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Question Number 6374 by sanusihammed last updated on 25/Jun/16

$$Showthat$$$$\stackrel{-}{a}\times \left(b\times c\right)=(\stackrel{-}{a}.\stackrel{-}{c})\stackrel{-}{b}-(\stackrel{-}{a}.\stackrel{-}{b})\stackrel{-}{c}$$

Commented by FilupSmith last updated on 25/Jun/16

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{difference}\mathrm{between}$$$$a\mathrm{and}\overline{a}?$$

Commented by prakash jain last updated on 25/Jun/16

$$\mathrm{I}\mathrm{think}\mathrm{he}\mathrm{meant}\mathrm{vector}:$$$$\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{c}$$$$\mathbf{a}\times \left(\mathbf{b}\times \mathbf{c}\right)=(\mathbf{a}\cdot \mathbf{c})\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{c}$$

Commented by FilupSmith last updated on 25/Jun/16

$$\mathrm{note}:$$$$\mathit{t}=<t_1,t_2,...,t_g>=\left[\begin{array}{c}{t}_1\\ t_2\\ ⋮\\ t_g\end{array}\right],\mathit{t}\in {\mathbb{R}}^g$$$$\mathrm{I}\mathrm{will}\mathrm{be}\mathrm{using}\mathrm{this}\mathrm{notation}\mathrm{for}\mathrm{simplicity}.$$$$\mathit{a}=<a_1,...,a_n>$$$$\mathit{b}=<b_1,...,b_n>$$$$\mathit{c}=<c_1,...,c_n>$$$$\mathit{a},\mathit{b},\mathit{c}\in {\mathbb{R}}^n$$$$$$$$\mathit{a}\times \mathit{b}=\parallel \mathit{a}\parallel \parallel \mathit{b}\parallel \sin \left(\theta \right)\mathit{n}\to \mathrm{vector}$$$$\mathit{a}\bullet \mathit{b}=\parallel \mathit{a}\parallel \parallel \mathit{b}\parallel \cos \left(\theta \right)\to \mathrm{scalar}$$$$\parallel \mathit{t}\parallel =\sqrt{t_1^2+t_2^2+...+t_n^2}\to \mathrm{scalar}$$$$\mathrm{where}\mathit{n}\mathrm{is}\mathrm{the}\mathrm{unit}\mathrm{vector}\mathrm{perpendicular}$$$$\mathrm{to}\mathit{a}\mathrm{and}\mathit{b}.$$$$$$$$\mathrm{Question}:$$$$\mathit{a}\times \left(\mathit{b}\times \mathit{c}\right)=(\mathit{a}\bullet \mathit{c})\mathit{b}-(\mathit{a}\bullet \mathit{b})\mathit{c}$$$$\mathrm{because}\mathit{a}\bullet \mathit{b}\mathrm{and}\mathit{a}\bullet \mathit{c}\mathrm{are}\mathrm{scalars},$$$$\mathrm{multiplying}\mathrm{them}\mathrm{by}\mathrm{a}\mathrm{vector}\mathrm{makes}$$$$\mathrm{a}\mathrm{new}\mathrm{vector}.$$$$e.g.5⟨2,1⟩=⟨10,5⟩$$$$\mathrm{continue}$$

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