Chapter 6, Problem 83RE

To determine

The angle $\theta$ between the vectors.

Answer to Problem 83RE

The angle $\theta$ between the vectors is $\frac{11\pi }{12}$

Explanation of Solution

Given info:

  $\begin{array}{l}u=\cos \left(\frac{7\pi }{4}\right)i+\sin \left(\frac{7\pi }{4}\right)j\\ v=\cos \left(\frac{5\pi }{6}\right)i+\sin \left(\frac{5\pi }{6}\right)j\end{array}$

Formula used:

If $\theta$ is the angle between two nonzero vectors u and v then

  $\cos \theta =\frac{u.v}{\Vert u\Vert \Vert v\Vert }$

Calculation:

Given that

  $\begin{array}{l}u=\cos \left(\frac{7\pi }{4}\right)i+\sin \left(\frac{7\pi }{4}\right)j\\ v=\cos \left(\frac{5\pi }{6}\right)i+\sin \left(\frac{5\pi }{6}\right)j\\ \cos \theta =\frac{u.v}{\Vert u\Vert \Vert v\Vert }=\frac{\left(\cos \left(\frac{7\pi }{4}\right)i+\sin \left(\frac{7\pi }{4}\right)j\right).\left(\cos \left(\frac{5\pi }{6}\right)i+\sin \left(\frac{5\pi }{6}\right)j\right)}{\sqrt{{\cos }^2\left(\frac{7\pi }{4}\right)+{\sin }^2\left(\frac{7\pi }{4}\right)}.\sqrt{{\cos }^2\left(\frac{5\pi }{6}\right)+{\sin }^2\left(\frac{5\pi }{6}\right)}}\\ \cos \theta =\frac{\cos \left(\frac{7\pi }{4}\right).\cos \left(\frac{5\pi }{6}\right)+\sin \left(\frac{7\pi }{4}\right)\sin \left(\frac{5\pi }{6}\right)}{1}\\ \cos \theta =\cos \left(\frac{7\pi }{4}-\frac{5\pi }{6}\right)\\ \theta =\frac{7\pi }{4}-\frac{5\pi }{6}=\frac{22\pi }{24}\\ \theta =\frac{11\pi }{12}\end{array}$

Conclusion:

Thus,theangle $\theta$ between the vectors is $\frac{11\pi }{12}$