If →a,→b,→c are three unit vectors, then |→a−→b|2+|→b−→c|2+|→c−→a|2 does not exceed
A
6
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B
8
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C
4
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D
9
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Solution
The correct option is D9 We know that, |→a+→b+→c|2≥0 ⇒|→a|2+|→b|2+|→c|2+2(→a.→b+→b.→c+→c.→a)≥0 ⇒1+1+1+2(→a.→b+→b.→c+→c.→a)≥0 ∵|→a|=|→b|=|→c|=1 ⇒2(→a.→b+→b.→c+→c.→a)≥−3
We know, |→a−→b|2+|→b−→c|2+|→c−→a|2=2(|→a|2+|→b|2+|→c|2−→a.→b−→b.→c−→c.→a) ⇒|→a−→b|2+|→b−→c|2+|→c−→a|2≤2(1+1+1)−(−3) ⇒|→a−→b|2+|→b−→c|2+|→c−→a|2≤9