MathematicsHard202×since 2002Q5814Let a→\overrightarrow aa and b→\overrightarrow bb be two non-zero vectors perpendicular to each other and ∣a→∣=∣b→∣|\overrightarrow a | = |\overrightarrow b |∣a∣=∣b∣. If ∣a→×b→∣=∣a→∣|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |∣a×b∣=∣a∣, then the angle between the vectors (a→+b→+(a→×b→))\left( {\overrightarrow a + \overrightarrow b + \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)(a+b+(a×b)) and a→{\overrightarrow a }a is equal to :Asin−1(16){\sin ^{ - 1}}\left( {{1 \over {\sqrt 6 }}} \right)sin−1(61)Bcos−1(12){\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right)cos−1(21)Csin−1(13){\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)sin−1(31)Dcos−1(13){\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)cos−1(31)Check answerSkip