Vectors Lines and Planes

Introduction to Vectors Line and Planes

Vector Lines

1) r  =  a + tb  ,  t  `in`  R

Cartesian Form : Let OXYZ be a right handed rectangular Cartesian coordinate framework . Let i , j , k be unit vectors along the positive X , Y , Z axes respectively .

We wish to find in Cartesian form the equation of the straight line L passing through a given point A (a1 , a2 , a3 ) and having direction cosines (b1 ,  b2 , b3) . We denote that L is the straight line passing through the point A with the position vector  a = a1i + a2j + a3k and parallel to the vector  b = b1i + b2j + b3k .

Hence P(x , y , z) is a point on L  `hArr`  `EE`   t  `in`  R such that  xi + yj - zk  =  OP   =  a + tb .

`hArr`  `EE`   t  `in`  R such that  xi + yj + zk   =  (a1 + tb1) i +  (a2 + tb2) j  +  (a3 + tb3) k

`hArr`  `EE`   t  `in`   R  such that   xi + yj - zk   =   a1 + tb1  ,  y  =  a2 + tb2  ,  z  =  a3 + tb3

`hArr`  x - a1  :  y - a2  :  z - a3  : : b1 : b2 : b3          ---------------->(1)

(1) is known as the Cartesian equation of a straight line in point - Direction cosines form .   It is usually expressed as  `(x-a_1)/b_1=(y-a_2)/b_2=(z-a_3)/b_3=t`    ,    `tin`

Since  kb1 , kb2 , kb3 , k  `!=` 0  are the direction ratios of the line the equation of a straight line in point - Direction ratios are  `(x-a_1)/(kb_1)=(y-a_2)/(kb_2)=(z-a_3)/(kb_3)`

2) r  =  ( 1 - t ) a + tb  ,  t  `in`  R

Cartesian Form:

Let OXYZ  be a right handed rectangular Cartesian coordinate framework . Let i , j , k be unit vectors along the positive X , Y, Z axes respectively . Let OP = r = (x , y , z) .

We wish to find the cartesian form of the equations of the straight line passing through two given points A(a1 , a2 , a3) and B(b1 , b2 , b3)

Let a = (a1 , a2 , a3) and  b  =  (b1 , b2 , b3)

P(x , y , z) is a point on L  `hArr` xi + yj + zk  =   OP  =  (1=i)a + tb for some  t  `in`  R .

`hArr`  (x-a1)i + (y-a2)j + (z-a3)k    =   t[ (b1-a1)i + (b2-a2)j + (b3-a3)k ]  ,    t  `in`  R

`hArr`  (x-a1) :  (y-a2) : (z-a3)  : :  (b1-a1) : (b2-a2) : (b3-a3)   --------------->(2)

(2) is known as the Cartesian form of the straight line passing through given points (a1 , a2 , a3) and (b1 , b2 , b3)  . It is usually written in form

`(x-a_1)/(b_1-a_1)=(y-a_2)/(b_2-a_2)=(z-a_3)/(b_3-a_3)`    =  `t`     ,    t  `in`   R

Properties of Vectors Lines and Planes

Properties

The vector equation of the plane passing through a poi A(a) and parallel to two non-collinear vectors b and c  is   r = a + sb + tc  ;   s , t  `in`  R
The vector equation of the plane passing through three non-collinear points A(a) , B(b)( , C(c)  is  r =  (1 - s - t)a + sb + tc  ;  s , t `in`  R
Let A and B  be the distinct points and c be a vector such that AB and C are non-collinear . Then the vector equation of the plane passing through A and B and parallel to c is  r  =  (1 - s)a + sb + tc   ;  s , t  `in`  R
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Solved Problems on Vectors Lines and Planes

1) Write the vector equation of the straight line passing through the points (2i + j + 3k) , (-4i + 3j - k) . Write also the Cartesian form of the equation .

Solution : If the position vectors of A and B with respect to O are OA = a , OB = b then the vector equation of a straight line passing through A and B is

r   =  (1 - t)a + tb  ,  t is real

Hence the vector equation of the straight line passing through the given points is

r   =  (1 - t) (2i+ j + 3k)  +  t(-4i + 3j - k)

Since r = (x , y , z) , a = (2 , 1 , 3) and b = (-4 , 3 , -1) the Cartesian form of the equation is

`(x-2)/(-4-2)=(y-1)/(3-1)=(z-3)/(-1-3)`

`(x-2)/(-6)=(y-1)/(2)=(z-3)/(-4)`

2) Find the vector equation of the plane passing through the points (1 , -2 , 5) , (0 , -5 , -1) and (-3 , 5 , 0)

Solution : If the position vectors of A , B and C with respect to O are OA = a , OB = b , OC = c then the vector equation of a straight line passing through A,B and C  is     (1 - s - t)a + sb + tc

Hence the vector equation of the straight line passing through the given three points is

r   =   (1 - s - t)(i - 2j + 5k)  +  s(-5j - k) + t(-3i + 5j)

Quartile Example

Introduction to quartile example:

The median calculate the central point of a distribution, bottom on the order, just about half of the information set falls under the median and half falls over the median.  The middle of the division is successful method to use. For the median, a usual measure of spreading can obtain from the lesser and higher quartile. The quartile contains the upper quartile and also the lower quartile. Let us see about the topic to quartile(upper quartile and lower quartile) example help are given below in the contents.

Example Problems to Solve the Method Quartile Example

let us see about the topic quartile example we can see bout different method to solve quartile (upper, lower, meadian and interquartile and range),

Example problem1:-using solve the method quartile example

Calculate the meadian, lesser quartile, higher quartile, interquartile and range of the given information of the following sequence.

15, 50, 31, 40, 18, 13, 24, 19, 25, 45, 12, 20.

Solution:

Initial, arrange the data in ascending order:

Given:  12, 13, 15, 18, 19, 20, 24, 25, 31, 40, 45, 50

Step :1

12, 13, 15 ,18,  19,  20,| 24, 25,  31, 40,45, 50

Lower quartile       Median         Upper quartile

Step :2 Find the meadian for lower quartile

Lower quartile = `(15+33)/2` = 16.5

Step :3

Median = `(20+24)/2` = 22

Step :4 Find the meadian for upper quartile

Upper quartile = `(31+40)/2` = 35.5

Step: 5

Interquartile range = Upper quartile – lower quartile

= 35.5 – 16.5 = 19

Step: 6

Range = largest value – smallest value

= 50 – 12 = 38

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Example Problem2:-using to Solve the Method Quartile Example

Estimate the median, lower quartile, upper quartile, interquartile and range of the given information of the following sequence.

15, 45, 35, 25, 65, 12, 30, 9, 20, 55, 95, 70.

Solution:

Initial, arrange the data in ascending order:

Given data:  9, 12, 15, 20, 25, 30, 35, 45, 55, 65, 70, 95

Step: 1

9, 12, 15, 20, 25, 30, | 35, 45, 55, 65, 70, 95

Lower quartile       Median              Upper quartile

Step :2 Find the meadian for lower quartile

Lower quartile = `(15 +20)/2` =17.5

Step :3

Median = `(30+35)/2` = 32.5

Step :4 Find the meadian for upper quartile

Upper quartile = `(55+65)/2` = 60

Step :5

Interquartile range = Upper quartile – lower quartile

= 60 – 17.5 = 77.5

Step :6

Range = largest value – smallest value

= 95 – 5 = 90

Triple Vector Product

In math, 3d vector is a vector with 3 dimensions. For example, 2`veci` + 2`vecj` + 4`veck` is a 3d vector. Cross product of a 3d vector gives resultant vector perpendicular to the two given vectors. A general formula is available for finding the cross product of 3d vectors. Let us study the explanation of 3d vector cross product with examples.

Let us take the following Triple Vector Product

a = a1 `veci ` + a2 `vecj` + a3 `veck ` and b = b1 `veci` + b2 `vecj` + b3 `veck`

Now, the cross product of the above two 3D vectors is as follows.

a x b = `|[veci,vecj, veck],[a_1,a_2, a_3],[b_1, b_2, b_3]|`

a x b = `veci` (a2b3 - a3b2) - `vecj` (a1b3 - a3b1) + `veck`(a1b3 - a3b1)

By using the above general description, we can find the cross product of two given 3D vectors.Looking out for more help on Algebra Tutor in algebra by visiting listed websites.

Derivation of cross product:

a x b = (a1 b1) `veci` + (a1 b2) `veci ` `vecj` + (a1 b3) `veci` `veck ` +

(a2 b1) `vecj ` `veci` + (a2 b2) `vecj` `vecj` + (a2 b3) `vecj` ` veck` +

(a3 b1) `veck`` veci` + (a3 b2)` veck` `vecj` + (a3 b2) `veck` `veck`

Some terms related to cross product:

`veci` x `vecj` = `veck`

`vecj` x `vec k` =` veci`

`veck` x `veci` =` vecj`

`vecj` x `veci` = -`veck`

`veck` x `vecj` = - `veci`

`veci ` x `veck` = -`vecj`

Based on the above terms, the formula of cross product becomes,

a x b = `veci` (a2b3 - a3b2) - `vecj` (a1b3 - a3b1) + `veck`(a1b3 - a3b1)

Inverse of Function Solver

Introduction to inverse function solver

If  ' f ' is a relation from A to B , then the relation  { (b,a) : (a,b) `in` f } is denoted as f-1 .

If ' f ' is a relation from A to B , then  (f   -1)-1  =  f   .

Theorem : If   f : A `|->` B is one-one , then f -1 is bijection from f(A) to A .

Proof : Let  f : A `|->` B be one-one .

Clearly f  -1  is a relation from   f(A) to A .

Let  b `in` f(A)  . then there exists  a `in` A such that   f(a) = b . Since f is one-one , a is the only element of A such that  f(a) = b . Thus given b `in` f(A) there is a  unique element a in A such that  (a,b) `in` f . Hence given b `in` f(A) there is unique element a `in` A such that  (b,a) `in` f-1 . Hence f-1 is function  from f(A) to A  and  f-1 (b) = a if and only  if f(a) = b . Clearly f-1 is an onto function . If  b1 , b2 `in` f(A) and f-1(b1) = f-1(b2) = a  say , then b1 = f(a) = b2 . Thus f-1 is one to one . Therfore  f-1 : f(A) `|->` A is bijection .

Definition : If  f : A `|->` B  is a bijection then the relation f -1 : { (b,a) : (a,b) `in` f } is a function from B to A and is called the inverse function of f .

Let us see some problems in inverse of function solver.

Examples on Inverse Function Solver
Problem: Find the inverse function of   f(x) = 4x +7 ?

Solution :   Given  f(x)  =  4x  + 7

Let   y  =  f(x)    `rArr`   x  =  f-1(y) .

`rArr`    y   =  4x   +   7  
`rArr`     x   =  `(y-7)/(4)`
`rArr`     f-1(y)   =  `(y-7)/(4)`

`:.`   f-1(x)  =  `(x-7)/(4)`

Answer :  f-1  =  `(x-7)/(4)` .

Problem: Find the inverse function of   f(x) = e4x+7  ?

Solution : Given  f(x) = e4x+7

Let    y  =  f(x)   `rArr`   x  =  f-1(y)

`rArr`    y   =  e4x+7  
`rArr`    log y =  4x + 7
`rArr`     x  =  `(logy-7)/(4)`    =  f-1(y)

f-1(x)  =  `(logx-7)/(4)` 

Answer : f-1(x)   =  `(logx-7)/(4)`

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Solved Problems on Inverse Function Solver

1) If A = { 1 , 2 , 3 } , B = { a , b , c } and  f = { (1,a) , (2,b) , (3,c) } , then find the inverse function of f ?

Solution : Given    A = { 1 , 2 , 3 } , B = { a , b , c }

Also given the function  f = { (1,a) , (2,b) , (3,c) }

Now , inverse of f  i.e. f-1  =  { (a,1) , (b,2) , (c,3) }

and f-1 : B `|->` A is also a bisection .

Answer :   f-1  =  { (a,1) , (b,2) , (c,3) }

2) If A = { 1 , 2 , 3 } , B = { a , b , c, d }  and g = { (1,b) , (2,a) , (3,c) } . Find the inverse function ?

Solution : Given        A = { 1 , 2 , 3 } , B = { a , b , c, d }  and g = { (1,b) , (2,a) , (3,c) }

Here g-1 does not exist .

g : A `|->` B  is one to one but not onto .  g-1 is bisection from { a, b c }  to A .

Absolute Value Inequalities

Introduction :

Normally absolute value means it is nothing but the values without considering it sign. Here we are going to solve the absolute values for inequalities. If we solve the absolute value inequalities we will get two values. Using this we value we will find the limitations of the inequalities. We will see some example problems for absolute value inequalities. Normally absolute value of |a| = `+-` a

Problems Based on Absolute Values Inequalities.

Example 1 for absolute value inequalities:

Find the absolute values of the inequalities. |x - 2| >= 10

Solution:

Given inequality is |x - 2| `gt= ` 10

To find the absolute value for the inequality is

(x – 2) `gt=` 10 ………… (1) And –(x – 2) `gt=` 10 …………………. (2)

Equation 1:

(x – 2) `gt= ` 10 ………… (1)

Add +2 on both sides

x - 2 + 2 `gt=` 10 +2

x `gt=` 12

Equation 2:

-(x – 2) `gt=` 10

-x + 2 `gt=` 10

Add -2 on both sides.

-x + 2 – 2 `gt=` 10 – 2

-x `gt=` 8

Divide by -1.

x `lt=` -8

So x lies between -8 `gt=` x `gt=` 12

We will see some more example problems based on absolute value inequalities.


Example 2 for Absolute Value Inequalities:

Find the absolute values of the inequalities. |x - 5| >= 2

Solution:

Given inequality is |x - 5| `gt=` 2

To find the absolute value for the inequality is

(x – 5) `gt= ` 2 ………… (1) And –(x – 5) `gt= ` 2 …………………. (2)

Equation 1:

(x – 5) `gt= ` 2 ………… (1)

Add +5 on both sides

x - 5 + 5 `gt= ` 2 + 5

x `gt= ` 7

Equation 2:

-(x – 5) `gt= ` 2

-x + 5 `gt= ` 2

Add -5 on both sides.

-x + 5 - 5 `gt= ` 2 - 5

-x `gt= ` -3

Divide by -1.

x `lt= ` 1

So x lies between 7 `gt= ` x `gt= ` 1                                            

These are some of the examples for absolute value inequalities. Here we will treat the inequalities like absolute values. Normally the difference between the inequalities and linear equation is less than and greater than sign.

Add and Subtract Matrices

Introduction about matrices:

Matrix has a list of data. In matrix is a rectangular arrangement of the elements. The elements are shown in the rows and the columns. Array elements are put in the parenthesis or square brackets. In matrix is represented by capital letters for example A, B, C…… We can do addition, subtraction and multiplication in matrices. In this article we shall discuss about the addition and subtraction process of matrix.

Addition of Matrices:

If we add two matrices add the first element of first matrix with first element of second matrices repeat the process for all the elements. An addition matrix is similar to normal addition.

If we add two matrices they are in the same order. For example matrix A present in 3 `xx` 3 structure the matrix B also present in 3 `xx` 3 structure else ( if the matrix B present in 2 `xx` 3 form or etc) we cannot add the matrices.

Example sums for addition of matrices:

Example:

Add the following two matrices

A= `[[2,-3],[-5,4]]`

B= `[[8,2],[-1,7]]`

A + B=?

Solution:

Given

A= `[[2,-3],[-5,4]]`

B= `[[8,2],[-1,7]]`

A + B= `[[2,-3],[-5,4]]`+ `[[8,2],[-1,7]]`

A + B= `[[2+8,-3+2],[-5-1,4+7]]`

A + B= `[[10,-1],[-6,11]]`

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Subtraction of Matrices:

Subtraction of matrices are some more different from normal substitution. Subtraction of matrices follows some rules. The following steps are used to subtract the matrices.

Step 1:

First create the negative matrix for subtrahend matrix.

Step 2:

Add the negative matrix (previous step result) with next matrix

For example if we do the process A-B matrix first discover the negative of B matrix and then add –B matrix with A matrix  like A + (-B)

Example Problems for Subtraction of Matrices:

Example:

Subtract the following two matrices

A = `[[-5,-8],[4,6]]`

B= `[[2,5],[-8,-7]]`

A-B=?

Solution:

Given

A = `[[-5,-8],[4,6]]`

B= `[[2,5],[-8,-7]]`

Step 1:

First create the negative matrix for subtrahend matrix.

- B = - `[[2,5],[-8,-7]]`

- B = `[[-2,-5],[8,7]]`

Step 2:

Add the negative matrix (previous step result) with next matrix

A + (-B) =  `[[-5,-8],[4,6]]` + `[[-2,-5],[8,7]]`

A + (-B) = `[[-5 + (-2),-8+(-5)],[4+8,6+7]]`

A + (-B) = `[[-7,-13],[12,13]]`

Graphing Parabola Inequalities

Introduction for graphing parabola inequalities:
Parabola is a conic sections, the intersection of a right circular conical surface and a plane to a generating straight line of that surface.

It is a polynomial function of degree two. A function f : R ? R defined by  the f(x) = ax2 + bx + c, where a, b, c ? R, a ? 0 is called a quadratic function. The graph of a quadratic function is always a parabola inequalities  f(x) > ax2 +  2 , f(x) < ax2 + 2

The graphing for parabola inequalities is shown below.

Example Problem for Graphing Parabola Inequalities:

Ex 1:  Draw the graph of inequalities y > x2

Sol :  Draw x-axis and y-axis on the graph sheet. Mark the scale on x-axis 1 cm = 1 unit and  y-axis 1 cm = 2 units. Assigning value for x = –4 to 4 and we get the corresponding y values:

Plot the points (–4,16), (–3,9), (–2,4),(–1, 1), (0,0), (1,1), (2,4), (3,9), (4,16) on the graph sheet and join the points by smooth curve. This curve is called the parabola y >x2

x  -4  -3  -2 -1  0  1  2  3  4

y  16  9   4    1   0  1  4  9  16

graph of inequalities y > x2

Ex 2:    Draw the graph of  inequalities   y < x2 – 2x – 3.

Sol :    Draw x-axis and y-axis on the graph sheet and mark the scales on x-axis

1 cm = 1 unit and on y-axis 1 cm = 2 units. Assign values x = –4 to 5 and calculate the

Corresponding y values

We plot the points (–4,21), (–3,12), (–2,5), (–1,0), (0, –3), (1, –4), (2, –3), (3,0), (4,5) and

(5,12) on the graph sheet and join these points by a smooth curve. We get a required graph of the parabola y < x2 – 2x – 3

X   -4   -3   -2   -1   0    1   2   3   4  5

Y   21  12   5    0   -3   -4  -3  0   5  12

inequalities   y < x2 – 2x – 3

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Practice Problem for Graphing Parabola Inequalities:

Draw the graph of inequalities   y > x2-1
Draw the graph of  inequalities   y  <  3x2- 4x + 5
3.  Draw the graph of  inequalities   y  <  5x2- 4x + 9
4.  Draw the graph of  inequalities   y > x2+4