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: 39


$$ (1,5,8)$$

Work Step by Step

The parametric equations of the line are given by $$ x=1, \quad y=1+2t, \quad z=4t.$$ Now, substituting in the equation of the plane $$ x+y+z=14\Longrightarrow 1+1+2t+4t=14\\ \Longrightarrow 6t=12\Longrightarrow t=2. $$ So the point of intersection is given by $$(1,1+2(2),4(2)=(1,5,8)$$
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