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For any vector $\bar r,\bar {i} \times (\bar {r} \times \bar {i}) \div \bar {j} \times (\bar {r} \times \bar {j})+ \bar k \times (\bar {r} \times {\bar k})=$

$(a)\;\bar {0} \qquad(b)\;2 \bar {r} \qquad(c)\;3 \bar {r} \qquad (d)\;4 \bar {r}$

1 Answer

$(b)\;2 \bar {r}$
answered Nov 7, 2013 by pady_1
 

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