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 and between a→and c→ varies in [π/6, 2π/3]. Then maximum value of cosθ1+3cosθ2 is

|a→+b→+3c→|2=16 ⇒|a→|2+|b→|2+9|c→|2+1cos⁡θ1+6cos⁡θ2+6cos⁡θ3=16,θ3∈[π/6,2π/3] or 2cos⁡θ1+6cos⁡θ2=5−6cos⁡θ3,θ3∈[π/6,2π/3] 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 and between a→and c→ varies in [π/6, 2π/3]. Then maximum value of cosθ1+3cosθ2 is

|a→+b→+3c→|2=16 ⇒|a→|2+|b→|2+9|c→|2+1cos⁡θ1+6cos⁡θ2+6cos⁡θ3=16,θ3∈[π/6,2π/3] or 2cos⁡θ1+6cos⁡θ2=5−6cos⁡θ3,θ3∈[π/6,2π/3] or cos⁡θ1+3cos⁡θ2max=4

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\vec{a}, \vec{b} and \vec{c}$ 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 $θ_{2}$ and between $\vec{a}$ and $\vec{c}$ varies in $[π / 6, 2 π / 3] .$ Then maximum value of ${cosθ}_{1} + 3 {cosθ}_{2}$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

22

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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