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: 6


$$ y= \frac{1}{2}.$$

Work Step by Step

Since $ n=\langle 0,1,0 \rangle $ and $(x_{0},y_{0},z_{0})=(-5,\frac{1}{2},\frac{1}{2})$ then the equation of the plane in scalar form is $ a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$ That is, $$0(x-3)+(y-\frac{1}{2})+0(z-\frac{1}{2})=0$$ and simplifying it, we get $$ y=\frac{1}{2}$$ Hence the equation is given by $$ y= \frac{1}{2}.$$
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