MathematicsMedium202×since 2002Q5730Let a→=2i^+j^−2k^\overrightarrow a = 2\widehat i + \widehat j - 2\widehat ka=2i+j−2k and b→=i^+j^\overrightarrow b = \widehat i + \widehat jb=i+j. If c→\overrightarrow cc is a vector such that a→. c→=∣c→∣,∣c→−a→∣=22\overrightarrow a .\,\overrightarrow c = \left| {\overrightarrow c } \right|,\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2a.c=c,c−a=22 and the angle between (a→×b→)(\overrightarrow a \times \overrightarrow b )(a×b) and c→\overrightarrow cc is π6{\pi \over 6}6π, then the value of ∣(a→×b→)×c→∣\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right|(a×b)×c is :A23{2 \over 3}32B4C3D32{3 \over 2}23Check answerSkip