a) Gọi \(H\) là trung điểm \(AB \Rightarrow SH \bot \left( {ABCD} \right)\) (vì \(\left( {SAB} \right) \bot \left( {ABCD} \right)\))
Tam giác \(SAB\) đều cạnh \(a\) nên \(SH = \dfrac{{a\sqrt 3 }}{2}\)
\({S_{ABMN}} = {S_{ABCD}} - {S_{CMN}} - {S_{ADN}}\)
\( = {a^2} - \dfrac{1}{2}\dfrac{a}{2}.\dfrac{a}{2} - \dfrac{1}{2}.a.\dfrac{a}{2}\) \( = \dfrac{{5{a^2}}}{8}\)
Suy ra \({V_{S.ABMN}} = \dfrac{1}{3}SH.{S_{ABMN}} = \dfrac{1}{3}.\dfrac{{a\sqrt 3 }}{2}.\dfrac{{5{a^2}}}{8} = \dfrac{{5{a^3}\sqrt 3 }}{{48}}\)
b) Gán hệ trục tọa độ Oxyz như hình vẽ với $O\equiv H$, $Ox\equiv HN$, $Oy\equiv HA$, $Oz\equiv HS$
$\Rightarrow A(0;\dfrac{a}{2};0)$; $P(\dfrac{a}{2};\dfrac{a}{4};\dfrac{a\sqrt3}{4})$
$M(\dfrac{a}{2};-\dfrac{a}{2};0)$; $K(0;-\dfrac{a}{4};\dfrac{a\sqrt3}{4})$
$\Rightarrow\vec{AP}=(\dfrac{a}{2};-\dfrac{a}{4};\dfrac{a\sqrt3}{4})$
$\vec{MK}=(-\dfrac{a}{2};\dfrac{a}{4};\dfrac{a\sqrt3}{4})$
$\vec{AM}=(\dfrac{a}{2};-a;0)$
$d(AP;MK)=\dfrac{|[\vec{AP},\vec{MK}].\vec{AM}|}{|[\vec{AP},\vec{MK}]|}=\dfrac{4a\sqrt3}{5}$
