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For any three vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ show that $\overrightarrow{a}\times(\overrightarrow{b}+\overrightarrow{c})+\overrightarrow{b}\times(\overrightarrow{c}+\overrightarrow{a})+\overrightarrow{c}\times(\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{0}$

1 Answer

  • $ \overrightarrow a \times \overrightarrow b = -(\overrightarrow b \times \overrightarrow a)\: or \: \overrightarrow b \times \overrightarrow a = -(\overrightarrow a \times \overrightarrow b)$ and $ \overrightarrow a \times ( \overrightarrow b + \overrightarrow c) = \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c$
$ LHS = \overrightarrow a \times ( \overrightarrow b+\overrightarrow c) + \overrightarrow b \times ( \overrightarrow c+\overrightarrow a) + \overrightarrow c \times ( \overrightarrow a+\overrightarrow b)$
$ = (\overrightarrow a \times \overrightarrow b)+(\overrightarrow a \times \overrightarrow c)+(\overrightarrow b \times \overrightarrow c) + ( \overrightarrow b \times \overrightarrow a)+(\overrightarrow c \times \overrightarrow a)+(\overrightarrow c \times \overrightarrow b)$
$ = (\overrightarrow a \times \overrightarrow b)-(\overrightarrow c \times \overrightarrow a)+(\overrightarrow b \times \overrightarrow c)-(\overrightarrow a \times \overrightarrow b) + (\overrightarrow c \times \overrightarrow a)-(\overrightarrow b \times \overrightarrow c)=\overrightarrow 0 = RHS $
answered Jun 5, 2013 by thanvigandhi_1

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