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: 25


$$ x+z=3.$$

Work Step by Step

Since the normal vector $ n $ is $\langle 1,0,1\rangle $ and, using the fact that the plane passes through $(-2,-3,5)$, the equation of the plane is given by $$(x+2)+0(y+3)+(z-5)=0\Longrightarrow x+z-3=0$$ Hence the equation is given by $$ x+z=3.$$
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