Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry
Chapter 7
Vectors
7.8
Vector Equation of a Straight Line
Chapter 10
7.8
2
Three Dimensional Coordinates Geometry
10.1
Basic Formulas
5
10.2
Equations of Straight Lines
10.3
Plane and Equation of a Plane
11
10.4
Coplanar Lines and Skew Lines
22
5
Vector Equation of a Straight Line
  
t : scalar parameter
r  a  tc ,

a : position v ector of a fixed point on the straight line

c : direction vector

r : position v ector of any point on straight line
Remark
 
 
r  a  t (b  a )
Example
Let A  (8,7,0) and B  (2,1,3) .
(a) Find the equation of the straight line AB .
(b) Find the perpendicular distance from the point P(4,7,9) to the line AB .
Find also the foot of perpendicular.
Remark
In above example (b), the distance from P to AB may also be found directly without
calculating the foot of perpendicular. The method is outlined as follows:
By referring to Figure,
PR  AP sin  
AB  AP
AB
Since
Example
By finding the foot of perpendicular from the point P(10,1,13) to the line,
L : r  i  5k  t (4i  5 j ) , find the equation of straight line passing through P and
perpendicular to L , find the perpendicular distance from P to L .
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Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry
Three Dimensional Co-ordinate Geometry
10.1
Basic Formula
The Distance Between Two Points
Distance between A( x1 , y1 , z1 ) and B( x2 , y2 , z 2 ) is
( x1  x2 ) 2  ( y1  y 2 ) 2  ( z1  z 2 ) 2 .
Section Formula
Let P( x, y, z ) divide the joint of A( x1 , y1 , z1 ) and B( x2 , y2 , z 2 ) in the ratio
AP m

PB n
 mx  nx1 my2  ny1 mz2  nz1 
,
,
The Coordinate of the point P is  2

mn
mn 
 mn
10.2
Equations of Straight Lines

  
In vector form, the equation of straight line is r  a  tc , where r is the position vector of any point in the


line, a is fixed point on line and c is direction vector of line.
If r  ( x, y, z ) , a  ( x1 , y1 , z1 ) , c  (a, b, c) , we have

 
xi  yj  zk
=




 
x1i  y1 j  z1 k  t (ai  bj  ck )
=



( x1  ta)i  ( y1  tb) j  ( z1  tc)k
  
Since i , j , k are basis vectors in R 3 , we have
x 

y 
z 

x1
y1
 ta
 tb
z1
 tc
or
x  x1 y  y1 z  z1


a
b
c
Parametric Form of a Straight Line
The equation of the straight line passing through the point ( x1 , y1 , z1 ) and with direction vector ( a, b, c )
can be expressed in the form of
 x  at  x1

 y  bt  y1 where t is a parameter.
 z  ct  z
1

This is called the parametric form of the straight line.
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Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry
Symmetric Form of a Straight Line
The equation of the straight line passing through the point ( x1 , y1 , z1 ) and with direction vector ( a, b, c )
and is
x  x1 y  y1 z  z1


a
b
c
and this is called the symmetric form of the straight line.
General Form of a Straight Line
The equation of a straight line can be written as a linear system
 A1 x  B1 y  C1 z  D1

 A2 x  B2 y  C 2 z  D2
 0
 0
which is called the general form of a straight line.
If given two points P1 ( x1 , y1 , z1 ) , P2 ( x2 , y2 , z 2 ) , the equation of straight line becomes
x 

y 
z 

Example
x1
 t ( x2  x1 )
y1
z1
 t ( y 2  y1 )
 t ( z 2  z1 )
or
x  x1
y  y1
z  z1


x2  x1 y 2  y1 z 2  z1
Find the equation of the line joining the points ( 2,0,3) and (4,1,2) .
S1
x 

Let L1 :  y 
z 

x1
y1
z1
1l1
 1m1
 1n1

and
x 

L2 :  y 
z 

x2
y2
z2
2 l 2
  2 m2
  2 n2

To find the intersection point of line L1 and L2
we solve
 x1

 y1
z
 1
 1l1

x2
 2 l 2
 1m1
 1n1


y2
z2
  2 m2
  2 n2
i.e. find 1 and 2 .
Note
After finding 1 and 2 is any two equations, 1 and 2 must put into the 3rd equation in
order to test whether it is satisfied or not.
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Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry
S2
Distance of a point P( x0 , y0 , z 0 ) from the line
x  x1 y  y1 z  z1


l
m
n
FIND P ' .
Let P ' be ( x1  l , y1  m , z1  n ) .
Direction vector of
PP' ( x1  l  x0 , y1  m  y0 , z1  n  z 0 )
Direction vector of line (l , m , n)
( x1  l  x0 , y1  m  y0 , z1  n  z 0 )  ( l , m , n )  0
As  is formed, P ' can be determined and so d  PP'
Theorem
Given
L1 :
x  x1 y  y1 z  z1


l1
m1
n1
L2 :
x  x2 y  y 2 z  z 2


l2
m2
n2
and
L1 // L2
 Their direction vectors are parallel 
Remark
L1  L2
 l1l2  m1m2  n1n2  0
10.3
Plane and Equation of Plane
l1 m1 n1


l 2 m2 n2
A vector perpendicular to (or orthogonal to) a plane is a normal vector o that plane.
In Figure, n is a normal vector of the plane ( ) .
Normal vector of a plane is not unique, for if n is a normal vector, then an (a is
any non-zero real number) is also a normal vector.
Let P0 ( x0 , y0 , z 0 ) be a fixed point and P( x, y, z ) be any point on it.
Set n  ( A , B , C ) i.e. A, B, C are given.

P0 P  n  0
( Vector Form )
We have ( x  x0 , y  y0 , z  z 0 )  ( A, B, C )  0

A( x  x0 )  B( y  y0 )  C ( z  z 0 )  0
( Normal Form )
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Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry
Remark
The general form of plane equation is Ax  By  Cz  D  0 .
Furthermore, if three points are given, Pi ( xi , yi , zi ) i  1,2,3 .
 A( x  x1 )  B( y  y1 )  C ( z  z1 )  0

We have  A( x1  x2 )  B( y1  y 2 )  C ( z1  z 2 )  0
 A( x  x )  B( y  y )  C ( z  z )  0
2
3
2
3
2
3

 n  ( A, B, C )  0  The system has non-trivial solution of A, B, C .
x  x1
y  y1
z  z1
Hence, x1  x2
y1  y 2
y 2  y3
z1  z 2  0 . It is an equation of plane.
z 2  z3
x 2  x3
Example
( 3 Point Form )
Find the equation of the plane passing through the points P(2,4,3) , Q(4,1,9) and R(0,1,6) .
Find also its distance from the origin.
The perpendicular distance between a point and a plane
Theorem
The perpendicular distance between a point P( x1 , y1 , z1 ) and a plane
 : Ax  By  Cz  D  0 is
d
Proof
Ax1  By1  Cz1  D
A2  B 2  C 2
Let P0 ( x0 , y0 , z 0 ) be any point on the plane ( ) .
Ai  Bj  Ck is a vector normal to the plane ( ) .
The unit vector n normal to the plane ( ) is n 
Ai  Bj  Ck
.
A2  B 2  C 2
The perpendicular distance d between the point P and the plane is equal to the
magnitude of the projection of P0 P on n .
Therefore
d
=
=
=
=
P0 P  n
( x1  x0 )i  ( y1  y0 ) j  ( z1  z 0 )k 
Ai  Bj  Ck
A2  B 2  C 2
A( x1  x0 )  B( y1  y0 )  C ( z1  z 0 )
A2  B 2  C 2
Ax1  By1  Cz1  Ax0  By 0  Cz0
A2  B 2  C 2
But, D   Ax0  By 0  Cz 0 , since P0 ( x0 , y0 , z 0 ) lies on the plane.

d
Ax1  By1  Cz1  D
A2  B 2  C 2
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Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry
Example
Find the perpendicular distance between two parallel planes
( 1 ) : x  y  2 z  6 and ( 2 ) : 2x  2 y  4z  5  0 .
Solution
Take a point P(0,0,3) on ( 1 ) .
The required distance is just the perpendicular distance between P and ( 2 ) .
i.e. d
2 0  2 0  43  5
=
2 2  (2) 2  4 2
=
17
6 units.
12
Angles Between Two planes
and  2 : A2 x  B2 y  C2 z  D2  0
Given 2 planes  1 : A1 x  B1 y  C1 z  D1  0
The angle between two planes is  and  , which are a pair of supplementary angles and
n1  n2  n1 n2 cos
cos
( A1 , B1 , C1 )  ( A2 , B2 , C 2 )
=
( A1  B1  C1 )( A2  B2  C 2 )
2
2
2
2
2
2
=
Remark
(a)  1 //  2
 n1  t n2 ,

(b)  1   2
t : scalar
A1 B1 C1


t
A2 B2 C 2
 n1  n2  0

A1 A2  B1 B2  C1C2  0
Equation of Plane Containing Two Given Lines
Given two lines
L1 :
x  x1 y  y1 z  z1


l1
m1
n1
L2 :
x  x2 y  y 2 z  z 2


l2
m2
n2
The normal vector of the required plane
n
=
=
n
(l1 , m1 , n1 )  (l2 , m2 , n2 )
i
l1
j
m1
k
n1
l2
m2
n2
=
(m1n2  m2 n1 ) i  (l1n2  l2 n1 ) j  (l1m2  l2 m1 ) k
=
(m1n2  m2 n1 ,  l1n2  l2 n1 , l1m2  l2 m1 )
Page 6
Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry

The equation of the plane
Example
Find the equation of the plane containing two intersecting lines.
L1 :
x  2 y 1
z
x  2 y 1 z




and L2 :
3
4
2
1
3
2
Example
3 x
Solve 
6 x
4y
 2y
 2z
z
 1
 0
Alternatively,
consider k  n  n
Solution
From the above examples we conclude that the
intersection of two planes is a line.
Family of Planes
Given two planes
 1 : A1 x  B1 y  C1 z  D1  0
 2 : A2 x  B2 y  C2 z  D2  0
The family of planes is any plane containing the line of intersection  1 and  2 .
 : A1 x  B1 y  C1 z  D1  k ( A2 x  B2 y  C2 z  D2 )  0 , where k is a constant.
Example
x  2 y  z  4
Find the equation of the plane containing the line 
and passing
x  6 y  5z  0
the point (1,1,2) .
Example
Example
 x  2z  4
Find the equation of the plane containing the line L1 : 
and parallel to
y  z  8
x3 y 4 z 7


the line L2 :
.
2
3
4
(a) The position vector of a point P( x, y, z ) is given by r  xi  yj  zk .
In Figure, P0 ( x0 , y0 , z 0 ) is a point on the plane  : r  n  d .
The line  : r  r0  ta, where t is a real scalar and r0  x0 i  y0 j  z 0 k , passing
through P0 and does not lie on  .
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Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry
an 

n  where t
Show that the projection of  on  is given by r  r0  t  a 
nn 

is a real scalar.
(b) Consider the lines  1 : r  3i  6 j  2k  t (2i  3 j  k )
and
 2 : r  10i  19 j  2k  t (8i  19 j  4k )
and the plane
 : r  (4i  j  2k )  4
(i)
Let A and B be the points at which  intersects  1 and  2 respectively.
Find the coordinates of A and B and show that AB is perpendicular to
both  1 and  2 .
(ii) Show that the projections of  1 and  2 on  are parallel.
Theorem
1 :
Two given planes
y  y1 z  z1
x  x1 y  y1


and  2 :
.
A
B
B
C
Prove that the equation of any plane through the line of intersection of  1 and  2 must
contain a line L :
Proof
x  x1 y  y1 z  z1


A
B
C
The equation of plane through the line of intersection of  1 and  2 is
B( x  x1 )  A( y  y1 )  k (C( y  y1 )  B( z  z1 )  0
 (*)
Normal Vector of (*) n1  ( B ,  A  kC ,  Bk ) .
Direction vector of line L : n2  ( A , B , C )
n1  n2  0

(*) is parallel to line L .
Since (*) and L pass through the point ( x1 , y1 , z1 ) .

10.4
(*) contains L .
Coplanar Lines and Skew Lines
Coplanar Lines
Definition
Two lines are said to be Coplanar if there exists a plane that contains both lines.
Two lines are Coplanar  they must be either parallel or they intersect.
Theorem
Two lines ( L1 ) :
x  x2 y  y 2 z  z 2
x  x1 y  y1 z  z1




and ( L2 ) :
a1
b1
c1
a2
b2
c2
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Advanced Level Pure Mathematics
Three Dimensional Co-ordinate Geometry
are coplanar if and only if
Example
x1  x2
y1  y 2
z1  z 2
a1
a2
b1
b2
c1
c2
 0   (*)
Show that the two lines
L1 :
x  2 y 1 z
x 1 y  2 z  3




and L2 :
1
2
3
4
1
2
are coplanar.
Skew Lines
Two straight lines are said to be Skew if they are non-coplanar i.e. neither do they intersect nor are they
being parallel.
To find the shortest distance between them, we have to find the common perpendicular to both lines
first. The method is illustrated by the following example.
Example
It is given that the two lines
L1 :
x  5 y z 1
x2 y4 z
 


and L2 :
1
2
1
1
1
1
are non-coplanar. Find the shortest distance between them.
Example
Consider the line L :
x 1 y  2 z


and the plane  : x  y  z  0 .
2
1
2
(a) Find the coordinates of the point where L intersects  .
(b) Find the angle between L and  .
Page 9