If →a,→b,→c are unit vectors, then |→a−2→b|2+|→b−2→c|2+|→c−2→a|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 B21 Given expression, |→a−2→b|2+|→b−2→c|2+|→c−2→a|2=5(|→a|2+|→b|2+|→c|2)−4(→a⋅→b+→b⋅→c+→c⋅→a)=7(|→a|2+|→b|2+|→c|2)−2(|→a|2+|→b|2+|→c|2+2→a⋅→b+2→b⋅→c+2→c⋅→a)=7×3−2[(→a+→b+→c)2]=21−2X
where, X=|→a+→b+→c|2
So, for maximum value of the expression : X=0
So, |→a−2→b|2+|→b−2→c|2+|→c−2→a|2≤21−2×0≤21