HOW TO SOLVE DETERMINANTS USING ELEMENTARY TRANSFORMATIONS

How to Prove Determinants using elementary transformations 

In this post we shall discuss Short trick of elementary transformation,Solving Determinants using elementary transformations,define elementary transformation, elementary transformation class 12, elementary row transformation questions.

By this method we have  to reduce maximum elements of specific Rows or column to zero, so that we can solve it easily

To solve the determinants using elementary transformations , Let us suppose L H S = △

As we can see that 'a' is common in 1st Row , 'b' is common in 2nd Row and 'c' is common in 3rd row ,
Therefore Taking a ,b ,c common from R_1  , R_2  , and  R₃       respectively

If we add R_1    to   R_2  and   R₁   to  R_{3   then we get zero in 1st column, so}  Operating  R_{1  →} R₁   + R₂   and    R_{3  →} R₁   + R₃  

_{As we have received maximum possible  zero in 1st column Therefore Expanding along} C₁  

△= (-a)×[(0)-(2c×2b)]
△ _{= abc{-a(-4bc)}}

△ = 4a²b²c² 

Hence the proof


Watch this video for Understanding Elementary transformations

PROBLEM

Proof:- Put L H S of determinant to Δ

Operating R₁ ➡️xR₁  , R₂ ➡️ yR₂ and R₃➡️ zR₃

Taking common xyz from C₃

Operating  R₂ ➡️ R₁ - R_2  and  R₃ ➡️ R₁ - R3

Expanding along  C₁ 

Δ = (x² - y² )( x³ - z³ ) - (x³ - y³ )(x² - z² )

Δ = (x- y )(x+ y)(x- z )( x² + z²  + xz ) - (x- y )( x² + y²  + xy ) (x - z )(x +z )

Δ = (x- y )(x- z )[(x+ y)( x² + z²  + xz ) -( x² + y²  + xy ) (x +z )]

Cancelling the same colour terms in the previous line ,then we have 

 Δ = (x- y )(x- z )[xz²   + yz² - y²x - y²z  ]

Arranging  terms in Squared Bracket  in such a way that the term containing z² must be at 1st and 3rd position and the term containing y² must be at 2nd and 4th position .

Δ = (x- y )(x- z )[(yz² - y²z) +( xz²  - y²x)]

Δ = (x- y )(x- z )[yz(z - y) + x(z²  - y²)]

Δ = (x- y )(x- z )[yz(z - y) + x(z - y)(z+ y)]

Δ = (x- y )(x- z )(z - y)[yz + x(z+ y)]

Δ = (x- y )(x- z )(z - y)[yz + xz+ xy]

Taking -1 common from (x- z )(z - y) in previous line ,

Δ = (x- y )(y- z )(z - x)[yz + xz+ xy]

Hence the   proof

Also Read   HOW TO FIND THE SLOPE OF LINE  AX + BY = C ,  SLOPE OF A  LINE

Final Words

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HOW TO FIND AREA BOUNDED BY TWO CIRCLES , INTEGRATION OF AREA UNDER CURVE

HOW TO FIND AREA OF TWO CIRCLES INTERSECTING EACH OTHERS,

Here we are going to discuss how to find common area of two circles which are overlapping or intersecting at two points with the help of an example

 Let us consider two circles whose equations are  given below
x² + y² =  1^{2                   .......................(1)}
( x - 1 )² + y² = 1²          .....................(2)

Let us draw these circles in coordinate planes, We can compare these equations with standard form of circle to find  the coordinate of  centre of both the circles are (0,0) and (1,0) respectively and radius of both the circles are 1.

If these two circles intersect with each other then we have to find their point/s of intersection.
To find points of intersection subtracting equation (1) from eq (2) , we get 

    ( X - 1 )² + Y² - X² - Y² =  1^2 - 1² 
⇒( X - 1 )²  - X² =  0
 ⇒X ²  +1 ²  - 2×(1)×(X) - X² =  0

⇒1 - 2X = 0

⇒ X = 1/2 ,
Now to find the values of y put the value of x in equation # 1

(1/2)² + Y² =  1²     
   Y² = 1- (1/4) = 3/4

   Y = 土⇃(3/4)
Therefore two points of intersection are B(1/2 , ⇃(3/4)) , C ( (1/2 , -⇃(3/4))

To understand better the solution of  this problem watch this  video 

Required area = shaded  Area , 

we can divide shaded area into four equal parts , As each parts is symmetrical , Therefore to find shaded area  it is sufficient to find the area of any one of four part and then then multiply it with 4.

Hence  Required area = 4 area OBLO = 4 Area BALB
To avoid tedious calculations choose 2nd part to integrate
I= Required area = 4 Area BALB    ------------- (3)

After simplification , we have

    ALSO READ        HOW TO INTEGRATE INTEGRAL WITH SQUARE ROOT IN NUMERATOR     

My previous post HOW TO FIND AREA OF THE CIRCLE  WHICH IS INTERIOR TO THE PARABOLA

  Final words 

Thanks for visiting this website and spending your valuable time to read this post regarding how to find area bounded by two circles .If you liked this post , do share it with your friends to benefit them also we shall meet in next post , till then bye and take care....

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HOW TO FIND AREA OF THE CIRCLE WHICH IS INTERIOR TO THE PARABOLA

HOW TO FIND AREA OF THE CIRCLE  WHICH IS INTERIOR TO THE PARABOLA

Area Under Curves

Let  us write two equations of circle and parabola respectively
4x²+ 4y²  = 9 ------------------- (1) 

and x²      = 4y    -------------------(2)

Reducing (1) to standard form by dividing 4 .we get 
x²+ y²  = (3/2)²   
Ist of all draw figures of both the circle and the parabola in cartesian plane.

As it can be seen from figure both  curves intersect each other at two points say A and A' . 
Next we have to find these two coordinates points of intersection . Solving (1) and (2) to find the values of x and y 
Putting  the value of ' x² ' from (2) in (1) we get 

4(4y)+ 4y² = 9

16y + 4y² - 9 = 0

 4y² - 16y -9 = 0

        

 y = (-16+20)/8  and (-16-20)/8
y = 1/2  and -9/2

So Rejecting the -ve value of y ,because when we put negative value (-9/2) in eq (2) , we shall have two complex values of "x" which are not acceptable.
so only put positive value (1/2) of   'y'  in (2) we get two real values of  'x' such that    x= 土⇃2,
Now we can write coordinate M(⇃2,0) and N (-⇃2,0)

Required Area = Shaded area
                         = 2 × Area OBAO 
Note this step carefully ↑

                   

Multiplying every terms with 2 which is written at  beginning  of the previous line.

Putting the values of upper and lower limits of x

    ALSO READ        HOW TO INTEGRATE INTEGRAL WITH SQUARE ROOT IN NUMERATOR     

  Final words 

Thanks for visiting this website and spending your valuable time to read this post.If you liked this post , do share it with your friends to benefit them also we shall meet in next post , till then bye and take care......

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HOW TO FIND THE ANGLE BETWEEN TWO LINES

HOW TO FIND THE ANGLE BETWEEN TWO LINES WHEN THE EQUATIONS OF GIVEN LINES ARE  IN CARTESIAN  FORMS

In this post we shall study How to find the angle between two lines ,angle between two lines vectors in Cartesian form, angle between two lines in 3d, angle between two lines calculator, angle between two lines coordinate geometry,derivation of angle between two lines

Problem 1

Consider two lines whose equations are given in cartesian form as

Note that direction Ratios of 1st line is (1 , 2 , 3) and the direction Ratios of 2nd line is (2 , 3 , 4).

Note that Direction Ratios of  any line are those numbers written is the denominator in standard form of the equation of the line

we know that if a_1,b_1,c_1   and    a_2,b_2,c_{2,  are Direction Ratios of line} L_{1  and Line} L<sub>2 .

If  θ be the angle between two lines ,Then this angle  can be formulated as follows</sub>

So putting the values of direction ratios of both the lines in (3) ;

we  get 

cos θ = [2 + 6 + 12]/sqrt[ 1+ 4+ 9]×sqrt[ 4 +9 +16 ]

cos θ = [20]/√[ 14×29 ];

cos θ = 10/√203

∴ θ = cos^-1 (10/√203)

Hence  cos^-1 (10/√203) is the angle between two lines

How to find angle between two  line in vector and cartesian   forms

 

Problem 2

Let us consider these two lines in cartesian form

Since these equations of lines are not in standard form  , in order to reduce these equations to standard form we shall have to make the coeffs of x, y , z unity.

In 1st part of equation (1) divide the num and den by 2 , and in 2nd part divide num and den by 3. 

Similarly divide 2nd part of  equation (2) by 5. we can rewrite these equations as follows

After cancellation and simplification , we get equations of lines in standard form

Here Direction Ratios of 1st Line are  (2,1 ,3)  and Direction Ratios of 2nd Line are (1 ,2 , 4) so using the formula

Putting the values of direction ratios of both the lines in above formula ,we get

cos θ = [2 + 2 + 12]/sqrt[ 4+ 1+ 9]×sqrt[ 1 +4 +16 ]

cos θ = [16]/sqrt[14×21];

cos θ = 16/√294

∴ θ = cos^-1 (16/√294)

Hence  cos^-1 (16/√294) is the angle between two lines

Problem 3

How to prove that the given lines are parallel to each others, 

To test whether given lines are parallel to each other , just check their direction ratios , id they are proportional to each others , then The given lines would be parallel to each other.

If the Direction ratios of 1st line is (1,2,5) and Direction ratios of 2nd  line is (2,4,10)  . If we take 2 common from direction ratios of 2nd lines then we have same direction Ratios as that of 1st line hence These lines are parallel o each other.

Problem 4

How to prove that the given lines are Perpendicular  to each others. If we calculate the angle between any two lines and it comes out to be 0 ( ZERO ) then  these would be perpendicular to each others.

E.g  If the D. R. of 1st line are ( 3, -1 , 3 )

And     D. R. of 2nd Line are  ( -2 ,3 ,3 ).

The cosine of angle between these  two line is zero

cos θ = 0

Then  θ = 90° 

⇒ Lines are Perpendicular to each others

Also Read previous posts 
How to Find perpendicular distance between skew lines

How to find slope of line ax+by=c


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To find the angle between two lines

x = y = z and

x = y = -z 


Want to check the solution of this problem ?

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Thanks for your  precious time to  read this post regarding how to find the angle between two lines in Cartesian form.

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HOW TO FIND THE PERPENDICULAR DISTANCE BETWEEN TWO SKEW LINES AND PARALLEL LINES

Shortest distance between two parallel lines,perpendicular distance between two parallel lines,shortest distance between two skew lines Cartesian form,shortest distance formula in 3d,distance between two non parallel lines,shortest distance between two parallel lines

The Shortest Distance Between Skew Lines

The shortest distance between the lines is the distance which is perpendicular to both the lines given as compared to any other lines that joins these two skew lines.

Vector Form

We shall consider two skew lines  L1 and L2 and  we are to calculate the distance between them. The equations of the given lines are:

Here  vector  and vector  are the vectors through which line (1) and (2) passes and and   are the vectors which are parallel to lines    L1 and L2  respectively. 

Then perform the following steps

(1)                 Calculate  -  

(2 )                Calculate   ×

(3)                Calculate    |  ×  | 

(4 )                Calculate ( − ) . (   ×  )

and if the dot product of these two vectors come out to be negative then take its absolute value as distance can not be a negative quantity  .

(5 )                Put all these values in the formula given below and  the value so calculated is the shortest distance between two skew Lines.

 ⃗⃗

Problem : How to find  the shortest distance between two skew lines in vector form whose equations are given by

  Now write the values of    ,, and 

Now find out the difference of and 

After solving this determinant and little simplification we get ,

The magnitude of this vector 

                             = √(169+64+4)=√(237)

Now find Dot Product    ( -  )  and  (   ×  ) ,

                                                    = (0)(-13)+(-6)(8)+(5)(-2)

                                                    = -48 -10

                                                    = -58

Taking the absolute value of

|   (− )  .  (   ×  ) | = 58

Now putting all these values in SD Formula written above , we can have

SD = 58/(√237) Units

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The Shortest Distance Between Parallel Lines

Consider two parallel lines

Here  vector  and  vector  are the vectors through which line (1) and (2) passes and    is the vector which is parallel to both lines    L1 and L2  respectively. Now perform the following steps 

( 1 )                 Calculate  -  

( 2 )                Calculate    | | 

( 3 )                Calculate ( -  )     ×  ,

     

Put all these values in the formula given below and the value so calculated is the shortest distance between two Parallel Lines, and if it comes to be negative then take its absolute value as distance can not be negative

Problem 2 : How to find  the shortest distance between two Parallel lines in vector form whose equations are given by

As we know these are parallel line because both these equations are parallel to same vector  -2i +3J+5k.

Now write the values of    ,, and  as follows

Now find out the difference of and 

Now  Calculate ( -  )     ×  ,

     After solving this determinant and  simplification we get ,            

                  

 

The magnitude of this vector is   

= √2069

Similarly the magnitude of vector    is √38

  Put all these values in the formula given below and the value so calculated is the shortest distance between two Parallel Lines, and if it comes to be negative then take its absolute value as distance can not be negative

SD = √(2069 /38) Units

At Last

Thanks for giving your precious time to read this post which include shortest distance between two lines in 3d pdf,shortest distance between two parallel lines,perpendicular distance between two parallel lines,shortest distance between two skew lines cartesian form,shortest distance between two points,shortest distance formula in 3d,distance between two non parallel lines,distance between two lines calculator,shortest distance between two parallel lines

Read previous Post How to find slope of line ax+by= c
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