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Physics Helpline
L K Satapathy
3D Geometry QA 5
Physics Helpline
L K Satapathy
3 D Geometry QA 5
Question: The distance of the point ( 1 , 0 , 2 ) from the point of intersection of the
line and the plane x – y + z =16 , is
( ) 2 14 ( ) 8 ( ) 3 21 ( ) 13a b c d
Answer :
12 2 . . . (1)
3 4 12
yx z  
12 2
3 4 12
yx z  
A(1,0,2)
d
B
L

The situation is shown in the figure
The given line is L , whose equation is
The given plane is  , whose equation is 16 . . . (2)x y z  
Let the given point be A whose coordinates are ( 1 , 0 , 2 )
12 2
3 4 12
yx zLet 
   
 Coordinates of any point on the line is (3 2 , 4 1, 12 2)    
Physics Helpline
L K Satapathy
3 D Geometry QA 5
Correct option = (d)
Since B lies on the plane x – y + z =16 , we have
(3 2) (4 1) (12 2) 16       
(3 2 , 4 1,12 2) (5 , 3 ,14)      
2 2 2
(5 1) (3 0) (14 2)d      
 The required distance between the points ( 1 , 0 , 2 ) and ( 5 , 3 , 14 ) is
 For some value of  , coordinates of B =(3 2 , 4 1, 12 2)    
Let the point of intersection of the line and the plane be B
 Coordinates of B
16 9 144 169 13    
11 11 1    
Physics Helpline
L K Satapathy
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3D Geometry QA 5

  • 1. Physics Helpline L K Satapathy 3D Geometry QA 5
  • 2. Physics Helpline L K Satapathy 3 D Geometry QA 5 Question: The distance of the point ( 1 , 0 , 2 ) from the point of intersection of the line and the plane x – y + z =16 , is ( ) 2 14 ( ) 8 ( ) 3 21 ( ) 13a b c d Answer : 12 2 . . . (1) 3 4 12 yx z   12 2 3 4 12 yx z   A(1,0,2) d B L  The situation is shown in the figure The given line is L , whose equation is The given plane is  , whose equation is 16 . . . (2)x y z   Let the given point be A whose coordinates are ( 1 , 0 , 2 ) 12 2 3 4 12 yx zLet       Coordinates of any point on the line is (3 2 , 4 1, 12 2)    
  • 3. Physics Helpline L K Satapathy 3 D Geometry QA 5 Correct option = (d) Since B lies on the plane x – y + z =16 , we have (3 2) (4 1) (12 2) 16        (3 2 , 4 1,12 2) (5 , 3 ,14)       2 2 2 (5 1) (3 0) (14 2)d        The required distance between the points ( 1 , 0 , 2 ) and ( 5 , 3 , 14 ) is  For some value of  , coordinates of B =(3 2 , 4 1, 12 2)     Let the point of intersection of the line and the plane be B  Coordinates of B 16 9 144 169 13     11 11 1    
  • 4. Physics Helpline L K Satapathy For More details: www.physics-helpline.com Subscribe our channel: youtube.com/physics-helpline Follow us on Facebook and Twitter: facebook.com/physics-helpline twitter.com/physics-helpline