The document discusses determining the angle between two lines and finding the point of intersection between two lines in 3D space. It provides the equations and process for finding the angle between two lines given their direction ratios. It also outlines the steps to find the point of intersection, which involves setting the coordinates of a point on each line equal to determine values for lambda and mu, and substituting those values back into one of the line equations. An example problem demonstrates finding the intersection point of two lines.

1. Physics Helpline
L K Satapathy
3D Geometry Theory 3

2. Physics Helpline
L K Satapathy
Angle between two lines :
Straight Lines in Space
1b
r
O

L1
X
Y
Z

L2
2b
r
Consider two lines L1 and L2 shown in the figure
which are parallel to the given vectors
 The equations are :
Where and are position vectors of the given points [ not shown in the figure
as they are not necessary for our derivation]
 The angle () between the lines = the angle between the vectors
1 1r a b  [ Parallel to ]1b
2 2r a b  [ Parallel to ]2b
1 2b and b
2a1a
1 2 1 2. . cosb b b b  1 2
1 2
.
cos
.
b b
b b
 
1 2b and b
3 D Geometry Theory 3

3. Physics Helpline
L K Satapathy
Let the direction ratios of the two lines are
Straight Lines in Space
1 1 1 2 2 2( , , ) ( , , )a b c and a b c
We will use the following vector results :
1 2 1 2 1 2 1 2
2 2 2 2 2 2
1 2 1 1 1 2 2 2
.
cos
.
b b a a b b c c
b b a b c a b c

 
  
   
1 2 1 2 1 2 1 2( ) .iii b b a a bb c c  
1 1 1 1 2 2 2 2
ˆ ˆˆ ˆ ˆ ˆb a i b j c k and b a i b j c k      
2 2 2
1 1 1 1( )i b a b c  
2 2 2
2 2 2 2( )ii b a b c  
3 D Geometry Theory 3

4. Physics Helpline
L K Satapathy
Straight Lines in Space
2
2 1 2 1 2 1 2
2 2 2 2 2 2
1 1 1 2 2 2
( )
sin 1 cos 1
( )( )
a a bb c c
a b c a b c
 
 
    
   
2 2 2 2 2 2 2
1 1 1 2 2 2 1 2 1 2 1 2
2 2 2 2 2 2
1 1 1 2 2 2
( )( ) ( )
( )( )
a b c a b c a a bb c c
a b c a b c
      

   
2 2 2
1 2 2 1 1 2 2 1 1 2 2 1
2 2 2 2 2 2
1 1 1 2 2 2
( ) ( ) ( )
( )( )
a b a b b c b c c a c a
a b c a b c
    

   
(i) When the two lines are PERPENDIULAR : cos = 0
1 2 1 2 1 2 0a a bb c c   
(ii) When the two lines are PARALLEL : Their direction ratios are proportional.
1 1 1
2 2 2
a b c
a b c
  
3 D Geometry Theory 3

5. Physics Helpline
L K Satapathy
Straight Lines in Space
Point of intersection of two lines :
Consider two lines whose equations are
1 1 1
1 1 1
. . . (1)
x x y y z z
a b c
  
 
2 2 2
2 2 2
. . . (2)
x x y y z z
a b c
  
 
1 1 1
1 1 1
x x y y z z
a b c

  
  
1 1 1 1 1 1( , , ) . . . (3)x a y b z c      Coordinates of any point on line (1)
2 2 2
2 2 2
x x y y z z
a b c

  
  
2 2 2 2 2 2( , , ) . . . (4)x a y b z c      Coordinates of any point on line (2)
3 D Geometry Theory 3

6. Physics Helpline
L K Satapathy
Straight Lines in Space
Since the point of intersection lies on both lines , we have
1 1 1 1 1 1 2 2 2 2 2 2( , , ) ( , , )x a y b z c x a y b z c           
Equating the x , y and z coordinates , we get
1 1 2 2x a x a    1 1 2 2y b y b    1 1 2 2z c z c   
In the above equations , the only unknown quantities are  and 
We can solve for  and  using any two equations
The two lines intersect , only if these values of  and  satisfy the third equation
Then we can find the point of intersection by putting the value of  in equation (3)
Or by putting the value of  in equation (4)
For better understanding of the method we will solve an example , as follows
3 D Geometry Theory 3

7. Physics Helpline
L K Satapathy
Straight Lines in Space
Question : Prove that the line through A (0 , – 1 , – 1 ) and B (4 , 5 , 1) intersects
the line through C (3 , 9 , 4) and D (– 4 , 4 , 4) . Also find their point of intersection.
Answer :Equation of the line through A (0 , – 1 , – 1 ) and B (4 , 5 , 1) is given by
0 1 1 1 1
( )
4 0 5 1 1 1 4 6 2
x y z x y z
say
    
     
  
 Coordinates of any point on AB = (4 , 6 1, 2 1)   
Equation of the line through C (3 , 9 , 4) and D (– 4 , 4 , 4) is given by
3 9 4 3 9 4
4 3 4 9 4 4 7 5 0
x y z x y z     
    
     
 Coordinates of any point on CD = (7 3 , 5 9 , 4)  
3 9 4
( )
7 5 0
x y z
say
  
   
3 D Geometry Theory 3

8. Physics Helpline
L K Satapathy
Straight Lines in Space
(4 , 6 1, 2 1) (7 3 , 5 9 , 4)         
The point of intersection lies on both lines AB and CD.
4 7 3 . . . ( )i  
2 1 4 2 5 . . . ( )iii    
6 1 5 9 6 5 10 . . . ( )ii        
Equating the coordinates :
5
( )
2
iii   ( ) 10 7 3 1i       
5
( ) 6 5 1 10 15 15
2
ii         Lines AB and CD intersect.
Putting  = 1 , we get (7 3 , 5 9 , 4) (10 , 14 , 4) [ ]Ans   
Also putting , we get5
2
 
5 5 5
(4 , 6 1 , 2 1) (10, 14, 4)
2 2 2
[ ]Ans     
3 D Geometry Theory 3

9. Physics Helpline
L K Satapathy
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