   Chapter 11.5, Problem 86E

Chapter
Section
Textbook Problem

# Finding the Distance Between Two Parallel Planes In Exercises 85-88, verify that the two planes are parallel, and find the distance between the planes. 4 x − 4 y + 9 z = 7 4 x − 4 y + 9 z = 18

To determine

To prove: The two planes 4x4y+9z=7 and 4x4y+9z=18 are parallel and also calculate the distance between the two planes if they are parallel.

Explanation

Formula used:

The distance between a point and a plane is:

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

The general form of the equation of the plane is:

ax+by+cz+d=0

Proof:

The provided pair of planes will be parallel if their normal vectors are proportional to each other, that is

n2=kn1

Here, n1 is the normal vector of the first plane, n2 is the normal vector of the second plane and k is any real number.

So, the equation of the first plane is:

4x4y+9z=7

And, the equation of the second plane is:

4x4y+9z=18

The coordinates of the normal vector of planes will be the coefficients of x,y,z. Then, the normal vector of the first plane is,

n1=4,4,9

And, the normal vector of the second plane is,

n2=4,4,9

Since, both vectors are same. Then, find the value of k as,

n2=kn14,4,9=k4,4,9k=1

So, the value of k is 1.

Hence, the provided planes are parallel.

Calculation:

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.

So, from first equation of the plane,

4x4y+9z=7

Substitute y=0,z=0 in the equation of the first plane Then,

4x4y+9z=74x4×0+9×0=74x0+0=7x=74

So, the value of x is,

x=74

Thus, a point on the first plane is,

(x0,y0,z0)=(74,0,0)

Now, the general form of the equation of the plane is given as,

ax+by+cz+d=0

Comparing this equation with the equation of second plane, then the value of d is 18

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