Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 491: 44

Answer

$P = (12,-15,28)$

Work Step by Step

$l_1: (x.y.z) = (2,0,3)+n(2,-3,5)$ $l_2: (x.y.z) = (4,1,-4)+r(-1,2,-4)$ Let $P = (x,y,z)$. The x-coordinate of $l_1$ and $l_2$ must both equal $x$: $x = 2+n(2) = 4+r(-1)$ $2n = 2-r$ $n = \frac{2-r}{2}$ The y-coordinate of $l_1$ and $l_2$ must both equal $y$: $y = 0+n(-3) = 1+r(2)$ $-3n = 1+2r$ $n = \frac{-1-2r}{3}$ We can equate the two expressions of $n$ to find $r$: $\frac{2-r}{2} = \frac{-1-2r}{3}$ $6-3r = -2-4r$ $r = -8$ We can use $l_2$ to find $P$: $P = (x,y,z) = (4,1,-4)+r(-1,2,-4)$ $P = (4,1,-4)+(-8)(-1,2,-4)$ $P = (4+8,1-16,-4+32)$ $P = (12,-15,28)$
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