Question Number 6374 by sanusihammed last updated on 25/Jun/16
$${Show}\:{that} \\ $$$$\overset{−} {{a}}\:×\:\left({b}\:×\:{c}\right)\:=\:\left(\overset{−} {{a}}\:.\:\overset{−} {{c}}\right)\overset{−} {{b}}\:\:−\:\:\left(\overset{−} {{a}}\:.\:\overset{−} {{b}}\right)\overset{−} {{c}} \\ $$
Commented by FilupSmith last updated on 25/Jun/16
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{between} \\ $$$${a}\:\mathrm{and}\:\bar {{a}}\:? \\ $$
Commented by prakash jain last updated on 25/Jun/16
$$\mathrm{I}\:\mathrm{think}\:\mathrm{he}\:\mathrm{meant}\:\mathrm{vector}: \\ $$$$\overset{\rightarrow} {{a}}×\left(\overset{\rightarrow} {{b}}×\overset{\rightarrow} {{c}}\right)=\left(\overset{\rightarrow} {{a}}\centerdot\overset{\rightarrow} {{c}}\right)\overset{\rightarrow} {{b}}−\left(\overset{\rightarrow} {{a}}\centerdot\overset{\rightarrow} {{b}}\right)\overset{\rightarrow} {{c}} \\ $$$$\boldsymbol{\mathrm{a}}×\left(\boldsymbol{\mathrm{b}}×\boldsymbol{\mathrm{c}}\right)=\left(\boldsymbol{\mathrm{a}}\centerdot\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{b}}−\left(\boldsymbol{\mathrm{a}}\centerdot\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{c}} \\ $$
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](Q6389.png)
$$\mathrm{note}: \\ $$$$\boldsymbol{{t}}=<{t}_{\mathrm{1}} ,\:{t}_{\mathrm{2}} ,\:...,\:{t}_{{g}} >=\begin{bmatrix}{{t}_{\mathrm{1}} }\\{{t}_{\mathrm{2}} }\\{\vdots}\\{{t}_{{g}} }\end{bmatrix},\:\:\:\boldsymbol{{t}}\in\mathbb{R}^{{g}} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{be}\:\mathrm{using}\:\mathrm{this}\:\mathrm{notation}\:\mathrm{for}\:\mathrm{simplicity}. \\ $$$$\boldsymbol{{a}}=<{a}_{\mathrm{1}} ,\:...,\:{a}_{{n}} > \\ $$$$\boldsymbol{{b}}=<{b}_{\mathrm{1}} ,\:...,\:{b}_{{n}} > \\ $$$$\boldsymbol{{c}}=<{c}_{\mathrm{1}} ,\:...,\:{c}_{{n}} > \\ $$$$\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\in\mathbb{R}^{{n}} \\ $$$$ \\ $$$$\boldsymbol{{a}}×\boldsymbol{{b}}=\parallel\boldsymbol{{a}}\parallel\parallel\boldsymbol{{b}}\parallel\mathrm{sin}\left(\theta\right)\boldsymbol{{n}}\:\:\:\rightarrow\:\mathrm{vector} \\ $$$$\boldsymbol{{a}}\bullet\boldsymbol{{b}}=\parallel\boldsymbol{{a}}\parallel\parallel\boldsymbol{{b}}\parallel\mathrm{cos}\left(\theta\right)\:\:\:\:\:\:\:\rightarrow\:\mathrm{scalar} \\ $$$$\parallel\boldsymbol{{t}}\parallel=\sqrt{{t}_{\mathrm{1}} ^{\mathrm{2}} +{t}_{\mathrm{2}} ^{\mathrm{2}} +...+{t}_{{n}} ^{\mathrm{2}} }\:\:\:\:\:\:\rightarrow\:\mathrm{scalar} \\ $$$$\mathrm{where}\:\boldsymbol{{n}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\boldsymbol{{a}}\:\mathrm{and}\:\boldsymbol{{b}}. \\ $$$$ \\ $$$$\mathrm{Question}: \\ $$$$\boldsymbol{{a}}×\left(\boldsymbol{{b}}×\boldsymbol{{c}}\right)=\left(\boldsymbol{{a}}\bullet\boldsymbol{{c}}\right)\boldsymbol{{b}}−\left(\boldsymbol{{a}}\bullet\boldsymbol{{b}}\right)\boldsymbol{{c}} \\ $$$$\mathrm{because}\:\boldsymbol{{a}}\bullet\boldsymbol{{b}}\:\mathrm{and}\:\boldsymbol{{a}}\bullet\boldsymbol{{c}}\:\:\mathrm{are}\:\mathrm{scalars}, \\ $$$$\mathrm{multiplying}\:\mathrm{them}\:\mathrm{by}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{makes}\: \\ $$$$\mathrm{a}\:\mathrm{new}\:\mathrm{vector}. \\ $$$${e}.{g}.\:\:\:\mathrm{5}\langle\mathrm{2},\:\mathrm{1}\rangle\:=\:\langle\mathrm{10},\:\mathrm{5}\rangle \\ $$$$\mathrm{continue} \\ $$