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: 7


$$ z= 2.$$

Work Step by Step

Since $ n=\langle 0,0,1 \rangle $ and $(x_{0},y_{0},z_{0})=(6,7,2)$ 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, $$0(x-3)+0(y-\frac{1}{2})+(z-2)=0$$ and simplifying it, we get $$ z=2$$ Hence the equation is given by $$ z= 2.$$
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