Đáp án:
\(\dfrac{{6\sqrt 7 a}}{7}\)
Giải thích các bước giải:
Trong \(\left( {SBC} \right)\) kẻ \(SH \bot BC\,\,\left( {H \in BC} \right) \Rightarrow SH \bot \left( {ABC} \right)\).
Xét tam giác vuông SBH có:
\(\begin{array}{l}BH = SB.\cos {30^0} = 2a\sqrt 3 .\dfrac{{\sqrt 3 }}{2} = 3a\\SH = SB.\sin {30^0} = 2a\sqrt 3 .\dfrac{1}{2} = a\sqrt 3 \end{array}\)
\( \Rightarrow HC = BC - BH = a\).
Ta có:
\(BH \cap \left( {SAC} \right) = C \Rightarrow \dfrac{{d\left( {B;\left( {SAC} \right)} \right)}}{{d\left( {H;\left( {SAC} \right)} \right)}} = \dfrac{{BC}}{{HC}} = \dfrac{{4a}}{a} = 4\)
\( \Rightarrow d\left( {B;\left( {SAC} \right)} \right) = 4d\left( {H;\left( {SAC} \right)} \right)\).
Trong (ABC) kẻ \(HE \bot AC\,\,\left( {E \in AC} \right)\), trong (SHE) kẻ \(HK \bot SE\,\,\left( {K \in SE} \right)\).
Ta có:
\(\begin{array}{l}\left\{ \begin{array}{l}AC \bot HE\\AC \bot SH\end{array} \right. \Rightarrow AC \bot \left( {SHE} \right) \Rightarrow AC \bot HK\\\left\{ \begin{array}{l}HK \bot SE\\HK \bot AC\end{array} \right. \Rightarrow HK \bot \left( {SAC} \right)\\ \Rightarrow d\left( {H;\left( {SAC} \right)} \right) = HK\end{array}\)
Xét \(\Delta CEH\) và \(\Delta CBA\) có:
\(\begin{array}{l}\angle CEH = \angle CBA = {90^0}\\\angle C\,\,chung\\ \Rightarrow \Delta CEH \sim \Delta CBA\,\,\left( {g.g} \right)\\ \Rightarrow \dfrac{{HE}}{{AB}} = \dfrac{{HC}}{{AC}}\\ \Rightarrow \dfrac{{HE}}{{3a}} = \dfrac{a}{{\sqrt {{{\left( {3a} \right)}^2} + {{\left( {4a} \right)}^2}} }}\\ \Rightarrow HE = \dfrac{{3a}}{5}\end{array}\)
Vì \(SH \bot \left( {ABC} \right) \Rightarrow SH \bot HE \Rightarrow \Delta SHE\) vuông tại H.
Áp dụng định lí Pytago trong tam giác vuông SHE có:
\(\begin{array}{l}\dfrac{1}{{H{K^2}}} = \dfrac{1}{{S{H^2}}} + \dfrac{1}{{H{E^2}}} = \dfrac{1}{{{{\left( {a\sqrt 3 } \right)}^2}}} + \dfrac{1}{{{{\left( {\dfrac{{3a}}{5}} \right)}^2}}} = \dfrac{{28}}{{9{a^2}}}\\ \Rightarrow HK = \dfrac{{3\sqrt 7 a}}{{14}}\end{array}\)
Vậy \(d\left( {B;\left( {SAC} \right)} \right) = 4HK = \dfrac{{6\sqrt 7 a}}{7}\).
