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If $\overrightarrow{a} \times(\overrightarrow{b}\times\overrightarrow{c})=(\overrightarrow{a}\times\overrightarrow{b})\times \overrightarrow{c}$ for non-coplanar vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ then

\[\begin{array}{1 1}(1)\overrightarrow{a} parallel\; to \overrightarrow{b}&(2)\overrightarrow{b} parallel\; to\overrightarrow{c}\\(3)\overrightarrow{c} parallel\; to \overrightarrow{a}&(4)\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\end{array}\]

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$\overrightarrow{a} \times(\overrightarrow{b}\times\overrightarrow{c})=(\overrightarrow{a} . \overrightarrow{c}) \overrightarrow{b}- (\overrightarrow{b}. \overrightarrow{c})\overrightarrow{a}$
$(\overrightarrow{a} \times \overrightarrow{b} ) \times\overrightarrow{c}=(\overrightarrow{a} . \overrightarrow{c}) \overrightarrow{b}- (\overrightarrow{b}. \overrightarrow{c})\overrightarrow{a}$
Given $\overrightarrow{a} \times(\overrightarrow{b}\times\overrightarrow{c})=(\overrightarrow{a}\times\overrightarrow{b})\times\overrightarrow{c}$
$(\overrightarrow{a} . \overrightarrow{c}) \overrightarrow{b}- (\overrightarrow{b}. \overrightarrow{c})\overrightarrow{a}=(\overrightarrow{a} . \overrightarrow{c}) \overrightarrow{b}- (\overrightarrow{b}. \overrightarrow{c})\overrightarrow{a}$
$-(\overrightarrow{a} . \overrightarrow{b})\overrightarrow{c}=-( \overrightarrow{b}.\overrightarrow{c})\overrightarrow{a}$
$\overrightarrow{c}= \large\frac{\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}$$\overrightarrow{a}$
$\overrightarrow{c} = m \overrightarrow{a} $ where $m= \large\frac{\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}$$\overrightarrow{a}$
Hence 3 is the correct answer.
answered May 12, 2014 by meena.p
 

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