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Question
Prove that, for any three vectors \[\vec{a} , \vec{b} , \vec{c}\] \[\left[ \vec{a} + \vec{b} , \vec{b} + \vec{c} , \vec{c} + \vec{a} \right] = 2 \left[ \vec{a} , \vec{b} , \vec{c} \right]\].
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Solution
We have:
\[\left[ \vec{a} + \vec{b} , \vec{b} + \vec{c} , \vec{c} + \vec{a} \right]\]
\[ = \left( \vec{a} + \vec{b} \right) . \left[ \left( \vec{b} + \vec{c} \right) \times \left( \vec{c} + \vec{a} \right) \right]\]
\[ = \left( \vec{a} + \vec{b} \right) . \left[ \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a} \right] (\text { By distributive law })\]
\[ = \left( \vec{a} + \vec{b} \right) . \left[ \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a} \right] ( \because \vec{c} \times \vec{c} = 0)\]
\[ = \vec{a} . \left( \vec{b} \times \vec{c} \right) + \vec{b} . \left( \vec{b} \times \vec{c} \right) + \vec{a} . \left( \vec{b} \times \vec{a} \right) + \vec{b .} \left( \vec{b} \times \vec{a} \right) + \vec{a} . \left( \vec{c} \times \vec{a} \right) + \vec{b .} \left( \vec{c} \times \vec{a} \right)\]
\[ = \left[ \vec{a} , \vec{b} , \vec{c} \right] + \left[ \vec{b} , \vec{b} , \vec{c} \right] + \left[ \vec{a} , \vec{b} , \vec{a} \right] + \left[ \vec{b} , \vec{b} , \vec{a} \right] + \left[ \vec{a} , \vec{c} , \vec{a} \right] + \left[ \vec{b} , \vec{c} , \vec{a} \right] \]
\[ = \left[ \vec{a} , \vec{b} , \vec{c} \right] + \left[ \vec{b} , \vec{c} , \vec{a} \right] ( \because \text { scalar triple product with two equal vectors is } 0) \]
\[ = \left[ \vec{a} , \vec{b} , \vec{c} \right] + \left[ \vec{a} , \vec{b} , \vec{c} \right] \left( \because \left[ \vec{b} , \vec{c} , \vec{a} \right] = \left[ \vec{a} , \vec{b} , \vec{c} \right] \right)\]
\[ = 2\left[ \vec{a} , \vec{b} , \vec{c} \right]\]
Hence,
\[\left[ \vec{a} + \vec{b} , \vec{b} + \vec{c} , \vec{c} + \vec{a} \right] = 2\left[ \vec{a} , \vec{b} , \vec{c} \right]\]
