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If $\overrightarrow{a}\times(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{b} \times (\overrightarrow{c} \times \overrightarrow{a})+\overrightarrow{c} \times (\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{x}\times\overrightarrow{y} $ then

\[\begin{array}{1 1}(1)\overrightarrow{x}=\overrightarrow{0}&(2)\overrightarrow{y}=\overrightarrow{0}\\(3)\overrightarrow{x} and \overrightarrow{y} \;are \;parallel&(4)\overrightarrow{x}=\overrightarrow{0} or\overrightarrow{y}=\overrightarrow{0} or\overrightarrow{x} and \overrightarrow{y} \;are \;parallel\end{array}\]

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$\overrightarrow{a}\times(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{b} \times (\overrightarrow{c} \times \overrightarrow{a})+\overrightarrow{c} \times (\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{x}\times\overrightarrow{y} $
but $\overrightarrow{a}\times(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{b} \times (\overrightarrow{c} \times \overrightarrow{a})+\overrightarrow{c} \times (\overrightarrow{a}\times \overrightarrow{b})=0 $
$\overrightarrow{x}+\overrightarrow{y}=\overrightarrow{0} $
$\overrightarrow{x}=\overrightarrow{0} or\overrightarrow{y}=\overrightarrow{0} or\overrightarrow{x} and \overrightarrow{y} \;are \;parallel$
Hence 4 is the correct answer.
answered May 5, 2014 by meena.p
 

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