ad1

HOW TO FIND DISTANCE BETWEEN TWO POINTS IN PLANE AND SPACE


Let us find out how to calculate the distance between 2 points in plane , distance between two points in plane ,distance between two points in 3d space, shortest distance between two points in 3d ,distance between two points in space , what is distance between two points in 3d formula , How to calculate distance in the 3d euclidean space.

HOW  TO FIND DISTANCE BETWEEN TWO POINTS IN PLANE 


Distance between two points A(x1,y1) and B(x2,y2) is given by 
HOW  TOP FIND DISTANCE BETWEEN TWO POINTS IN PLANE AND SPACE





Problem 1


Suppose we have two points A(1 , 2 ) and B (3 , 4) in the plane the distance between them can be calculated as follows:-
1  . 1st of all take (x1y ) as  (1 , 2 ) and (x2y2)  as (3 , 4)  then
2.   Take the differences of x coordinates and y coordinates 
3.   Then take the square of the differences of these coordinates 
4.   After that take the sum of these squares and in the last step 
5.   Take the square root of this sum obtained in the previous step .

Then |AB| = √{(3 - 1)2 +(4 - 2)2 }
|AB| = √{(2)2 +(2)2 }
|AB| = √{4 + 4 }
|AB| = √8
|AB| = 2 √2  Units


Problem 2


Suppose we have two points A(-3 , 6 ) and B (-5 , 2) in the plane the 
distance between them can be calculated as follows:-

1  . In the 1st step  take (x1y ) as  (-3 , 6 ) and (x2y2)  as (-5 , 2)  then

2.   Take the differences of x coordinates and y coordinates 

3.   Then take the square of the differences of these coordinates 
4.   After that take the sum of these squares and in the last step 
5.   Take the square root of this sum obtained in the previous step .

Then |AB| = √{(-5 - (-3))2 +(2 - 6)2 }
 |AB| = √{(-5 + )2 +(2 - 6)2 }
|AB| = √{(-2)2 +(- 4)2 }
|AB| = √{4 + 16 }
|AB| = √20
|AB| = 4√5  Units

Problem 3




Suppose we have two points A(-3 , 8 ) and B (-6 , 12) in the plane
 the distance between them can be calculated as follows:-

1 .  1st of all  take (x1y ) as  (-3 , 8 ) and (x2y2)  as (-6 , 12) 
2.   Take the differences of x coordinates and y coordinates 

3.   Then take the square of the differences of these coordinates 

4.   After that take the sum of these squares and in the last step 

5.   Take the square root of this sum obtained in the previous step .


Then |AB| = √{(-6 - (-3))2 +(12 - 8)2 }
 |AB| = √{(-6 + )2 +(12 - )2 }
|AB| = √{(-3)2 +(4)2 }
|AB| = √{9 + 16 }
|AB| = √25  ,  As the square root of 25 is 5 
|AB| = 5 Units

Problem 4


How to show that A(1,9) ,B(-2,0) ,C(3,15) are collinear points in the plane .
before we proceed Let us know
Collinear points :- Three or more points are said to be collinear 
points if they lies in the  same line .

We shall show these points collinear by the following formula
|AB|+|BC|=|AC| , or by using the formula that sum of distances of 1st two distances in a line is equal to the whole distance. So

|AB| = {(-2 - 1)2 +(0 - 9)2 }
|AB| = {(-3)2 +(0 - 9)2 }
|AB| = {9 + 81}
|AB| = √90
|AB| = √{9×10}
|AB| = 3√10 ----------- (1)


|BC| ={(3 - (-2))2 +(15 - 0)2 }
|BC| ={(3 + 2))2 +(15)2 }
|BC| ={(3 + 2))2 +(15)2 }
|BC| ={25 +225 }
|BC| =250
|BC| =√{25×10}
|BC| = 510 ----------- (2)
 ,

|AC| ={(3 - 1)2 +(15 - 9)2 }
|AC| ={4 + 36 }
|AC| =√40
|AC| =√{4×10}
|AC| = 2√10 ----------- (3)

Therefore from (1) , (2) and (3)

|AB| + |AC|=|BC|

3√10 +2√10 = 510
And also point B is common in this case, so A,B and C are collinear points 

DISTANCE BETWEEN TWO POINTS IN SPACE



Distance between two points P(x1,y1,z1) and Q(x2,y2,z1) is given by 
HOW  TOP FIND DISTANCE BETWEEN TWO POINTS IN PLANE AND SPACE





Problem 1

Suppose we have two points A(-3 , -3, 7 ) and B (-5 , 7, -9) in the plane the distance between them can be calculated as follows:-
1  . 1st take (x1y1 ,z1 ) as  (-3 , -3, 7 ) and (x2y2 ,z2)  as (-5 , 7, -9)  then
2.   Take the differences of x coordinates , y coordinates and z coordinates 
3.   Then take the square of the differences of these coordinates 
4.   After that take the sum of these squares and in the last step 
5.   Take the square root of this sum obtained in the previous step .


Then |AB| = {(-5 - (-3))2 + (7 -  (-3))2 (-9 - 7)2}
 |AB| = √{(-5 + )2 + (7 + 4)2 + (-9 - 7)2}
 |AB| = √{(-2)2 + (11)2 + (-16)2}
|AB| = √{4 + 121 + 256 }
|AB| = √381  Units


Problem 2

Suppose we have two points A(√2 , √3, 3 ) and B (2√2 , 3√3 , 6) in the plane the distance between them can be calculated as follows:-
1  . 1st take (x1y1 ,z1 ) as (√2 , √3, 3 ) and (x2y2 ,z2)  as (2√2 , 3√3 , 6)  then
2.   Take the differences of x coordinates , y coordinates and z coordinates 
3.   Then take the square of the differences of these coordinates 
4.   After that take the sum of these squares and in the last step 
5.   Take the square root of this sum obtained in the previous step .



Then |AB| = √{(2√2  - √2 )2 + (3√3 -  √3)2 (6 - 3)2}
 |AB| = √{(√2 )2 + (2√3)2 + (3)2}
 |AB| = √{2 + 12 + 9}
 |AB| = √23   Units

Problem 3

Show that A(-2,3,5), B(1,2,3), C(7,0,-1) are collinear points


 In order to show these points collinear . we apply the following
 formula.

|AB|+ |BC|= |AC|
|AB| = {(1 - (-2))2 + (2 - 3)2 (3 - 5)2}
|AB |= {(1 + 2)2 + ( - 1)2 (- 2)2}
|AB |= {32 + 1 +  4 }
|AB| = {9+ 1 +  4 }
|AB| = 14  --------------(1)

|BC| = {(7 - 1)2 + (0 - 2)2 (-1 - 3)2}
|BC| = {(6)2 + ( 2)2 (-4)2}
|BC| = 36 + 4+16}
|BC| = 36+ 4+16}
|BC| = 56
|BC| = 214 ----------------(2)

|AC| = {(7 - (-2))2 + (0 - 3)2 (-1 - 5)2}
|AC| = {(9)2 + ( - 3)2 (-6)2}
|AC| = {81 + 9 + 36}
|AC| = 126
|AC| = 314 ----------------(3)
From (1), (2) and (3) we say that
Since |AB|+|BC|=|AC|

       √14 + 2√14 = 3√14

As B is common point , Therefore A,B and C are collinear points

Also read     Solving systems of linear equations

Conclusion 



Thanks for spending your precious time to read this post, In this post we discussed how to calculate the distance between 2 points in plane , distance between two points in plane ,distance between two points in 3d space, shortest distance between two points in 3d ,distance between two points in space , what is distance between two points in 3d formula , How to calculate distance in the 3d euclidean space.

Share:

Ad3