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Let \mathrm{}\vec{a}, \vec{b}, \vec{c}be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and \mathrm{}(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=168$,then $|\vec{a}|+|\vec{b}|+|\vec{c}| is equal to :

Option: 1

10


Option: 2

14


Option: 3

16


Option: 4

18


Answers (1)

best_answer

\begin{aligned} &\therefore All \: of\: \vec{a} \times \vec{b}, \vec{b} \: \times \vec{c}\: and\: \vec{c} \times \vec{a} \: are\\ &\therefore(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=|a||b| \sin 120^{\circ} \cdot |b| |c| \sin 120^{\circ}\\ &=\frac{3}{4}|a||b|^{2}|c|\\ &\therefore \text{Required expression}\\ &\Rightarrow \mathrm{\frac{3}{4}\left(|a||b|^{2}|c|+|a||b||c|^{2}+|a|^{2}|b||c|\right)=168}\\ &\mathrm{\Rightarrow|a||b||c|(|a|+|b|+|c|)=224 }\\ &\mathrm{\Rightarrow|a|+|b|+|c|=16}\\ &\therefore \text{option (c) } \end{aligned}

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