   Chapter 11.5, Problem 42E

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 ( 3 , − 1 , 2 ) , ( 2 , 1 , 5 ) , and ( 1 , − 2 , − 2 ) .

To determine

To calculate: The equation of the plane that through the points (3,1,2),(2,1,5) and (1,2,2).

Explanation

Given:

The plane passes through the points (3,1,2),(2,1,5) and (1,2,2).

Formula used:

The equation of the plane in standard form is given by

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

Cross product of the two vectors is:

n=u×v

Calculation:

To determine equation of a plane:

The point through which it is passing

And the normal vector to the plane should be known.

Three choices for the point but no normal vector is provided.

To obtain a normal vector, use the cross product of vectors u and v extending from the point (3,1,2) to the points (2,1,5) and (1,2,2). The component forms of u and v are

Consider the first point as:

(x1,y1,z1)=(3,1,2)

The second point as:

(x2,y2,z2)=(2,1,5)

And the third point as:

(x3,y3,z3)=(1,2,2)

Now, find the vectors,

For the points (3,1,2) and (2,1,5) then the vector u is,

u=[(x2x1),(y2y1),(z2z1)]=23,1+1,52=1,2,3

For the points (3,1,2) and (1,2,2) then the vector v is,

v=[(x3x1),(y3y1),(z3z1)]=(13),(2+1),(22)=2,1,4.

Consider the normal vector is n

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