   Chapter 11.5, Problem 85E

Chapter
Section
Textbook Problem

# Finding the Distance Between Two Parallel PlanesIn Exercises 91–94, verify that the two planes are parallel and find the distance between the planes. x − 3 y + 4 z = 10 x − 3 y + 4 z = 6

To determine

To calculate: For the given plane, find the distance between the provided planes x3y+4z=10, x3y+4z=6 if they are parallel.

Explanation

Given:

The equation of the first plane is:

x3y+4z=10

The equation of the second plane is:

x3y+4z=6

Formula used:

The general form of the equation of the plane is:

ax+by+cz+d=0

The distance between a point and a plane is:

D=|ax0+by0+cz0+d|a2+b2+c2

Calculation:

As we know given pair of planes will be parallel if their normal vectors are proportional to each other,

Therefore,

n2=kn1

n1 — Normal vector of the first plane,

n2 —Normal vector of the second plane

k —is any real number.

Hence,

The equation of the first plane is:

x3y+4z=10

And the equation of the second plane is:

x3y+4z=6

Further the coordinates of the normal vector of planes will be the coefficients of x,y,z.

So,

The normal vector of the first plane is,

n1=1,3,4

The normal vector of the second plane is,

n2=1,3,4

Since, from the above calculation both vectors are same. Then,

Find the value of k,

n2=kn11,3,4=k1,3,4k=1

So, the value of k is 1.

Thus, the provided planes are parallel.

Now, to find the distance between two parallel planes, obtain a point on the first plane by substituting any two coordinates as zero an obtaining value of the third coordinate

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