Three Distance Theorem

The three-distance theorem states that there are at most three distinct gaps between consecutive elements in the set of fractional parts of the first n multiples of any real number. In a real world of three - dimensions, we must extend our knowledge of geometry ti three dimensional space.

To find the distance between two points whose co-ordinates are given. Let P( x1 , y1 , z1 ) and Q ( a2, b2, c2 ) be the two given points. Through P and Q draw planes parallel to the co-ordinate planes to form a rectangular box whose one diagonal is PQ. Geometrically to find the distance PQ is the computing of the length of the diagonal PQ of the box ( as shown in the figure) by means of Pythagoras theorem.



Since MQ is perpendicular to the plane PAMB and PM lies in the plane so MQ is perpendicular to PM `=>` `angle` PMQ = 900 .

in triangle PMQ, `angle` PMQ = 900 , therefore by pythagoras theorem we get,

PQ2 = PM2 + MQ2 ........(1)

since, AM is perpendicular to AP, so `angle` MAP = 900

In triangle AMP, `angle` MAP = 900 , therefore, by pythagoras theorem

PM2 = AP2 + AM2 .....(2)

From (1) and (2), we get

PQ2 = AP2+ AM2+ MQ2 ......(3)

Clearly from the figure

AP = x2 - x1 , AM = y2 -y1 , MQ = z2-z1

Substituting these values in (3) we get

PQ2 = (x2 - x1 ) + ( y2 - y1 ) + (z 2- z1)

PQ = `sqrt(( x_2-x_1)+ ( y_2-y_1) + (z_2-z_1))`

Distance from origin: The distance of P(x,1y,1z1) from the origin o(0,0,0) is `sqrt((x_1-0)^2 +( y_2 - 0)^2 + (z_2 - 0)^2)` = `sqrt((x_1)^2 + (y_1)^2 + (z_1)^2)`



Example Problems on three Distance Theorem:

Problem1: Find ' a' so that the distance between ( -2,1,-3) and ( 1, 3, -6) be 7 units?

Solution:

Let A, B be ( -2,1,3) ,( 1, 3, -6) respectively then

`=>` `sqrt((a-2)^2 + (3-1)^2 + (-6 + 3 )^2)` = 7

`=>` a2 + a + 4 + 4 + 9 = 49 `=>` a2 + 4a - 32 = 0

`=>` (a-4) (a+ 8) = 0 `=>` a= 4 , -8

Problem2: Find the distance between the points A(1,2,3) from he point B(3, 6, 8)?

Solution: The distance

`sqrt((3-1)^2 +(6-3)^2 + ( 8-3)^2)`

AB = `sqrt(2^2 + 3 ^2 + 5^2)`

AB = `sqrt(4 + 9 + 25)`

AB = `sqrt(38)` <br>

Practice Problems on three Distance Theorem

Problem2: Find the distance of the point P( 4, 6, 8) from the point Q ( 9,5,2)? ( Answer: 3`sqrt(7)` )

problem3: By using the distance formula prove that the points A( 3, -5, 1), B ( -1, 0 , 8) and C( 7, -10, -6) are collinear ("Answer : The sum of two side is equal to the third side so the points are collinear)

Dihedral Angle Calculator

Dihedral angle is the one of the special kind of angles in mathematical geometry.  Dihedral angle is an angle that forms between two plans.  Usually a plane is the three dimensional flat surface.  The points in the plane take the form of (x, y, z).  In this topic we are going to study about how to calculate the dihedral angle when the points of the plane are given through the dihedral angle calculator.

Explanation about Dihedral Angle through Calculator

Important Guidelines:

The following steps are the important guide lines to calculate the dihedral angle.

• First we have to choose the three points on the plane 1 and then choose the three points on the plane 2

• Enter the (x, y, Z) values for each point on the dihedral angle calculator we have to get the equation of the plane 1 and plane 2 separately

• The equation of the plane 1 is taking the form of A1x + B1y + C1z + D1 =0 and then the equation of the plane 2 is taking the form of A2x + B2y + C2z + D2 =0

Now we have to calculate the dihedral angle through the following formula,

cos`alpha` = `((A_1)(A_2) + (B_1)(B_2) + (C_1) (C_2))/ (sqrt((A_1)^2 +(B_1)^2 +(C_1)^2) * sqrt((A_2)^2 +(B_2)^2 +(C_2)^2))`

This formula is used for our manual calculation.

Example Problem on Calculate Dihedral Angle via Calculator

Calculate the dihedral angle for the plane 1 and plane 2.  The points on the plane 1 are (1, 2, 3) and (3, 2, 1) and then (2, 1, 3) and the points on the plane 2 are (4, 1, 2) and (1, 4, 4) and then (2, 4, 2)

Solution:

Now we have to substitute the values of the points in the following dihedral Angle Calculator,



This calculator gives the distance between each points and equation of the each plane and then the angle in between the planes in both degrees and radians.  These are the main usage of the dihedral angle calculator.

Distributive Law Definition

The distributive law is placed in set theory. The distributive law is the piece of sets that used to learn about the sets. Group of items is called the sets. Set theory have some operations union, intersection, difference, complement and performs some law.The set theory is used to select the number of objects simultaneously. Here we will see about the distributive law definition with examples.

Examples - Distributive Law Definition:

Now we are going to solve the examples for distributive law definition.

AU (B∩C) = (AUB) ∩ (AUC)

A∩ (BUC) = (A∩B) U (A∩C)

Example 1

Get the solution of the given sets by using distributive law. A={ a,b,c,d } B={ b,c,g } and C={ b,c,k,m }.

Solution:

The given sets are A={a,b,c,d} B={b,c,g} and C={b,c,k,m}.

Distributive law for set is as follows,

i)AU (B∩C)=(AUB) ∩(AUC)

ii)A∩(BUC)=(A∩B) U (A∩C)

i)AU (B∩C)=(AUB) ∩(AUC)

First find the solution left hand side.

AU (B∩C)

First we can solve the inner bracket set.

B∩C

Intersection means we choose the common elements from the set B and C.

So B∩C =   B= {b, c, g} ∩ C= {b, c, k, m}.

B∩C = {b, c}

Now find the AU (B∩C)

Grouping the values of A and B∩C

AU (B∩C) = {a, b, c, d} U (b, c)

= {a, b, c, d}

So AU (B∩C) = {a, b, c, d} -------------- (1)

(AUB) ∩ (AUC)

Groping the values of A and B sets.

A= {a, b, c, d} B= {b, c, g}

AUB = A= {a, b, c, d} U B= {b, c, g}

= {a, b, c, d, g}

Joining the values of A and C sets.

AUC:

A= {a, b, c, d} U C= {b, c, k, m}.

= {a, b, c, d, k, m}

(AUB) ∩ (AUC):

AUB ∩AUC = {a, b, c, d, g} ∩ {a, b, c, d, k, m}

(AUB) ∩ (AUC) = {a, b, c, d} ------------- (2)

So AU (B∩C) = (AUB) ∩ (AUC)

ii) A∩ (BUC) = (A∩B) U (A∩C)

A={ a,b,c,d } B={ b,c,g } and C={ b,c,k,m }.

Now solve the left side.

A∩ (BUC)

BUC:

= {b, c, g} U {b, c, k, m}

BUC = {b, c, g, k, m}

A∩ (B U C):

A= {a, b, c, d} ∩ {b, c, g, k, m}

= {b, c} -------------- (1)

(A∩B) U (A∩C)

A∩B= {a, b, c, d} ∩ B= {b, c, g}

= {b, c}

A∩C= A={ a,b,c,d } ∩ C={ b,c,k,m }.

= {b, c}

(A∩B) U (A∩C) = {b, c}

A∩ (BUC) = (A∩B) U (A∩C)


Example 2 – Distributive Law Definition:

A= {2, 7, 9} B= {5, 10, 11} C= {5, 8, 10}

What is AU (B∩C)

(B∩C):

= {5, 10, 11} ∩ {5, 8, 10}

= {5, 10}

AU (B∩C):

= {2, 7, 9} U {5, 10}

= {2, 7, 9, 5, 10}

These are examples for distributive law definition.

That’s all about distributive law definition.

Resultant of three Vectors

This article is about resultant of three vectors. Resultant of three vectors is vectors that results from adding two or more vectors together. There are two ways to calculate the resultant vector. There are many different websites to help the students on obtaining the resultant of three vectors. Tutor vista is the famous website to provide help on finding the resultant of three vectors with the help of highly qualified tutors. Below are some of the problems regarding resultant of three vectors.

Methods - Resultant of three Vectors:

There are two different methods to find the resultant vector.



The head to tail method:

The head to tail method is represented with the arrow mark. The arrow mark is the head and the tail is with out the arrow mark. Place the two vectors such that the head of the vectors joins the tail of another vector. Draw the resultant vector as shown in the figure above. To find the resultant vectors we use the Pythagorean theorem. There is also another method to calculate the resultant vector which is studied in college grade.

Parallelogram Method:

The another method is parallelogram method. To use this parallelogram method it is very important to be well know with trigonometry basics.


Example Problems - Resultant of three Vectors

vector addition is also the resultant of 3 vectors.

Example 1: The vector a = 3 and vector b = 4. Find the resultant vector using Pythagorean theorem.

Solution

We know that Pythagorean theorem is a2+b2 = c2

So given a =3 and b = 4

So the resultant vector c2 = a2 + b2

c =` sqrt (a^2 + b^2)`

= `sqrt (3^2 + 4^2)`

=` sqrt (9+16)`

= `sqrt(25)`

c = 5

So the resultant vector is 5


Example 2: Add 5`veci` +4`vecj`+3`veck` with 2`veci`+1`vecj`+0`veck`

Solution

Here two set of vectors are given. we have to perform the sum of two vectors by adding the magnitudes alone

5`veci` + 4`vecj` +3`veck`
2`veci` + 1`vecj` +0`veck`
----------------------
7`veci` + 5`vecj` + 3`veck`
-----------------------
Example 3: `veca =4vecp + 3vecq+2vecr` , `vec b = 5vecp +2vecq+vecr` and `vecc = 2 vecp + vecq + 4 vecr`
Solution
Resultant vector is `veca + vecb + vecc`
`4 vecp + 3vecq+2 vecr`
`5 vecp + 2vecq+ vecr`
`2 vecp + vecq + 4 vecr`
---------------------------------
`11 vecp +6 vecq + 7 vecr`
---------------------------------

Various Types of Graphs

In mathematics, a graph is an abstract representation of a set of objects where some pairs of the objects are connected by links. The interconnected objects are represented by mathematical abstractions called vertices, and the links that connect some pairs of vertices are called edges. Typically, a graph is depicted in diagrammatic form as a set of dots for the vertices, joined by lines or curves for the edges. (Source: Wikipedia)

Detail Explanation of Various Types of Graphs

various types of graphs:

Line graphs: A line graph is the way of to representing the two pieces of information, which is usually related with respect to each other. This is more useful when comparisons are needed.



Pie Charts: A pie chart or (circle graphs) are normally used in showcasing a wholesome quantity; we have to show that how in this whole quantities are broken into parts.



Bar Charts: Bar chart is a type of chart, which has labeled horizontal or vertical bars showing a piece of information and an axis



Area Graphs are used to view the changes, how it get changed with respect to time. An area graph shows the contribution of to each data series in this form of a picture.




Another Type of the Problem for Various Types of Graphs


Coordinate type graph:

Plot the various types of graphs in this equation

1. y=x+6

2. y=x2

Solution

Problem 1: Make a graph with this equation

y =x+6

Substitute x=-2,-1, 0,1,2,3

x=-2 y= (-2)+6

By solving it we get

y=4

The ordered pair is (-2, 4)

x=-1 y= (-1)+6

By solving it we get

y=5

The ordered pair is (-1,5)

Similarly

x=0, we get y= 6

x=1, we get y=7

x=2, we get y=8

The co ordinate’s points are (-2,4),(-1,5),(0,6),(1,7), and (2,8)

Plot the table use this various coordinates

 

x	y
-2	4
-1	5
0	6
1	7
2	8

Problem 2: Make a graph with this equation

y =x2

Substitute x=-2,-1, 0,1,2,3

x=-2 y= (-2)2

By solving it we get

y=4

The ordered pair is (-2, 4)

x=-1 y= (-1)2

By solving it we get

y=1

The ordered pair is (-1,1)

Similarly

x=0, we get y= 0

x=1, we get y=1

x=2, we get y=4

The co ordinate’s points are (-2,4),(-1,1),(0,0),(1,1), and (2,4)

Plot the table use this various coordinates

x	y
-2	4
-1	1
0	0
1	1
2	4

Plot the graph for this various tables

Cube Root Formula

In math, a cube root of a value, represents `root(3)(x)` or x1/3, is a value a such that a3 = x. In general the cube root formula is defined as a value x are the numbers y which satisfy the following formula. By using this formula we can solve the cube root problems. Let, see some of the examples of cube root. This article very much helpful to you to solve the problem about the cube root. The cube root problem is used on many places and it is one of the basic mathematical operation.

Examples of Cube Root Formula:

All real numbers contains perfectly one real cube root and a duo of complex conjugate roots, and every nonzero complex values contains three distinct complex cube roots.

Example problem1:

Find the cube root of 8.

Solution:

The given problem can be written as: `root(3)(8)=2` since `2*2*2=2^3=8.`

Example problem2:

Find the cube root of 1.

Solution:

The given problem can be written as: `root(3)(1)=1` since `1*1*1=1^3=1.`

Example problem3:

Find the cube root of 27.

Solution:

The given problem can be written as: `root(3)(27)=3` since `3*3*3=3^3=27.`

Example problem4:

Find the cube root of -8.

Solution:

The given problem can be written as: `root(3)(-8)=-2` since `-2xx-2xx-2=-2^3=-8.`

Example problem5:

Find the cube root of -792.

Solution:

The given problem can be written as: `root(3)(-792)=-9.25213002`

Since, `-9.25213002xx-9.25213002xx-9.25213002=-9.25213002^3=-792.`

Practice Problems of Cube Root Formula:

Problem 1:

Find the cub root of 81.

Solution:

4.32674871

Problem 2:

Find the cub root of 100.

Solution:

4.64158883

Problem 3:

Find the cub root of 125.

Solution:

5

Problem 4:

Find the cub root of 343.

Solution:

7

Problem 5:

Find the cub root of 1000.

Solution:

10

Percentage to Grade Conversion

The percentage of the number is a method of showing the number with the denominator 100 as fraction. The percentage of the number is represented by the symbol “%” or “pct”. Percentage shows the relation of the two quantities, the first quantity is associated with the second quantity. So, first quantity should be larger than zero

Grades are the measurements which are standardized for understanding the academic level of the students. Percentage and its grades are listed below.

Percentage to Grade Conversion Table

Percent to grade point table
Percent	Grade points	Letter grade
0.1	0	F
0.25	0	F
0.26	0	F
0.27	0	F
0.28	0	F
0.29	0	F
0.3	0	F
0.31	0	F
0.32	0	F
0.33	0	F
0.34	0	F
0.35	0	F
0.36	0	F
0.37	0	F
0.38	0	F
0.39	0	F
0.4	0	F
0.41	0	F
0.42	0	F
0.43	0	F
0.44	0.1	F
0.45	0.2	F
0.46	0.3	F
0.47	0.4	F
0.48	0.5	F
0.49	0.6	F
0.5	0.7	D-
0.51	0.8	D-
0.52	0.9	D
0.53	1	D
0.54	1.1	D
0.55	1.2	D+
0.56	1.3	D+
0.57	1.4	D+
0.58	1.5	C-
0.59	1.6	C-
0.6	1.7	C-
0.61	1.8	C-
0.62	1.9	C
0.63	2	C
0.64	2.1	C
0.65	2.2	C+
0.66	2.3	C+
0.67	2.3	C+
0.68	2.4	C+
0.69	2.4	C+
0.7	2.5	B-
0.71	2.5	B-
0.72	2.6	B-
0.73	2.6	B-
0.74	2.7	B-
0.75	2.7	B-
0.76	2.8	B-
0.77	2.8	B-
0.78	2.9	B
0.79	2.9	B
0.8	3	B
0.81	3	B
0.82	3.1	B
0.83	3.1	B
0.84	3.2	B+
0.85	3.2	B+
0.86	3.3	B+
0.87	3.3	B+
0.88	3.4	B+
0.89	3.4	B+
0.9	3.5	A-
0.91	3.5	A-
0.92	3.6	A-
0.93	3.6	A-
0.94	3.7	A-
0.95	3.7	A-
0.96	3.8	A-
0.97	3.8	A-
0.98	3.9	A
0.99	3.9	A
1	4	A

Example to Percentage to Grade Conversion:

Example 1:

John scored the marks in five subjects 87, 90, 88, 76 and 65. The test is conducted for 100 marks in each subject. Find the grade of the john.

Solution:

Total marks of the test = 5 `xx ` 100 = 500

Total marks of the john = 87 + 90 + 88 + 76 + 65 = 406

Percentage of the marks scored by john = `406/500` `xx` 100 = 81.2% or `81.2/100` = .812

Therefore john got B grade


Example 2:

Martin scored the marks in five subjects 85, 95, 65, 75 and 55. The test is conducted for 100 marks in each subject. Find the grade of the Martin.

Solution:

Total marks of the test = 5 `xx` 100 = 500

Total marks of the Martin = 85 + 95 + 65 + 75 + 55 = 375

Percentage of the marks scored by Martin = `375/500` `xx` 100 = 75% or `75/100` = .75

Therefore Martin got B- grade

Practice Problems to Percentage to Grade Conversion:

Problem 1:

Paul scored the marks in five subjects 85, 98, 58, 67 and 94. The test is conducted for 100 marks in each subject. Find the grade of the Paul.

Solution is, Paul got B grade

Problem 2:

Steve scored the marks in five subjects 76, 86, 96, 66 and 56. The test is conducted for 100 marks in each subject. Find the grade of the Steve.

Solution is, Steve got B- grade