   Chapter 12.5, Problem 76E

Chapter
Section
Textbook Problem

# Find equations of the planes that are parallel to the plane x + 2y − 2z = 1 and two units away from it.

To determine

To show: The equations of the parallel planes, which are parallel to the plane x+2y2z=1 and two units away from it.

Explanation

Solution:

The equations of the parallel planes which are parallel to the plane x+2y2z=1 and two units away from it are x+2y2z=7 and x+2y2z=5.

As the planes are parallel to the plane x+2y2z=1, the normal vectors of the parallel planes are the scalar multiple of normal vector of the plane x+2y2z=1.

Write the normal vector from the plane x+2y2z=1 as follows.

n=1,2,2

The normal vector of one of the parallel planes is written as follows.

n1=t1,2,2=t,2t,2t

Here,

t is the scalar parameter.

Write the expression of plane equation of one of the parallel with the use of normal vector as follows.

tx+2ty2tz=k

Here,

k is the constant parameter.

The expression is also written as follows.

t(x+2y2z)=kx+2y2z=kt

Consider the term kt as c.

x+2y2z=c

x+2y2zc=0 (1)

Rewrite the expression x+2y2z=1 as follows.

x+2y2z1=0 (2)

The equations (1) and (2) are parallel planes.

Formula used:

Write the expression to find the distance between the parallel planes ax+by+cz+d1=0 and ax+by+cz+d2=0 as follows

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