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Question

If a,b,c are unit vectors such that a+b+c=0 and (a,b)=π3 then |a×b|+|b×c|+|c×a|=

A
32
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B
0
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C
332
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D
1
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Solution

The correct option is C 332
|¯a|=¯b=|¯c|=1
angle between (¯a,¯c=π3)
angle between (¯b,¯c=π3)
¯aׯb+¯bׯc+|¯cׯa|
=|¯a|¯bsin60°+¯b|¯c|sin60°+|¯c||¯a|sin60°=3×32=332

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