## Precalculus (6th Edition) Blitzer

Vector $\mathbf{v}$ can be written as $\mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j}$.
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 $\sqrt{13}\cos 56.30\mathbf{i}+\sqrt{13}\sin 56.30\mathbf{j}$.