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 =
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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INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 273
CA
2
= (3 – 25)
2
+ (6 + 41)
2
+ (9 – 5)
2
= 484 + 2209 + 16 = 2709
We find that CA
2
+ AB
2
≠ BC
2
.
Hence, the triangle ABC is not a right angled triangle.
Example 6 Find the equation of set of points P such that PA
2
+ PB
2
= 2k
2
, where
A and B are the points (3, 4, 5) and (–1, 3, –7), respectively.
Solution Let the coordinates of point P be (x, y, z).
Here PA
2
= (x – 3)
2
+ (y – 4)
2
+ ( z – 5)
2
PB
2
= (x + 1)
2
+ (y – 3)
2
+ (z + 7)
2
By the given condition PA
2
+ PB
2
= 2k
2
, we have
(x – 3)
2
+ (y – 4)
2
+ (z – 5)
2
+ (x + 1)
2
+ (y
– 3)
2
+ (z + 7)
2
= 2k
2
i.e., 2x
2
+ 2y
2
+ 2z
2
– 4x – 14y + 4z = 2k
2
– 109.
EXERCISE 12.2
1. Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1)
(iii) (–1, 3, – 4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3).
2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
3. Verify the following:
(i) (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.
4. Find the equation of the set of points which are equidistant from the points
(1, 2, 3) and (3, 2, –1).
5. Find the equation of the set of points P, the sum of whose distances from
A (4, 0, 0) and B (– 4, 0, 0) is equal to 10.
12.5 Section Formula
In two dimensional geometry, we have learnt how to find the coordinates of a point
dividing a line segment in a given ratio internally. Now, we extend this to three dimensional
geometry as follows:
Let the two given points be P(x
1
, y
1
, z
1
) and Q (x
2
, y
2
, z
2
). Let the point R (x, y, z)
divide PQ in the given ratio m : n internally. Draw PL, QM and RN perpendicular to
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MATHEMATICS
the XY-plane. Obviously PL || RN || QM and feet
of these perpendiculars lie in a XY-plane. The
points L, M and N will lie on a line which is the
intersection of the plane containing PL, RN and
QM with the XY-plane. Through the point R draw
a line ST parallel to the line LM. Line ST will
intersect the line LP externally at the point S and
the line MQ at T, as shown in Fig 12.5.
Also note that quadrilaterals LNRS and
NMTR are parallelograms.
The triangles PSR and QTR are similar.
Therefore,
1
2
PR SP SL PL NR PL
QR QT QM TM QM NR
m z z
n z z
= = = = =
– – –
– – –
This implies
2 1
mz nz
z
m n
=
+
+
Similarly, by drawing perpendiculars to the XZ and YZ-planes, we get
2 1 2 1
my +ny mx +nx
y = and x=
m+ n m+ n
Hence, the coordinates of the point R which divides the line segment joining two points
P (x
1
, y
1
, z
1
) and Q (x
2
, y
2
, z
2
) internally in the ratio m : n are
+
+
+
+
+
+
nm
nzmz
nm
nymy
nm
nxmx
121212
,,
If the point R divides PQ externally in the ratio m : n, then its coordinates are
obtained by replacing n by – n so that coordinates of point R will be
−
−
−
−
−
−
nm
nzmz
nm
nymy
nm
nxmx
121212
,,
Case 1 Coordinates of the mid-point: In case R is the mid-point of PQ, then
m : n = 1 : 1 so that
x
=
2
and
2
,
2
212121
zz
z
yy
y
xx
+
=
+
=
+
.
These are the coordinates of the mid point of the segment joining P (x
1
, y
1
, z
1
)
and Q (x
2
, y
2
, z
2
).
Fig 12.5
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INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 275
Case 2 The coordinates of the point R which divides PQ in the ratio k : 1 are obtained
by taking
m
k
n
=
which are as given below:
+
+
+
+
+
+
k
zkz
k
yky
k
xxk
1
,
1
,
1
121212
Generally, this result is used in solving problems involving a general point on the line
passing through two given points.
Example 7 Find the coordinates of the point which divides the line segment joining
the points (1, –2, 3) and (3, 4, –5) in the ratio 2 : 3 (i) internally, and (ii) externally.
Solution (i) Let P (x, y, z) be the point which divides line segment joining A(1, – 2, 3)
and B (3, 4, –5) internally in the ratio 2 : 3. Therefore
2(3) + 3(1) 9
= =
2 + 3 5
x
,
2(4) + 3(–2) 2
= =
2 + 3 5
y
,
2(–5) + 3(3) –1
= =
2 + 3 5
z
Thus, the required point is
−
5
1
,
5
2
,
5
9
(ii) Let P (x, y, z) be the point which divides segment joining A (1, –2, 3) and
B (3, 4, –5) externally in the ratio 2 : 3. Then
2(3) + (–3)(1)
= = – 3,
2 + (–3)
x
2(4) + (–3)(–2)
= = – 14
2 + (–3)
y
,
2(–5) + (–3)(3)
= =19
2 + (–3)
z
Therefore, the required point is (–3, –14, 19).
Example 8 Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6)
and (14, 0, –2) are collinear.
Solution Let A (– 4, 6, 10), B (2, 4, 6) and C(14, 0, – 2) be the given points. Let the
point P divides AB in the ratio k : 1. Then coordinates of the point P are
+
+
+
+
+
−
1
106
,
1
64
,
1
42
k
k
k
k
k
k
Let us examine whether for some value of k, the point P coincides with point C.
On putting
2 – 4
= 14
+ 1
k
k
, we get
3
2
k
= −
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MATHEMATICS
When
3
2
k
= −
, then
3
4( ) 6
4 6
2
0
3
1
1
2
k
k
− +
+
= =
+
− +
and
3
6( ) 10
6 10
2
2
3
1
1
2
k
k
− +
+
= = −
+
− +
Therefore, C (14, 0, –2) is a point which divides AB externally in the ratio 3 : 2 and is
same as P.Hence A, B, C are collinear.
Example 9 Find the coordinates of the centroid of the triangle whose vertices are
(x
1
, y
1
, z
1
), (x
2
, y
2
, z
2
) and (x
3
, y
3
, z
3
).
Solution Let ABC be the triangle. Let the coordinates of the vertices A, B,C be
(x
1
, y
1
, z
1
), (x
2
, y
2
, z
2
) and (x
3
, y
3
, z
3
), respectively. Let D be the mid-point of BC.
Hence coordinates of D are
+++
2
,
2
,
2
323232
zzyyxx
Let G be the centroid of the triangle. Therefore, it divides the median AD in the ratio 2 : 1.
Hence, the coordinates of G are
2 3 2 3 2 3
1 1 1
2 2 2
2 2 2
2 1 2 1 2 1
x x y y z z
x y z
, ,
+ + +
+ + +
+ + +
or
1 2 3 1 2 3 1 2 3
3 3 3
x x x y y y z z z
, ,
+ + + + + +
Example 10 Find the ratio in which the line segment joining the points (4, 8, 10) and
(6, 10, – 8) is divided by the YZ-plane.
Solution Let YZ-plane divides the line segment joining A (4, 8, 10) and B (6, 10, – 8)
at P (x, y, z) in the ratio k : 1. Then the coordinates of P are
+
−
+
+
+
+
1
810
,
1
108
,
1
64
k
k
k
k
k
k
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INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 277
Since P lies on the YZ-plane, its x-coordinate is zero, i.e.,
4 + 6
= 0
+ 1
k
k
or
2
3
k
= −
Therefore, YZ-plane divides AB externally in the ratio 2 : 3.
EXERCISE 12.3
1. Find the coordinates of the point which divides the line segment joining the points
(– 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2 : 3 internally, (ii) 2 : 3 externally.
2. Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio
in which Q divides PR.
3. Find the ratio in which the YZ-plane divides the line segment formed by joining
the points (–2, 4, 7) and (3, –5, 8).
4. Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and
1
C 0 2
3
, ,
are collinear.
5. Find the coordinates of the points which trisect the line segment joining the points
P (4, 2, – 6) and Q (10, –16, 6).
Miscellaneous Examples
Example 11 Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) and
D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
Solution To show ABCD is a parallelogram we need to show opposite side are equal
Note that.
AB =
222
)31()22()11( −−+−−+−−
=
4 16 16
+ +
= 6
BC =
222
)12()23()12( +++++
=
9259 ++
=
43
CD =
222
)26()37()24( −+−+−
=
616164 =++
DA =
222
)63()72()41( −+−+−
=
439259 =++
Since AB = CD and BC = AD, ABCD is a parallelogram.
Now, it is required to prove that ABCD is not a rectangle. For this, we show that
diagonals AC and BD are unequal. We have
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MATHEMATICS
AC =
3111)32()23()12(
222
=++=−+−+−
BD =
155498125)16()27()14(
222
=++=+++++
.
Since AC ≠ BD, ABCD is not a rectangle.
A
Note We can also show that ABCD is a parallelogram, using the property that
diagonals AC and BD bisect each other.
Example 12 Find the equation of the set of the points P such that its distances from
the points A (3, 4, –5) and B (– 2, 1, 4) are equal.
Solution If P (x, y, z) be any point such that PA = PB.
Now
222222
)4()1()2()5()4()3( −+−++=++−+− zyxzyx
or
222222
)4()1()2()5()4()3(
−+−++=++−+−
zyxzyx
or 10
x
+ 6y – 18z – 29 = 0.
Example 13 The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates
of A and B are (3, –5, 7) and (–1, 7, – 6), respectively, find the coordinates of the
point C.
Solution Let the coordinates of C be (x, y, z) and the coordinates of the centroid G be
(1, 1, 1). Then
x
+
−
=
3
1
3
1
, i.e.,
x
= 1;
y
−
+
=
5
7
3
1
, i.e., y = 1;
z
+
−
=
7
6
3
1
, i.e., z = 2.
Hence, coordinates of C are (1, 1, 2).
Miscellaneous Exercise on Chapter 12
1. Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and
C (– 1, 1, 2). Find the coordinates of the fourth vertex.
2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0)
and (6, 0, 0).
3. If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6),
Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.
4. Find the coordinates of a point on y-axis which are at a distance of 5
2
from
the point P (3, –2, 5).
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INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 279
5. A point R with x-coordinate 4 lies on the line segment joining the points
P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.
[Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given
by
+
+
+
−
+
+
1
410
,
1
3
,
1
28
k
k
kk
k
].
6.If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the
set of points P such that PA
2
+ PB
2
= k
2
, where k is a constant.
Summary
®
In three dimensions, the coordinate axes of a rectangular Cartesian coordinate
system are three mutually perpendicular lines. The axes are called the x, y
and z-axes.
®
The three planes determined by the pair of axes are the coordinate planes,
called XY, YZ and ZX-planes.
®
The three coordinate planes divide the space into eight parts known as octants.
®
The coordinates of a point P in three dimensional geometry is always written
in the form of triplet like (x, y, z). Here x, y and z are the distances from the
YZ, ZX and XY-planes.
®
(i) Any point on x-axis is of the form (x, 0, 0)
(ii) Any point on y-axis is of the form (0, y, 0)
(iii) Any point on z-axis is of the form (0, 0, z).
®
Distance between two points P(x
1
, y
1
, z
1
) and Q (x
2
, y
2
, z
2
) is given by
2 2 2
2 1 2 1 2 1
PQ
( x x ) ( y y ) ( z z )
= − + − + −
®
The coordinates of the point R which divides the line segment joining two
points P (x
1
y
1
z
1
) and Q (x
2
, y
2
, z
2
) internally and externally in the ratio m : n
are given by
2 1 2 1 2 1
mx nx my ny mz nz
, ,
m n m n m n
+ + +
+ + +
and
2 1 2 1 2 1
mx – nx my – ny mz – nz
, ,
m – n m – n m – n
,
respectively.
®
The coordinates of the mid-point of the line segment joining two points
P(x
1
, y
1
, z
1
) and Q(x
2
, y
2
, z
2
) are
1 2 1 2 1 2
2 2 2
x x y y z z
, ,
+ + +
.
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MATHEMATICS
®
The coordinates of the centroid of the triangle, whose vertices are (x
1
, y
1
, z
1
)
(
x
2
, y
2
, z
2
) and (x
3
, y
3
, z
3
), are
1 2 3 1 2 3 1 2 3
+ + + + + +
3 3 3
x x x y y y z z x
, ,
.
Historical Note
Rene’ Descartes (1596–1650), the father of analytical geometry, essentially dealt
with plane geometry only in 1637. The same is true of his co-inventor Pierre
Fermat (1601-1665) and La Hire (1640-1718). Although suggestions for the three
dimensional coordinate geometry can be found in their works but no details.
Descartes had the idea of coordinates in three dimensions but did not develop it.
J.Bernoulli (1667-1748) in a letter of 1715 to Leibnitz introduced the three coor-
dinate planes which we use today. It was Antoinne Parent
(1666-1716), who gave a systematic development of analytical solid geometry
for the first time in a paper presented to the French Academy in 1700.
L.Euler (1707-1783) took up systematically the three dimensional coordinate ge-
ometry, in Chapter 5 of the appendix to the second volume of his “Introduction
to Geometry” in 1748.
It was not until the middle of the nineteenth century that geometry was extended
to more than three dimensions, the well-known application of which is in the
Space-Time Continuum of Einstein’s Theory of Relativity.
—
v
v
v
v
v
—
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