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: 50


$\approx 67.62^{\circ} $ or $1.18$ radians

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 $ax+by=c$ ${\bf n_{1}}=(12){\bf i} + (5){\bf j}=\langle 12,5\rangle$ is perpendicular to $12x+5y=1$ ${\bf n_{2}}=(2){\bf i} + (-2){\bf j}=\langle 2,-2\rangle$ is perpendicular to $2x-2y=3.$ $\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$ $=\displaystyle \cos^{-1}(\frac{24-10}{\sqrt{144+25}\cdot\sqrt{4+4}})$ $=\displaystyle \cos^{-1}(\frac{14}{13\cdot 2\sqrt{2}})\approx 67.62^{\circ} $ or $1.18$ radians
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