MathematicsHard202×since 2002Q5820Let a→\overrightarrow aa and b→\overrightarrow bb be the vectors along the diagonals of a parallelogram having area 222\sqrt 222. Let the angle between a→\overrightarrow aa and b→\overrightarrow bb be acute, ∣a→∣=1|\overrightarrow a | = 1∣a∣=1, and ∣a→ . b→∣=∣a→×b→∣|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |∣a.b∣=∣a×b∣. If c→=22(a→×b→)−2b→\overrightarrow c = 2\sqrt 2 \left( {\overrightarrow a \times \overrightarrow b } \right) - 2\overrightarrow bc=22(a×b)−2b, then an angle between b→\overrightarrow bb and c→\overrightarrow cc is :Aπ4{\pi \over 4}4πB−-− π4{\pi \over 4}4πC5π6{{5\pi } \over 6}65πD3π4{{3\pi } \over 4}43πCheck answerSkip