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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

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

$[\overrightarrow{a} + \overrightarrow{b},\overrightarrow{b} + \overrightarrow{c},\overrightarrow{c} +\overrightarrow{a}]$
$\qquad=2 [\overrightarrow{a} , \overrightarrow{b},\overrightarrow{c} ]$
$8=2 [\overrightarrow{a} , \overrightarrow{b},\overrightarrow{c} ]$
$[\overrightarrow{a} , \overrightarrow{b},\overrightarrow{c} ]=\large\frac{8}{2}$
$\qquad=4$
Hence 1 is the correct answer.