For-vectors-v-and-u-where-v-u-R-n-What-is-the-angle-between-v-and-u-both-are-from-origin-

Question Number 6875 by FilupSmith last updated on 01/Aug/16

For vectors v and u where :

v, u \in R^{n}

What is the angle between v and u ?

(both are from origin)

Commented by prakash jain last updated on 01/Aug/16

\cos θ = \frac{v \cdot u}{v u}

Commented by FilupSmith last updated on 01/Aug/16

How can we show this ?

Commented by FilupSmith last updated on 01/Aug/16

Is this correct ?

c^{2} = a^{2} + b^{2} - 2 a b \cos θ

\cos θ = \frac{c^{2} - a^{2} - b^{2}}{- 2 a b}

c = Distance between v and u

c =∥ v - u ∥= \sqrt{∥ v ∥^{2} + ∥ u ∥^{2} - 2 v \cdot u} p r o o f f o r t h i s n e e d e d

a =∥ v ∥= \sqrt{v_{1}^{2} + \dots + v_{n}^{2}} = \sqrt{\underset{t = 1}{\sum^{n}} v_{t}^{2}}

b =∥ u ∥= \sqrt{u_{1}^{2} + \dots + u_{n}^{2}} = \sqrt{\underset{t = 1}{\sum^{n}} u_{t}^{2}}

\cos θ = - \frac{∥ v - u ∥^{2} - ∥ v ∥^{2} - ∥ u ∥^{2}}{2 ∥ v ∥∥ u ∥}

\cos θ = - \frac{{(\sqrt{∥ v ∥^{2} + ∥ u ∥^{2} - 2 v \cdot u})}^{2} - ∥ v ∥^{2} - ∥ u ∥^{2}}{2 ∥ v ∥∥ u ∥}

\cos θ = - \frac{∥ v ∥^{2} + ∥ u ∥^{2} - 2 v \cdot u - ∥ v ∥^{2} - ∥ u ∥^{2}}{2 ∥ v ∥∥ u ∥}

\cos θ = - \frac{- 2 v \cdot u}{2 ∥ v ∥∥ u ∥}

\cos θ = \frac{2 v \cdot u}{∥ v ∥∥ u ∥} = \frac{2 \underset{t = 1}{\sum^{n}} v_{t} u_{t}}{\sqrt{(\underset{t = 1}{\sum^{n}} v_{t}^{2}) (\underset{t = 1}{\sum^{n}} u_{t}^{2})}}