Answer
$\mathbf{v}=-6\sqrt{2}\mathbf{i}-6\sqrt{2}\mathbf{j}$
Work Step by Step
Therefore,
$\begin{align}
& \mathbf{v}=\parallel v\parallel \cos {{225}^{\circ }}\mathbf{i}+\parallel v\parallel \sin {{225}^{\circ }}\mathbf{j} \\
& =\left[ 12\times \left( -\frac{1}{\sqrt{2}} \right) \right]\mathbf{i}+\left[ 12\times \left( -\frac{1}{\sqrt{2}} \right) \right]\mathbf{j} \\
& =-6\sqrt{2}\mathbf{i}-6\sqrt{2}\mathbf{j}
\end{align}$
