}=\frac{a^{3}\sqrt{3}}{4}\)
Ta có \(CA^{2}=CH.CB\Rightarrow CH=\frac{CA^{2}}{CB}=\frac{(a\sqrt{3})^{2}}{2a}=\frac{3a}{2}.\) Từ đó
\(\frac{d(B;(SAC))}{d(H;(SAC))}=\frac{CB}{CH}=\frac{2a}{\frac{3a}{2}}=\frac{4}{3}\Rightarrow d(B;(SAC))=\frac{4}{3}d(H;(SAC))\)
Hạ \(HE\perp AC,HK\perp SE.\) Ta có \(HE\perp AC,HS\perp AC\Rightarrow AC\perp (SHE)\Rightarrow AC\perp HK\)
Từ đó \(\left\{\begin{matrix} HK\perp AC\\HK\perp SE \end{matrix}\right.\Rightarrow HK\perp (SAC).\) Do đó \(d(B;(SAC))=\frac{4}{3}d(H;(SAC))=\frac{4}{3}HK\)
Ta có: \(\frac{HE}{AB}=\frac{CH}{CB}=\frac{3}{4}\Rightarrow HE=\frac{3}{4}AB=\frac{3a}{4}.\) Từ đó
\(HK^{2}=\frac{HS^{2}.HE^{2}}{HS^{2}+HE^{2}}=\frac{(\frac{3a}{4})^{2}.(\frac{3a}{2})^{2}}{(\frac{3a}{4})^{2}+(\frac{3a}{2})^{2}}=\frac{9a^{2}}{20}\Rightarrow HK=\frac{3a\sqrt{5}}{10}\)
\(d(B;(SAC))=\frac{4}{3}HK=\frac{4}{3}.\frac{3a\sqrt{5}}{10}=\frac{2a\sqrt{5}}{5}\)
