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Question

If a,b,c are unit vectors, then |a2b|2+|b2c|2+|c2a|2 does not exceed ?

A
18
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B
21
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C
27
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D
30
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Solution

The correct option is B 21
Given expression,
|a2b|2+|b2c|2+|c2a|2=5(|a|2+|b|2+|c|2)4(ab+bc+ca)=7(|a|2+|b|2+|c|2)2(|a|2+|b|2+|c|2+2ab+2bc+2ca)=7×32[(a+b+c)2]=212X
where, X=|a+b+c|2
So, for maximum value of the expression : X=0
So, |a2b|2+|b2c|2+|c2a|2212×021

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