## Three Dimensional Geometry

(i) Common solids and Nets of Solids All solids are three dimensional

This is because they have length and width, this makes them to occupy space.

Examples

(i) Common solids and Nets of Solids All solids are three dimensional

This is because they have length and width, this makes them to occupy space.

Examples

This can be best demonstrated by the following nets and their respective solids.

Geometric properties of common solids

Solids have length, area and volume since measurements on them are said to be three dimensional. A vertex on a solid is a point where three or more edges meet. An edge is a line where faces meet. A plane / face is a set of points in a flat surface and extends indefinitely in all directions. In many solids some face, line or line and faces are parallel while others are not.

Example

Consider the triangular prism below.

Solution

a) . Line(s) parallel to (i). AB is DC (ii). BC are AD and FE (iii). BF is CE b) Face parallel to ABF is DCE c) Line of intersection of ABCD and ABF is AB. d) Point of intersection of (i) Line DA and face ABF is A (ii) Line DC and face BCEF is C

1.Pythagorus theorem

This theorem states that in any right-angled triangle the sum of the squares on the two shorter sides equals to the square of the hypothenuse; i.e. AB2 + BC2 = AC2 ( Height 2 + Base 2 = Hypothenuse2)

Example: Calculate the length of the side KM shown below given that the triangle is right angled at M.

Solution:

(KM)2
+ (ML)2
= (KL)2

letting KM = y cm

y2 + 52
= 132

y2 + 25
= 169

y 2 =
169 - 25

y 2 =
144

y = 12 cm

TRIGONOMETRIC RATIOS

There are three main Basic trigonometric rations
in a right angled triangle. These are sine, cosine and the tangent.

Example 1
Calculate the size of the angle θ
in the figure below.

Solution

The side 5 is on the opposite side of the angle
θ and side 7
is the adjacent side: the ratio tan θ
will be used.

Tan θ = Opp/
adj = 5/ 7

Tan θ = 0.7143 (look for the angle whole
tan = 0.7143

θ = 35.54o

Example 2

Find the size of the angle
θ shown below.

Solution

In this case the ratio sin is 10 be used since
the lines given are opposite and hypotenuse.

Sine θ = Opp/
Hyp = 3/ 8

Sin θ = 0.375

θ = 22.02

Example 3

Find the length of the side shown with label y
on the figure show below given that the images in right angled at B.

Solution

The relationship between the side 12cm and the
angle 300 given is that so the side is the adjacent side and y is the
hypotenuse of the of the triage therefore we use the ratio cosine

i.e. Cos θ = Adj/Hyp ; Cos 30o
= 12/y

Y = 12/Cos 30o ; y = 12/0.866

Y = 13.85 cm

SINE AND COSINE RULE

The sine and cosine rules are used to solve for the sides and angles of
triangle which are not right angles.

The sine rule: consider the non-right
angled triangle ABC below.

The sine Rule will given by

a/Sin A = b/ = c/sin C = 2R where R is the Radius of the circumcircle of
the triangle.

The sine Rule is used to

(i) Determine the Radius of the circumcircle

(ii) Solve for a triangle completely given two

a) Angles and a side

b) Sides and the angle opposite on of the sides

Example

Solve completely the triangle XYZ in which X =
42o ; x
= 8.4 cm and y = 10.2 cm

Solution :

The triangle is drawn as shown below

Using sine rule

x /Sin x = y/Sin y = z/sin z = 2R

But we do not have R therefore

8.4/Sin 42 = 10.2/Sin y = 8.4 Sin y =10.2 sin 42

Sin Y = (10.2 x sin 42)/8.4

Sin y = 0.8125 - y = 54.34

Therefore :

Angle XYZ = Angle Y = 54.34o

Angle XZY = 180 - (42 + 54.34) = 180 - 96.34

Angle XZY = Angle z = 83.66o

Using sine rule Again implies that Z/Sin Z = x / Sin x

Z = 8.4/Sin 83.66 = 8.4/sin 42 which implies that

Z = ( 8.4 sin 83.66)/ Sin 42

Z = 12.48 cm

**NB:** The triangle has been solved completely since
all its angles and sides are now known

The Cosine rule

By use of the non-right angled triangle shown below

The Cosine rule

a2 = b2
+ c2 - 2bc cos A

Or

b2 = a2
+ c2 - 2ac Cos B

c2 + a2
+ b2 - 2ab Cos B

All these alternative forms of cosine Rule are used to find the :

(i) The third side of a triangle given two sides and an included angle

(ii) Three angle given three sides

Example

Solve completely for the following triangle
given that P = 6.9cm, Q = 12.3cm and R = 7.6cm

Solution

The triangle is drawn and labeled as shown below

By use of the cosine rule

p2 = q2 + r2 - 2qr cos P

6.92 = (12.3)2 + (7.6)2 - 2 (12.3) (7.6) Cos P

47.61 = 151.29 + 57.76 -186.96 cos P

47.61 = 209.05 - 186.96 Cos P

-161.44/-186.96 = -186.96 Cos P

Cos P = -161.44/-186.96

Cos P = 0.8635

P = 30.29o

By use of cosine Rule still

R2 = p2
+ q2 - 2(p)(q) Cos R

(7.6)2 =6.92
+ 12.32 - 2(6.9) 912.30 Cos R

57.76 = 47.61 + 151.29 - 169.74 Cos R

57.76 = 198.90 - 169.74 Cos R

141.14 = -169.74 Cos R

Cos R = -141.14/ -169.74 which implies that Cos R = 0.8315

R = 33.75o

From the triangle PQR then therefore Angle Q can be obtained by

Q = 180 - (30.29 + 33.75)

Q =180 - 64.04

Q = 115.96o

The triangle has been solved completely since all its angles and
sides are now known.

OBJECTIVE

#### By the end of the lesson you should be able to:

1. State the geometry properties of common solids.

2. Identify the projection of a line on to a plane.

3. Identify skew lines.

4. Calculate the length between two points in three dimensional geometry.

5. Identify and calculate the angle between

i. Two lines

ii. A line and a plane

iii. Two planes.

THREE DIMENSIONAL GEOMETRY

In our everyday life we are in touch with 3
dimensional geometry. Every object one sees and touches has 3 dimension:
Length, width and height. Look at any room and items in it. These can be
described in 3 dimension.

3 dimensional geometry is important in many
careers such as architecture.

Identifying the projections of a line onto a
plane

Projections

If the plane WXYZ is lit from an external source
as shown the shadow of line AB forms on the plane as AC and is known as
the projection of the line AB onto the plane, WXYZ.

NOTE:

Point B is projected on C.

The angle between the line AB and AC is taken to be the angle between AB
and plane WXYZ.

In general, the angle between a line and a plane is the angle between
the line and its projection on the plane.

Example

Skew Lines

The lines HG and AD do not intersect and are not
parallel. These lines do not lie on the same plane such lines are skew
Lines.

The angle between the skew lines is found by translating one of the lines to the plane containing the other.

To get the angle between HG and AD
translate line HG to plane ABCD such that it lies at line AB on the
plane, then the angle between the two is angle DAB = 90o.

Example

Use the figure below and name

Solution

(i) Skew lines to GB are AD, DC, HE, EF
(blink lines GB, AD, DC, HE, EF)

(ii) Angle FEG or CDB

(iii) Angle EDB and angle GBD

Calculating Lengths And Angles In A Solid

In three dimensional geometry unknown lengths and angles can be determined by solving tringles (i.e. determining the unknown sides and angles of triangles as illustrated in the background information)

Calculating the length of a line

Example

In the figure below ABCDEFGH is a cuboid and K
is the mid-point of EH

Calculate the length of lines

(i) DH

(ii) AH

(iii) KD

Solution

By use of Pythagoras theorem

(i) (DH)2 = (DC)2 + ( CH)2

but AC = AB = 8 cm

Therefore,

(DH)2 =
82 + 42

(DH)2 =
64 + 16

(DH)2 =
80

DH = 8.944 cm

(ii) To find AH. We must consider triangle
ACH

Therefore,

we first work out the line AC

(AC)2
= (AB)2
+ (BC)2

(AC)2 =
82 + 62

(AC)2 =
64 + 36 = 100

AC = 10 cm

Therefore, (AH)2
= (AC)2
+ (CH)2

(AH)2 =
102 + 42
= 116

AH = 10.77 cm

(iii)To find KD we must consider triangle KGD
Right angled at G

For Right angled triangle HGK

(KG)2
= (HG)2
+ (AK)2

(KG)2 =
82 + 32

(KG)2 =
64 + 9

(KG)2 =
73

KG = 8.544 cm2

From Right angled triangle KGD

(KD)2
= (KG)2
+ (GD)2

(KD)2 =
8.5442 +
42

(KD)2 =
73 + 16

(KD)2 =
89

KD = 9.433 cm2

Calculating The Length And Angles Between

(a) Two Lines

The figure below shows a wedge and its given
measurements.

Solution

(i) Lines AF and BD are skewed lines. Therefore,
we must translate (move) line BD to where AE is, to form angle EAF or
move line AF to where BC is, to form angle DBC.

Therefore , moving line BD

(BD)2
= (DC)2
+ (BC)2

(BD)2 =
82 + 62

(BD)2 =
64 + 36 = 100

BD = 10 cm

To calculate angle EAF,

Use the trigonometric ratio

SineO
= 0.8

angle EAF = O
= 53.13o
(the angle between lines AF and BD)

(ii)Angle between line AD and DF

We must first calculate the lengths of lines AD and DF

Therefore , (AD)2
= (DB)2
+ (AB)2

(AD)2 =
102 + 122

(AD)2 =
100 + 144

(AD)2 =
244

AD = 15.62 cm

(DF)2 =
(DF)2 +
(CF)2

(DF)2 =
82 + 122

(DF)2 = 64 + 144

(DF)2 =
208

DF = 14.42 cm

By use of trigonometry (SOHCAHTOA) the angle required is the angle ADF

Tan O = 0.4161;

O = 22.59o
(this is the angle between line AD and DF)

(iii)Angle between line FD and ED

The angle between line FD and ED is angle FDE.

By use of trigonometry

Cos O = 0.8322

O= 33.68o
(this is the angle between line FD and ED)

Angles between a line and a plane

Example

The figure along side is a square based pyramid.
The triangular faces of the pyramid are isosceles triangles of slanting
edges 10 cm each.

(i) Find the angle between VA and plane ABCD

(ii) If m is the mid-point of line BC find the angle between VM and the
plane ABCD.

(AC)2
= (AB)2
+ (BC)2

(AC)2 =
82 + 82

(AC)2 =
64 + 64

(AC)2 =
128

AC = 11.32 cm

AD =1/2 AC

Therefore, AD = 5.66 cm

CosO=
0.566;

O =
55.53O
(this is the angle between lines VA and plane ABCD)

(ii) Since M is the mid-point of BC then MC = BM = 4 cm.

(VM)2 = (VC)2
+ (CM)2

(VM)2 = 102
- 42

(VM)2 = 100 - 16

(VM)2 = 84

VM = 9.165 cm

Angle between VM and the plane ABCD is angle VMO(blink angle between
line VM and plane ABCD)

CosO
= 0.4364;

O =
64.12O

NOTE:

An angle between a line and a plane is the size of the angle between the
line and its projection on that given plane.

Angle between a plane and another plane

Example

The figure below is a cuboid 8.5 cm long, 6 cm
wide and 3.5 cm high.

(i) Plane KLQP and MNPQ

(ii) KLQP and KLMN

(iii) KMQS and MNPQ

Solution

(i) The angle between planes KLQP and MNPQ is
given by (formed on line PQ)

(ii) The angle between planes KLQP and KLMN
is angle formed on line KL (intersection of the plane)

(iii) Angle between planes KMQS and MNPQ is the angle formed on
the line QM (intersection of the planes)

Three Dimensional Geometry

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