Chapter 6.2, Problem 19E

Solution

To Find:

The angle between the given vectors using an algebraic method. Approximate the exact answers using a calculator when needed.

  ${135}^{°}$ .

Given:

The vectors $u=\left(2\cos \frac{\pi }{4}\right)i+\left(2\sin \frac{\pi }{4}\right)j$ and $v=\left(\cos \frac{3\pi }{2}\right)i+\left(\sin \frac{3\pi }{2}\right)j$ .

Concepts Used:

If $u=ai+bj$ then $\left|u\right|=\sqrt{a^2+b^2}$ .

The dot product of the vectors $u=u_{1}i+u_{2}j$ and $v=v_{1}i+v_{2}j$ is defined as:

  $u\cdot v=u_1{v}_1+u_2{v}_2$

The smaller angle $\theta$ between two vectors $u$ and $v$ is given by ${\cos }^{-1}\left(\frac{u\cdot v}{\left|u\right|\left|v\right|}\right)$

Calculations:

Determine the angle between the vectors $u$ and $v$ .

  

  

  

  $={\cos }^{-1}\left(\frac{\left(\frac{2}{\sqrt{2}}\right)\left(0\right)+\left(\frac{2}{\sqrt{2}}\right)\left(-1\right)}{\sqrt{{\left(\frac{2}{\sqrt{2}}\right)}^2+{\left(\frac{2}{\sqrt{2}}\right)}^2}\sqrt{{\left(0\right)}^2+{\left(-1\right)}^2}}\right)$

  $={\cos }^{-1}\left(\frac{-\frac{2}{\sqrt{2}}}{\sqrt{4}\sqrt{1}}\right)$

  $={\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

  ${\text{=135}}^{°}$

Conclusion:

The angle between the given vectors is ${135}^{°}$ .