## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 600: 60

#### Answer

a) $z$ ; b) $x$; c) $y$

#### 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 xy-plane, that is, $(x,y,0)$ is given by $\sqrt{(x-x)^2+(y-y)^2+(z-0)^2}=\sqrt{z^2}=z$ 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 yz-plane, that is, $(0,y,z)$ is given by $\sqrt{(x-0)^2+(y-y)^2+(z-z)^2}=\sqrt{x^2}=x$ 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 zx-plane, that is, $(x,0,z)$ is given by $\sqrt{(x-x)^2+(y-0)^2+(z-z)^2}=\sqrt{y^2}=y$ Hence, a) $z$ ; b) $x$; c) $y$

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.