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Question

# If ¯a,¯b and ¯c are unit vectors, then ∣∣¯a−¯b∣∣2+∣∣¯b−¯c∣∣2+|¯c−¯a|2 does not exceed

A
4
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B
9
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C
8
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D
6
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Solution

## The correct option is D 9|→a−→b|2+|→a−→b|2+|→a−→b|2=2(a2+b2+c2)−2abcosθ−2bcosϕ−2acosγNow a=b=c=1Hence the above equation, reduces to2(3)−2(cosθ+cosϕ+cosγ)=2(3−(cosθ+cosϕ+cosγ))=kNow k is maximum for cosθ=cosϕ=cosγ=−1However this is not possible, since the above implies →a is anit-parallel to →bAnd →b is anti-parallel to →cAnd →c is anti-parallel to →a but this is not possible, since →a antiparallel to →b and →b antiparallel to →c indicates that →a||→c.Thus the above option is rules out.Consider that all the three vectors form an equilateral triangle.Then, cosθ=cosϕ=cosγ=−12, or the vectors are at 1200.Hence, we get k =2(3−(−32))=2(92)=9Thus kmax=9.Hence, |→a−→b|2+|→a−→b|2+|→a−→b|2≤9

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