Cho hình chóp S.ABCD có đáy ABCD là hình chữ...

C

- Hướng dẫn giải

Đáp án C

 Ta có: $\overrightarrow{SC}.\overrightarrow{BD}=\left(\overrightarrow{SA}+\overrightarrow{AC}\right).\overrightarrow{BD}=\overrightarrow{SA}.\overrightarrow{BD}+\overrightarrow{AC}.\overrightarrow{BD}=\overrightarrow{AC}.\overrightarrow{BD}$

          $=AC.BD.\cos \widehat{DOC}=A{C}^2.\frac{O{D}^2+O{C}^2-D{C}^2}{2OD.OC}$ 

          $=A{C}^2.\frac{O{D}^2+O{C}^2-D{C}^2}{2O{C}^2}=2\left(2O{C}^2-D{C}^2\right)$

           $=2\left(\frac{5a^2}{2}-a^2\right)=3a^2$ 

Do đó: $\cos \left(\overrightarrow{SC},\overrightarrow{BD}\right)=\frac{\overrightarrow{SC}.\overrightarrow{BD}}{SC.BD}=\frac{3a^2}{3a.a\sqrt{5}}=\frac{1}{\sqrt{5}}$ 

Vậy $\cos \left(SC,BD\right)=\left|\cos \left(\overrightarrow{SC},\overrightarrow{BD}\right)\right|=\frac{1}{\sqrt{5}}$