Câu hỏi 3

\(HC=\sqrt{B{{H}^{2}}+B{{C}^{2}}}=\sqrt{{{a}^{2}}+{{a}^{2}}}=a\sqrt{2}\)

Ta có \(\left( SC;\left( ABCD \right) \right)=\left( SC;HC \right)=\widehat{SCH}={{60}^{0}}\)

Xét tam giác vuông SHC có \(SH=HC.\tan 60=a\sqrt{2}.\sqrt{3}=a\sqrt{6}\)

Ta có:

\(\begin{align}  AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}=\sqrt{4{{a}^{2}}+{{a}^{2}}}=a\sqrt{5} \\  SB=\sqrt{S{{H}^{2}}+H{{B}^{2}}}=\sqrt{6{{a}^{2}}+{{a}^{2}}}=a\sqrt{7} \\ \end{align}\)

Ta có:

\(\begin{align}  \overrightarrow{SB}.\overrightarrow{AC}=\left( \overrightarrow{SH}+\overrightarrow{HB} \right).\overrightarrow{AC}=\underbrace{\overrightarrow{SH}.\overrightarrow{AC}}_{\overrightarrow{0}}+\overrightarrow{HB}.\overrightarrow{AC}=\overrightarrow{HB}.\overrightarrow{AC} \\  \Rightarrow \overrightarrow{SB}.\overrightarrow{AC}=HB.AC.\cos \left( HB;AC \right)=HB.AC.\cos \widehat{BAC}=HB.AC.\frac{AB}{AC}=a.2a=2{{a}^{2}} \\ \end{align}\)

Lại có \(\overrightarrow{SB}.\overrightarrow{AC}=SB.AC.\cos \left( SB;AC \right)\Rightarrow \cos \left( SB;AC \right)=\frac{\overrightarrow{SB}.\overrightarrow{AC}}{SB.AC}=\frac{2{{a}^{2}}}{a\sqrt{7}.a\sqrt{5}}=\frac{2}{\sqrt{35}}\)

Chọn A.