   Chapter 12.5, Problem 39E

Chapter
Section
Textbook Problem

# Find an equation of the plane.39. The plane that passes through the point (1, 5, 1) and is perpendicular to the planes 2x + y −2z = 2 and x + 3z = 4

To determine

To find: An equation of the plane that passes through the point (1,5,1) and perpendicular to the planes 2x+y2z=2 and x+3z=4.

Explanation

Formula used:

The expression to find the equation of the plane through the point P0(x0,y0,z0) with normal vector n=a,b,c is as follows.

a(xx0)+b(yy0)+c(zz0)=0 (1)

The normal vector (n) to the plane is the cross product of two direction vectors in the plane.

The normal vector of the perpendicular plane is the direction vector which is parallel to the required plane.

Calculation:

Write the normal vector from the first perpendicular plane 2x+y2z=2 and consider it as vector a.

a=2,1,2

Write the expression of second perpendicular plane.

x+3z=4

Rearrange the expression as follows.

x+(0)y+3z=4

Write the normal vector from the second perpendicular plane x+(0)y+3z=4 and consider it as vector b.

b=1,0,3

As both the vectors a and b lie on the plane, the cross product of a and b is orthogonal to the plane and is considered as the normal vector to the required plane.

Find the normal vector (n).

n=a×b

Substitute 2,1,2 for a and 1,0,3 for b,

n=2,1,2×1,0,3

Calculate the cross product as follows

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