\(\vec{AB} ⊥\vec{AC}\Rightarrow \vec{AB}.\vec{AC} = 0\)
\(\vec{AC}.\vec{CB} =- \vec{CA}\). \(\vec{CB}\)
Ta có: \(CB= \sqrt{AB^2+AC^2}\)\(=\sqrt{a^2+a^2}=a\sqrt2\); \(\widehat{C} = 45^0\) vì \(\Delta ABC\) là tam giác vuông cân tại \(A.\)
Vậy \(\vec{AC}.\vec{CB} = -\vec{CA}. \vec{CB}\)\(= -|\vec{CA}|. |\vec{CB}|. cos45^0\)
\(= - a.a\sqrt 2 .{{\sqrt 2 } \over 2} = - {a^2}.\)
