   Chapter 11.5, Problem 56E

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. ( − 5 , 1 , − 3 ) ,     ( 2 , − 1 , 6 )

To determine

To calculate: The equation of the plane through the given points (5,1,3) and (2,1,6).

Explanation

Given:

Plane contains all the points that are equidistant from the points (5,1,3) and (2,1,6).

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 (5,1,3) and Q (2,1,6)

Thus, the normal vector of required plane will be:

n=2(5),11,6(3)=7,2,9

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

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