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