Đáp án:

b) $\left( {SI,\left( {ABCD} \right)} \right) = \arctan \left( {\dfrac{3}{{\sqrt {10} }}} \right)$

Giải thích các bước giải:

Kẻ $AE\bot BP=E; AH\bot SE=H$

a) Ta có:

$\left\{ \begin{array}{l}
\left( {SAB} \right) \cap \left( {SAD} \right) = SA\\
\left( {SAB} \right) \bot \left( {ABCD} \right)\\
\left( {SAD} \right) \bot \left( {ABCD} \right)
\end{array} \right. \Rightarrow SA \bot \left( {ABCD} \right)$

b) Ta có:

$\dfrac{{AI}}{{CI}} = \dfrac{{AB}}{{CP}} = 2 \Rightarrow d\left( {A,\left( {SBP} \right)} \right) = 2d\left( {C,\left( {SBP} \right)} \right) \Rightarrow d\left( {A,\left( {SBP} \right)} \right) = a$

Lại có:

$\begin{array}{l}
\left\{ \begin{array}{l}
AE \bot BP\\
SA \bot BP
\end{array} \right. \Rightarrow BP \bot \left( {SAE} \right) \Rightarrow BP \bot AH\\
\left\{ \begin{array}{l}
AH \bot BP\\
AH \bot SE
\end{array} \right. \Rightarrow AH \bot \left( {SBP} \right)
\end{array}$

$ \Rightarrow H$ là hình chiếu của $A $ trên $SE$

$ \Rightarrow d\left( {A,\left( {SBP} \right)} \right) = AH \Rightarrow AH = a$

Ta có:

$\begin{array}{l}
\widehat {ABE} = \widehat {CPB}\left( {slt} \right) \Rightarrow \sin \widehat {ABE} = \sin \widehat {CPB} = \dfrac{{CB}}{{BP}} = \dfrac{{CB}}{{\sqrt {B{C^2} + C{P^2}} }} = \dfrac{a}{{\sqrt {{a^2} + {a^2}} }} = \dfrac{1}{{\sqrt 2 }}\\
\Delta ABE;\widehat {AEB} = {90^0};\sin \widehat {ABE} = \dfrac{1}{{\sqrt 2 }}\\
 \Rightarrow AE = AB.\sin \widehat {ABE} = 2a.\dfrac{1}{{\sqrt 2 }} = a\sqrt 2 
\end{array}$

Xét $\begin{array}{l}
\Delta SAE;\widehat {SAE} = {90^0};AH \bot SE = H;AE = a\sqrt 2 ;AH = a\\
 \Rightarrow \dfrac{1}{{A{H^2}}} = \dfrac{1}{{S{A^2}}} + \dfrac{1}{{A{E^2}}} \Rightarrow \dfrac{1}{{S{A^2}}} = \dfrac{1}{{{a^2}}} - \dfrac{1}{{{{\left( {a\sqrt 2 } \right)}^2}}} = \dfrac{1}{{2{a^2}}} \Rightarrow SA = a\sqrt 2 
\end{array}$

Ta có:

$SA\bot (ABCD)$$\to A$ là hình chiếu của $S$ trên $(ABCD)$

$ \Rightarrow \left( {SI,\left( {ABCD} \right)} \right) = \left( {SI,AI} \right) = \widehat {SIA}$

$\Delta ABC;\widehat {ABC} = {90^0},AI = \dfrac{2}{3}AC = \dfrac{2}{3}\sqrt {A{B^2} + B{C^2}}  = \dfrac{2}{3}\sqrt {{{\left( {2a} \right)}^2} + {a^2}}  = \dfrac{{2a\sqrt 5 }}{3}$

Xét $\begin{array}{l}
\Delta SAI;\widehat {SAI} = {90^0};SA = a\sqrt 2 ;AI = \dfrac{{2a\sqrt 5 }}{3}\\
 \Rightarrow \tan \widehat {SAI} = \dfrac{{SA}}{{AI}} = \dfrac{{a\sqrt 2 }}{{\dfrac{{2a\sqrt 5 }}{3}}} = \dfrac{3}{{\sqrt {10} }}\\
 \Rightarrow \widehat {SAI} = \arctan \left( {\dfrac{3}{{\sqrt {10} }}} \right)\\
 \Rightarrow \left( {SI,\left( {ABCD} \right)} \right) = \arctan \left( {\dfrac{3}{{\sqrt {10} }}} \right)
\end{array}$

Vậy $\left( {SI,\left( {ABCD} \right)} \right) = \arctan \left( {\dfrac{3}{{\sqrt {10} }}} \right)$