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For any vector $\overrightarrow{a}$ Prove that $\overrightarrow{i} \times (\overrightarrow{a}\times\overrightarrow{i})+ \overrightarrow{j} \times (\overrightarrow{a}\times\overrightarrow{j})+ \overrightarrow{k } \times(\overrightarrow{a}\times\overrightarrow{k})=\overrightarrow{2a}$

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  • If $ \overrightarrow a = a_1\overrightarrow i + a_2 \overrightarrow j+a_3 \overrightarrow k,\: \: \overrightarrow b = b_1 \overrightarrow i+b_2\overrightarrow j + b_3 \overrightarrow k$ then $ \overrightarrow a.\overrightarrow b = a_1b_1+a_2b_2+a_3b_3$
  • Vector triple product $(\overrightarrow a \times \overrightarrow b) \times \overrightarrow c=(\overrightarrow a.\overrightarrow c)\overrightarrow b-(\overrightarrow b.\overrightarrow c)\overrightarrow a$ $\overrightarrow a \times (\overrightarrow b \times \overrightarrow c)= (\overrightarrow a.\overrightarrow c)\overrightarrow b - (\overrightarrow a.\overrightarrow b)\overrightarrow c$
Let $ \overrightarrow a = a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k$
LHS $ = \overrightarrow i \times (\overrightarrow a \times \overrightarrow i) + \overrightarrow j \times (\overrightarrow a \times \overrightarrow j)+\overrightarrow k \times (\overrightarrow a \times \overrightarrow k)$
$ = (\overrightarrow i.\overrightarrow i)\overrightarrow a-(\overrightarrow i.\overrightarrow a)\overrightarrow i+(\overrightarrow j.\overrightarrow j)\overrightarrow a-(\overrightarrow j.\overrightarrow a)\overrightarrow j+(\overrightarrow k.\overrightarrow k)\overrightarrow a-(\overrightarrow k.\overrightarrow a)\overrightarrow k$
$ = \overrightarrow a-a_1\overrightarrow i+\overrightarrow a-a_2\overrightarrow j+\overrightarrow a-a_3\overrightarrow k$
$ = 3\overrightarrow a-(a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k)=3\overrightarrow a-\overrightarrow a=2\overrightarrow a =$ RHS


answered Jun 9, 2013 by thanvigandhi_1
edited Jun 23, 2013 by thanvigandhi_1

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