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319Chapter
Complex Numbers 319
7
CONTENTS
System of Co-ordinates
7.1 Co-ordinates of a point in space
7.2 Distance formula
7.3 Section formulas
7.4 Triangle
7.5 Volume of tetrahedron
7.6 Direction cosines and Direction ratio
7.7 Projection
7.8 Angle between two lines
The Straight Line
7.9 Straight line in space
7.10 Equation of line passing through two given points
7.11 Changing unsymmetrical form to symmetrical form
7.12 Angle between two lines
7.13 Reduction of cartesian form of the equation of a line to
vector form and vice-versa
7.14 Intersection of two lines
7.15 Foot of perpendicular from a point Pierre Fermat
7.16 Shortest distance between two lines
7.17
The Plane
Definition of plane and its equations
Rene' Descartes (1596-1650 A.D.), the father of
7.18 Equation of plane passing through the given point analytical geometry, essentially dealt with plane
7.19 Foot of perpendicular from a point to a given plane geometry only in 1637. The same is true of his
7.20 Angle between two planes coinventor Pierre Fermet (1601-1665 A.D.)
7.21
7.22
Equation of planes bisecting angle between two given planes
Image of a point in a plane Descartes had the idea of co-ordinates in three
7.23 Coplaner lines dimensions but did not develop it.
Line and Plane J.Bernoulli (1667-1748 A.D.) in a letter of 1715
7.24 Equation of plane through a given line A.D. to Leibnitz introduced the three co-ordinate
7.25 Transformation from unsymmetric from of the equation planes which we use today. It was Antoinne
of line to the symmetric form Parent (1666-1716 A.D.), who gave a systematic
7.26 Intersection point of a line and plane development of analytical solid geometry for the
7.27 Angle between line and plane first time in a paper presented to the French
7.28 Projection of a line on a plane Academy in 1700 A.D.
Sphere L.Euler (1707-1783 A.D.) took up systematically
7.29 General equation of sphere the three dimensional co-ordinate geometry.
7.30 Equation of sphere of various types It was not until the middle of the nineteenth
7.31 Section of a sphere by a plane century that geometry was extended to more than
7.32 Condition of tangency of a plane to a sphere three dimensions, the well-known application of
7.33 Intersection of straight line and a sphere which is in the Space-Time Continuum of
7.34 Angle of intersection of two spheres Einstein's Theory of Relativity.
Assignment (Basic and Advance Level)
Answer Sheet of Assignment
,320
320 Three Dimensional Co-ordinate Geometry
System of Co-ordinates
7.1 Co-ordinates of a Point in Space.
(1) Cartesian Co-ordinates : Let O be a fixed point, known as origin and let OX, OY and OZ be three mutually
perpendicular lines, taken as x-axis, y-axis and z-axis respectively, in such a way that they form a right-handed system.
Z
C E
k
F j P(x, y, z)
O Y
B
i
A D
X
The planes XOY, YOZ and ZOX are known as xy-plane, yz-plane and zx-plane respectively.
Let P be a point in space and distances of P from yz, zx and xy-planes be x, y, z
Z
respectively (with proper signs), then we say that co-ordinates of P are (x, y, z). Y
Also OA = x, OB = y, OC = z. O
X X
The three co-ordinate planes (XOY, YOZ and ZOX) divide space into eight parts and
these parts are called octants. Y Z
Signs of co-ordinates of a point : The signs of the co-ordinates of a point in three dimension follow the convention
that all distances measured along or parallel to OX, OY, OZ will be positive and distances moved along or parallel to OX,
OY, OZ will be negative.
The following table shows the signs of co-ordinates of points in various octants :
Octant OXYZ OXYZ OXYZ OXYZ OXYZ OXYZ OXYZ OXYZ
co-ordinate
x + – + – + – + –
y + + – – + + – –
z + + + + – – – –
(2) Other methods of defining the position of any point P in space :
(i) Cylindrical co-ordinates : If the rectangular cartesian co-ordinates of P are (x, y, z), then those of N are (x, y, 0)
and we can easily have the following relations : x = u cos, y = u sin and z = z.
Hence, u 2 x 2 y 2 and tan 1 (y / x ) . Z Y
Cylindrical co-ordinates of P (u, , z) P(x, y, z)
(u, , z)
r (r, , )
O X
X
u
N
Z
, Three Dimensional Co-ordinate Geometry 321
(ii) Spherical polar co-ordinates : The measures of quantities r, , are known as spherical or three dimensional
polar co-ordinates of the point P. If the rectangular cartesian co-ordinates of P are (x, y, z) then
z = r cos, u = r sin x = u cos = r sin cos, y = u sin = r sin sin and z = r cos
u x 2 y2 y
Also r 2 x 2 y 2 z 2 and tan ; tan
z z x
Note : The co-ordinates of a point on xy-plane is (x, y, 0), on yz-plane is (0, y, z) and on zx-plane is (x, 0,
z)
The co-ordinates of a point on x-axis is (x, 0, 0), on y-axis is (0, y, 0) and on z-axis is (0, 0, z)
Position vector of a point : Let i , j, k be unit vectors along OX, OY and OZ respectively. Then position
vector of a point P(x, y, z) is OP x i y j zk .
7.2 Distance Formula.
(1) Distance formula : The distance between two points A(x 1 , y 1 , z 1 ) and B(x 2 , y 2 , z 2 ) is given by
AB [(x 2 x 1 )2 (y 2 y 1 )2 (z 2 z 1 )2 ]
(2) Distance from origin : Let O be the origin and P(x, y, z) be any point, then OP (x 2 y 2 z 2 ) .
(3) Distance of a point from co-ordinate axes : Let P(x, y, z) be any point in the space. Let PA, PB and PC
be the perpendiculars drawn from P to the axes OX, OY and OZ respectively.
Z
C
Then, PA (y 2 z 2 ) P(x,y,z)
PB (z 2 x 2 ) O A
X
B
PC (x y )
2 2
Y N
Example: 1 The distance of the point (4, 3, 5) from the y-axis is [MP PET 2003]
(a) 34 (b) 5 (c) 41 (d) 15
Solution: (c) Distance x 2 z 2 16 25 41
Example: 2 The points (5, –4, 2), (4, –3, 1), (7, –6, 4) and (8, –7, 5) are the vertices of [Rajasthan PET 2002]
(a) A rectangle (b) A square (c) A parallelogram (d) None of these
Solution: (c) Let the points be A(5, –4, 2), B(4, –3, 1), C(7, –6, 4) and D(8, –7, 5).
AB 1 1 1 3 , CD 1 1 1 3 , BC 9 9 9 3 3 , AD 9 9 9 3 3
Length of diagonals AC 4 4 4 2 3 , BD 16 16 16 4 3
i.e., AC BD
Hence, A, B, C, D are vertices of a parallelogram
7.3 Section Formulas.
, 322 Three Dimensional Co-ordinate Geometry
(1) Section formula for internal division : Let P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) be two points. Let R be a point
on the line segment joining P and Q such that it divides the join of P and Q internally Z P(x1,y1,z1)
m1
in the ratio m1 : m 2 . Then the co-ordinates of R are R(x,y,z)
m 2
r1
m1 x 2 m 2 x 1 m1 y 2 m 2 y1 m1 z 2 m 2 z1 Q(x2,y2,z2)
r
, , . r
2
m1 m 2 m1 m 2 m1 m 2 Y
X
O
(2) Section formula for external division : Let P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) be two points, and let R be a
point on PQ produced, dividing it externally in the ratio m 1 : m 2 (m 1 m 2 ) . Then the co-ordinates of R are
m1 x 2 m 2 x 1 m1 y 2 m 2 y1 m1 z 2 m 2 z1
, , .
m1 m 2 m1 m 2 m1 m 2
Note : Co-ordinates of the midpoint : When division point is the mid-point of PQ then ratio will be
x x 2 y1 y 2 z1 z 2
1 : 1, hence co-ordinates of the mid point of PQ are 1 , , .
2 2 2
Co-ordinates of the general point : The co-ordinates of any point lying on the line joining points
kx x 1 ky 2 y 1 kz 2 z 1
P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) may be taken as 2 , , , which divides PQ in
k 1 k 1 k 1
the ratio k : 1. This is called general point on the line PQ.
Example: 3 If the x-co-ordinate of a point P on the join of Q (2, 2, 1) and R (5, 1, –2) is 4, then its z-co-ordinate is
[Rajasthan PET 2003]
(a) 2 (b) 1 (c) –1 (d) –2
5 k 2 k 2 2k 1 5k 2 2(2) 1
Solution: (c) Let the point P be , , . ∵ Given that 4 k 2 z-co-ordinate of P 1
k 1 k 1 k 1 k 1 2 1
7.4 Triangle.
(1) Co-ordinates of the centroid
(i) If (x 1 , y 1 , z 1 ), (x 2 , y 2 , z 2 ) and (x 3 , y 3 , z 3 ) are the vertices of a triangle, then co-ordinates of its centroid are
x 1 x 2 x 3 y1 y 2 y 3 z1 z 2 z 3
, , .
3 3 3
(ii) If (x r , y r , z r ) ; r = 1, 2, 3, 4, are vertices of a tetrahedron, then co-ordinates of its centroid are
x 1 x 2 x 3 x 4 y1 y 2 y 3 y 4 z1 z 2 z 3 z 4
, , .
4 4 4
(iii) If G (, , ) is the centroid of ABC, where A is ( x 1 , y 1 , z 1 ) , B is (x 2 , y 2 , z 2 ) , then C is
(3 x 1 x 2 , 3 y 1 y 2 , 3 z 1 z 2 ) .
(2) Area of triangle : Let A(x 1 , y 1 , z 1 ) , B(x 2 , y 2 , z 2 ) and C(x 3 , y 3 , z 3 ) be the vertices of a triangle, then
y1 z1 1 x1 z1 1 x1 y1 1
1 1 1
x y2 z2 1 , y x2 z2 1 , z x2 y2 1
2 2 2
y3 z3 1 x3 z3 1 x3 y3 1
Now, area of ABC is given by the relation 2x 2y 2z . A(x1,y1,z1)
B(x2,y2,z2) C(x3,y3,z3)
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