vMathematics is both the queen and the hand-maiden of

all sciences – E.T. BELLv

12.1 Introduction

You may recall that to locate the position of a point in a

plane, we need two intersecting mutually perpendicular lines

in the plane. These lines are called the coordinate axes

and the two numbers are called the coordinates of the

point with respect to the axes. In actual life, we do not

have to deal with points lying in a plane only. For example,

consider the position of a ball thrown in space at different

points of time or the position of an aeroplane as it flies

from one place to another at different times during its flight.

Similarly, if we were to locate the position of the

lowest tip of an electric bulb hanging from the ceiling of a

room or the position of the central tip of the ceiling fan in a room, we will not only

require the perpendicular distances of the point to be located from two perpendicular

walls of the room but also the height of the point from the floor of the room. Therefore,

we need not only two but three numbers representing the perpendicular distances of

the point from three mutually perpendicular planes, namely the floor of the room and

two adjacent walls of the room. The three numbers representing the three distances

are called the coordinates of the point with reference to the three coordinate

planes. So, a point in space has three coordinates. In this Chapter, we shall study the

basic concepts of geometry in three dimensional space.*

*

For various activities in three dimensional geometry one may refer to the Book, “A Hand Book for

designing Mathematics Laboratory in Schools”, NCERT, 2005.

Leonhard Euler

(1707-1783)

12Chapter

INTRODUCTION TO THREE

DIMENSIONAL GEOMETRY

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INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 269

12.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space

Consider three planes intersecting at a point O

such that these three planes are mutually

perpendicular to each other (Fig 12.1). These

three planes intersect along the lines X′OX, Y′OY

and Z′OZ, called the x, y and z-axes, respectively.

We may note that these lines are mutually

perpendicular to each other. These lines constitute

the rectangular coordinate system. The planes

XOY, YOZ and ZOX, called, respectively the

XY-plane, YZ-plane and the ZX-plane, are

known as the three coordinate planes. We take

the XOY plane as the plane of the paper and the

line Z′OZ as perpendicular to the plane XOY. If the plane of the paper is considered

as horizontal, then the line Z′OZ will be vertical. The distances measured from

XY-plane upwards in the direction of OZ are taken as positive and those measured

downwards in the direction of OZ′ are taken as negative. Similarly, the distance

measured to the right of ZX-plane along OY are taken as positive, to the left of

ZX-plane and along OY′ as negative, in front of the YZ-plane along OX as positive

and to the back of it along OX′ as negative. The point O is called the origin of the

coordinate system. The three coordinate planes divide the space into eight parts known

as octants. These octants could be named as XOYZ, X′OYZ, X′OY′Z, XOY′Z,

XOYZ′, X′OYZ′, X′OY′Z′ and XOY′Z′. and denoted by I, II, III, ..., VIII , respectively.

12.3 Coordinates of a Point in Space

Having chosen a fixed coordinate system in the

space, consisting of coordinate axes, coordinate

planes and the origin, we now explain, as to how,

given a point in the space, we associate with it three

coordinates (x,y,z) and conversely, given a triplet

of three numbers (x, y, z), how, we locate a point in

the space.

Given a point P in space, we drop a

perpendicular PM on the XY-plane with M as the

foot of this perpendicular (Fig 12.2). Then, from the point M, we draw a perpendicular

ML to the x-axis, meeting it at L. Let OL be x, LM be y and MP be z. Then x,y and z

are called the x, y and z coordinates, respectively, of the point P in the space. In

Fig 12.2, we may note that the point P (x, y, z) lies in the octant XOYZ and so all x, y,

z are positive. If P was in any other octant, the signs of x, y and z would change

Fig 12.1

Fig 12.2

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MATHEMATICS

accordingly. Thus, to each point P in the space there corresponds an ordered triplet

(x, y, z) of real numbers.

Conversely, given any triplet (x, y, z), we would first fix the point L on the x-axis

corresponding to x, then locate the point M in the XY-plane such that (x, y) are the

coordinates of the point M in the XY-plane. Note that LM is perpendicular to the

x-axis or is parallel to the y-axis. Having reached the point M, we draw a perpendicular

MP to the XY-plane and locate on it the point P corresponding to z. The point P so

obtained has then the coordinates (x, y, z). Thus, there is a one to one correspondence

between the points in space and ordered triplet (x, y, z) of real numbers.

Alternatively, through the point P in the

space, we draw three planes parallel to the

coordinate planes, meeting the x-axis, y-axis

and z-axis in the points A, B and C, respectively

(Fig 12.3). Let OA = x, OB = y and OC = z.

Then, the point P will have the coordinates x, y

and z and we write P (x, y, z). Conversely, given

x, y and z, we locate the three points A, B and

C on the three coordinate axes. Through the

points A, B and C we draw planes parallel to

the YZ-plane, ZX-plane and XY-plane,

respectively. The point of interesection of these three planes, namely, ADPF, BDPE

and CEPF is obviously the point P, corresponding to the ordered triplet (x, y, z). W

e

observe that if P (x, y, z) is any point in the space, then x, y and z are perpendicular

distances from YZ, ZX and XY planes, respectively.

A

Note The coordinates of the origin O are (0,0,0). The coordinates of any point

on the x-axis will be as (x,0,0) and the coordinates of any point in the YZ-plane will

be as (0, y

, z).

Remark The sign of the coordinates of a point determine the octant in which the

point lies. The following table shows the signs of the coordinates in eight octants.

Table 12.1

Fig 12.3

I II III IV V VI VII VIII

x + – – + + – – +

y + + –

– + + – –

z + + + + – – – –

Octants

Coord

inates

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INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 271

Example 1 In Fig 12.3, if P is (2,4,5), find the coordinates of F.

Solution

For the point F, the distance measured along OY is zero. Therefore, the

coordinates of F are (2,0,5).

Example 2 Find the octant in which the points (–3,1,2) and (–3,1,– 2) lie.

Solution

From the Table 12.1, the point (–3,1, 2) lies in second octant and the point

(–3, 1, – 2) lies in octant VI.

EXERCISE 12.1

1. A point is on the

x

-axis. What are its y-coordinate and z-coordinates?

2. A point is in the XZ-plane. What can you say about its y-coordinate?

3. Name the octants in which the following points lie:

(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5),

(–3, –1, 6) (– 2, – 4, –7).

4. Fill in the blanks:

(i) The x-axis and y-axis taken together determine a plane known as_______.

(ii) The coordinates of points in the XY-plane are of the form _______.

(iii) Coordinate planes divide the space into ______ octants.

12.4 Distance between Two Points

We have studied about the distance

between two points in two-dimensional

coordinate system. Let us now extend this

study to three-dimensional system.

Let P(x

1

, y

1

, z

1

) and Q ( x

2

, y

2

, z

2

)

be two points referred to a system of

rectangular axes OX, OY and OZ.

Through the points P and Q draw planes

parallel to the coordinate planes so as to

form a rectangular parallelopiped with one

diagonal PQ (Fig 12.4).

Now, since ∠PAQ is a right

angle, it follows that, in triangle PAQ,

PQ

2

= PA

2

+ AQ

2

... (1)

Also, triangle ANQ is right angle triangle with ∠ANQ a right angle.

Fig 12.4

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MATHEMATICS

Therefore AQ

2

= AN

2

+ NQ

2

... (2)

From (1) and (2), we have

PQ

2

= PA

2

+ AN

2

+ NQ

2

Now PA = y

2

– y

1

, AN = x

2

– x

1

and NQ = z

2

– z

1

Hence PQ

2

= (x

2

– x

1

)

2

+ (y

2

– y

1

)

2

+ (z

2

– z

1

)

2

Therefore PQ =

2

12

2

12

2

12

)()()( zzyyxx −+−+−

This gives us the distance between two points (x

1

, y

1

, z

1

) and (x

2

, y

2

, z

2

).

In particular, if x

1

= y

1

= z

1

= 0, i.e., point P is origin O, then OQ =

2

2

2

2

2

2

zyx ++

,

which gives the distance between the origin O and any point Q (x

2

, y

2

, z

2

).

Example 3 Find the distance between the points P(1, –3, 4) and Q (– 4, 1, 2).

Solution The distance PQ between the points P (1,–3, 4) and Q (– 4, 1, 2) is

PQ =

222

)42()31()14( −+++−−

=

41625 ++

=

45

=

3 5

units

Example 4 Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear.

Solution We know that points are said to be collinear if they lie on a line.

Now, PQ =

14419)53()32()21(

222

=++=−+−++

QR =

1425616436)31()20()17(

222

==++=−−+−+−

and PR =

14312636981)51()30()27(

222

==++=−−+−++

Thus, PQ + QR = PR. Hence, P, Q and R are collinear.

Example 5 Are the points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the vertices

of a right angled triangle?

Solution By the distance formula, we have

AB

2

= (10 – 3)

2

+ (20 – 6)

2

+ (30 – 9)

2

= 49 + 196 + 441 = 686

BC

2

= (25 – 10)

2

+ (– 41 – 20)

2

+ (5 – 30)

2

= 225 + 3721 + 625 = 4571

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