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 684: 22



Work Step by Step

Since the plane is parallel to the plane $$ x+y+z=3$$ then its normal vector $ n $ is $\langle 1,1,1\rangle $ and, using the fact that the plane passes through $(4,1,9)$, the equation of the plane is given by $$(x-4)+(y-1)+(z-9)=0\Longrightarrow x+y+z-14=0$$ Hence the equation is given by $$x+y+z=14.$$
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