   Chapter 11.5, Problem 48E

Chapter
Section
Textbook Problem

# Finding an Equation of a PlaneIn Exercises 45–56, find an equation of the plane with the given characteristics.The plane passes through the point ( 2 , 2 , 1 ) and contains the line given by x 2 = y − 4 − 1 = z .

To determine

To calculate: The equation of the plane passing through the point (2,2,1).

Explanation

Given:

The plane passes through the point (2,2,1) and contains the line is:

x2=y41=z

Formula used:

The standard equation of the plane is given by:

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

Calculation:

To determine equation of a plane:

The point through which it is passing

And the normal vector to the plane should be known.

The points lying on the line will also lie on plane

As the plane contains the two provided lines,

The symmetric equation of a line is given by

xx1a=yy1b=zz1c

Here, x1,y1,z1 are the coordinates of the point through which line is passing and a,b,c are the direction numbers of parallel vectors of line.

The equation of the provided line is:

x2=y41=z

So, the point on the line is (x1,y1,z1)=(0,4,0) and the parallel vector to the line is v=a,b,c=2,1,1

Now, the normal vector to the plane can be obtained by the cross product of u and v.

u is extending from the point (0,4,0) to the point (2,2,1) and v is the parallel vector to the provided line

### 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

#### In Exercises 7-10, solve for x or y. (62)2+(2y)2=52

Calculus: An Applied Approach (MindTap Course List)

#### Find the derivative of the function. g(x)=ex2x

Single Variable Calculus: Early Transcendentals, Volume I

#### What is the partial fraction form of ?

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