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Question

If a,b,c are unit vectors, then the value of a2b2+b2c2+|c2a|2 does not exceed

A
12
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B
15
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C
18
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D
21
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Solution

The correct option is C 21
|a2b|2+|b2c|2+|c2a|2

=a2+4b24ab+b2+4c24bc+c2+4a24ca

=(a2+b2+c2)+4(a2+b2+c2)4(ab+bc+ca)

=5(a2+b2+c2)4(ab+bc+ca)

=5(3)4(32)

=15+6

=21

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