\(BP=\dfrac{1}{3}AB\Rightarrow BP=\dfrac{1}{2}AP\)
\(\Rightarrow d\left(B;\left(SPC\right)\right)=\dfrac{1}{2}d\left(A;\left(SPC\right)\right)\)
Trong tam giác APC, kẻ \(AH\perp CP\Rightarrow CP\left(SAH\right)\)
Trong tam giác vuông SAH, kẻ \(AK\perp SH\Rightarrow AK\perp\left(SPC\right)\Rightarrow AK=d\left(A;\left(SPC\right)\right)\)
\(AP=\dfrac{2}{3}AB=\dfrac{2a}{3}\Rightarrow CP=\sqrt{AP^2+AC^2-2AP.AC.cos60^0}=\dfrac{a\sqrt{7}}{3}\)
Áp dụng định lý hàm sin:
\(\dfrac{AP}{sin\widehat{ACP}}=\dfrac{CP}{sinA}\Rightarrow sin\widehat{ACP}=\dfrac{AP.sin60^0}{CP}=\dfrac{\sqrt{21}}{7}\)
\(\Rightarrow AH=AC.sin\widehat{ACP}=\dfrac{a\sqrt{21}}{7}\)
\(\dfrac{1}{AK^2}=\dfrac{1}{AH^2}+\dfrac{1}{SA^2}\Rightarrow AK=\dfrac{SA.AH}{\sqrt{SA^2+AH^2}}=\dfrac{2a\sqrt{93}}{31}\)
\(\Rightarrow d\left(B;\left(SPC\right)\right)=\dfrac{1}{2}AK=\dfrac{a\sqrt{93}}{31}\)
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