If vectors →a,→band →c are unit vectors such that →a+→b+→c=0 then the value of (→a.→b+→b.→c+→c.→a) is
A
12
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B
−12
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C
32
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D
−32
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Solution
The correct option is D−32 Given →a,→band →c are three unit vectors|→a|=|→b|=|→c|=1
We know →a+→b+→c=0
Squaring both sides, (→a+→b+→c)2=0 |→a|2+|→b|2+|→c|2+2(→a.→b+→b.→c+→c.→a) 1+1+1+2(→a.→b+→b.→c+→c.→a)=0 2(→a.→b+→b.→c+→c.→a)=−3 (→a.→b+→b.→c+→c.→a)=−32