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Question

If a, b and c are unit vectors, then the maximum value of |ab|2+|bc|2+|ca|2 is

A
1
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B
4
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C
9
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Solution

The correct option is C 9
|ab|2+|bc|2+|ca|2=2(a2+b2+c2)2(a.b+b.c+c.a)=2×32(a.b+b.c+c.a)=6(a+b+c)2+a2+b2+c2=9|a+b+c|2

The minimum value of |a+b+c| is 0. Hence, the maximum value of |ab|2+|bc|2+|ca|2 is 9 - 0 which is 9.

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