   Chapter 11.5, Problem 55E

Chapter
Section
Textbook Problem

# Finding an Equation of a PlaneIn Exercises 57–60, find an equation of the plane that contains all the points that are equidistant from the given points. ( − 3 , 1 , 2 ) ,     ( 6 , − 2 , 4 )

To determine

To calculate: The equation of the plane which contains all the points that are equidistant from the points (3,1,2), and (6,2,4).

Explanation

Given:

Plane contains all the points that are equidistant from the points (3,1,2), and (6,2,4).

Formula used:

The standard equation of the plane is given by:

a(xx1)+b(yy1)+c(zz1)=0

Calculation:

The normal vector of the required plane will be

the line joining the two provided points P(3,1,2), and (6,2,4)

Thus, the normal vector of required plane will be:

n=6(3),21,42=9,3,2

And, as the required plane is equidistant from these two points. So, the plane lies at the mid-point of the line joining the points P and Q

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Inequalities Solve the inequality graphically. 143. 4x 3 x2

Precalculus: Mathematics for Calculus (Standalone Book)

#### Graph each equation: y=3x

Elementary Technical Mathematics

#### In Problems 1-20, evaluate the improper integrals that converge.

Mathematical Applications for the Management, Life, and Social Sciences

#### For f(x) = ex x, f(x) = a) ex x b) ex 1 c) e1 x d) e1 1

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

#### The area of the parallelogram at the right is: 26

Study Guide for Stewart's Multivariable Calculus, 8th 