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:
Let angle between $\vec{a}$ and $\vec{b}$ be $\theta$.
$|\vec{a}+\vec{b}|=\sqrt{1+1+2 \cos \theta}=2\left|\cos \frac{\theta}{2}\right| \quad[\because|a|=|b|=1]$
Similarly, $|\vec{a}-\vec{b}|=2\left|\sin \frac{\theta}{2}\right|$
So, $\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|=2\left[\sqrt{3}\left|\cos \frac{\theta}{2}\right|+\left|\sin \frac{\theta}{2}\right|\right]$
$\because$ Maximum value of $(a \cos \theta+b \sin \theta)=\sqrt{a^{2}+b^{2}}$
$\therefore$ Maximum value $=2 \sqrt{(\sqrt{3})^{2}+(1)^{2}}=4$