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

$$\mathrm{For}\mathrm{vectors}\mathit{v}\mathrm{and}\mathit{u}\mathrm{where}:$$$$\mathit{v},\mathit{u}\in {\mathbb{R}}^n$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathit{v}\mathrm{and}\mathit{u}?$$$$\left(\mathrm{both}\mathrm{are}\mathrm{from}\mathrm{origin}\right)$$

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

$$\cos \theta =\frac{\mathit{v}\cdot \mathit{u}}{vu}$$

Commented by FilupSmith last updated on 01/Aug/16

$$\mathrm{How}\mathrm{can}\mathrm{we}\mathrm{show}\mathrm{this}?$$

Commented by FilupSmith last updated on 01/Aug/16

$$\mathrm{Is}\mathrm{this}\mathrm{correct}?$$$$c^2=a^2+b^2-2ab\cos \theta$$$$\cos \theta =\frac{c^2-a^2-b^2}{-2ab}$$$$c=\mathrm{Distance}\mathrm{between}\mathit{v}\mathrm{and}\mathit{u}$$$$c=\parallel \mathit{v}-\mathit{u}\parallel =\sqrt{\parallel \mathit{v}{\parallel }^2+\parallel \mathit{u}{\parallel }^2-2\mathit{v}\cdot \mathit{u}}proofforthisneeded$$$$a=\parallel \mathit{v}\parallel =\sqrt{v_1^2+...+v_n^2}=\sqrt{\underset{t=1}{\sum ^n}{v}_t^2}$$$$b=\parallel \mathit{u}\parallel =\sqrt{u_1^2+...+u_n^2}=\sqrt{\underset{t=1}{\sum ^n}{u}_t^2}$$$$$$$$\cos \theta =-\frac{\parallel \mathit{v}-\mathit{u}{\parallel }^2-\parallel \mathit{v}{\parallel }^2-\parallel \mathit{u}{\parallel }^2}{2\parallel \mathit{v}\parallel \parallel \mathit{u}\parallel }$$$$\cos \theta =-\frac{{\left(\sqrt{\parallel \mathit{v}{\parallel }^2+\parallel \mathit{u}{\parallel }^2-2\mathit{v}\cdot \mathit{u}}\right)}^2-\parallel \mathit{v}{\parallel }^2-\parallel \mathit{u}{\parallel }^2}{2\parallel \mathit{v}\parallel \parallel \mathit{u}\parallel }$$$$\cos \theta =-\frac{\parallel \mathit{v}{\parallel }^2+\parallel \mathit{u}{\parallel }^2-2\mathit{v}\cdot \mathit{u}-\parallel \mathit{v}{\parallel }^2-\parallel \mathit{u}{\parallel }^2}{2\parallel \mathit{v}\parallel \parallel \mathit{u}\parallel }$$$$\cos \theta =-\frac{-2\mathit{v}\cdot \mathit{u}}{2\parallel \mathit{v}\parallel \parallel \mathit{u}\parallel }$$$$\cos \theta =\frac{2\mathit{v}\cdot \mathit{u}}{\parallel \mathit{v}\parallel \parallel \mathit{u}\parallel }=\frac{2\underset{t=1}{\sum ^n}{v}_t{u}_t}{\sqrt{\left(\underset{t=1}{\sum ^n}{v}_t^2\right)\left(\underset{t=1}{\sum ^n}{u}_t^2\right)}}$$

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