a) Ta có : \(OA=\dfrac{1}{2}AC=\dfrac{a\sqrt{2}}{2}\left(t/c.hình.vuông\right)\)
\(\Delta SOA:\left(Pitago\right)\)
\(SO=\sqrt{SA^2-OA^2}=\sqrt{\dfrac{6a^2}{9}-\dfrac{2a^2}{4}}=\sqrt{\dfrac{6a^2}{36}}=\dfrac{a\sqrt{6}}{6}\Rightarrow Sai\)
b) \(tan\left(\widehat{SA;\left(ABCD\right)}\right)=tan\widehat{SAO}=\dfrac{SO}{OA}=\dfrac{\dfrac{a\sqrt{6}}{6}}{\dfrac{a\sqrt{2}}{2}}=\sqrt{3}\)
\(\Rightarrow\left(\widehat{SA;\left(ABCD\right)}\right)=30^o\RightarrowĐúng\)
c) Ta có :
\(\left(P\right)//BD\Rightarrow HK//BD\left(HK\in\left(P\right)\right)\)
Ta lại có : \(BD\perp AC\left(t/c.hình.vuông\right);BD\perp SO\left(SO\perp\left(ABCD\right)\right)\)
\(\Rightarrow BD\perp\left(SAC\right)\)
mà \(HK//BD\left(cmt\right)\)
\(\Rightarrow HK\perp\left(SAC\right)\RightarrowĐúng\)
