## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 685: 58

#### Answer

$$\theta=94.79^o.$$

#### Work Step by Step

Since $n_1=\langle 1,-3,1\rangle$ and $n_2=\langle 2,0,-3\rangle$, then we have \begin{align*} \cos \theta &=\frac{n_1 \cdot n_2}{\|n_1\|\|n_2\|}\\ &=\frac{2+0-3}{\sqrt{1+9+1}\sqrt{4+0+9}}\\ &=\frac{-1}{\sqrt{11}\sqrt{13}}. \end{align*} That is $$\theta=94.79^o.$$

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