Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
SlideShare a Scribd company logo

3D Geometry Theory 4

•Download as PPTX, PDF•
•
Lakshmikanta Satapathy
Lakshmikanta Satapathy

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 of 10
Physics Helpline L K Satapathy 3D Geometry Theory 4
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
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
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
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
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
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        
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
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
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

More Related Content

3D Geometry Theory 4

  • 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 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