JEE Main 202117 Mar, Morning shiftMathematicsHardVector Algebra184× since 2002If a→=αi^+βj^+3k^\overrightarrow a = \alpha \widehat i + \beta \widehat j + 3\widehat ka=αi+βj+3k, b→=−βi^−αj^−k^\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat kb=−βi−αj−k and c→=i^−2j^−k^\overrightarrow c = \widehat i - 2\widehat j - \widehat kc=i−2j−k such that a→ . b→=1\overrightarrow a \,.\,\overrightarrow b = 1a.b=1 and b→ . c→=−3\overrightarrow b \,.\,\overrightarrow c = - 3b.c=−3, then 13((a→×b→) . c→){1 \over 3}\left( {\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c } \right)31((a×b).c) is equal to _____________.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
