University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 600: 59

Answer

a) $\sqrt{y^2+z^2}$; b) $\sqrt{x^2+z^2}$ c) $\sqrt{x^2+y^2}$

Work Step by Step

a) The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Here, the distance from point $P(x,y,z)$ to the x-axis, that is, $(x,0,0)$ is given by $\sqrt{(x-x)^2+(y-0)^2+(z-0)^2}=\sqrt{y^2+z^2}$ b) The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Here, the distance from point $P(x,y,z)$ to the y-axis, that is, $(0,y,0)$ is given by $\sqrt{(x-0)^2+(y-y)^2+(z-0)^2}=\sqrt{x^2+z^2}$ c) The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Here, the distance from point $P(x,y,z)$ to the z-axis, that is, $(0,0,z)$ is given by $\sqrt{(x-0)^2+(y-0)^2+(z-z)^2}=\sqrt{x^2+y^2}$
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