Answer
$\mathbf{v}=3\sqrt{3}\mathbf{i}+3\mathbf{j}$
Work Step by Step
Consider the following vector $\mathbf{v}$,
$\mathbf{v}=a\mathbf{i}+b\mathbf{j}$ (I)
The components $a$ and $b$ can be expressed in terms of the magnitude of vector $\mathbf{v}$ and direction angle $\theta $ as,
$\begin{align}
& a=\parallel \mathbf{v}\parallel \cos \theta \\
& b=\parallel \mathbf{v}\parallel \sin \theta \\
\end{align}$
Substitute the value of $a,b$ and $\theta $ in equation (I) to get,
$\begin{align}
& \mathbf{v}=\parallel v\parallel \cos {{30}^{\circ }}\mathbf{i}+\parallel v\parallel \sin {{30}^{\circ }}\mathbf{j} \\
& =\left( 6\times \frac{\sqrt{3}}{2} \right)\mathbf{i}+\left( 6\times \frac{1}{2} \right)\mathbf{j} \\
& =3\sqrt{3}\mathbf{i}+3\mathbf{j}
\end{align}$
