## Precalculus (6th Edition) Blitzer

The angle between $\mathbf{v}$ and $\mathbf{w}$ is ${{30}^{{}^\circ }}$
Vector $\mathbf{v=}2\cos \frac{4\pi }{3}\mathbf{i}+2\sin \frac{4\pi }{3}\mathbf{j}$ can be written as, \begin{align} & \mathbf{v=}2\cos \frac{4\pi }{3}\mathbf{i}+2\sin \frac{4\pi }{3}\mathbf{j} \\ & =2\left( -\frac{1}{2} \right)\mathbf{i}+2\left( -\frac{\sqrt{3}}{2} \right)\mathbf{j} \\ & =-\mathbf{i}-\sqrt{3}\mathbf{j} \end{align} And, Vector $\mathbf{w=}3\cos \frac{3\pi }{2}\mathbf{i}+3\sin \frac{3\pi }{2}\mathbf{j}$ can be written as, \begin{align} & \mathbf{w=}3\cos \frac{3\pi }{2}\mathbf{i}+3\sin \frac{3\pi }{2}\mathbf{j} \\ & =3\left( 0 \right)\mathbf{i}+3\left( -1 \right)\mathbf{j} \\ & =0\mathbf{i}-3\mathbf{j} \end{align} Now, let the angle between $\mathbf{v}$ and $\mathbf{w}$ be $\theta$ such that the angle between the vectors $\mathbf{v}$ and $\mathbf{w}$ can be obtained using the formula $\theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right)$ as, \begin{align} & \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( -\mathbf{i}-\sqrt{3}\mathbf{j} \right)\cdot \left( 0\mathbf{i}-3\mathbf{j} \right)}{\left( \sqrt{{{\left( -1 \right)}^{2}}+{{\left( -\sqrt{3} \right)}^{2}}} \right)\left( \sqrt{{{\left( 0 \right)}^{2}}\mathbf{+}{{\left( 3 \right)}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( -1 \right)\cdot \left( 0 \right)+\left( -\sqrt{3} \right)\cdot \left( -3 \right)}{\left( \sqrt{4} \right)\left( \sqrt{9} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{0+3\sqrt{3}}{6} \right) \end{align} Solve ahead to get the result as, \begin{align} & \theta ={{\cos }^{-1}}\left( \frac{0+3\sqrt{3}}{6} \right) \\ & ={{\cos }^{-1}}\left( \frac{\sqrt{3}}{2} \right) \\ & ={{30}^{{}^\circ }} \end{align} Hence, the angle, in degrees, between $\mathbf{v}$ and $\mathbf{w}$ is ${{30}^{{}^\circ }}$.