If a → and b → are unit vectors, then the greatest value of 3 | a → + b → | + | a…

Let angle between $\overrightarrow{\mathrm{a}} \text{and} \overrightarrow{\mathrm{b}}=\alpha$.
So, $\sqrt{3}|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|+|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{3}\sqrt{2+2\mathrm{cos\alpha }}+\sqrt{2-2\mathrm{cos\alpha }}$
$=\sqrt{3}\sqrt{2\times 2{\cos }^2\frac{\alpha }{2}}+\sqrt{2\times 2{\sin }^2\frac{\alpha }{2}}$
$=2\left(\sqrt{3}\left|\cos \frac{\alpha }{2}\right|+\left|\sin \frac{\alpha }{2}\right|\right)$

As $\alpha \in \left[0,\mathrm{\pi }\right]\Rightarrow \frac{\alpha }{2}\in \left[0,\frac{\mathrm{\pi }}{2}\right]$

So, given expression is

$=2\left(\sqrt{3}\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}\right)$
Thus, maximum value is$=2\sqrt{{\left(\sqrt{3}\right)}^2+1^2}=2\times 2=4$