Practicing Success
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: |
$[\vec{a}, \vec{b}, \vec{c}]$ $2[\vec{a}, \vec{b}, \vec{c}]$ Zero None of these |
$2[\vec{a}, \vec{b}, \vec{c}]$ |
${[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]}$ $=(\vec{a}+\vec{b}) . ((\vec{b}+\vec{c}) \times(\vec{c}+\vec{a}))$ $=(\vec{a}+\vec{b}) . (\vec{b}+\vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{a})$ $=[\vec{a} \vec{b} \vec{c}]+[\vec{b} \vec{c} \vec{a}]$ $=2[\vec{a} \vec{b} \vec{c}]$ Hence (2) is correct answer. |