   Chapter 12.5, Problem 60E

Chapter
Section
Textbook Problem

# Find symmetric equations for the line of intersection of the planes.z = 2x − y − 5, z = 4x + 3y − 5

To determine

To find: The symmetric equations for the line of intersection of the planes z=2xy5 and z=4x+3y5.

Explanation

Formula used:

The symmetric equations of a line through the point (x0,y0,z0) and parallel to the direction vector a,b,c is,

xx0a=yy0b=zz0c (1)

Calculation:

Write the equation of first plane as follows.

2xyz=5 (2)

Write the equation of second plane as follows.

4x+3yz=5 (3)

Consider a point on the line of intersection P0(x0,y0,z0) such that the point satisfies the equations of both the planes z=2xy5 and z=4x+3y5.

Set z=0 and solve the two equations of the intersecting planes.

Substitute 0 for z in equation (2),

2xy(0)=52xy=5

y=2x+5 (4)

Substitute 0 for z in equation (3),

4x+3y(0)=5

4x+3y=5 (5)

Substitute equation (4) in (5) and obtain the x- and y-coordinates as follows.

x=2y=1

Therefore, the point on the line of intersection P0(x0,y0,z0) is (2,1,0).

The parallel direction vector is the cross product of normal vectors of both the planes.

v=n1×n2

The expression is also written as follows.

v=|ijka1b1c1a2b2c2| (6)

The normal vector (n1) of first plane.

n1=2ijk

The normal vector of first plane is also written as follows.

n1=2,1,1

The normal vector (n2) of second plane.

n2=4i+3jk

The normal vector of second plane is also written as follows

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