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: 46

Answer

$(x,y,z) = (12,10,0)+n(-3,5,1)$

Work Step by Step

$x+3z = 12$ $y-5z = 10$ The normal vectors to the planes are $a = (1,0,3)$ and $b = (0,1,-5)$ respectively. We can find $a \times b$: $a \times b = (0-3, 0-(-5), 1-0) = (-3, 5, 1)$ Then $(-3,5,1)$ is the direction vector of the line of intersection. We can let $z=0$ to find a point on the line of intersection: $x+3(0) = 12$ $x = 12$ $y-5(0) = 10$ $y = 10$ Therefore, the point $(12,10,0)$ is on the line of intersection. The can write the equation for the line of intersection: $(x,y,z) = (12,10,0)+n(-3,5,1)$
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