1
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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
3
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Solution

## The correct option is C 3√32|→a|=|→b|=|→c|=1=> position vectors A,B,C of →a,→b,→c lie on a circle of radii 1, and origin as centre.Also, centroid of triangle formed by position vectors of →a,→b,→c = →a+→b+→c3=0=> origin is both circumcentre and centroid of the triangle=> Triangle ABC is an equilateral triangle with origin as centroid/circumcentre.=> angle between each of the →a,→b,→c is 2π3=> |→a×→b|=|→a||→b|sin(2π3)=√32Similarly, |→b×→c|=|→c×→a|=√32Thus ∣∣→a×→b∣∣+∣∣→b×→c∣∣+|→c×→a|=3√32

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