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April 18, 2014

April 18, 2014

Posted by **manpreet** on Friday, November 23, 2012 at 4:32pm.

- dps -
**Steve**, Friday, November 23, 2012 at 6:35pmwithout loss of generality, we can let

a = <1,0>

b = <cosθ,sinθ>

a+b = <(1+cosθ),sinθ>

|a+b|^2 = (1+cosθ)^2 + sin^2 θ

if a+b is a unit vector,

|a+b|^2 = 1

1+2cosθ+cos^2θ + sin^2θ = 1

1+2cosθ = 0

cosθ = -1/2

θ = 2pi/3

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