HOW TO FIND DISTANCE BETWEEN TWO POINTS IN PLANE AND SPACE

How to calculate the distance between 2 points in plane , distance between two points in plane , 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 

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 (x1, y1  ) as  (1 , 2 ) and (x2, y2)  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 (x1, y1  ) as  (-3 , 6 ) and (x2, y2)  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 + 3 )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 (x1, y1  ) as  (-3 , 8 ) and (x2, y2)  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 + 3 )² +(12 - 8 )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| = 5√10 ----------- (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 = 5√10

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 

Problem 1

Suppose we have two points A(-3 , -4, 7 ) and B (-5 , 7, -9) in the plane the distance between them can be calculated as follows:-
1  . 1st take (x1, y1 ,z1,  ) as  (-3 , -3, 7 ) and (x2, y2 ,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 -  (-4))² + (-9 - 7)²}
 |AB| = √{(-5 + 3 )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 (x1, y1 ,z1,  ) as (√2 , √3, 3 ) and (x2, y2 ,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| = 2√14 ----------------(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| = 3√14 ----------------(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.

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HOW TO PROVE TRIGONOMETRIC IDENTITIES || TRIGONOMETRY

Proof of trigonometric identities , trigonometric identities problems, proving trigonometric identities formulas,these trigonometric identities of class 10, fundamental trigonometric identities,trigonometric identities class 11 and its formation with the help of some examples.

How to prove Identity

cos 6x = 32cos⁶ x - 48.cos⁴ x   + 18.cos² x  - 6.cos² x  - 1

Proof

1st of all  rewrite 3x as 3.2x

L.H.S. = cos 6x =  cos (3.2x) 

Now using the result cos 3θ = 4cos³ θ - 3 cos θ  -----(1)

Replacing θ as 2x in (1), we get 

L.H.S. = 4cos3 2x - 3 cos 2x  -----------(2)

Now using the result  1+ cos 2θ = 2 cos2 θ 

                                   ⇒ cos 2θ = 2 cos2 θ -1

Replacing cos 2x = 2 cos2 x -1 in (2), we get 

L.H.S.= 4 {2cos² x -1}3 - 3 {2 cos² x  -1}

Now using the result {a - b }³ = {a}³ - { b }³  -  3{a }² .b   + 3(a). b²

  cos 6x    = 4[ {2cos² x  }³ - { 1 }³  -  3{2cos² x  }² .1   +3.(2cos² x) .1² ] - 3 . {2 cos² x  -1}

Taking the product of powers to simplify it

cos 6x  =   4[ 8cos⁶ x  - 1 - 12cos⁴ x  + 6cos² x]  - 3{2cos² x-1}

Multiply by 4 in 1st term and multiply by -3 in 2nd term

 cos 6x  = 32cos⁶ x  - 4 - 48cos⁴ x  + 24cos² x  - 6cos² x + 3

Adding the like powers terms and arranging in descending order

cos 6x   = 32cos⁶ x - 48cos⁴ x  + 18cos² x  - 6cos² x  - 1

Hence the Proof

Prove the Identity 

tan (2x) =  2tan x  1 - tan² x 

Proof

We know that 

tan (A+B) =  tan A +  tan B1 - tan A tan B 

Put A = B  = x in above formula . then it becomes

tan (x+x) =  tan x +  tan x1 - tan x tan x 

tan (2x) =  2tan x  1 - tan² x 

Hence the Proof

Prove that sin 2x = 2sin x cos x

Proof

As we know that sin (A + B) = sin A cos B + cos A sin B..  ...(1)

Put A = B  = x in ...   (1)

sin (x + x) = sin x cos x + cos x sin x

sin (2x) = sin x cos x +  sin x cos x

sin (2x) = 2 sin x cos x

Hence the Proof

Prove that cos 2x = cos² x - sin² x

Proof

As we know that cos (A + B) = cos A cos B - sin A sin B..  ...(1)

Put X = A = B in (1) , we get

cos (x + x) = cos x cos x - sin x sin x
cos 2x = cos² x - sin² x      


Hence the Proof

Prove that cos 4x = 8 cos⁴ x - 8 cos² x + 1

Proof    

 Using the result 

1+cos 2θ = 2cos² θ

cos 2θ = 2cos² θ -1 -------------(1)

Replacing θ with 2x in eq (1)

1+ cos 4x = 2cos² 2x

cos 4x = 2cos² 2x -1

Again using  cos 2θ = 2cos² θ -1

cos 4x = 2(2cos² x -1)² -1

It is the square of 2cos² x -1

cos 4x = 2(2cos² x -1)² -1

cos 4x = 2(4cos⁴ x +1 - 4cos² x) -1

cos 4x = 8cos⁴ x +2 - 8cos² x -1

cos 4x =  8cos⁴ x - 8cos² x +1

Hence the Proof

What is the value of sin3x?

To find the value of sin 3x ,  use this formula which contain sin (A+B)

therefore sin (A+B) = sin A cos B cos A sin B——-(1)

put A = 2x and B = x in (1)

then Sin 3x = sin 2x cos x + cos 2x sin x

As we know that cos 2x = 1 - 2sin³ x and sin 2x = 2 sin x cos x

Sin 3x = Sin (2x+x)

Sin 3x = sin 2x cos x + cos 2x sin x

sin 3x = (2 sin x cos x) cos x + (1 - 2sin³ x ) sin x

sin 3 x = 2 sin x cos² x + sin x -  2sin³ x

As we know that cos² x = 1- sin² x

sin 3x= 2 sin x (1-sin³ x) + sin x - 2sin³ x

sin 3x = 2 sin x -2 sin³ x + sin x - 2sin³ x

sin 3x = 3 sin x - 4 sin³ x

Similarly we can prove that cos 3x= 4 cos³ x - 3 cos x

For learning and memorising more trigonometric formulas

visit here for  EASY TRIGONOMETRY PART 1

and EASY TRIGONOMETRY PART 2

Conclusion

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