We know that
#norm(B-A)^2 = normA^2+normB^2-2 normA norm B cos theta#
and also
#norm(B-A)-norm(A) = normB-norm(B-A)# or
#2norm(B-A) = normA+normB#
and after substitution
#costheta= (normA^2+normB^2-norm(B-A)^2)/(2 normA norm B ) =3/8(norm(A)^2+norm(B)^2)/(normA norm B)-1/4#
Now making #normB = lambda norm A#
#cos theta = 3/8(1+lambda^2)/lambda -1/4# which has a minimum at #lambda = 1# giving #cos theta = 1/2# and then
#theta = pi/3#
then, the value of #n# is #3#.
Concluding, #OAB# is an equilateral triangle.
