Let $\vec{a}, \vec{b}, \vec{c}$ be any three vectors. Then $[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]$ is always equal to:
Answer & explanation
Correct answer: option 2
${[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]}$
$=(\vec{a}+\vec{b}) . ((\vec{b}+\vec{c}) \times(\vec{c}+\vec{a}))$
$=(\vec{a}+\vec{b}) . (\vec{b}\times\vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{a})$
$=[\vec{a} \vec{b} \vec{c}]+[\vec{b} \vec{c} \vec{a}]$
$=2[\vec{a} \vec{b} \vec{c}]$
Hence (2) is correct answer.