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: 21



Work Step by Step

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