JEE Mathematics/ Lakshmikanta Satapathy/ Three dimensional Geometry part 4/ Perpendicular distance of a point from a line/ Image of a point in a line

1. Physics Helpline
L K Satapathy
3D Geometry Theory 4

2. Physics Helpline
L K Satapathy
Perpendicular distance of a point from a line :
Straight Lines in Space
Let the coordinates of the given point P = ( ,  , )
P ( ,  , )
Q BA
And the equation of the given line AB is
1 1 1
. . . (1)
x x y y z z
a b c
  
 
Where (a , b , c) are the direction ratios of the line AB
Let Q be the foot of the perpendicular from P on the line AB
1 1 1x x y y z z
Let
a b c

  
  
1 1 1( , , )x a y b z c      Coordinates of any point on AB
1 1 1( , , )x a y b z c            Direction ratios of line PQ
1 1 1( , , ) . . . (2)x a y b z c     For some value of  , coordinates of Q
3 D Geometry Theory 4

3. Physics Helpline
L K Satapathy
Straight Lines in Space
1 1 1( ) ( ) ( ) 0a x a b y b c z c              
Line PQ is perpendicular to line AB [ using for  lines ]
2 2 2
1 1 1 0ax a a by b b cz c c              
2 2 2
1 1 1( ) ( ) ( ) ( )a b c a x b y c z           
1 1 1
2 2 2
( ) ( ) ( )
( )
a x b y c z
a b c
  

    
 
 
Putting the value of  in equation (2) , we find the coordinates of Q.
Knowing the coordinates of P and Q , we find
the perpendicular distance PQ using the relation
2 2 2
2 1 2 1 2 1( ) ( ) ( )d x x y y z z     
For better understanding of the method we will solve an example , as follows
1 2 1 2 1 2 0a a bb c c  
3 D Geometry Theory 4

4. Physics Helpline
L K Satapathy
Straight Lines in Space
Question : Find the length of the perpendicular from the point (1 , 2 , 3 ) to the line
Answer : The coordinates of the given point P = (1 , 2 , 3)
6 7 7
3 2 2
x y z  
 

P (1 , 2 , 3)
Q BA
6 7 7
3 2 2
x y z
Let 
  
  

(3 6, 2 7, 2 7)       Coordinates of any point on AB
Let Q be the foot of the perpendicular from P on the line
For some value of  , coordinates of Q (3 6, 2 7, 2 7)      
 Direction ratios of line PQ (3 6 1, 2 7 2, 2 7 3)         
(3 5, 2 5, 2 4)      
3 D Geometry Theory 4

5. Physics Helpline
L K Satapathy
Straight Lines in Space
3(3 5) 2(2 5) 2( 2 4) 0         
Line PQ is perpendicular to line AB [ using for  lines ]1 2 1 2 1 2 0a a bb c c  
 Coordinates of Q
 Distance PQ
9 15 4 10 4 8 0        
17 17 0 1      
(3 6, 2 7, 2 7)      
[(3)( 1) 6, (2)( 1) 7, ( 2)( 1) 7]        (3, 5, 9)
We have , P = (1 , 2 , 3) and Q = (3 , 5 , 9)
2 2 2
(3 1) (5 2) (9 3)     
4 9 36 49 7 [ ]uni Ansts    
3 D Geometry Theory 4

6. Physics Helpline
L K Satapathy
Straight Lines in Space
P ( ,  , )
Q
BA
R (, , )
Image of a point in a line :
Let the coordinates of the given point P = ( ,  , )
And the equation of the given line AB is
1 1 1x x y y z z
a b c
  
 
Where (a , b , c) are the direction ratios of the line.
If R ( ,  , ) be the image of the point P in the line AB, then
(i) The line PR is perpendicular to the line AB
(ii) Distance PQ = distance QR [ Q is the foot of the perpendicular from P on AB ]
(i) First we find the coordinates of Q [ foot of the perpendicular ]
Method :
(ii) Then we find the coordinates of R such that Q is the mid point of PR
3 D Geometry Theory 4

7. Physics Helpline
L K Satapathy
Straight Lines in Space
1 1 1x x y y z z
Putting
a b c

  
  
and proceeding as in the pervious section, we get the coordinates of the point Q .
Let the coordinates of Q ( , , )o o o  
Also the coordinates of
Since Q is the mid point of PR , we have
, ,
2 2 2
o o o
     
  
    
  
2 , 2 , 2o o o                
Putting the values , we get the coordinates ( ,  , ) of R .
For better understanding of the method we will solve an example , as follows
3 D Geometry Theory 4
( , , ) & ( , , )P R        

8. Physics Helpline
L K Satapathy
Straight Lines in Space
P (5 , 9 , 3)
Q
BA
R (, , )
Question : Find the image of the point (5 , 9 , 3) in the line
Answer : The coordinates of the given point P = (5 , 9 , 3)
1 2 3
2 3 4
x y z  
 
Let Q be the foot of the perpendicular from P to the line
1 2 3
2 3 4
x y z
Let 
  
  
(2 1, 3 2, 4 3)      Coordinates of any point on AB
For some value of  , coordinates of Q
 Direction ratios of line PQ
(2 1, 3 2, 4 3)     
(2 1 5, 3 2 9, 4 3 3)        
(2 4, 3 7, 4 )    
3 D Geometry Theory 4

9. Physics Helpline
L K Satapathy
Straight Lines in Space
Line PQ is perpendicular to line AB [ using for  lines ]1 2 1 2 1 2 0a a bb c c  
2(2 4) 3(3 7) 4(4 ) 0       
4 8 9 21 16 0 29 29 0 1              
 Coordinates of Q (2 1, 3 2, 4 3) (3, 5, 7)      
If the coordinates of the image point R = ( ,  , ) , then
Also the coordinates of P =(5 , 9 , 3)
5 9 3
3 , 5 , 7
2 2 2
      
  
6 5 1 , 10 9 1 , 14 3 11             
 The coordinates of the image point R = (1 , 1 , 11) [ Ans ]
3 D Geometry Theory 4

10. Physics Helpline
L K Satapathy
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