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: 8


$$ x-z=-12$$

Work Step by Step

Since $n=\langle 1,0,-1 \rangle$ and $(x_{0},y_{0},z_{0})=(4,2,-8)$ 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, $$(x-4)+0(y-2)-(z+8)=0$$ and simplifying it, we get $$ x-z=-12$$ Hence the equation is given by $$ x-z=-12.$$
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