Chapter 9 - Section 9.4 - Vectors in Three Dimensions - 9.4 Exercises - Page 658: 35

$100.9^{\circ}$

$\textbf{u}\cdot\textbf{v}=(0\times1)+(1\times2)+(1\times-3)= -1$ $u=\sqrt {1^{2}+1^{2}}=\sqrt 2$ $v=\sqrt {1^{2}+2^{2}+(-3)^{2}}=\sqrt {14}$ $\theta= \cos^{-1}(\frac{\textbf{u}\cdot\textbf{v}}{uv})=\cos^{-1}(\frac{-1}{\sqrt 2\times\sqrt {14}})= 100.9^{\circ}$