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Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

Find an equation of the plane.

40. The plane that passes through the line of intersection of the planes xz = 1 and y + 2z = 3 and is perpendicular to the plane x + y − 2z = 1

To determine

To find: An equation of the plane that passes through the line of intersection of the planes xz=1 and y+2z=3 , and perpendicular to the plane x+y2z=1 .

Explanation

Formula:

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

Write the normal vector from the plane xz=1 .

Rearrange the plane xz=1 and write the normal vector.

x+(0)yz=1

n1=1,0,1

Write the normal vector from the plane y+2z=3 .

Rearrange the plane y+2z=3 and write the normal vector.

(0)x+y+2z=3

n2=0,1,2

The cross product of the normal vectors n1 and n2 is the direction vector (a) and it is parallel to the required plane.

Find the direction vector a from the normal vectors of two intersected planes.

a=n1×n2

Substitute 1,0,1 for n1 and 0,1,2 for n2 ,

a=1,0,1×0,1,2

Calculate the cross product as follows.

a=|ijk101012|=i[(0)(2)(1)(1)]j[(1)(2)(0)(1)]+k[(1)(1)(0)(0)]=i(0+1)j(20)+k(10)=i2j+k

a=1,2,1

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

Write the normal vector from the perpendicular plane x+y2z=1 and consider it as vector b .

b=1,1,2

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

Find the normal vector (n) .

n=a×b

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

n=1,2,1×1,1,2

Calculate the cross product as follows

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Chapter 12 Solutions

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