Proving the Formula for the Angle Between Two Vectors

[Mathispoweru]

@khanshamim1 It is a property of dot products. For any vector v.v = ||v||^2. Try with with for example v={4, 3}. Use this property and do it the long way by determining the magnitude and then squaring it.

[Emc]

Thank You so much for this. This is by far the clearest and easiest to follow prove I've seen. I was wondering where that came from.

[franarcuri]

Thanks!!!! I needed this. Really good explanation!!!

[ashly_gonsalves]

i was like when is he gonna finish combining and separating but at the end i'm left mind blown. thnxxxx

[sir.k]

great!every step was clear to me, thanks alot

[rickmonarch]

let's say V - U = K we can do this, since subtraction doesn't change the dimensions and structure of a vector.
so we end up with :
IIKII^2 = (K).(K) So now I show that both equal something.
since IIKII = (sum as i goes 1->N ki^2)^(1/2) // it's a p norm when p=2
since the sum is a constant and that is on the "(1/2)" squaring it just cancels it and doesn't change anything further.
So let's square both sides.
IIKII^2 = ((sum as i goes 1->N ki^2)^(1/2))^2
since (a^b)^c = a^(b*c)

( I ) IIKII = sum as i goes 1->N ki^2 // that's exactly what the dot product does. So let's see.

K.K = sum as i goes 1->N ( ki * ki )
And since a*a=a^2

( II ) K.K = sum as i goes 1->N k^2

Since the right side of the equations state the same, the left sides must be equal.