Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 482: 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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.