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: 30


The planes are parallel.

Work Step by Step

Since the normal vector to the plane $2x-4y-z=3$ is $\langle2,-4,-1 \rangle $ and the normal vector to the plane $-6x+12y+3z=1$ is $\langle -6,12,3 \rangle $, one can easily see that $$\langle -6,12,3 \rangle=-3\langle2,-4,-1 \rangle,$$ That is, the normal vectors are parallel and hence the planes are parallel.
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