First slide

Dot product or scalar product of vectors

Question

$\vec{a}, \vec{b} and \vec{c}$ 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 $\cos θ_{1} + 3 \cos θ_{2} is$

Moderate

A

3
B

4
C

$2 \sqrt{2}$
D

6

Solution

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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App