Đáp án:
$d(M;(SBC)) =\dfrac{a\sqrt3}{3}$
Giải thích các bước giải:
Ta có:
$CB\perp AB$
$SA\perp CB\quad (SA\perp (ABC))$
$\Rightarrow CB\perp (SAB)$
$\Rightarrow CB\perp SB$
Xét $(SBC)$ và $(ABC)$ có:
$\begin{cases}(SBC)\cap (ABC)=CB\\SB\perp BC;\, SB\subset (SBC)\\AB\perp CB;\, AB\subset (ABC)\end{cases}$
$\Rightarrow \widehat{((SBC);(ABC))}=\widehat{SBA}=30^o$
$\Rightarrow \begin{cases}SA = AB.\tan30^o = a\\SB = \dfrac{AB}{\cos30^o}=2a\end{cases}$
$\Rightarrow S_{SBC}=\dfrac{1}{2}SB.BC =a^2$
Ta có:
$V_{S.ABC}=\dfrac{1}{3}S_{ABC}.SA =\dfrac{1}{6}AB.BC.SA =\dfrac{a^3\sqrt3}{6}$
Ta lại có:
$V_{S.ABC}=V_{A.SBC}=\dfrac{1}{3}S_{SBC}.d(A;(SBC))$
$\Rightarrow d(A;(SBC)) = \dfrac{3V_{S.ABC}}{S_{SBC}}=\dfrac{\dfrac{a^3\sqrt3}{2}}{a^2}=\dfrac{a\sqrt3}{2}$
Mặt khác:
$\dfrac{d(M;(SBC))}{d(A;(SBC))}=\dfrac{CM}{CA}=\dfrac{2}{3}$
$\Rightarrow d(M;(SBC)) =\dfrac{2}{3}d(A;(SBC))=\dfrac{a\sqrt3}{3}$
