由题意,|i|=|j|=1,=π/3,故：i·j=|i|*|j|*cos=1/2
a=2i+j,b=-3i+2j,故：a·b=(2i+j)·(-3i+2j)=-6|i|^2+2|j|^2-3i·j+4i·j=-6+2+1/2=-7/2
而：|a|^2=(2i+j)·(2i+j)=4|i|^2+|j|^2+4i·j=5+2=7,故：|a|=sqrt(7)
|b|^2=(-3i+2j)·(-3i+2j)=9|i|^2+4|j|^2-12i·j=13-6=7,故：|b|=sqrt(7)
故：cos=a·b/(|a|*|b|)=(-7/2)/7=-1/2,故：=2π/3