3D Geometry Theory 4
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The document discusses the perpendicular distance from a point to a line in three-dimensional geometry, detailing the coordinates involved and the mathematical equations required for the calculation. It provides an example problem regarding finding the perpendicular distance from a specific point to a line, demonstrating the method and calculations step-by-step. Additionally, it explains how to find the image of a point relative to a line and includes further examples for clarity.
![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](https://image.slidesharecdn.com/geometry3d-160410041509/85/3D-Geometry-Theory-4-3-320.jpg)
![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](https://image.slidesharecdn.com/geometry3d-160410041509/85/3D-Geometry-Theory-4-5-320.jpg)
![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](https://image.slidesharecdn.com/geometry3d-160410041509/85/3D-Geometry-Theory-4-6-320.jpg)
![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](https://image.slidesharecdn.com/geometry3d-160410041509/85/3D-Geometry-Theory-4-9-320.jpg)
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