Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
10
Calculus II
Chapter 7
7.8
Chapter 10
Vectors
Vector Equation of a Straight Line
2
Three Dimensional Coordinates Geometry
10.1
Basic Formulas
6
10.2
Equations of Straight Lines
6
10.3
Plane and Equation of a Plane
14
10.4
Coplanar Lines and Skew Lines
28
Prepared by K. F. Ngai
Page 1
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
7.8
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
 
 
r  a  t (b  a )
Remark
or
Example
r  (1  t )a  t b
Find the vector equation of the straight line  1 , in the direction of i  2 j  2k and passing
through the point with position vector (1,2,3) .
Solution
Example
Find the vector equation of the straight line through the point (3,5,4) in the direction of
i  j  k . Find also the point on this line which has 4i as one component vector of its
position vector.
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Find the equation of the line joining the points A( 1,2,6) and B(4,8,3) .
Find the coordinates of the point of intersection of this line and the x-y plane and the ratio in
which x-y plane divides AB .
Solution
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.
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
The line joining two points P (1,8,1) and Q(4,4,2) meets the xz  and yz  planes
respectively at R and S . Find the coordinates of R and S and the ratios in which they
divide PQ .
Solution
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
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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 .
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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) .
Solution
Example
Find the equation of the line which passes through (1,3,2) and intersects the line
x  2 y z 1
 
1
3
2
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
 1 is a line passing through A(1,0,3) and B(2,4,2) ,  2 is another line passing
through C (2,4,1) and D(6,8,3) . Find, in degrees, the acute angle between  1 and  2 .
Solution
Example
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 angle between two line is  
Example
Find the parametric form of a straight line L which passes through the point (1,1,1) and
parallel to the straight line L1 :
x3 y 4 z 2


. Show also that this line is perpendicular
2
1
3
 x  3t  2

to the straight line L2 :  y  6t  1 .
z  4

Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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.
Example
Find the intersection point of the lines
x4 y 5 z 9
x2
y
4 z




and
.
1
1
2
1
2
1
Find the intersection point of the lines
x2 y 3 z 6
x  4 y  6 z  11




and
.
1
2
3
2
3
5
Solution
Example
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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'
Example
Find the perpendicular distance from the point P(4,7,9) to the line L :
x  2 y 1 z  2


.
2
2
1
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
(a) Find the vector equation of the straight line  1 , in the direction of i  2 j  2k and
passing through the point with position vector (1,2,3) .
(b) Find a vector parallel to the straight line  2 with vector equation
r  (t  1)i  (2t  2) j  (3t  6)k
where t is a scalar parameter.
(c) Determine whether  1 meets  2 ; if so, find the point of intersection of  1 meets  2 .
(d) Find the angle between  1 meets  2 .
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
If the foot of the perpendicular from the point P(4,7,9) to the line L :
x  2 y 1 z  3


2
2
1
is Q , find the coordinates of Q. Hence, find the perpendicular distance from P to L .
Solution
Example
Consider the two straight lines L1 :
x 1 y  2 z  3
x 1 y  2 z  6




and L2 :
.
1
2
2
1
2
3
Find the point of intersection of L1 and L2 .
Find also the acute angle between L1 and L2 .
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Let a  i  3 j  k , b  3i  6 j , c  2i  4 j  3k be the position vectors of the points A ,
B and C respectively.
(a) Find the equation of the line L , which passes through A and B .
(b) Find the shortest distance from C to L .
Solution
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 
L1  L2
 l1l2  m1m2  n1n2  0
l1 m1 n1


l 2 m2 n2
Proof
Remark
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
10.3
Plane and Equation of Plane
A vector perpendicular to (or orthogonal to) a plane is a normal vector to 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 a n (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

Remark
A( x  x0 )  B( y  y0 )  C ( z  z 0 )  0
( Normal Form )
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 .
 Ax1

 Ax2
 Ax
 3
 By1
 Cz1
D 0
 By 2
 By 3
 Cz 2
 Cz3
D 0
D 0

 A( x  x1 )  B( y  y1 )  C ( z  z1 )  0

 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


 x  x1

 x1  x2
x  x
3
 2
y  y1
y1  y 2
y 2  y3
z  z1  A   0 
   
z1  z 2  B    0 
z 2  z 3  C   0 
 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
( 3 Point Form )
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
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.
Solution
Example
Find the equation of the plane passing through the point (1,2,3) and parallel to the plane
x  3 y  4 z  3 . Find also its distance from the origin.
Solution
Example
Find the equation of the plane containing the line
x3 y 3 z


and the origin.
1
2
1
Solution
Example
Find the equation of the plane passing through the origin and the point (3,1,2) and parallel
to the line
x 1 y 1 z 1


.
2
2
1
Solution
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Page 15
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Find the equation of the plane containing the point P(2,3,4) and perpendicular to both of the
planes  1 : 2x  y  2z  8  0 and  2 : x  2 y  3z  7  0 .
Solution
AL03-II-6
Let  be the plane containing (2,1,0) , (1,0,1) and (3,0,1) . Suppose L is the straight line
passing through A(0,0,2) and perpendicular to  . Find
(a) the equation of  ,
(b) the coordinates of the point of intersection of L and  ,
(c) the distance from A to  .
Ans:
(a)
y  z 1  0
1 3
(b) (0, , )
2 2
(c)
2
2
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Find the equation of the plane which contains the origin and the line
x 1
z 3
 y2
.
2
2
Solution
Example
Find the coordinates of the point at which the line joining the points (3,1,4) and (2,6,1)
meets the plane 2 x  y  3z  3 .
Solution
Example
Find the equations of the line which contains the point (2,3,4) and is parallel to the line of
intersection of the planes x  y  2 z  1 and 2 x  3 y  z  3  0 .
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Find the equation of the plane which contains the line 6 x  3 y  9  2 z  10 and is at right
angles to the plane 2 x  7 y  3 z  1 .
Solution
Example
Find the equation of the plane containing the parallel lines
L1 :
x 1 y 1 z  2
x y 1 z  3



and L2 : 
.
1
2
3
1
2
3
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Find the equations of the following planes.
(a) Passing through the points (3,1,0), (2,8,3) and (1,3,2) .
(b) Having x  intercept =  3 and perpendicular to the line joining the points (5,1,4)
to (1,1,7) .l
(c) Contains the line x  y  2  6 z  6 and parallel to the line x  1  2 y  12 z.
(d) Contains the lines
x
y 3 z 5

2
3
and
x 1 y 1 z  2


.
2
5
3
Solution
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Page 19
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Find a formula in order to find the distance from a fixed point P( x0 , y0 , z 0 ) to the plane
Ax  By  Cz  D  0 .
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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
P0 P  n
=
d
( 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
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)  4
2
2
2
=
17
6 units.
12
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Find the equations of the two planes which are parallel to the plane
( ) : 3x  6 y  2 z  14  0
and are 5 units away from the point P(2,1,3) .
Solution
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
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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

Example
(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 )
The equation of the plane
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
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Alternatively,
Example
3 x
Solve 
6 x
4y
 2y
 2z
z
 1
 0
consider k  n  n
Solution
From the above examples we conclude that the
intersection of two planes is a line.
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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) .
Solution
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
Solution
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Page 25
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
(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  .
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 )
 : r  (4i  j  2k )  4
and the plane
(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.
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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 ) .

(*) contains L .
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Page 27
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
10.4
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
are coplanar if and only if
x1  x2
y1  y 2
z1  z 2
a1
a2
b1
b2
c1
c2
 0   (*)
Proof
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Page 28
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Show that the two lines
L1 :
x  3 y  2 z 1
x 1 y  2 z  6




and L2 :
2
5
3
4
1
2
are coplanar and intersect.
Solution
Example
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.
Solution
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Page 29
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
If the lines
x2 y4 z4


1
p
1
and
x y 3 z 2


1
1
q
are
coplanar
and
perpendicular to each other, find p and q .
Solution
Example
Show that the lines
x  2 y  3 z  4 x  3 y 1 z 1




,
are coplanar.
2
1
3
1
3
2
Find the intersection and find the equation of the plane containing them.
Solution
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Page 30
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
(a) Show that the two lines L1 :
x 1 y  2 z  3
x 1 y 1 z 1




and L2 :
1
2
1
3
2
1
are non-coplanar.
(b) Find a straight line passing through the origin and intersecting each of the lines L1
and L2 .
Solution
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Page 31
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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.
Solution
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Page 32
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
x  y  0
Solve L1 : 
y  z  0
 x  1  2t

and L2 :  y  1
.
z  1  t

Solution
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  .
Solution
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Page 33
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
Example
Let L1 be the line of intersection of the planes x  y  z  1 and x  y  z  5 , and
L2 be the line of passing through (1,1,1) and intersecting L1 at right line.
(a) Find a parametric equation of L1 .
(b) Find the coordinates of the point of intersection of L1 and L2 , and a parametric
equation of L2
Solution
Example
Find the image of the line
x 1 y  2 z  3


in the plane 3x  2 y  5 z  24 .
2
1
6
Solution
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Page 34
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
AL-97I-11
(a) Let m, n be vectors in R 3 . Show that
m  m m  n
2
  m  n
(i) det
 mn nn 
(ii) (n  n)m  (m  n)n  n  (m  n)
(b) Two planes (r  a)  m  0 and (r  b)  n  0 intersect in a Line L , where
a, b, m, n are constant vectors and r is any position vector R 3 . Express the real
numbers  and  in terms of a, b, m and n such that the point represented
by the position vector p  m  n lies on the line L .
Show that
p  ( a  m)
n  ( m  n)
mn
2
 (b  n)
m  ( n  m)
mn
2
Solution
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Page 35
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
AL-98II-4
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  .
Ans:
(a)
( 1,3,2)
(b)

 1 
 cos 1 

2
 3
Solution
AL-94II-3
Find the equations of the straight line which satisfies the following two conditions:
(i)
passing through the point (4,2,3) ,
(ii) parallel to the planes x  y  z  10  0 and x  2 y  0
Ans:
x4 y2 z3


2
1
1
Solution
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
AL-93II-2
Find the equation of the plane passing through the line of intersection of the planes
x  y  z  1  0 and x  4 y  3z  0
and parallel to the straight line
Ans:
x  1  3 y  3( z  1) .
 x  2y  z  2  0
Solution
AL-92II-3
If the lines
x2 y4 z4
x y 3 z 2
and




1
p
1
1
1
q
are coplanar and perpendicular to each other, find p and q .
Ans:

 p  2
or

q

1


1

 p2

1
q  
2

Solution
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Page 37
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
AL-91II-11
Consider the lines
L1 :
x2 y 3 z 3


1
2
3
L2 :
and
x  4 y  6 z  11


2
3
5
(a) Prove that L1 and L2 are non-coplanar．
(b) (i)
Find the equation of the plane  containing L1 and parallel to
L2 ．
(ii) Find the equation of the plane  ' containing L2 and perpendicular to  .
(c) (i)
Find the point S at which L1 intersects  ' ．
(ii) Find the equations of line through S and perpendicular to both L1 and L2 .
Ans:
(b) (i)
x yz20
(ii) 8 x  7 y  z  1  0
(c) (i)
(1,1,0)
(ii)
x 1 y 1 z  0


1
1
1
Solution
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Page 38
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
AL-90II-6
Find the equation of the plane containing the line ( L) :
x 1 y 1 z  2


and the
3
2
2
point A(1,1,3)
Ans:
2 x  3 y  6 z  13
AL-89II-12 (a)
The position vector of a point Rx , y , z  is given by z  xi  yj  zk ．In the
figure﹐ R0 x0 , y0 , z 0  is a point on the plane  : r．n=  ．
The line  : r=r 0 + t a﹐where r 0 = x0 i  y0 j  z 0 k ﹐passes through R0 and does not
lie on  ．
an 

n , t  R ．
Show that the projection of  on  is given by ': r=r 0 + t  a 
nn 

 x  1  2t

(b) Consider the lines  1 :  y  3  3t
 z  1 t

 x  2  8t

﹐t  R and  2 :  y  19t
 z  2  4t

﹐t  R
and the plane  1 : 4 x  y  2 z  4  0 ．
(i) Let P1 and P2 be the points at which  1 intersects  1 and  2
respectively．
Find P1 and P2 and show that the line segment P1 P2 is perpendicular
to both  1 and  2 ．
(ii) Show that the projections of  1 and  2 on  1 are parallel．
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Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
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Page 40
Three Dimensional Co-ordinate Geometry
Advanced Level Pure Mathematics
AL-87II-6
(a)
 x  a1  p1t

Let  1 :  y  b1  q1t
 z  c rt
1
1

x  a2  p2t

and  2 :  y  b2  q 2 t be two given lines．Suppose  1
z c r t
2
2

and  2 intersect．
(i)
a1  a 2
Show that b1  b2
p1
q1
p2
q2  0
c1  c 2
r1
r2
(ii) If  1 and  2 are distinct﹐find a vector normal to the plane containing  1 and
2．
Hence﹐or otherwise﹐obtain the equation of this plane．
(b) Consider the lines
 x  pt

L1 :  y  qt ,
 z  rt

 x  qt

L2 :  y  rt and
 z  pq

 x  rt

L3 :  y  pt
 z  qt

where p , q and r are distinct and non-zero．Find the equation of a plane
containing L1 and perpendicular to the plane which contains L2 and L3 when
(i) pq  qr  rp  0
(ii) pq  qr  rp  0
Solution
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Page 41