## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 684: 10

#### Answer

$$3x+2y+z=3, \quad 3x+2y+z=2.$$ (Other answers are possible.)

#### Work Step by Step

The equation of any plane is given by $$a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0.$$ Since $n=\langle 3,2,1\rangle$, then we can arbitrarily choose the planes as follows $$3(x-1)+2(y-0)+(z-0)=0, \quad 3(x-0)+2(y-1)+(z-0)=0.$$ Note that we could have chosen any other vector, as long as does not pass through $(0,0,0)$. By simplification, we have $$3x+2y+z=3, \quad 3x+2y+z=2.$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.