Answer
The vector is $6\mathbf{i}+6\sqrt{3}\mathbf{j}$.
Work Step by Step
A non-zero vector $\mathbf{v}$ is expressed as $\mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j}$ if the magnitude and direction angle from positive x-axis are $\left\| \mathbf{v} \right\|\text{ and }\theta $,
Here, $\left\| \mathbf{v} \right\|=12$ and $\theta =60{}^\circ $.
So, $\mathbf{v}$ can be expressed as:
$\begin{align}
& \mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j} \\
& =12\cos 60{}^\circ \mathbf{i}+12\sin 60{}^\circ \mathbf{j} \\
& =12\left( \frac{1}{2} \right)\mathbf{i}+12\left( \frac{\sqrt{3}}{2} \right)\mathbf{j} \\
& =6\mathbf{i}+6\sqrt{3}\mathbf{j}
\end{align}$
So, the vector $\mathbf{v}$ is $6\mathbf{i}+6\sqrt{3}\mathbf{j}$.
