Question

Cho lăng trụ đều ABC.A'B'C' có AB =1, AA' = \(\sqrt{2}\) . Tính góc giữa AB' và BC' 

Answer 1

\(AB'=BC'=\sqrt{1+2}=\sqrt{3}\)

\(\overrightarrow{AB'}.\overrightarrow{BC'}=\left(\overrightarrow{AA'}+\overrightarrow{AB}\right)\left(\overrightarrow{BC}+\overrightarrow{BB'}\right)=AA'^2+\overrightarrow{AB}.\overrightarrow{BC}=2+1.1.cos120^0=\dfrac{3}{2}\)

\(\Rightarrow cos\left(AB';BC'\right)=\dfrac{\left|\overrightarrow{AB'}.\overrightarrow{BC'}\right|}{AB'.BC'}=\dfrac{1}{2}\)