vector{BA}=(3;-3;0)
vector{BC}=(-1;-3;4)
vector{BA}*vector{BC}=3*(-1)+(-3)*(-3)+0*4=-3+9+0=6
|vector{BA}|=sqrt(3^2+(-3)^2)=sqrt(18)
|vector{BA}|=sqrt((-1)^2+(-3)^2+4^2)=sqrt(26)
cos ∠ BAC=[m]\frac{6}{\sqrt{18}\cdot \sqrt{26}}=\frac{1}{\sqrt{13}}[/m]
∠ BAC=arccos[m]\frac{1}{\sqrt{13}}[/m]