   Chapter 11.5, Problem 83E

Chapter
Section
Textbook Problem

# Finding the Distance Between a Point and a PlaneIn Exercises 87–90, find the distance between the point and the plane. ( 2 , 8 , 4 ) 2 x + y + z = 5

To determine

To calculate: For the given Point and plane, find the distance between the provided point (2,8,4) and the provided plane 2x+y+z=5.

Explanation

Given:

The equation of the plane is,

2x+y+z=5

And the provided point is (2,8,4).

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:

By using the equation of the plane.

Let us first, determine the normal vector n to the provided plane

2x+y+z=5

Now since, the coordinates of normal vector n are the coefficients of x,y and z terms in the equation of the plane.

So, the normal vector is,

n=a,b,c=2,1,1

Where, a,b and c are the direction numbers of the normal vector n.

Further now from the above,

values of direction numbers are,

a=2,b=1 and c=1.

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

ax+by+cz+d=0

Let us compare this equation from the equation of the given plane. Then the value of d is 5.

Presently, discover the distance between a point and the plan

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