B=( 1/2, √3/2), that is |b|= 1.

a b=√3/2-√3/2=0

1

M⊥n, that is, m n = (a+(tanq 2-3) b) (-ma+tanqb).

=-m|a|^2+tanQ(tanQ^2-3)|b|^2

=-4m+tanQ(tanQ^2-3)=0

Namely: m = tanq (tanq 2-3)/4.

Namely: m = f (q) = tanq (tanq 2-3)/4, Q∈(-π/2, π/2).

2

Q∈[-π/6, π/3], namely: tanQ∈[-√3/3, √3]

Order: t=tanQ, then: t∈[-√3/3, √3]

Namely: g (t) = t (t 2-3)/4 = (t 3-3t)/4.

g'(t)=3(t^2- 1)/4

G'(t)=0, then: t= 1 or-1 (truncation)

1 & lt； G'(t) > when t≤√3; 0, in which case g(t) is increasing function.

-√3/3≤t & lt； When 1, g' (t) < 0, g(t) is a decreasing function.

So when t= 1, g(t) takes the minimum value: gmin=- 1/2.

That is, the minimum value of f(Q): fmin=gmin=- 1/2.

At this time, tanQ= 1, that is, Q=π/4.