Given vector $\vec{A} = 2\hat{i} + 3\hat{j}$,the angle between $\vec{A}$ and the $y$-axis is:

The velocity of a particle having a magnitude of $10 \ m/s$ in the direction of $60^{\circ}$ with the positive $X$-axis is:

$A$ vector is represented by $3\,\hat i + \hat j + 2\,\hat k$. Its length in the $XY$ plane is

The angles which the vector $\vec{A} = 3\hat{i} + 6\hat{j} + 2\hat{k}$ makes with the coordinate axes are

Two forces of magnitude $P$ and $Q$ acting at a point have a resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. The angle between the forces is:

The resultant of these forces $\overrightarrow{OP}, \overrightarrow{OQ}, \overrightarrow{OR}, \overrightarrow{OS}$ and $\overrightarrow{OT}$ is approximately $\ldots \ldots \text{N}$.
[Take $\sqrt{3}=1.7, \sqrt{2}=1.4$. Given $\hat{i}$ and $\hat{j}$ are unit vectors along $x, y$ axes.]