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 cos⁡θ1+3cos⁡θ2 is

|a→+b→+3c→|2=16⇒ |a→|2+|b→|2+9|c→|2+2cos⁡θ1+6cos⁡θ2+6cos⁡θ3=16,θ3∈[π/6,2π/3] or 2cos⁡θ1+6cos⁡θ2=5−6cos⁡θ3 or cos⁡θ1+3cos⁡θ2max=4

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 cos⁡θ1+3cos⁡θ2 is

|a→+b→+3c→|2=16⇒ |a→|2+|b→|2+9|c→|2+2cos⁡θ1+6cos⁡θ2+6cos⁡θ3=16,θ3∈[π/6,2π/3] or 2cos⁡θ1+6cos⁡θ2=5−6cos⁡θ3 or cos⁡θ1+3cos⁡θ2max=4

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$\vec{a}, \vec{b} and \vec{c}$ are unit vectors such that $| \vec{a} + \vec{b} + 3 \vec{c} | = 4$ Angle between $\vec{a} and \vec{b} is θ_{1}$ between $\vec{b} and \vec{c} is θ_{2}$ $\vec{a} and \vec{c} varies [π / 6, 2 π / 3]$ .Then the maximum value of $\cos θ_{1} + 3 \cos θ_{2} is$

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$2 \sqrt{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} | \vec{a} + \vec{b} + 3 \vec{c} |^{2} = 16 \\ \Rightarrow | \vec{a} |^{2} + | \vec{b} |^{2} + 9 | \vec{c} |^{2} + 2 \cos θ_{1} + 6 \cos θ_{2} + 6 \cos θ_{3} = 16, θ_{3} \in [π / 6, 2 π / 3] \\ or 2 \cos θ_{1} + 6 \cos θ_{2} = 5 - 6 \cos θ_{3} \\ or {(\cos θ_{1} + 3 \cos θ_{2})}_{max} = 4 \end{array}$