a→,b→ and c→ are unit vectors such that |a→+b→+3c→|=4 Angle between a→ and b→ is θ1 between b→ and c→ is θ2 a→ and c→ varies [π/6,2π/3] .Then the maximum value of cosθ1+3cosθ2 is
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a
3
b
4
c
22
d
6
answer is B.
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Detailed Solution
|a→+b→+3c→|2=16⇒ |a→|2+|b→|2+9|c→|2+2cosθ1+6cosθ2+6cosθ3=16,θ3∈[π/6,2π/3] or 2cosθ1+6cosθ2=5−6cosθ3 or cosθ1+3cosθ2max=4