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: 45


$45^{\circ}$, 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 $ax+by=c$ ${\bf n_{1}}=(3){\bf i} + (1){\bf j}=\langle 3,1\rangle$ is perpendicular to $3x+y=5.$ ${\bf n_{2}}=(2){\bf i} + (-1){\bf j}=\langle 2,-1\rangle$ is perpendicular to $2x-y=4.$ $\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$ $=\displaystyle \cos^{-1}(\frac{6-1}{\sqrt{9+1}\cdot\sqrt{4+1}})$ $=\displaystyle \cos^{-1}(\frac{5}{\sqrt{50}})$ $=\displaystyle \cos^{-1}(\frac{5}{5\sqrt{2}})$ $=\displaystyle \cos^{-1}(\frac{1}{\sqrt{2}})=45^{\circ}$, or $\pi/4$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.