write the equation of a plane containing the lines?

Question

x = y = (4-z)/4 2x=2-y=z

whats the best way to do this i am stuck

Answer 1

Write the equation of a plane containing the lines:
s = x/1 = y/1 = (4 - z)/4
t = x/(1/2) = (2 - y)/1 = z/1

First we need to see if the lines are coplanar. They are if they are parallel or if they intersect.

First Plane
x = s
y = s
z = 4 - 4s

Second Plane
x = t/2
y = 2 - t
z = t

The lines are clearly not parallel. We need to check for a point of intersection.

x = y
t/2 = 2 - t
3t/2 = 2
t = 4/3

x = s = t/2
s = 2/3

Plugging in we get:

First Plane
x = s = 2/3
y = s = 2/3
z = 4 - 4s = 4/3

Second Plane
x = t/2 = 2/3
y = 2 - t = 2/3
z = t = 4/3

The lines have a point in common so they intersect.
The point is P(2/3, 2/3, 4/3).


Take the two directional vectors of the line. They are also directional vectors of the plane.

u = <1, 1, -4>
v = <1/2, -1, 1>

Any non-zero multiple of v will also be a directional vector. Multiply by 2.

v = <1, -2, 2>

The normal vector of the plane will be perpendicular to both u and v. Take the cross product.

n = u X v = <1, 1, -4> X <1, -2, 2> = <-6, -6, -3>

Any non-zero multiple of n will also be a normal vector to the plane. Divide by -3.

n = <2, 2, 1>

With the normal vector of the plane and a point
P(2/3, 2/3, 4/3) we can write the equation of the plane.

2(x - 2/3) + 2(y - 2/3) + 1(z - 4/3) = 0
2x - 4/3 + 2y - 4/3 + z - 4/3 = 0
2x + 2y + z - 4 = 0