Let →A,→B and →C be unit vectors. Suppose that →A⋅→B=→A⋅→C=0 and that the angle between →B and →C is π6 then →A=
A
±2(→B×→C)
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B
±(→B×→C)
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C
±2(→B+→C)
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D
±(→B+→C)
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Solution
The correct option is A±2(→B×→C) Since, →A⋅→B=0,→A⋅→C=0 Hence, (→B+→C)⋅→A=0 So →A is perpendicular to (→B+→C)⋅→A is a unit vector perpendicular to the plane of vectors →B and →C. →A=→B×→C∣∣→B×→C∣∣ ∣∣→B×→C∣∣=∣∣→B∣∣∣∣→C∣∣sinπ6=1×1×12=12 ∴→A=→B×→C∣∣→B×→C∣∣=±2(→B×→C)