   Chapter 11.5, Problem 51E

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 points ( 1 , − 2 , − 1 ) and ( 2 , 5 , 6 ) and is parallel to the x-axis.

To determine

To calculate: The equation of the plane passes through the point (1,2,1), and (2,5,6), also it is parallel to the x-axis.

Explanation

Given: The plane passes through the point (1,2,1), and (2,5,6), also it is parallel to the x-axis.

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.

Two choices for the point but there is no normal vector provided.

Since, the required plane is parallel to x-axis, so the normal vector will be obtained by the cross product of u 1,0,0, and the vector v which extends from the point (2,5,6) to the point (1,2,1).

So, component form of v will be:

v=21,5(2),6(1)=1,7,7

So, it follows that a vector normal to the required plane is:

n=u×v=|ijk1001<

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