Câu 1:
Tam giác ABC đều (gt). \(\Rightarrow\) \(\widehat{ABC}\) \(=60^o\) (Tính chất tam giác đều).
Ta có: \(\left(\overrightarrow{BA};\overrightarrow{BC}\right)=\) \(\widehat{ABC}\) \(=60^o.\)
Câu 2:
Xét tam giác ABC cân tại A có: \(\widehat{BAC}\) \(=120^o\) (gt).
\(\Rightarrow\)\(\widehat{ABC} = \) \(\widehat{ACB}\) \(=30^o.\)
Vẽ \(\overrightarrow{BD}=\overrightarrow{AC.}\) \(\Rightarrow BD\) // \(AC.\) \(\Rightarrow\) \(\widehat{ACB}\) \(=\widehat{CBD}\) \(=30^o.\)
Ta có: \(\left(\overrightarrow{BA};\overrightarrow{AC}\right)=\left(\overrightarrow{BA};\overrightarrow{BD}\right)=\) \(\widehat{ABD}\) \(=\widehat{ABC} + \widehat{CBD} =\) \(30^o+30^o=60^o.\)
Câu 3:
ABCD là hình vuông (gt). \(\Rightarrow\) \(\widehat{BAC} = \)\(\dfrac{90^o}{2}=45^o\) và \(AC=a\sqrt{2}.\)
Ta có: \(\overrightarrow{AB}.\overrightarrow{AC}=\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|.cosBAC=a.a\sqrt{2}.cos45^o=a^2.\)
