2. Finding the Distance Between Two Skew Lines by Projection

Theorem: Distance between two skew lines is equal to the distance between their projections on the plane perpendicular to one of them.

Proof: Given two skew lines a and b with common perpendicular NM, and a plane α such that b ⊥ α.

Let M₁ be intersection of α and b, and a₁ be projection of a on α (Figure 2.37).

So, we need to prove that the length of NM is equal to the distance between M₁ and a₁.

Let N₁ ∈ a₁ be the projection of N on α .

Since b ⊥ α, b ⊥ M₁N₁.

For the common perpendicular, we can write NM ⊥ b.

So, both NM and N₁M₁ are perpendicular to the same line b.

Therefore, NM // N₁M₁.

On the order hand, MM₁ ⊥ α and NN₁ ⊥ α . So, MM₁ // NN₁ , which yields the result that NMM₁N₁ is a rectangle. So we have NM = N₁M₁.

Now, we will show that N₁M₁ is the distance between a₁ and M₁. It can be done just by showing that N₁M₁ ⊥ a₁.

Since MN ⊥ a, ∠MNK = 90°.

The projection of angle MNK on α is ∠M₁N₁K₁.

Since ∠MNK= 90° and MN // M₁N₁ we can write ∠M₁N₁K₁= 90°, which means that M₁N₁ ⊥ a₁.

So, we can conclude that the distance between a and b is equal to the distance between M₁ and a₁.

Example 65: Given an equilateral ΔABC with a side of m units. If ΔA₁B₁C₁ is the corresponding projection of ΔABC on a parallel plane, find the distance between AA₁ and B₁C.

S olution: We just need to find a plane perpendicular to one of skew lines AA₁ and B₁C.

Here, AA₁ is perpendicular to ΔA₁B₁C₁.

We find the projections of AA₁ and B₁C on (A₁B₁C₁).

A₁ = Proj (AA₁)

B₁C₁ = Proj (B₁C)

The answer of our question will be the distance from A₁ to B₁C₁ which is the height to B₁C₁ in ΔA₁B₁C₁.

ΔABC is equilateral. So its projection on a parallel plane will be a congruent triangle with a side of m units.

Thus, A₁H = . So, the distance between AA₁ and B1C is units.

Example 66: Given parallelogram ABCD and point K not lying in the plane of ABCD. If DK ⊥ (ABCD), AB = 6 cm, AD = 8 cm, DK = 3 cm, and ∠BAD = 30°,

a) find the distance between DK and AB.

b) find the distance between BK and CD.

S olution:

a) DK ⊥ (ABCD), so we can take ABCD as projection plane.

Proj_ABC DK = D and Proj_ABC AB = AB.

Therefore, distance between DK and AB will be distance between D and AB, i.e. height DH of parallelogram ABCD.

m(∠BAD) = 30° , so DH = = 4 cm.

Hence, the distance betweem DK and AB is DH = 4 cm.

b ) DH ⊥ DC and DK ⊥ DC, so DC ⊥ (DKH).

Proj_(DKH) DC = D.

By the three perpendiculars theorem,

BH ⊥ (DKH), so Proj_(DKH) KB = KH.

Now we need to find the distance from D to KH.

Since ΔKDH is a right triangle with sides 3, 4, 5, this distance is

Example 67: Given a square ABCD and O intersection point of its diagonals. Line segment MO is perpendicular to the plane of the square. MO = cm and one side of the square is 4 cm. Find the distance

a) between AB and MO.

b) between BD and MC.

S olution: a) (ABC) is a plane perpendicular to MO.

Proj_(ABC) AB = AB and Proj_(ABC) MO = O.

Hence, the distance between MO and AB is the distance from O to AB, that is OK.

b ) We know that, in a square the diagonals are perpendicular to each other.

So, BD ⊥ AC.

BD ⊥ AC and BD ⊥ MO, so BD ⊥ (MOC).

Therefore, we can take plane MOC as the projection plane.

Proj_(MOC) BD = O, Proj_(MOC) MC = MC.

S o, the distance between BD and MC is the distance between point O and MC.

In ΔMOC, ∠MOC = 90°.

Draw OT ⊥ MC. We need to find OT.

MC² = MO² + OC²; MC² = ( )² + ( )² = 16; MC = 4 cm.

Finally, MO ⋅ OC = MC ⋅ OT and

Check Yourself 16

1. ABCDA₁B₁C₁D₁ is a cube with one side is 6 cm. Find the distance between AC and DC₁.

2 . In the adjacent figure, point P is not lying in the plane of ΔABC. Point A₁ ∈ AP such that .

Point C₁ ∈ CP such that A₁C₁ // AC. PH is perpendicular to (ABC).

If PH = 6 cm, find the distance between lines A₁C₁ and

BC.

3. Given a square ABCD with a side of 1 cm. Line segment MB is perpendicular to the plane of square ABCD. If MB = 1 cm, find the distance between AC and MD.

4. Given two parallel planes α and β and line segments AB and CD such that A and C are in α, B and D are in β, and AB ⊥ α . Find the distance between AB and CD if

AB = 20 cm, CD = 25 cm, AC = 14 cm, and BD = 13 cm.

Answers

1. cm 2. 4 cm 3. cm 4.

EXERCISES 3

A. Types of Projection

1. ABC is a triangle in a plane and A₁B₁C₁ is its projection in another plane. Given that AA₁ = 6 cm, BB₁ = 9 cm and CC₁ = 12 cm. Find the distance between the centroids of these triangles.

2. Two intersecting planes α and  are given. Line d is their intersection. Given that a line n is perpendicular to α and line m is parallel to n. What can we say about m and d?

3. An equilateral triangle ABC is given in a plane α.

a. Can we always say that its projection is also an equilateral triangle?

b. If plane  is passing through AB then can we say that projection of ABC on  is equilateral?

c. At which condition the projection of ABC in plane  is an equilateral triangle?

4. ABCD is a square in plane α. Plane  passes through AB. The distance between the intersection point of diagonals of ABCD and plane  is 7 cm. Find the distance between point D and plane .

5. Given that AB and CD are two line segments in space. Both of AB and CD are parallel to line d. Distance between AB and d is 12 cm, CD and d is 6 cm. If both of the lengths of AB and CD are 20 cm and length of projection of AB in line d is 10 cm find the length of the projection of line CD in d?

6. An isosceles right triangle has legs 12 cm. The hypothenuse of this triangle is parallel to a line d. What is the length of the projection of this triangle in line d?

7. Can we say that the projection of a right angle in a plane is also a right angle if none of the arms of the angle are parallel to given plane?

8 . In the figure the lines d₁ and d₂ are perpendicular to each other. The projection of point S on the plane of lines d₁ and d₂ is point A. The points K, L, M and N are equidistant from A. Given that,

KM = 24 cm and KS = 13 cm.

Find A(KNS) + A(KMS).

9. In the figure G is centroid of equailateral triangle ABC. R is an exterior point and RG (ABC). Given that,

A B = 12 cm and

RG = 4 cm.

Find RC.

1 0. In the figure the planes P and Q intersected alond the line d. Point A is in plane P and AB d. T is the projection of point A in plane Q. Given that,

AT = 4 cm,

TB = 3 cm and

BC = 5 cm.

Find AC.

1 1. In the figure the lines d₁, d₂, d₃ are perpendicular to the parallel planes P and T. The three lines are equidistant from each other. Distance between the planes P and T is 16 cm.

If R is the midpoint of AK and AK = find the distance from S to R.

B. Using Projection in Calculations

12. In the figure ABCD and BCLK are two congruent squares.

Given that AB BK and DL = 8 cm.

F ind DK.

13. In the plane α, consider the points B, C and E so that BC = 9 cm, CE = 12 cm and BE = 15 cm. Let A be a point not in plane α such that AB ⊥ α and AB = 36 cm.

a) Find the angle between line CE and plane ABC.

b) Find the tangent of the angle between planes ACE and BCE.

1 4. In the figure E₁ and E₂ are two intersecting planes and their intersection is the line BC. The points K and P are on the planes E₁ and E₂, respectively. Given that,

KC BC,

PC BC,

KC = PC = cm and

KP = 9 cm.

Find the angle between the planes E₁ and E₂.

15. In the figure B, C, D are in the plane E. The propejction of point A in the plane E is D. Given that,

B D DC,

BD = DC = 6 cm and

AD = 8 cm.

Find the sine of the dihedral angle between the plane E and the plane (ABC).

16*. Given ABC and ABD two equilateral triangles included in planes α and β respectively. If α ⊥ β, calculate the angle between lines AB and CD.

17*. In triangle ABC we have BC = 10 cm and

cos B ⋅ cos C = cos A + sin B ⋅ sin C .

Given a point O not in the plane of triangle ABC with the property OA = OB = OC = 13 cm.

Find the cosine of the angle between line OA and plane ABC.

18. In the figure the dihedral angle between P and Q is 120. The intersection of the planes is line d. Line k is equidistant from both of the planes and 10 cm away from the line d. Find the distance between k and plane Q.

1 9. The dihedral angle between P and Q is 60. There is a circe in plane P with radius 4 cm. What is the area of the projection of this circle in plane Q?

20. In the figure ABCD is a rhombus. PA is perpendicular to the plane of ABCD. The distance from A to BC is 4 cm and DC = 5 cm. If PA = cm find the area of PAC.

21. In the figure two planes are parallel to each other.

The line d make 60 angle with these planes. If the distance between the intersection points of line d and the planes is 12 cm find the distance between these two planes.

2 2. In the figure ABC is an equilateral triangle and BDC is an isosceles right triangle. Given that DC = cm

and dihedral angle between the planes of (ABC) and (BDC) is 150. Find AD.

2 3. In the figure BCDE is a rectangle and ABC is an isosceles triangle. The projection of point A is on the intersection point of the diagonals of BCDE. Given that,

BC = 30 cm and the projection of AK on plane (BCDE) is 20 cm. Find the sum of the lengths of diagonals of the rectangle.

24. Decide whether a trihedral angle can be constructed in which the face angles are respectively;

a) 70°, 80°, and 100°

b) 25°, 30°, and 75°

c) 70°, 115°, and 190°

d) 120°, 140°, and 160°

25. Can a trihedral angle be constructed with the following dihedral angles?

a) 130°, 25°, 75°

b) 30°, 105°, 35°

c) 60°, 80°, 100°

d) 60°, 190°, 140°

26*. A right triangle ABC is projected onto a plane α which is parallel to BC and passes through vertex A, so that projections AB' and AC' of sides AB and AC have the lengths 3 cm and 5 cm and cos ∠B'AC' = . Calculate the distance from point B to plane α .

27. In the figure line AB cut the plane P by a 30 angle. If the length of AB is 20 cm find the length of its projection on the plane P.

2 8. In the figure P is a plane and it is passing through the center of given circle. The circle is inclined for 45. The area of the projection of the circle is cm².

Find the radius of the circle.

29. Area of a rectangle in a plane P is a. If the area of the projection of this rectangle on another plane Q is find the angle between these two planes.

3 0. In the figure ABCD is a rectangle and PCD is a triangle. Point T is the center of rectangle and the projection of point P in the plane of (ABCD). The dihedral angle between the planes of ABCD and PCD is 60.

Given that A(PCD) = 10 cm².

Find A(ABCD).

3 1. In the adjacent figure, a cube is given with one side is m units.

Find distance between

a) B1D and D1C1.

b) BC1 and AC.

c) B1C and BD1. (in terms of m)

32. Given a right trapezoid ABCD where ∠A = ∠B = 90°. Line segment KA is perpendicular to the plane of ABCD. AD = b and AB = a with ∠C = α.

Find the distance between

a) AK and BC

b) KD and BC

c) AK and CD

3 3. In the adjacent figure, ABC is an equilateral triangle with a side of 6 cm. A₁B₁C₁ is the

corresponding projection of ΔABC on a parallel plane. AA₁= BB₁ = CC₁ = 6 cm and they are perpendicular to plane ABC. Find the distance

between AA₁ and B₁C.

34. Through vertex B of ΔABC drawn a line c, perpendicular to the plane of ΔABC. Find the distance between line c and AC if AC = 25 cm, BC = 15 cm, and ∠ABC = 90°.

35. Given a square ABCD and O the intersection point of its diagonals. Line segment MO is perpendicular to the plane of the square and MO = units. If one side of the square is 2a units, find:

a) the distance between AB and MO

b) the distance between BD and MC in terms of a.

36. In a plane α given a circle with center O and radius r. Through point C on the circle drawn a line c, perpendicular to plane α . Line d, lying in plane α, tangent to the given circle at point A. What is the distance between lines d and c if ∠AOC = 120° in terms of r?

37. Through vertex A of square ABCD with side length m drawn a plane α perpendicular to AC. Find the distance between BD and a line c which is lying in α and not parallel to BD, in terms of m.