Answer
Vector $\mathbf{v}$ can be written as $\mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j}$.
Work Step by Step
Let $\mathbf{v}$ be any non-zero vector given as $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$.
In order to express the vector in terms of magnitude and direction, first, calculate $\left\| \mathbf{v} \right\|$
$\left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ …… (1)
Calculate the direction angle $\theta $ measured from the positive x-axis to $\mathbf{v}$ as follows:
$\cos \theta =\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
$\theta ={{\cos }^{-1}}\left( \frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)$ …… (2)
Then vector $\mathbf{v}$ can be written as
$\mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j}$ …… (3)
Example: Information:
Let vector $\mathbf{v}=2\mathbf{i}+3\mathbf{j}$.
Calculate $\left\| \mathbf{v} \right\|$
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
& =\sqrt{{{2}^{2}}+{{3}^{2}}} \\
& =\sqrt{13}
\end{align}$
Calculate $\theta $
$\begin{align}
& \theta ={{\cos }^{-1}}\left( \frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right) \\
& ={{\cos }^{-1}}\left( \frac{2}{\sqrt{13}} \right) \\
& =56.30
\end{align}$
Put the above value in equation (3) to get
$\begin{align}
& \mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j} \\
& =\sqrt{13}\cos 56.30\mathbf{i}+\sqrt{13}\sin 56.30\mathbf{j}
\end{align}$
The vector $\mathbf{v}$ is $\sqrt{13}\cos 56.30\mathbf{i}+\sqrt{13}\sin 56.30\mathbf{j}$.
