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Question

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=0a.(a+b+c)=a.0a.a+a.b+a.c=0|a|2+a.b+a.c=0(thereforea.a=|a|2)a.b+a.c=1....equation(1)(since|a|=1)Now,a+b+c=0b.(a+b+c)=b.0b.a+b.b+b.c=0a.b+b2+b.c=0a.b+b.c=16....equation(2)(b.b=|b|2)and(a.b=b.a)Alsoa+b+c=0c(a+b+c)=c.0a.c+b.c+c.c=0(Sincea.b=b.a)

a.c+b.c+|c|2=0 (c.c=|c|2)

a.c+b.c=4 ----- Equation (3)

On adding 1,2,3

2(a.b+b.c+c.a)=1164

(a.b+b.c+c.a)=212

λ=212

Hence option B is the correct answer.

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