3. Mutual Positions of a Plane and a Line

Rule: A line and a plane in the space may have the following three positions relative to each other.

1. The line is lying in the plane.

2. The line intersects the plane at a unique point.

3. The line is parallel to the plane.

Related with the mutual positions of a line and a plane in space we can write the following theorems and the conclusions:

Theorems:

1. If a line is parallel to another line lying in a plane, the line will be parallel to the plane.

2. If a line is parallel to a plane, in this plane there are lines parallel to the given line.

3. Two lines parallel to the same line are parallel.

4. Two angles with respectively parallel arms in the same direction are congruent.

5. If one of two parallel lines intersects a plane, the other intersects too.

Proofs:

1 . Let d be a line parallel to another line m lying in plane α like in the following figure.

Since d and m are parallel lines they lie in a plane λ by rule b of determination of a plane.

α and λ are intersecting planes along line m by axiom 5.

So, if dand α intersect each other, the intersection point must be on m by axiom 5.

This contradicts with the parallelity of m and d.

So d and α have no common point, in other words they are parallel.

2 . Let d be a line parallel to a given plane α and A be any point in β (Figure 1.8).

Then d and A determine a plane β by 3 rule of determination of a plane.

β and α have a common point, that is A. So, they have a common line by axiom 5.

Let us name this line as m.

Both d and m are in β.

Since d has no common point with α, it can not intersect m.

So, d and m are parallel lines by the rules of plane geometry.

Therefore, a line is parallel to a plane if and only if it does not lie in the plane and it is parallel to a line lying in that plane.

L et d be a line parallel to a plane α and A₁ be a point in α (Figure 1.9). Then d and A₁ determine a plane β₁.

Let m₁ be the intersection of α and β₁. We proved that m₁ // d.

Let A₂ be a point in α but not on m₁. Line d and A₂ determine another plane β₂.

Let m₂ be the intersection of α and β₂. Then m₂ // d.

Now let us think about m₁ and m₂. They are both in α.

If they have a common point, this point will be a common point of planes β₁ and

β₂.

Since d is the intersection of β₁ and β₂ this common point must be on d.

This is impossible. Because d // m₁ and d //m₂.

So m₁ and m₂ can not have any common point. Therefore, they are parallel.

As a conclusion, if a line is parallel to a plane then in this plane there are infinitely many lines parallel to the given line and these lines are parallel to each other.

3 . Let m, n and d be three lines in space so that m // d and n // d

(Figure 1.10).

Since m and d are parallel they determine a plane α, and since n and d are parallel they determine another plane β by the rule 2 of determination of a plane.

Let A be a point on n.

Line m and point A determine a plane λ.

Since β and λ have a common point A they have a common line k by axiom 5.

We know that m // d. So, m is parallel to β by theorem 1.

Then lines d and k are two lines in plane β parallel to line m. So, k and d are parallel.

Through point A there can be drawn only one line parallel to d by the rules of plane geometry. So, n and k should be coincident lines.

Therefore, m and n are parallel lines.

4. Let ∠ABC and ∠A₁B₁C₁ be two angles with respectively parallel arms in the same direction.

Let M and N be any two points on

a rms BA and BC respectively.

On B₁A₁ and B₁C₁ take two points

M₁ and N₁ so that M₁B₁ = MB and

N₁B₁ = NB. (Figure 1.11)

Since BA // B₁A₁, BMM₁B₁ is a parallelogram.

So BB₁ // MM₁ and BB₁ = MM₁. (1)

Similarly BC // B₁C₁ and BNN₁B₁ is a parallelogram.

So BB₁ // NN₁ and BB₁ = NN₁ . (2)

From (1) and (2) we get NN₁ // MM₁ and NN₁ = MM₁.

So MNN₁M₁ is a parallelogram and MN = M₁N₁.

Then by S.S.S, MBN and M₁B₁N₁ are congruent.

That means ∠MBN = ∠M₁B₁N₁.

5. Let α be a plane, d and m be two parallel lines and d intersect α.

For m there are three positions: It lies in α or it is parallel to α or it intersects α.

1. If m is in α then d will be parallel to a line in α. So d is parallel to α. This is a contradiction.

2. If m is parallel to α then in α there will be a line (for example n) parallel to m. Since d // m and m // n, it can be concluded that d // n by theorem 3. For this case again d will be parallel to α.

Hence m intersects α.

Conclusions:

1. If one of two parallel lines is parallel to a plane the other is either in the plane or parallel to the plane.

2. If the corresponding arms of two angles are parallel and are in opposite directions, the angles are equal.

3. If the corresponding arms of two angles are parallel and if one of corresponding arms is in the same direction while the other is in opposite then the sum of the angles is 180°.

Example 8: Show that if one of two lines lies in a plane and the other intersects this plane at a point not on the first line then these lines are skew to each other.

S olution:

Let d be a line in a plane λ and m intersect λ at a point A. (Figure 1.14)

Assume that there is a plane β containing both m and d.

Then d and A will be in β.

So β and λ will be coincident.

However λ does not contain m.

So β can not contain m either.

Hence there is no plane containing both d and m. It means that they are skew.

E xample 9: In the figure, A is not in (BCD), B₁C₁ // BC, and C₁D₁ // CD.

If and the perimeter of BCD is

24 cm, find the perimeter of B₁C₁D₁ .

Solution:

Since B₁C₁ // BC, triangles AB₁C₁ and ABC are similar.

So .

Since .

So , BC = 3B₁C₁.

Moreover, since C₁D₁ // CD, AC₁D₁ ACD.

So , CD = 3C₁D₁.

Since , AB₁D₁ ABD .

Therefore, , BD = 3B₁D₁ .

P_BCD = 24 cm , BC + BD + CD = 24 cm,

3(B₁C₁ + B₁D₁ + C₁D₁) = 24 cm, .

Hence the perimeter of B₁C₁D₁ is 8 cm.

Example 10: Show that the midpoints of the sides of a skew quadrilateral

are the vertices of a parallelogram.

S olution: Let ABCD be a skew quadrilateral

and M, N, P, Q be the midpoints of sides AB, BC, CD, DA, respectively (Figure 1.16).

In DAC and BAC, QP // AC, MN // AC,

and by the rules of plane geometry.

So QP // MN by theorem 3 and QP = MN.

If we use the same logic in ABD and BCD we will obtain that QM // PN and .

Hence QMNP is a parallelogram.

E xample 11: In the figure, A₁C₁ // AC,

C₁B₁ // CB, and A₁B₁ // AB.

If A₁C₁ = 5 cm,

A₁B₁ = cm and

B₁C₁ = 7 cm, find the measure of angle DCB.

Solution:

Since ∠A₁C₁B₁ and ∠ACB are two angles with respectively parallel arms in the same direction, they are equal by theorem 4.

In A₁B₁C₁ by cosine theorem we get

,

,

and cos ∠C₁ = .

So ∠A₁C₁B₁ = 60°. Then ∠ACB = 60° and ∠DCB = 180° - 60°= 120°.

Example 12: One side of a rhombus ABCD is 4 cm. Sides AB and AD intersect

a plane α at points P and Q respectively.

AP = 1 cm and AQ = 3 cm are given.

a) Show that lines CB and CD intersect α.

b) If CB and CD intersect α at P₁ and Q₁, respectively, find the lengths of CP₁ and CQ₁.

Solution:

a) In a rhombus opposite sides are parallel.

So AB // CD and AD // BC (Figure 1.18).

If one of two parallel lines intersects

a plane, the other intersects too by theorem 5.

It is given that AB and AD intersect plane α.

Hence CD and CB intersect α too.

b) Points P₁, Q₁, P and Q are all intersections of planes α and (ABC) by axiom 5. So they are collinear.

Since AQ // P₁B, triangles APQ and BPP₁ are similar.

then BP₁ = 9 cm.

So CP₁ = CB + BP₁ = 4 + 9 = 13 cm.

On the other hand, since AP // DQ₁, we obtain APQ DQ₁Q.

Then

So CQ₁ = CD + DQ₁ = 4 + = cm.

Example 13: Show that when two parallel lines are intersected by a line, all

these three lines lie in the same plane.

Solution: Two parallel lines determine a plane.

The line intersecting these parallel lines have two points in common with this plane which are the intersection points.

So, it lies in this plane too by axiom 4.

Example 14: Show that through one of two skew lines, there can be drawn

a plane parallel to the other.

Solution: Let m and d be two skew lines and A be a point on d.

Through A let us draw line m' parallel to m.

d and m' determine a plane α by rule 4 of determination of a plane.

Since m // m' and m' is in α, m // α. (Figure 1.19)

C heck Yourself 4

1. In figure, A is not in (BCD), and P and Q are

the centroids of BCD and ACD, respectively.

Is PQ parallel to plane ABD?

2. In figure, d intersects plane α at point A.

B and C are in α.

For any three points P, Q , R on d, show that

if PB = PC and QB = QC , RB = RC.

3. ABCD is a trapezoid (AB // CD). AB is in a plane α and CD is in a plane β.

α and β are intersecting along the line m.

Show that MN // m where M and N are the midpoints of AD and BC respectively.

4. State the followings as true or false

a) Two lines parallel to the same plane are always parallel.

b) If two lines m and n are parallel to a given line d then m // n.

c) If the line d is parallel to one of two intersecting lines it is parallel to the plane that include the intersecting lines.

5. m and n are two skew lines.

S how that if both m and n are

parallel to planes α and β then α

and β are parallel.

6. Two points A and B are in a plane P. Point C is outside of plane P. We connect point C with A and B. M is the midpoint of CA and N is the midpoint of CB. Prove that MN is parallel to the plane P.

Answers

1. Yes, PQ // (ABD). Hint: Extend CP and CQ then use similarity.

2. Use congruency in triangles.

3. Hint: We know MN is median of the trapezoid and MN // AB // CD.

4. a) False b) True c) True

5. Hint: Use theorem 1 and 2.

6. Hint: Use theorem 1.