Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.7 Cylindrical and Spherical Coordinates - Exercises - Page 700: 72

Answer

The projection ${Q_1}$ of ${P_1}$ onto the $xy$-plane is ${Q_1} = \left( {1, - \sqrt 3 ,0} \right)$. The polar angle ${\theta _1}$ of ${Q_1}$ is $\frac{{5\pi }}{3}$. The projection ${Q_2}$ of ${P_2}$ onto the $xy$-plane is ${Q_2} = \left( { - 1,\sqrt 3 ,0} \right)$. The polar angle ${\theta _2}$ of ${Q_2}$ is $\frac{{2\pi }}{3}$

Work Step by Step

The projection ${Q_1}$ of ${P_1} = \left( {1, - \sqrt 3 ,5} \right)$ onto the $xy$-plane is obtained by setting the $z$-coordinate to zero. So, ${Q_1} = \left( {{x_1},{y_1}} \right) = \left( {1, - \sqrt 3 ,0} \right)$. Since ${x_1} = 1$ and ${y_1} = - \sqrt 3 $, ${Q_1}$ is in the fourth quadrant. The polar angle ${\theta _1}$ of ${Q_1}$ is obtained by solving: $\tan {\theta _1} = \frac{{ - \sqrt 3 }}{1} = - \sqrt 3 $ ${\theta _1} = - \frac{\pi }{3}$ ${\ \ }$ or ${\ \ }$ ${\theta _1} = 2\pi - \frac{\pi }{3} = \frac{{5\pi }}{3}$. Similarly, the projection ${Q_2}$ of ${P_2} = \left( { - 1,\sqrt 3 ,5} \right)$ onto the $xy$-plane is obtained by setting the $z$-coordinate to zero. So, ${Q_2} = \left( {{x_2},{y_2}} \right) = \left( { - 1,\sqrt 3 ,0} \right)$. Since ${x_2} = - 1$ and ${y_2} = \sqrt 3 $, ${Q_2}$ is in the second quadrant. The polar angle ${\theta _2}$ of ${Q_2}$ is obtained by solving: $\tan {\theta _2} = \frac{{\sqrt 3 }}{{ - 1}} = - \sqrt 3 $ ${\theta _2} = \frac{{2\pi }}{3}$
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.