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Question Number 6374 by sanusihammed last updated on 25/Jun/16
Show that  a^−  × (b × c) = (a^−  . c^− )b^−   −  (a^−  . b^− )c^−
$${Show}\:{that} \\ $$$$\overset{−} {{a}}\:×\:\left({b}\:×\:{c}\right)\:=\:\left(\overset{−} {{a}}\:.\:\overset{−} {{c}}\right)\overset{−} {{b}}\:\:−\:\:\left(\overset{−} {{a}}\:.\:\overset{−} {{b}}\right)\overset{−} {{c}} \\ $$
Commented by FilupSmith last updated on 25/Jun/16
what is the difference between  a and a^�  ?
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{between} \\ $$$${a}\:\mathrm{and}\:\bar {{a}}\:? \\ $$
Commented by prakash jain last updated on 25/Jun/16
I think he meant vector:  a^→ ×(b^→ ×c^→ )=(a^→ ∙c^→ )b^→ −(a^→ ∙b^→ )c^→   a×(b×c)=(a∙c)b−(a∙b)c
$$\mathrm{I}\:\mathrm{think}\:\mathrm{he}\:\mathrm{meant}\:\mathrm{vector}: \\ $$$$\overset{\rightarrow} {{a}}×\left(\overset{\rightarrow} {{b}}×\overset{\rightarrow} {{c}}\right)=\left(\overset{\rightarrow} {{a}}\centerdot\overset{\rightarrow} {{c}}\right)\overset{\rightarrow} {{b}}−\left(\overset{\rightarrow} {{a}}\centerdot\overset{\rightarrow} {{b}}\right)\overset{\rightarrow} {{c}} \\ $$$$\boldsymbol{\mathrm{a}}×\left(\boldsymbol{\mathrm{b}}×\boldsymbol{\mathrm{c}}\right)=\left(\boldsymbol{\mathrm{a}}\centerdot\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{b}}−\left(\boldsymbol{\mathrm{a}}\centerdot\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{c}} \\ $$
Commented by FilupSmith last updated on 25/Jun/16
note:  t=<t_1 , t_2 , ..., t_g >= [(t_1 ),(t_2 ),(⋮),(t_g ) ],   t∈R^g   I will be using this notation for simplicity.  a=<a_1 , ..., a_n >  b=<b_1 , ..., b_n >  c=<c_1 , ..., c_n >  a,b,c∈R^n     a×b=∥a∥∥b∥sin(θ)n   → vector  a•b=∥a∥∥b∥cos(θ)       → scalar  ∥t∥=(√(t_1 ^2 +t_2 ^2 +...+t_n ^2 ))      → scalar  where n is the unit vector perpendicular  to a and b.    Question:  a×(b×c)=(a•c)b−(a•b)c  because a•b and a•c  are scalars,  multiplying them by a vector makes   a new vector.  e.g.   5⟨2, 1⟩ = ⟨10, 5⟩  continue
$$\mathrm{note}: \\ $$$$\boldsymbol{{t}}=<{t}_{\mathrm{1}} ,\:{t}_{\mathrm{2}} ,\:…,\:{t}_{{g}} >=\begin{bmatrix}{{t}_{\mathrm{1}} }\\{{t}_{\mathrm{2}} }\\{\vdots}\\{{t}_{{g}} }\end{bmatrix},\:\:\:\boldsymbol{{t}}\in\mathbb{R}^{{g}} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{be}\:\mathrm{using}\:\mathrm{this}\:\mathrm{notation}\:\mathrm{for}\:\mathrm{simplicity}. \\ $$$$\boldsymbol{{a}}=<{a}_{\mathrm{1}} ,\:…,\:{a}_{{n}} > \\ $$$$\boldsymbol{{b}}=<{b}_{\mathrm{1}} ,\:…,\:{b}_{{n}} > \\ $$$$\boldsymbol{{c}}=<{c}_{\mathrm{1}} ,\:…,\:{c}_{{n}} > \\ $$$$\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\in\mathbb{R}^{{n}} \\ $$$$ \\ $$$$\boldsymbol{{a}}×\boldsymbol{{b}}=\parallel\boldsymbol{{a}}\parallel\parallel\boldsymbol{{b}}\parallel\mathrm{sin}\left(\theta\right)\boldsymbol{{n}}\:\:\:\rightarrow\:\mathrm{vector} \\ $$$$\boldsymbol{{a}}\bullet\boldsymbol{{b}}=\parallel\boldsymbol{{a}}\parallel\parallel\boldsymbol{{b}}\parallel\mathrm{cos}\left(\theta\right)\:\:\:\:\:\:\:\rightarrow\:\mathrm{scalar} \\ $$$$\parallel\boldsymbol{{t}}\parallel=\sqrt{{t}_{\mathrm{1}} ^{\mathrm{2}} +{t}_{\mathrm{2}} ^{\mathrm{2}} +…+{t}_{{n}} ^{\mathrm{2}} }\:\:\:\:\:\:\rightarrow\:\mathrm{scalar} \\ $$$$\mathrm{where}\:\boldsymbol{{n}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\boldsymbol{{a}}\:\mathrm{and}\:\boldsymbol{{b}}. \\ $$$$ \\ $$$$\mathrm{Question}: \\ $$$$\boldsymbol{{a}}×\left(\boldsymbol{{b}}×\boldsymbol{{c}}\right)=\left(\boldsymbol{{a}}\bullet\boldsymbol{{c}}\right)\boldsymbol{{b}}−\left(\boldsymbol{{a}}\bullet\boldsymbol{{b}}\right)\boldsymbol{{c}} \\ $$$$\mathrm{because}\:\boldsymbol{{a}}\bullet\boldsymbol{{b}}\:\mathrm{and}\:\boldsymbol{{a}}\bullet\boldsymbol{{c}}\:\:\mathrm{are}\:\mathrm{scalars}, \\ $$$$\mathrm{multiplying}\:\mathrm{them}\:\mathrm{by}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{makes}\: \\ $$$$\mathrm{a}\:\mathrm{new}\:\mathrm{vector}. \\ $$$${e}.{g}.\:\:\:\mathrm{5}\langle\mathrm{2},\:\mathrm{1}\rangle\:=\:\langle\mathrm{10},\:\mathrm{5}\rangle \\ $$$$\mathrm{continue} \\ $$

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