Solve this following

Question:

If $\vec{a}$ and $\vec{b}$ are unit vectors, then the greatest

value of $\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|$ is

Solution:

$\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|$

$=\sqrt{3}(\sqrt{2+2 \cos \theta})+\sqrt{2-2 \cos \theta}$

$=\sqrt{6}(\sqrt{1+\cos \theta})+\sqrt{2}(\sqrt{1-\cos \theta})$

$=2 \sqrt{3}\left|\cos \frac{\theta}{2}\right|+2\left|\sin \frac{\theta}{2}\right|$

$\leq \sqrt{(2 \sqrt{3})^{2}+(2)^{2}}=4$

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