( 717, #53 ) Find an equation of the plane that passes through the line of intersection of the planes
x + y - z = 2 and 2x - y + 3z = 1 and passes through the point ( -1, 2, 1 ).

Solution:

We must first find two other points on the plane. When z = 0:

x + y = 2

2x - y = 1

3x = 3 -> x = 1, y = 1

Thus the point is ( 1, 1, 0 ).

When x = 0:

y - z = 2 -> 3y - 3z = 6 -y + 3z = 1 -y + 3z = 1 2y = 7 -> y = 7/2, z = 3/2

Thus the point is ( 0, 7/2, 3/2 ).

Let a be the vector from ( -1, 2, 1 ) to ( 1, 1, 0 ):

a = < 1 - (-1), 1 - 2, 0 - 1 > = < 2, -1, -1 >

Let b be the vector from ( -1, 2, 1 ) to ( 0, 7/2, 3/2):

b = < 0 - (-1), 7/2 - 2, 3/2 - 1 > = < 1, 3/2, 1/ 2 >

The normal to the plane is:

The general equation for a plane is:

1 (x + 1 ) - 2 ( y - 2 ) + 4 ( z - 1 ) = 0

x - 2y + 4z = -1

by: nl