Scalar product of two vectors

a)
u+v=[(cosα)i+(sinα)j]+[(cosβ)i+(sinβ)j]
=(cosα+cosβ)i+(sinα+ sinβ)j

u-v=[(cosα)i+(sinα)j]-[(cosβ)i+(sinβ)j]
=(cosα-cosβ)i+(sinα-sinβ)j

(u+v)．(u-v)= [(cosα+cosβ)i+(sinα+ sinβ)j][(cosα-cosβ)i+(sinα-sinβ)j]
=(cosα+cosβ)(cosα-cosβ)+(sinα+ sinβ)(sinα-sinβ)
=(cos2α-cos2β)+ (sin2α-sin2β)
=(sin2α+ cos2α)- (sin2β+ cos2β)
=1-1
=0



Let the required angle be θ.

圖片參考：http://g.imagehost.org/0013/ScreenHunter_01_May_03_21_21.gif


cos2α+cos2β=(cosα-cosβ)2+(sinα-sinβ)2
1=( cos2α+cos2β-2cosαcosβ)+ (sin2α+sin2β-2sinαcosα)
1=( sin2α+cos2α)+ (sin2β+cos2β)-2(cosαcosβ+ sinαcosα)
1=1+1-2cos(α-β)
1=2cos(α-β)
1/2= cos(α-β)
1/2= cos(β-α)
β-α=π/3

(a) (u + v) and ( u - v) is equivalent to the diagonals of a parallelogram with u and v as the 2 sides.
Since magnitude of u = magnitude of v = 1, so it is a rhombus. Therefore, angle between the 2 diagonals = 90 degrees which is the property of a rhombus. That means angle between (u + v) and (u - v) is 90 degree.
(b) If magnitude of u = magnitude of (u -v), that means u, v and ( u-v) formed an equilateral triangle, so angle between the vectors = 60 degree.
That is beta - alpha = 60 degree.