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: 40

Answer

$$(2,-1,-1).$$

Work Step by Step

The parametric equations of the line are given by $$ x=2+t, \quad y=-1+2t, \quad z=-1-4t.$$ Now, substituting in the equation of the plane $$2x+y =3\Longrightarrow 4+2t-1+2t=3\\ \Longrightarrow 4t=0\Longrightarrow t=0. $$ So the point of intersection is given by $$(2,-1,-1).$$
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