Find an equation of the plane.

Question

The plane that passes through the line of intersection of the planes
x − z = 2 and y + 2z = 2
and is perpendicular to the plane
x + y − 4z = 4

Answer 1

First, we need to find the direction vector of the line of intersection of the planes x - z = 2 and y + 2z = 2. To do this, we can find two points on each plane and take their cross product:

For x - z = 2, we can choose (2, 0, 0) and (0, 0, -2):
(2, 0, 0) - (0, 0, -2) = (2, 0, 2)

For y + 2z = 2, we can choose (0, 2, 0) and (0, 0, -1):
(0, 2, 0) - (0, 0, -1) = (0, 2, 1)

Taking the cross product:
(2, 0, 2) x (0, 2, 1) = (-4, -2, 4)

Therefore, the direction vector of the line of intersection is (-4, -2, 4).

Next, we need to find the normal vector of the plane x + y - 4z = 4. We can read this off from the coefficients of x, y, and z: (1, 1, -4).

Since the plane we want is perpendicular to this normal vector, the dot product of their normal vectors should be zero:

(1, 1, -4) dot (-4, -2, 4) = -4 + (-2) + (-16) = -22

Therefore, we need to find the equation of the plane with normal vector (-4, -2, 4) and passing through the line of intersection of x - z = 2 and y + 2z = 2. To do this, we can choose a point on the line of intersection and use it as the point on the plane. For example, we can choose (2, 0, 0) since it satisfies x - z = 2. Then we can use the formula for the equation of a plane:

-4(x-2) - 2y + 4z = -8

Simplifying:

-4x + 2y + 4z = 0

This is our equation for the plane.