Ta có: \(\overrightarrow {AB} = \left( { - 1;3} \right);\overrightarrow {AC} = \left( {2; - 1} \right)\)
Vậy\(\cos \left( {AB,AC} \right) = \left| {\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right)} \right| = \frac{{\left| { - 1.2 + 3.\left( { - 1} \right)} \right|}}{{\sqrt {{{\left( { - 1} \right)}^2} + {3^2}} .\sqrt {{2^2} + {{\left( { - 1} \right)}^2}} }} = \frac{1}{{\sqrt 2 }} \Rightarrow \widehat {BAC} = {45^o}\)