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: 47


$30^{\circ}$ or $\pi/6$

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 $ax+by=c$ ${\bf n_{1}}=(\sqrt{3}){\bf i} + (-1){\bf j}=\langle\sqrt{3},-1\rangle$ is perpendicular to $\sqrt{3}x-y=-2$ ${\bf n_{2}}=(1){\bf i} + (-\sqrt{3}){\bf j}=\langle 1,-\sqrt{3}\rangle$ is perpendicular to $x-\sqrt{3}y=1.$ $\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$ $=\displaystyle \cos^{-1}(\frac{\sqrt{3}+\sqrt{3}}{\sqrt{3+1}\cdot\sqrt{1+3}})$ $=\displaystyle \cos^{-1}(\frac{2\sqrt{3}}{4})$ $=\displaystyle \cos^{-1}(\frac{\sqrt{3}}{2})=30^{\circ}$ or $\pi/6$
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