line and plane intersection

Question

In three-space, there are three possibilities for the intersection of a line and a plane.

1. (a) The line intersects the plane at a point.
2. (b) The line lies on the plane.
3. (c) The line is parallel to the plane and distinct.

Given the plane 2𝑥 − 3𝑦 + 4𝑧 − 12 = 0

1. give an example of a vector equation of a line for each
2. possible intersection
3. prove that it is the type of intersection you say it is.
4. (You can use algebra or geometric proofs to show that you have a valid example.)

Answer 1

1) The easiest example is (x, y, z) = (6 + 2t, - 3t, 4t) We have even defined the point of intersection: (6, 0, 0)

Note the direction of the line is the coefficients of the plane, so it is orthogonal to the plane

We chose 6 because 2x - 12 = 0 when x = 6

We can verify the single point of intersection by plugging (6 + 2t, - 3t, 4t) into 2x - 3y + 4z - 12 and getting

2(6 + 2t) -3(-3t) +4(4t) - 12 = 12 + 4t + 9t + 16t - 12 = 29t, which equals 0 only when t = 0

2) The line is on the plane: (x, y, z) = (3t, 2t, 3) Note that (3, 2) . (2, -3) = 0, and 4z - 12 = 0 when z = 3

We can verify by plugging into 2x - 3y + 4z - 12 and getting 2(3t) -3(2t) + 4(3) - 12 = 0t + 0

0t + 0 = 0

There is no dependence on t and the equality is valid, so it is true for all t.

3) The line is not on the plane: (x, y, z) = (3t, 2t, 4) This line is 1 above the line on the plane, where this line has a constant z.

Plugging into 2x - 3y + 4z - 12, we get 2(3t) -3(2t) + 4(4) - 12 = 4 + 0t = 4

4 is never equal to 0.

This line is not on the plane.