1. Determination of a Plane
By different ways we can define a plane. Now, let us define these ways.
Rule: We can define a plane in space by:
Three non-collinear points. (Axiom 2)
Two parallel lines.
A line and a point outside of this line.
Two intersecting lines.
Proofs
a. Axiom 2.
b. Let m and d be two parallel lines in space. Let α and β be two different planes containing both m and d.
Since α and β have common points, which are the points on m and d, they are intersecting planes.
By Axiom 5 this intersection must be a unique line.
But here the intersection is the union of two lines m and n.
There is a contradiction. So α and β are coincident.
c
.
Let
d
be
a line and A be a point not on line d.
Let us take points B and C on d. (Figure 1.2)
Since points A, B, C are three non-collinear points they determine a plane λ by axiom 2.
Then λ contains two points of line d. So d lies in λ by axiom 4.
Therefore line d and point A determine plane λ.
Namely, there is only one plane, which is λ, containing both point A and line d.
d
.
Let
d
and
m
be
two intersecting lines and A be their intersection point like in the
figure 1.3.
Besides A, let us take two points B and C, on
lines d and m, respectively.
Since A, B, C are three non-collinear
points, they determine a plane λ.
Then λ will contain both lines by axiom 4 since it has two points in common with each of the lines.
Rule:
From n non-collinear points in space we can draw C(n; 2) lines.
From n non-coplanar and non-collinear points in space we can draw C(n; 3) planes.
Example 3: a) From 6 non-collinear points in space how many lines can we draw?
b) From 6 non-coplanar points is space how many planes can we draw?
Solution:
a) C(6;
2) =
.
sb)
C(6;
3) =
.
Example 4: Show that all sides of a triangle are in the same plane.
Solution: Let ABC be a triangle.
Since AB and AC are intersecting lines they determine a plane α.
Since two points of line segment BC are in α, α contains all points on BC.
So, all sides of a triangle are in the same plane.
Example 5: A, B, C and D are four non-coplanar points. Can three of these
points be on the same line?
Solution: No.
Because if three of these points are on a line then for the fourth point we have two cases:
i) It can be on this line. Then there can be drawn infinitely many planes containing these four points.
ii) If it is not on this line then the line and the point not on this line determine a plane, and this plane contains all the given points.
In both cases the points will be coplanar. However it is given that points are non-coplanar.
So, three of them can not be collinear.
Check Yourself 2
1. How many planes can be determined by three parallel lines? (Write all possible cases)
2. How many planes can be determined by three intersecting lines? (Write all possible cases)
3. a) From 8 non-collinear points is space how many lines can we draw?
b) From 8 non-coplanar points is space how many planes can we draw?
4. Four points on a line are given. Can we determine a plane by using these points? Or how many planes can we determine?
5. We draw a line intersecting by two sides of a triangle. Can we say that this line must be included in the plane of the triangle? Why?
Answers
1. One or Three 2. One 3. a) 28 b) 56 4. No. Infinitely many
5. Yes, Proof can be shown by fourth axiom.