Question

Cho hình vuông ABCD. Tính :

\(\cos\left(\overrightarrow{AC},\overrightarrow{BA}\right);\sin\left(\overrightarrow{AC},\overrightarrow{BD}\right);\cos\left(\overrightarrow{AB},\overrightarrow{CD}\right)\)

Answer 1

\(cos\left(\overrightarrow{AC};\overrightarrow{BA}\right)=cos\left(\overrightarrow{AC};\overrightarrow{AB'}\right)=cos\widehat{CAB'}=cos135^o\)\(=\dfrac{\sqrt{2}}{2}\).
\(sin\left(\overrightarrow{AC};\overrightarrow{BD}\right)=sin90^o=1\) do \(AC\perp BD\).
\(cos\left(\overrightarrow{AB};\overrightarrow{CD}\right)=cos180^o=-1\) do hai véc tơ \(\overrightarrow{AB};\overrightarrow{CD}\) ngược hướng.

 

Answer 2