K. Nicole Clark Math 311-200

3/8/2004 Fulling

URL: people.tamu.edu/~knc8928/5.1.18.pdf

5.1.18

Consider affine subspaces in R³ (lines and planes that do not necessarily pass through the origin). Use Theorem 6 to classify the possible intersections and unions of

(a) a plane and a line;

(b) two planes.

HINT: If the sets intersect at all, you can choose the origin of coordinates to be a point in the intersection.

Solution

(a) A plane and a line can intersect at a point, the line can lie in the plane, or they can be parallel and not intersect at all. As is obvious, a line is one-dimensional and a plane is two-dimensional. We are going to assume the origin falls anywhere along the intersections.

Case I: A plane and a line intersect at a point.

Intersection: dim(LP) = 0 – this is because the intersection is at a point, which has a dimension of zero. (Having a point as an intersection is also called a direct sum).

Union: Using the equations from page 218 and Theorem 6 of Linearity, we know that

The second equation is used when the intersection is a direct sum, as defined above.

Thus, in this case,

Case II: A line lies in a plane.

Intersection: dim(LP) =1 - In this case, the intersection of the two will be simply the line itself, which has by definition a dimension of one.

Union:

= 1 + 2 -1 = 2

This is obvious from intuition as well, as the union of a plane and a line that lies in that plane will be the plane itself, which by definition has a dimension of two.

Case III: The line and the plane never intersect.

In this case, it is impossible for both the line and the plane to pass through the origin, thus Theorem 6 does not apply.

(b) Two planes can intersect at a line, lie in the same plane, or be parallel and not intersect at all. We are again assuming that the origin falls somewhere along the intersection

Case I: Two planes intersect along a line.

Intersection: - This makes sense, as the dimension of a line is one.

Union:

= 2 +2 – 1 = 3

Case II: Two planes lie in the same plane.

Intersection: This intersection is just a plane, having a dimension of two by definition.

2

Union:

= 2 +2 – 2 = 2

This makes sense, as the union of two planes lying in the same plane is just a plane.

Case III: The two planes never intersect.

In this case, it is impossible for both planes to pass through the origin, thus Theorem 6 does not apply.