A. Angle Between Two Lines

We know that, in space, lines can be coincident, parallel, intersecting or skew. We know finding the angle between two intersecting lines from the plane geometry. But how can we find the angle between the skew lines? In this case, on one of the skew lines we take a point and through this point we draw a line parallel to the other one. Then we get two intersecting lines and the angle between them will be the angle between the skew lines.

Note 1: If two lines are parallel, the angle between them is 0°. If two lines are perpendicular, the angle between them is 90°.

Note 2: Two lines form two different angles. By the angle between two lines, we mean the smaller one.

In Figure 2.12, we take α or 180 – α as the angle between d₁ and d₂ by finding the smaller one.

b. Angle Between a Line and a Plane

Angle between a line and a plane is the angle between the line itself and its projection on that plane.

In Figure 2.13, Proj_α d = d' and ∠AOA' is the angle between d and α.

Note: If a line is perpendicular to a plane, the angle between them is 90°. If the line is parallel to that plane, the angle between them is 0°.

Theorem: Between all angles formed by a line and all of the lines in a plane, the angle with the projection of the given line is the smallest one.

Proof: Let d be a given line, m its projection on plane α and n another line in α.

On line n, take line segment AN₁ = AM₁ i.e. equal to the projection of inclined line segment MA, where M₁ is the projection of point M on d.

Then two sides of triangles AM₁M and AN₁M are equal: side AM is common to both, and AM₁ and AN₁ are equal by construction.

But the third side MN₁ in triangle AN₁M is longer than the third side MM₁ in the triangle AM₁M (an inclined line is greater than a perpendicular).

Hence the opposite angle β in ΔAN₁M is greater than angle γ in ΔAM₁M.

Therefore, β > γ.

Conclusion:

The acute angle between a line lying in a plane and the projection of an inclined line on this plane is less than the angle between the line and the inclined line.

Refer to Figure 2.14. m is the projection of n on (AM₁M) and d is another line in (AM₁M). So the angle between m and n is smaller than the angle between n and d.

c. Angle Between Two Planes (Dihedral Angle)

Definition: (dihedral angle, faces, edges)

A dihedral angle is a figure formed by two half planes having a common line.

The two half planes are called as the faces and their intersection line as the edge of the dihedral angle.

A dihedral angle may be read by naming a point in one

face, the edge, and a point in the other face.

Definition: (plane angle of a dihedral angle)

A plane angle of a dihedral angle is an angle formed by two perpendiculars drawn to the same point on the edge.

In Figure 2.17, ∠POP₁ is a plane angle of dihedral angle ABCF.

There are infinitely many plane angles of a dihedral angle.

Theorem: Plane angles of a dihedral angle are equal to each other.

Proof:

In Figure 2.18, ∠POP₁ and ∠TO₁T₁ are two plane angles of dihedral angle ABCF.

Rays OP and O₁T lie on the same face of the dihedral angle and they are perpendicular to edge BC, so they are parallel in the same direction.

Similarly rays OP₁ and O₁T₁ are also parallel in the same direction.

Therefore, ∠POP₁ = ∠TO₁T₁.

Note 1: In our book we will use dihedral angle as plane angle of dihedral angle.

Note 2: If two planes are intersecting planes, the dihedral angle between them is the smaller one. If two planes are parallel, the dihedral angle between then is 0°.

Definition: (perpendicular and oblique planes)

If the angle between two planes is 90°, planes are said to be perpendicular, otherwise they are oblique planes.

Example 52: Given an equilateral triangle ABC with one side 8 cm and point O is its centroid. Through point O, a line segment OT is drawn that is perpendicular to the plane of ΔABC. The length of OT is 4 cm. Find the angle between the planes of ΔABC and ΔABT.

Solution: Triangle ABC is equilateral so its height CH will pass through point O.

Let us draw TA and TB and join T and H (Figure 2.21).

By the three perpendiculars theorem, TH will be perpendicular to AB.

Hence, the angle between planes of ΔABC and ΔABT

will be ∠THO. Now, let us calculate this angle.

cm,

cm,

then m(ÐTHO) = 60°.

Example 53: ABCD is a rhombus with ∠A = 60°. AB = 6 cm and BE ⊥ (ABC),

BE = cm. Find the angle between (AED) and (ABC).

Solution: First of all, let us draw BH ⊥ AD.

If we join E and H then EH will be perpendicular to AD by the three perpendiculars theorem.

∠EHB is the angle between planes AED and ABC. Now, let us calculate it.

In ΔAHB, BH = AB ⋅ sin60° = 6 ⋅ = cm.

In ΔEBH, BE = BH = cm.

So ΔEBH is an isosceles right triangle. Therefore, ∠EHB = 45°.

Example 54: The endpoints of line segment AB lie in the planes as in the figure. AB = 16 cm.

Two half planes make an angle of 120°.

AC and BD are perpendicular to intersection line of these two planes.

Find length of CD if AC = 7 cm and BD = 11 cm.

Solution: In plane β, we take CE = BD that is perpendicular to intersection line of planes α and β.

So, angle ACE will be the dihedral angle between α and β, BDCE will be a rectangle.

Now, let us join A and E.

AC ⊥ CD and CE ⊥ CD, so CD ⊥ (ACE).

Since CD // BE, BE ⊥ (ACE). Thus, ∠AEB = 90°.

Now, in ΔACE by applying the cosine theorem.

AE² = AC² + CE² – 2 ⋅ AC ⋅ CE ⋅ cos (∠ACE)

AE² = 7² + 11² – 2 ⋅ 7 ⋅ 11 ⋅ cos 120°

AE² = 247.

ΔAEB is a right triangle. By the Pythagorean theorem, we have

AE² + EB² = AB²; 247 + EB² = 16²; EB² = 9 and EB = 3 cm.

In rectangle BDCE, EB = CD. Therefore, CD = 3 cm.

Example 55: In the adjacent figure, triangle ABC is given. Point P is not lying in the plane of ΔABC.

AB = AC = AP = 4 cm.

BC = BP = CP = 6 cm.

Calculate:

a) the angle between lines AB and CP.

b) the cosine of the angle between planes ACP and BCP.

Solution:

a) In ΔBPC draw height BH of side PC.

Since ΔBPC is an equilateral triangle H is the midpoint of PC.

In ΔPAC, PA = AC and H is the midpoint of

PC. So AH ⊥ PC too. Hence PC ⊥ (AHB).

Since AB is a line in (AHB), PC ⊥ AB. That means the angle between PC and AB is 90°.

b) The angle between given faces is ∠AHB, So we need to find cos ∠AHB.

cm,

cm,

AB= 4 cm.

In ΔAHB by cosine theorem:

AB² = AH² + BH² – 2 · AH · BH · cos ∠AHB,

16 = 7 + 27 – 2 · · · cos ∠AHB, then .

Example 56: A rhombus ABCD has AB = cm. The projection of the rhombus on a plane α which contains diagonal AC is a square AB'CD' with an area of 2 cm². Find:

a. the area of the rhombus.

b. the angle between plane ABC and plane α.

Solution:

a. Since the area of AB′CD′ is 2 cm² one side of it is cm.

DD′ ⊥ α. So DD′ ⊥ AD′.

Then cm.

Let D′B′ intersect AC at O. Then O is the midpoint of AC.

So DB intersects AC at O too, and DO = OB.

DD′ ⊥ D′O. So cm.

So DB = 4 cm.

It is obvious that AC = · = 2 cm.

Therefore the area of rhombus is

b. The intersection of (ABC) and α is AC.

DD′ ⊥ α, D′O ⊥ AC.

So by three perpendiculars theorem DO ⊥ AC.

So the angle between (ABC) and α is ∠DOD′.

Then

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