## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 832: 59

#### Answer

$x=1,y-2=-z$

#### Work Step by Step

Given: $5x-2y-2z=1$ and $4x+y+z=6$ Here, $(x_0,y_0,z_0)=(1,1,1)$ and $a=\lt 0,-13,13\gt$ The symmetric equations are defined by: $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ Hence, the symmetric equations are: $x=1,y-2=-z$

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