Elementary Geometry for College Students (7th Edition)

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: 32

Answer

Since the two lines are not parallel and they do not intersect, the two lines are skew.

Work Step by Step

$l_1 : (x,y,z) = (-2,-2,-2)+n(1,2,3)$ $l_2 : (x,y,z) = (1,1,1)+r(1,-3,5)$ We can verify the ratio of each coordinate of the direction vectors of each line: $x-coordinates: \frac{1}{1} = 1$ $y-coordinates: \frac{2}{-3} = -\frac{2}{3}$ $z-coordinates: \frac{3}{5} = \frac{3}{5}$ Since the ratios are not the same for all three coordinates, the two lines are not parallel. Let's assume that the two lines intersect at the point $(a,b,c)$. Then: $-2+n = 1+r = a$ $n = r+3$ Then: $-2+2n = 1-3r = b$ $-2+2(r+3) = 1-3r$ $-2+2r+6 = 1-3r$ $5r = -3$ $r = -\frac{3}{5}$ We can find $n$: $n = r+3$ $n = -\frac{3}{5}+3$ $n = \frac{12}{5}$ However: $c = -2+3n = -2+3(\frac{12}{5}) = \frac{26}{5}$ $c = 1+5r = 1+5(-\frac{3}{5}) = -2$ Clearly, this is a contradiction. Therefore, the assumption that the two lines intersect is false. Since the two lines are not parallel and they do not intersect, the two lines are skew.
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