vector{AC}=(2-(-1);2-(-1);-2-2)=(3;3;-4)
|vector{AB}|=sqrt(1^2+2^2+(-2)^2)=sqrt(9)=3
|vector{AC}|=sqrt(3^2+3^2+(-4)^2)=sqrt(34)
Находим скалярное произведение
vector{AB}*vector{AC}=1*3+2*3+(-2)*(-4)=17
cos ∠ (vector{AB},vector{AC})=vector{AB}*vector{AC}/(|vector{AB}|*|vector{AC}|=
=17/(3*sqrt(34))