Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 714: 48


$45^{\circ}$ or $\pi/4$

Work Step by Step

The result of exercise 31 tells us that ${\bf n}=a{\bf i} + b{\bf j}$ is perpendicular to the lines $ax+by=c$ ${\bf n_{1}}=(1){\bf i} + (\sqrt{3}){\bf j}=\langle 1,\sqrt{3}\rangle$ is perpendicular to $x+\sqrt{3}y=1$ ${\bf n_{2}}=(1-\sqrt{3}){\bf i} + (1+\sqrt{3}){\bf j}=\langle 1-\sqrt{3},1+\sqrt{3}\rangle$ is perpendicular to $(1-\sqrt{3})x+(1+\sqrt{3})y=8.$ $\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$ $=\displaystyle \cos^{-1}(\frac{(1)(1-\sqrt{3})+\sqrt{3}(1+\sqrt{3})}{\sqrt{1+3}\cdot\sqrt{(1-\sqrt{3})^{2}+(1+\sqrt{3})}})$ $=\displaystyle \cos^{-1}(\frac{1-\sqrt{3}+\sqrt{3}+3}{2\cdot\sqrt{1-2\sqrt{3}+3+1+2\sqrt{3}+3}})$ $=\displaystyle \cos^{-1}(\frac{4}{2\cdot\sqrt{8}})$ $=\displaystyle \cos^{-1}(\frac{1}{\sqrt{2}})=45^{\circ}$ or $\pi/4$
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