   Chapter 11.5, Problem 81E

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. ( 0 , 0 , 0 ) 2 x + 3 y + z = 12

To determine

To calculate: For the given Point and plane, find the distance between the provided point (0,0,0) and the plane 2x+3y+z=12.

Explanation

Given:

The equation of the plane is,

2x+3y+z=12

And the provided point is (0,0,0).

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+3y+z=12

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=2,3,1

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=3 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 provided plane. Then the value of d is 12 Now, find the distance between a point and the plan

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