Three vectors |→a|,∣∣→b∣∣ and |→c| satisfy the condition →a+→b+→c=0. Evaluate the quantity λ=→a.→b+→b.→c+→c.→a, if |→a|=1,∣∣→b∣∣=4and|→c|=2.
A
212
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B
−212
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C
−172
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D
172
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Solution
The correct option is C−212 →a+→b+→c=0→a.(→a+→b+→c)=→a.0→a.→a+→a.→b+→a.→c=0|→a|2+→a.→b+→a.→c=0(therefore→a.→a=|→a|2)→a.→b+→a.→c=−1....equation(1)(since|→a|=1)Now,→a+→b+→c=0→b.(→a+→b+→c)=→b.0→b.→a+→b.→b+→b.→c=0→a.→b+∣∣→b∣∣2+→b.→c=0→a.→b+→b.→c=−16....equation(2)(→b.→b=|→b|2)and(→a.→b=→b.→a)Also→a+→b+→c=0→c(→a+→b+→c)=→c.0→a.→c+→b.→c+→c.→c=0(Since→a.→b=→b.→a)