A) 0
B) 2[a b c]
C) a + b + c
D) 3[a b c]
Correct Answer: A
Solution :
\[\because \]\[\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=(\mathbf{b}\,\mathbf{.}\,\mathbf{c})\mathbf{a}-(\mathbf{a}\,\mathbf{.}\,\mathbf{b})\mathbf{c}\] \\[\mathbf{a}\times (\mathbf{b}\times \mathbf{c})+\mathbf{b}\times (\mathbf{c}\times \mathbf{a})+\mathbf{c}\times (\mathbf{a}\times \mathbf{b})\] \[=(\mathbf{b}\,\mathbf{.}\,\mathbf{c})\mathbf{a}-(\mathbf{a}\,\mathbf{.}\,\mathbf{b})\mathbf{c}+(\mathbf{b}\,\mathbf{.}\,\mathbf{a})\mathbf{c}-(\mathbf{b}\,\mathbf{.}\,\mathbf{c})\mathbf{a}\]\[+(\mathbf{b}\,\mathbf{.}\,\mathbf{c})\mathbf{a}-(\mathbf{a}\,\mathbf{.}\,\mathbf{c})\mathbf{b}\] \[=0\], \[\{\because \,\,\,\,\mathbf{a}\,\mathbf{.}\,\mathbf{b}=\mathbf{b}\mathbf{.a}\,\,\,\,etc.\}\].
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