JEE Mathematics/ Lakshmikanta Satapathy/ 3D Geometry theory part 7/ Equation of a Plane through a given point and normal to given vector in vector and Cartesian forms

1. Physics Helpline
L K Satapathy
3D Geometry Theory 7

2. Physics Helpline
L K Satapathy
Theorem : Every first degree equation in x , y and z represents a plane.
A plane is a surface such that the line segment joining any two points on it , lies
completely on the surface i.e. every point on the line segment lies on the surface.
represents the general equation of a plane. . 0i e ax by cz d   
Proof : Consider the first degree equation in x , y and z given by
0 . . . (1)ax by cz d   
Consider two points on the surface represented by equation (1) whose coordinates are
1 1 1 2 2 2( , , ) ( , , )A x y z and B x y z 
 We have 1 1 1 0 . . . (2)ax by cz d   
And 2 2 2 0 . . . (3)ax by cz d   
The Plane
3 D Geometry Theory 7

3. Physics Helpline
L K Satapathy
The Plane
 We have to prove that 1 2 1 2 1 2
0
1 1 1
x x y y z z
a b c d
  
  
       
               
1 2 1 2 1 2
1
[ ( ) ( ) ( ) (1 )]
1
LHS a x x b y y c z z d   

       

1 1 1 2 2 2
1
[( ) ( )]
1
ax by cz d ax by cz d

       

Let a point P = (x , y , z) lie on the line segment AB , which divides it in ratio  : 1
1 2 1 2 1 2
( , , ) , , . . . (4)
1 1 1
x x y y z z
x y z
  
  
   
      
We need to prove that the point P lies on the surface represented by equation (1)
1
[(0) (0)] 0
1


  

[ Using equation (2) and (3) ]
 The equation represents a plane0ax by cz d   
3 D Geometry Theory 7

4. Physics Helpline
L K Satapathy
Equation of Plane passing through a given point and perpendicular to a given vector
The Plane
O
PA
N
a
r
Consider the plane shown in the figure .
It passes through a given point A having position vector a
And perpendicular to the vector N
Consider any point P on the plane having position vector r
AP is a line segment lying on the plane
AP r a  
 AP is perpendicular to N . 0AP N 
( ). 0 . . . (1)r a N  
Vector form :
[ Vector equation ]
3 D Geometry Theory 7

5. Physics Helpline
L K Satapathy
Equation of Plane passing through a given point and perpendicular to a given vector
The Plane
Let the coordinates of the given point A = 1 1 1
ˆˆ ˆa x i y j z k   
Cartesian form :
[ Cartesian equation ]
1 1 1( , , )x y z
And the coordinates of any point P = ( , , )x y z ˆˆ ˆr xi y j z k   
1 1 1
ˆˆ ˆ( ) ( ) ( )r a x x i y y j z z k       
Let the direction ratios of the normal N = ( , , )A B C ˆˆ ˆN Ai B j C k   
 Equation (1)  1 1 1
ˆ ˆˆ ˆ ˆ ˆ[( ) ( ) ( ) ]. [ ] 0x x i y y j z z k Ai B j C k       
1 1 1( ) ( ) ( ) 0 . . . (2)A x x B y y C z z      
3 D Geometry Theory 7

6. Physics Helpline
L K Satapathy
Question : Find the equation of the plane passing through the point (3 , – 3 ,1) and
normal to the line joining the points (3 , 4 , –1) and (2 , – 1 , 5) in cartesian form.
The Plane
Answer :
O
PA
N
a
r
D
C
The plane is normal to the line joining the points
ˆ ˆˆ ˆ ˆ ˆ(2 3) ( 1 4) (5 1) 5 6N CD i j k i j k            
Plane passing through the point (3 , – 3 ,1)
ˆˆ ˆ3 3a i j k   
Let the coordinates of any point P = ( , , )x y z
ˆˆ ˆr xi y j z k   
C (3 , 4 , –1) and D (2 , – 1 , 5) as shown .
3 D Geometry Theory 7

7. Physics Helpline
L K Satapathy
The Plane
ˆˆ ˆ( 3) ( 3) ( 1)r a x i y j z k       
( ). 0r a N 
ˆ ˆˆ ˆ ˆ ˆ[( 3) ( 3) ( 1) ].[ 5 6 ] 0x i y j z k i j k         
Equation of a plane in vector form is
( 1)( 3) ( 5)( 3) (6)( 1) 0x y z        
3 5 15 6 6 0x y z       
5 6 18 0x y z     
5 6 18 [0 ]Anx y z s    
3 D Geometry Theory 7

8. Physics Helpline
L K Satapathy
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