Đáp án:
\(\dfrac{\sqrt{210}}{15}a\)
Giải thích các bước giải:
Tính \(d(O;(SBC))\)
Từ O kẻ \(OK \perp BC\)
Ta có:
\(\left\{\begin{matrix} BC \perp OK
& & \\ BC \perp SO
& &
\end{matrix}\right.\)
\(\Rightarrow BC \perp (SOK)\)
Kẻ OF \(\perp SK\)
\(\left\{\begin{matrix} OF \perp BC
& & \\ OF \perp SK
& &
\end{matrix}\right.\)
\(\Rightarrow OF \perp (SBC)\)
Vậy \(d(O,(SBC))=OF\)
\(AC=\sqrt{a^{2}+a^{2}}=\sqrt{2}a\)
\(SO=\sqrt{4a^{2}-(\dfrac{\sqrt{2}}{2}a)^{2}}=\dfrac{\sqrt{14}}{2}a\)
\(OK=\dfrac{1}{2}BC=\dfrac{a}{2}\)
Ta có: \(\dfrac{1}{OF^{2}}=\dfrac{1}{SO^{2}}+\dfrac{1}{OK^{2}}\)
\(\Leftrightarrow OF=\dfrac{\sqrt{210}}{30}a\)
Ta có: \(AO \bigcap (SBC)=C\)
Ta có: \(\dfrac{AC}{OC}=\dfrac{d(A,(SBC))}{d(O,(SBC))}\)
\(\Leftrightarrow 2=\dfrac{d(A,(SBC))}{\dfrac{\sqrt{210}}{30}a}\)
\(\Leftrightarrow d(A,(SBC))=\dfrac{\sqrt{210}}{15}a\)
