Vector $\overrightarrow{A}$ is of length $2\text{cm}$ and is inclined at ${60}^{\circ }$ above the $x-$axis in the first quadrant. Vector $\overrightarrow{B}$ is of length $2\text{cm}$ and inclined at ${60}^{\circ }$ below the $x-$axis in the fourth quadrant. Then $(\overrightarrow{A}-\overrightarrow{B})$ is a vector of magnitude

The correct option is B $2\sqrt{3}\text{cm}$ along $(+ve)y-$ axis

Let us assume,
$\overrightarrow{R}=\overrightarrow{A}-\overrightarrow{B}$

$\overrightarrow{R}=(A\cos {60}^{\circ }\hat{i}+A\sin {60}^{\circ }\hat{j})-(B\cos {60}^{\circ }\hat{i}-B\sin {60}^{\circ }\hat{j})$

$\because |\overrightarrow{A}|=|\overrightarrow{B}|=2\text{cm}$

So, $R=(A\sin {60}^{\circ }+B\sin {60}^{\circ })\hat{j}=2\sqrt{3}\hat{j}$

Hence, $(\overrightarrow{A}-\overrightarrow{B})$ is $2\sqrt{3}\text{cm}$ along $(+ve)y-$ axis.

Therefore, option $\left(\text{B}\right)$ is correct.