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 483: 45

Answer

$(x,y,z) = (12,6,0)+n(-2,3,1)$

Work Step by Step

$x+2z = 12$ $y-3z = 6$ The normal vectors to the planes are $a = (1,0,2)$ and $b = (0,1,-3)$ respectively. We can find $a \times b$: $a \times b = (0-2, 0-(-3), 1-0) = (-2, 3, 1)$ Then $(-2,3,1)$ is the direction vector of the line of intersection. We can let $z=0$ to find a point on the line of intersection: $x+2(0) = 12$ $x = 12$ $y-3(0) = 6$ $y = 6$ Therefore, the point $(12,6,0)$ is on the line of intersection. The can write the equation for the line of intersection: $(x,y,z) = (12,6,0)+n(-2,3,1)$
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