## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 483: 43

#### Answer

$P = (5,6,5)$

#### Work Step by Step

$l_1: (x.y.z) = (2,3,-1)+n(1,1,2)$ $l_2: (x.y.z) = (7,7,2)+r(-2,-1,3)$ Let $P = (x,y,z)$. The x-coordinate of $l_1$ and $l_2$ must both equal $x$: $x = 2+n(1) = 7+r(-2)$ $n = 5-2r$ The y-coordinate of $l_1$ and $l_2$ must both equal $y$: $y = 3+n(1) = 7+r(-1)$ $n = 4-r$ We can equate the two expressions of $n$ to find $r$: $5-2r = 4-r$ $r = 1$ We can use $l_2$ to find $P$: $P = (x,y,z) = (7,7,2)+r(-2,-1,3)$ $P = (7,7,2)+(1)(-2,-1,3)$ $P = (7-2,7-1,2+3)$ $P = (5,6,5)$

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