$\overrightarrow{AB}=\left( -1;3 \right)$ ; $\overrightarrow{AC}=\left( 2;-1 \right)$
(AB) qua A(2;-1), nhận $\overrightarrow{{{n}_{AB}}}\left( 3;1\right)$ làm vecto pháp tuyến
$\Rightarrow$ (AB): 3(x-2)+(y+1)=0
(AB): 3x+y-5=0
(AC) qua A(2;-1), nhận $\overrightarrow{{{n}_{AC}}}\left(1;2 \right)$ làm vecto pháp tuyến
$\Rightarrow$ (AC): (x-2)+2(y+1)=0
(AC): x+2y = 0
$\cos \left( \overrightarrow{AB,}\overrightarrow{AC} \right)=\left| \cos \left( \overrightarrow{{{n}_{AB}}},\overrightarrow{{{n}_{AC}}} \right) \right|=\frac{\left| \overrightarrow{{{n}_{AB}}.}\overrightarrow{{{n}_{AC}}} \right|}{\left| \overrightarrow{{{n}_{AB}}} \right|.\left| \overrightarrow{{{n}_{AC}}} \right|}=\frac{\left| 3.1+1.2 \right|}{\sqrt{{{3}^{2}}+{{1}^{2}}}.\sqrt{{{1}^{2}}+{{2}^{2}}}}=\frac{5}{\sqrt{10}.\sqrt{5}}=\frac{5}{5\sqrt{2}}=\frac{\sqrt{2}}{2}$
$\Rightarrow \widehat{\left( \overrightarrow{AB,}\overrightarrow{AC} \right)}={{45}^{o}}$
$\Rightarrow \widehat{BAC}=\widehat{\left( \overrightarrow{AB,}\overrightarrow{AC} \right)}={{45}^{o}}$