If →a,→b and →c are unit vectors then, |→a−→b|2+|→b−→c|2+|→c−→a|2 can take the value of
A
3
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B
5
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C
12
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D
8
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Solution
The correct options are A 3 B 5 D 8 |→a−→b|2+|→b−→c|2+|→c−→a|2=a2+b2+b2+c2+c2+a2−2[→a.→b+→b.→c+→c.→a] =6−2[→a.→b+→b.→c+→c.→a]−−−(1) Since →a,→b&→c are unit vectors |→a+→b+→c|≥0 a2+b2+c2+2[→a.→b+→b.→c+→c−→a]≥0 3+2[→a.→b+→b.→c+→c.→a]≥0[→a.→b+→b.→c+→c.→a]≥−32−−−(2) By (1) & (2), |→a−→b|2+|→b−→c|2+|→c−→a|2≤6−2−32≤9