Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 685: 42



Work Step by Step

The parametric equations of the line are given by $$ x=1+4t, \quad y=9t, \quad z=-1+2t.$$ Now, substituting in the equation of the plane $$ x-z=6\Longrightarrow 1+4t+1-2t=6\\ \Longrightarrow 2t=4\Longrightarrow t=2. $$ So the point of intersection is given by $$(9,18,3).$$
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