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

S h o w t h a t

\bar{a} \times (b \times c) = (\bar{a} . \bar{c}) \bar{b} - (\bar{a} . \bar{b}) \bar{c}

Commented by FilupSmith last updated on 25/Jun/16

what is the difference between

a and \bar{a} ?

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

I think he meant vector :

\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}

\boldsymbol a \times (\boldsymbol b \times \boldsymbol c) = (\boldsymbol a \cdot \boldsymbol c) \boldsymbol b - (\boldsymbol a \cdot \boldsymbol b) \boldsymbol c

Commented by FilupSmith last updated on 25/Jun/16

note :

\boldsymbol t =< t_{1}, t_{2}, \dots, t_{g} >= [\begin{matrix} t_{1} \\ t_{2} \\ ⋮ \\ t_{g} \end{matrix}], \boldsymbol t \in R^{g}

I will be using this notation for simplicity .

\boldsymbol a =< a_{1}, \dots, a_{n} >

\boldsymbol b =< b_{1}, \dots, b_{n} >

\boldsymbol c =< c_{1}, \dots, c_{n} >

\boldsymbol a, \boldsymbol b, \boldsymbol c \in R^{n}

\boldsymbol a \times \boldsymbol b =∥ \boldsymbol a ∥∥ \boldsymbol b ∥ \sin (θ) \boldsymbol n \to vector

\boldsymbol a ∙ \boldsymbol b =∥ \boldsymbol a ∥∥ \boldsymbol b ∥ \cos (θ) \to scalar

∥ \boldsymbol t ∥= \sqrt{t_{1}^{2} + t_{2}^{2} + \dots + t_{n}^{2}} \to scalar

where \boldsymbol n is the unit vector perpendicular

to \boldsymbol a and \boldsymbol b .

Question :

\boldsymbol a \times (\boldsymbol b \times \boldsymbol c) = (\boldsymbol a ∙ \boldsymbol c) \boldsymbol b - (\boldsymbol a ∙ \boldsymbol b) \boldsymbol c

because \boldsymbol a ∙ \boldsymbol b and \boldsymbol a ∙ \boldsymbol c are scalars,

multiplying them by a vector makes

a new vector .

e . g . 5 ⟨ 2, 1 ⟩ = ⟨ 10, 5 ⟩

continue