Chapter 6.4, Problem 30E

To determine

To Find: -The angle $\theta$ between two vectors.

Answer to Problem 30E

The angle between two angle is $\frac{5\pi }{12}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}u=\cos \frac{\pi }{3}i+\sin \frac{\pi }{3}j\\ v=\cos \frac{3\pi }{4}i+\sin \frac{3\pi }{4}j\end{array}$

Formula used: -

  $\begin{array}{l}\cos \theta =\frac{u.v}{\left|\left|u\right|\right|\left|\left|v\right|\right|}\\ u.v=a_1{a}_2+b_1{b}_2\\ \left|\left|u\right|\right|=\sqrt{a_1{}^2+b_1{}^2}\\ \\ \cos A\cos B+\sin A\sin B=\cos \left(A-B\right)\end{array}$

  ${\cos }^{2}A+{\sin }^{2}A=1$

  $\begin{array}{l}\cos \theta =\frac{\left(\cos \frac{\pi }{4}\right)\left(\cos \frac{2\pi }{3}\right)+\left(\sin \frac{\pi }{4}\right)\left(\sin \frac{2\pi }{3}\right)}{\sqrt{{\left(\cos \frac{\pi }{4}\right)}^2+{\left(\sin \frac{\pi }{4}\right)}^2}\sqrt{{\left(\cos \frac{2\pi }{3}\right)}^2+{\left(\sin \frac{2\pi }{3}\right)}^2}}\\ \cos \theta =\frac{\cos \left(\frac{2\pi }{3}-\frac{\pi }{4}\right)}{\left(1\right)\left(1\right)}\\ \cos \theta =\cos \frac{5\pi }{12}\\ \theta =\frac{5\pi }{12}\end{array}$