calculezi analitic produsul u*v=(2i+j)*(i+m²j)=2+m²
Dar u*v=lul*lvlcosα unde α=<(u,v)
lul=l2+jl=√(2²+1)=√5
lvl=lm²+1l=√(m²+1²)
u*v=√5*(m²+1)cosα (2
Egalezi (1 si (2
2+m²=√5(m²+1)*cosα=>
cosα=(2+m²)/√5(m²+1)>0 pt ca e un raport de numere srict pozitive=>
cosα>0 Cosinusul e pozitiv in cadranul[0,π/2] deci
α∈(0,π/2)