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If$[\overrightarrow{a} + \overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c} + \overrightarrow{a}]=8$ then $[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$ is

\[\begin{array}{1 1}(1)4&(2)16\\(3)32&(4)-4\end{array}\]

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$[\overrightarrow{a} + \overrightarrow{b},\overrightarrow{b} + \overrightarrow{c},\overrightarrow{c} +\overrightarrow{a}]$
$\qquad=2 [\overrightarrow{a} , \overrightarrow{b},\overrightarrow{c} ]$
$8=2 [\overrightarrow{a} , \overrightarrow{b},\overrightarrow{c} ]$
$ [\overrightarrow{a} , \overrightarrow{b},\overrightarrow{c} ]=\large\frac{8}{2}$
$\qquad=4$
Hence 1 is the correct answer.
answered May 12, 2014 by meena.p
 

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