|a|=|b|=|a-b|
设a、b的夹角为α
|a-b|^2 = |a|^2 + |b|^2 - 2a.b
|a|^2= |a|^2 + |b|^2 - 2a.b
|a|^2 - 2|a|^2cosα =0
=> α = 60°
设a、a+b的夹角为β
consider
|a+b|^2 = (a+b).(a+b)
= |a|^2 +|b|^2 + 2a.b
= 3|a|^2
a.(a+b) = |a|^2 + |a||b|cos60°
= 3|a|^2/2 (1)
also
a.(a+b) = |a||a+b|cosβ
= √3 |a|^2cosβ (2)
equaling (1) and (2)
3|a|^2/2 = √3 |a|^2cosβ
cosβ = √3/2
β = 30°