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 - Preliminary Questions - Page 684: 1



Work Step by Step

Since the plane is parallel to $3x+4y-z=5$, then its normal vector will be $\langle 3,4,-1\rangle $, and since the plane passes through the origin, the equation of the plane is given by $$3(x-0)+4(y-0)-(z-0)=0\Longrightarrow 3x+4y-z=0.$$
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