Algebra is one of the building blocks of Mathematics in __IIT JEE examination__. In fact, in the preparation of JEE this is the starting point. Algebra is a very scoring and an easy portion in the Mathematics syllabus of JEE. Though Algebra begins with Sets and Relations but we seldom get any direct question from this portion. Functions can be said to be a prerequisite to Calculus and hence it is critical in __IIT JEE preparation__. Sequence and series is another section which is mixed with other concepts and then asked in the examination. Quadratic equations fetch direct questions too and are also easy to grasp. Binomial Theorem is also a marks fetching topic as the questions on this topic are quite easy. Permutations and Combinations along with Probability is the most important section in Algebra. IIT JEE exam fetches a lot of questions on them. Those who get good IIT JEE rank always do well in this section. Complex Numbers are also important as this fetches question in the IIT JEE exam almost every year. Matrices and Determinant mostly give direct question and there are no twist and turns in the questions based on them.

We cover some of these contents here in brief as they have been discussed in detail in the coming sections:

** Sets:** A set is a well-defined collection of distinct objects. The different members of a set are called the elements of the set. Moreover all the elements are unique.

Example: {1, 2, 3, 6} represents a set of numbers less than 10. Note that there is no repetition of elements in a set.

** Relations and Functions:** A relation is a set of ordered pairs. A

**function**is a relation for which each value from the set the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.

The first elements in the ordered pairs i.e. the x values form the domain while the second elements i.e. the y-values form the range. But, only the elements used by the relation are counted in the range.

The figure given below describes a mapping which shows a relation from set A to set B. As is clear from the figure, the relation consists of ordered pairs (1, 2), (3, 2), (5, 7), and (9, 8). The domain is the set {1, 3, 5, 9} and the range which is the dependent variable is the set {2, 7, 8}. Note that 3, 5, 6 are not a part of the range as they are not associated to any member of the first set.

**Example:** f(x) = x/2 is a function because for every value of x we get another value x/2, so f(2) = 1

f (3) = 3/2

f (-6) = -3.

**Quadratic Equations: **An equation of the form ax^{2} + bx +c = 0

where x represents the variable and a,b,c are constants with a not equal to zero.

If a = 0, the equation becomes linear and is no more a quadratic equation. An equation must have a second degree term in order to be a quadratic equation.

We give the formula of solving the quadratic equation:

The general form of quadratic equation is ax^{2}+bx+ c

Dividing by a, we obtain

x^{2} + b x/a = -c /a

(x+ b/2a)^{2} = -c/a + b^{2} / 4a^{2} = (b^{2}-4ac) / 4a^{2}

x+ b/2a = Â±âˆšb^{2}-4ac / 2a

Solving this for x we get

x = (-bÂ±âˆšb^{2}-4ac) / 2a

The above equation is called the quadratic formula.

**Binomial Theorem: **The binomial theorem is used for expanding binomial expressions (a+b) raised to any given power without direct multiplication. Mathematically a binomial theorem can be defined as the theorem that gives the expansion of any binomial raised to a positive integral power say n. Such an expansion contains (n+1) terms.

The general expression for it is

(* x* + *a*)* ^{n}* =

*x*+

^{n}*nx*

^{n-1}

*a*+ [

*n*(

*n*-1)/2]

*x*

^{n}^{-2}

*a*

^{2}+â€¦+ (

*)*

^{n}_{k}*x*

^{n}^{–}

*+ â€¦ +*

^{k}a^{k}*a*, where (

^{n}*) =*

^{n}_{k}*n*!/(

*n-k*)!

*k*!, the number of combinations of

*k*items selected from

*n.*

** Permutation:** Permutation is an ordered arrangement of the numbers, terms, etc., of a set into specified groups. The number of permutations of n objects taken r at a time is given by n! / (n-r)!

The permutations of a, b, and c, taken two at a time, are ab, ba, ac, ca, bc, cb.

** Combinations: **A combination is also a way selecting certain things out of a larger group. But here order does not matter unlike permutation.

For example: If we need to form combination of two out f given three balls as in the previous case, only three cases are possible. As in the previous figure the first two constitute the same case in combination. Similarly the middle two and the last two also represent the same case.

The number of combinations of n objects taken r at a time is given by n! / r! (n-r)!

**Complex Numbers: **A complex number is a combination of a real and a imaginary number. They are written as a+ib, where a and b are real and I is an imaginary number with value âˆš-1. Even 0 is also a complex number as 0 = 0 + 0i. Examples: 1+i, 2-3i, 6i, 3.

Trigonometry is a vital constituent of Mathematics in __IIT JEE examination__. Trigonometric functions and trigonometry ratios are some of the most imperative areas of trigonometry. Since the IIT JEE exam asks a good amount of questions on this topic, so getting the knack of the basic trigonometry can surely help an IIT JEE aspirant to smooth his way through the exam.

Trigonometry is the branch of Mathematics that deals with triangles and the relationships between the sides and angles. The trigonometry functions are universal in parts of pure mathematics and applied mathematics which also lay the groundwork for many branches of science and technology. The IIT JEE Trigonometry problems range from the trigonometry basics to the applications of trigonometry.

Some of the trigonometry questions are simply based on trigonometry formulae and are quite easy to crack while others may demand some trigonometry tricks. Thus IIT JEE trigonometry syllabus is a perfect blend of questions of all levels

Given below is the trigonometry table that illustrates the values of the functions at different angles:

It would be an added advantage if the aspirants could memorize all above trigonometry formulas but if not; they must at least grasp the major ones like:

The graphs also constitute a vital component in Trigonometry. A student who is well versed with the graphs of the major functions is able to tackle questions with ease. Some of the fundamental graphs are sketched below:

Enlisted below are some of the prime heads that come under trigonometry and are covered in the coming sections:

- Multiple and Sub-multiple Angles
- Trigonometric Equations
- Inverse circular Functions
- Trigonometric Inequality
- Trigonometric Functions
- Trigonometric Identities and Equations

**For more in depth knowledge refer**

**Illustration 1:**

Find the angles and sides indicated by the letters in the diagram. Give each answer correct to the nearest whole number.

Since in general, it is simpler to use the value of 30Â°, so we will have to consider the triangle with 30Â°. Here,

r/60 = tan 30Â°.

Hence, r = 60tan (30Â°) = 34.64101615…

So answer correct to the nearest whole number is r= 35.

Now, that we have obtained the value of r, we use the first triangle to evaluate s.

*r/s = tan(55Â°)*

so, 35/s = tan(55Â°)

so, 35/tan(55Â°) = s = 24.50726384…

**r = 35, s = 25**

**Illustration 2:**

Evaluate the value of (1 + cos Ï€/8) (1 + cos 3Ï€/8) (1 + cos 5Ï€/8) (1 + cos 7Ï€/8).

Solution: The given expression is

(1 + cos Ï€/8) (1 + cos 3Ï€/8) (1 + cos 5Ï€/8) (1 + cos 7Ï€/8)

= (1 + cos Ï€/8) (1 + cos 3Ï€/8) (1 – cos 3Ï€/8) (1 – cos Ï€/8)

= (1 â€“ cos^{2}Ï€/8) (1 – cos^{2}3Ï€/8)

= Â¼ [2 sin Ï€/8 sin 3Ï€/8]^{2}

= Â¼ [2 sin Ï€/8 cos Ï€/8]^{2}

= Â¼ [sin Ï€/4]^{2}

= 1/4. 1/2 = 1/8

**Illustration 3**: If cos (Î± â€“ Î²) = 1 and cos (Î± + Î²) = 1/e, where Î±, Î² âˆˆ [-Ï€, Ï€], then the values of Î± and Î² satisfying both the equations is/are

Solution: It is given in the question that cos (Î± â€“ Î²) = 1 and cos (Î± + Î²) = 1/e, where Î±, Î² âˆˆ [-Ï€, Ï€].

Now, cos (Î± â€“ Î²) = 1

Î± â€“ Î² = 0, 2Ï€, -2Ï€

Hence, Î± â€“ Î² = 0 (since Î±, Î² âˆˆ [-Ï€, Ï€])

So, Î± = Î²

Hence, cos 2Î± = 1/e

So, the number of solutions of above will be number of points of intersection of the curves y = cos 2Î± and y = 1/e

where Î±, Î² âˆˆ [-Ï€, Ï€]

It is quite clear that there are four solutions corresponding to four points of intersection P_{1}, P_{2}, P_{3} and P_{4}.

**Illustration 4**: If k = sin (Ï€/18) sin (5Ï€/18) sin (7Ï€/18), then what is the numerical value of k?

Solution: The value of k is given to be k = sin (Ï€/18) sin (5Ï€/18) sin (7Ï€/18).

Hence, k = sin 10Â° sin 50Â° sin 70Â°

= sin 10Â° sin (60Â° – 10Â°). sin (60Â° + 10Â°)

= sin 10Â° [sin^{2}60Â° – sin^{2}10Â°]

= sin 10Â° [(âˆš3/2)^{2} – sin^{2}10Â°]

= sin 10Â° [3/4 – sin^{2}10Â°]

= 1/4 [3sin 10Â° – 4sin^{3}10Â°]

= 1/4 x sin (3 x 10) (since, sin 3Î¸ = 3sinÎ¸- 4sin^{3} Î¸)

= 1/4 sin 30Â° = 1/8

Hence, the numerical value of k is 1/8

Co-ordinate Geometry is a method of analyzing geometrical shapes. Co-ordinate Geometry is one of the most scoring topics of the mathematics syllabus of IIT JEE and other engineering exams. Besides calculus, this is the only topic that can fetch you maximum marks. It is a vast topic and can further be divided into various parts like:

Â· Circle Â· Parabola Â· Ellipse Â· Hyperbola Â· Straight Lines

All these topics hold great importance from examination point of view but the Straight Line and the Circle are the most important. These topics together fetch maximum questions in the JEE and moreover they are a pre requisite to conic sections as well.

**The Coordinate Plane**

In Coordinate geometry, points are placed on the coordinate plane. The horizontal line is the x-axis while the vertical line is the y-axis. The point where they cross each other is called the origin.

A point’s location on the plane is given by two numbers, the first tells where it is on the x-axis and the second tells where it is on the y-axis. Together, they define a single, unique position on the plane.

In the figure below, we have plotted the point (20, 15). Note here that the order is important as the first in the pair always stands for the x coordinate. Sometimes these are also referred to as the rectangular coordinates.

We have listed some of the important facts here, but the rest have been covered in detail in the later sections:

__The Midpoint of a Line Joining Two Points__

__The Midpoint of a Line Joining Two Points__

The midpoint of the line joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is: [Â½(x_{1} + x_{2}), Â½(y_{1} + y_{2})] **Illustration:** Find the coordinates of the midpoint of the line joining (11, 2) and (3, 4). Midpoint = [Â½(11+ 3), Â½(2 + 4)] = (7, 3) You may refer the maths past papers to get an idea about the type of questions asked.

__The Gradient of a Line Joining Two Points__

__The Gradient of a Line Joining Two Points__

The gradient of a line joining two points is given by

y_{2}-y_{1 }âˆ• x_{2}-x_{1}

__Parallel and Perpendicular Lines__

__Parallel and Perpendicular Lines__

Two parallel lines have the same gradient while if two lines are perpendicular then the product of their gradient is -1. * Example* a) y = 4x + 1 b) y = –

^{1}/

_{4}x + 12 c) Â½y = 2x – 3 The gradients of the lines are 4, –

^{1}/

_{4}and 4 respectively. Hence, as stated above lines (a) and (b) are perpendicular, (b) and (c) are perpendicular and (a) and (c) are parallel. The figure below shows the various topics included in the conic sections:

Generally, conic section includes ellipse, parabola and hyperbola but sometimes circle is also included in conic sections. Circle can actually be considered as a type of ellipse. When a cone and a plane intersect and their intersection is in the form of a closed curve, it leads to the formation of circle and ellipse. As it is visible from the figure above, the circle is obtained when the cutting plane is parallel to the plane of generating circle of the cone. Similarly, in case the cutting plane is parallel to one generating line of the cone, the resulting conic is unbounded and is called parabola. The last case in which both the halves of the cone are intersected by the plane, produce two distinct unbounded curves called hyperbola. Conic section is a vital organ of coordinate geometry. It is easy to gain marks in this section as there are some standard questions asked from this section so they can be easily dealt with. Some topics no doubt are difficult but can be mastered with continuous practice. The importance of Co-ordinate Geometry lies in the fact that almost all the students who aspire to get high All India rank in IIT JEE, AIEEE, DCE, EAMCET and other engineering entrance examinations give a good emphasis on Co-ordinate Geometry.

Differential Calculus is one of the most important topics in the preparation of IIT JEE. This is the easiest part of Calculus and there is no doubt in the fact that it is scoring too. It is also important to attain proficiency in Differential Calculus as it is a prerequisite to the learning of Integral Calculus too. Differential Calculus is a branch of mathematical analysis which deals with the problem of finding the rate of change of a function with respect to the variable on which it depends. So, differential calculus is basically concerned with the calculation of derivatives for using them in problems involving non constant rates of change. Applications also include computation of maximum and minimum values of a function. The study of Differential Calculus includes **Functions, Sets and Relations** though they are considered to be a part of Algebra. **Limits, Continuity and Differentiability** are the favorite topics of those who have a bent towards Differential Calculus. It is not only an easy topic but also fetches direct question in the examination. A person who has already done a good practice of this chapter is also likely to do well in the next topic of **Differentiation**. The lifeline of Differential Calculus is basically the topics which include the application of Derivatives i.e.**Tangent and Normal **and** Maxima and Minima. **Differential calculus is closely related to integral calculus. In fact, differentiation is the reverse process of integration. Here we shall discuss the main heads that are counted under differential calculus. These topics have been discussed in detail in the coming sections: ** Relation: **A **relation **is a set of ordered pairs and is usually defined by a rule. The domain of a relation is the set of all first elements (usually *x* values) of its ordered pairs. The range of a relation is the set of all second elements (usually* y* values) of its ordered pairs. **Function: **A **function** is a relation for which each value from the set of the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair. Here is an example of a function:

It is an important example as students are likely to get confused as to whether it is a function or not. As stated in the definition, since each value of x is associated with exactly one value of y (though in this case that y is same for every x) hence it is a function.

Example: Consider f to be defined by {(0,0),(1,1),(2,2),(3,3),(1,2),(2,3),(3,1),(2,1),(3,2),(1,3)} This is a relation (not a function) since we can observe that 1 maps to 2 and 3, for instance. **Limit of a function: ** Suppose *f* : **R** â†’ **R** is defined on the real line and *p,L* âˆˆ **R**. Then we say that the limit of the function f is l if For every real *Îµ* > 0, there exists a real *Î´* > 0 such that for all real x, 0 < | *x* âˆ’ *p* | < *Î´* implies | *f*(*x*) âˆ’ *L* | < *Îµ*. Mathematically it is represented as **Example:** Find the limit of the function f(x) = (x^{2}-6x + 8) / x-4, as xâ†’5. **Solution:** The limit is 3, because *f* (5) = 3 and this function is continuous at *x* = 5. Such easy questions are not asked in the exam so it was just meant to clear the concept of limit. We now move on to a bit difficult question: **Example:** Find the limit of the function g(x) = âˆš(x-4) -3 / (x-13) as x approaches 13. **Solution:** In such a question you first need to reduce the function in simple form so that the computation of limit becomes simple.First rationalize the numerator and denominator by multiplying by its conjugate So, [âˆš(x-4) -3] / (x-13) x [âˆš(x-4) +3]/ [âˆš(x-4) +3] On multiplying and solving we get the result, 1/ âˆš( x-4) +3 Now since the terms have been simplified the limit can now be calculated by substituting the value of x as 13. Hence putting x=13 in the last equation we get the limit as 1/6. For more on limits, you may refer the video **Continuity of a function**: A function *y* = *f*(*x*) is continuous at point *x*=*a* if the following three conditions are satisfied: (1) *f*(*a*) is defined , (2) exists (i.e., is finite) , (3) A function is continuous when its graph is a single unbroken curve. This definition proves to be useful when it is possible to draw the graph of a function so that just by the graph the continuity of the function can be judged.

**Differentiation: **Differentiation is an operation that allows us to find a function that outputs the rate of change of one variable with respect to another variable. The derivative of a function at a chosen input value describes the rate of change of the function near that input value. The process of finding a derivative is called differentiation. Geometrically, the derivative at a point is the slope of the tangent line to the graph of the function at that point, provided that the derivative exists and is defined at that point.Since Differential Calculus is new to the students as they do not study it in their 10th standard examination, so they are advised to master the topic by practicing questions on Limits, Continuity and Differentiability. The preparation of Differential Calculus also gives another opportunity to prepare and revise the chapter on Functions, Sets and Relations. To get an idea about the types of questions asked you may also consult the Previous Year Papers.

Integral Calculus is the part of calculus that deals with integration and its application in the solution of differential equations and in determining areas or volumes etc. It involves problems like solution of area and volume. In the figure given above, integral calculus may be used to find out the area between the two curves F_{1} and F_{2}. It would not be an exaggeration if we call Calculus to be the castle of Mathematics and its most important part is the integral calculus. The topic is quite scoring and has a good weightage in exams like IIT JEE, AIEEE etc. but it can be mastered only through constant practice. Definite integrals and Indefinite integrals are the major tools which are used for applications under the topic of Area. Calculation of area is considered a hard topic but since it fetches many questions in JEE so it must be given due importance. **Indefinite Integral: ** An integral of the form i.e. without upper and lower limits is called an indefinite integral. Indefinite integrals are often written as where c is an arbitrary constant. Indefinite integrals are also called Antiderivatives. **Note**: The constant of integration is very important. In fact, the graphs of Antiderivatives of a function are vertical translations of each other, the location of each graph depending on the value of c. Consider the following example: The function *F*(*x*) = *x*^{3}/3 is an antiderivative of *f*(*x*) = *x*^{2}. As the derivative of constant is zero, *x*^{2} will have an infinite number of antiderivatives; such as (*x*^{3}/3) + 0, (*x*^{3}/3) + 7, (*x*^{3}/3) âˆ’ 42, (*x*^{3}/3) + 293 etc. Thus, all the antiderivatives of *x*^{2} can be obtained by changing the value of C in *F*(*x*) = (*x*^{3}/3) +C **Definite Integral**:Given a function that is continuous on the interval [*a,b*] we divide the interval into *n* subintervals of equal width, , and from each interval choose a point x_{1}* Then the **definite integral of ****f(x)**** from ****a**** to *** b* is

**Properties of Definite Integral:**We list below various important rules that form the basis of solving numerical in definite integral: (1) .The limits on any definite integral can be interchanged. You just need to add a minus sign on the integral while doing so. (2) .When the upper and lower limits coincide, the value of the integral is zero. (3), where

*c*is any number. A constant can be taken out of the integral sign in case of both definite and indefinite integral. (4).A definite integral can be broken into parts across a sum or difference. (5) where

*c*is any number. This property tells us how to integrate a function over adjacent intervals, [

*a,c*] and [

*c,b*]. Note that it is not necessary for c to be between a and b. (6) This property shows that as long as the function and limits are same, the variable used for integration does not make any difference.

**Illustration 1:**Let f(x) = x/(1+x

^{n})

^{1/n}for n â‰¥ 2 and g(x) = (fofofofofâ€¦ n times) (x). Then find the value of âˆ«x

^{n-2}g(x) dx.

**Solution:**The value of f(x) is given to be f(x) = x/(1+x

^{n})

^{1/n}Then ff(x) = f(x)/(1 + f(x)

^{n})

^{1/n}= x/(1+2x

^{n})

^{1/n}So, fff(x) = x/(1+3x

^{n})

^{1/n}Hence, g(x) = (fofofâ€¦..n times)(x) = x/(1+nx

^{n})

^{1/n}Hence, I = âˆ«x

^{n-2}g(x) dx = âˆ« x

^{n-1}dx/(1 + nx

^{n})

^{1/n}= 1/n

^{2}âˆ« n

^{2}x

^{n-1}dx/(1 + nx

^{n})

^{1/n}= 1/n

^{2 }âˆ« [d/dx (1 + nx

^{n})] / (1 + nx

^{n})

^{1/n}. dx Hence, I = [(1+ nx

^{n})

^{1-1/n}]/ n(n-1) + K.

**Illustration 2:**Evaluate the given expression âˆ« dx/ [ x

^{2}(x

^{4}+ 1)]

^{3/4}

**Solution:**Let I = âˆ« dx/ [x

^{2}(x

^{4}+ 1)]

^{3/4}= âˆ« dx/ x

^{2}x

^{3}(1+x

^{-4})

^{3/4}= âˆ« dx/ x

^{5}(1+x

^{-4}) Put 1+x

^{-4}= z Then -4x

^{-5}dx = dz Hence, I = âˆ« -dz/4z

^{3/4}= -1/4 âˆ« dz/z

^{3/4}= -z

^{1/4}+ c

**Illustration 3:**Evaluate âˆ« (x +1)/ x(1+xe

^{x})

^{2}dx

**.**

**Solution:**Let I

**=**âˆ« (x +1)/ x(1+xe

^{x})

^{2}dx

**=**âˆ« e

^{x}(x +1)/ xe

^{x}(1+xe

^{x})

^{2}dx Put (1+xe

^{x}) = t, then (e

^{x}+xe

^{x}) dx = dt Hence, I = âˆ« dt/ (t-1)t

^{2}= âˆ«[1/(t-1) â€“ 1/t â€“ 1/t

^{2}] dt = log |t-1| – log |t| +1/t + c = log | [(t-1)/t] | +1/t + c = log | xe

^{x}/ (1+xe

^{x})| + 1/ (1+xe

^{x}) + c Most of the students are weak in Integral calculus but this weakness must be converted into strength with regular practice. All the high rank holders in JEE have their expertise in Integral calculus.

Vectors constitute an important topic in the Mathematics syllabus of JEE. It is important to master this topic to remain competitive in the IIT JEE. It often fetches some direct questions too. The topic of Vectors is quite simple and it also forms the basis of several other topics. Various topics that have been covered in this chapter include: Â· Introduction Â· Addition of Two Vectors Â· Fundamental Theorem of Vectors Â· Orthogonal System of Vectors We now discuss some of these topics in brief as they have been covered in detail in the coming sections: **Length / Magnitude of a vector:** The **length** or **magnitude** of the vector **w **= (a, b, c) is defined as |w|= w= âˆša^{2}+b^{2}+c^{2} **Unit vectors:** A unit vector is a vector of unit length. A unit vector is sometimes denoted by replacing the arrow on a vector with a “^” or just adding a “^” on a boldfaced character (i.e., È‡ or **È‡**). Therefore, | **È‡| = 1.** Any vector can be made into a unit vector by dividing it by its length. **È‡ = u / |u|** Any vector can be fully represented by providing its magnitude and a unit vector along its direction. A vector can be written as u = u**È‡**

Base vectors and vector components: Base vectors represent those vectors which are selected as a base to represent all other vectors. For example the vector in the figure can be written as the sum of the three vectors u_{1}, u_{2}, and u_{3}, each along the direction of one of the base vectors e_{1}, e_{2}, and e_{3}, so that u= u_{1}+u_{2}+u_{3}

It is clear from the figure that each of the vectors u_{1}, u_{2} and u_{3} is parallel to one of the base vectors and can be written as a scalar multiple of that base. Let *u*_{1}, *u*_{2}, and *u*_{3} denote these scalar multipliers such that one has u_{1}= u_{1}e_{1} u_{2}=u_{2}e_{2} u_{3}=u_{3}e_{3}

The original vector u can now be written as u= u_{1}e_{1}+u_{2}e_{2}+u_{3}e_{3} Negative of a vector: **A negative vector is a vector that has the opposite direction to the reference positive direction.**

**A vector connecting two points: **

The vector connecting point *A *to point *B* is given by **r**= (x_{B}-x_{A}) i+ (y_{B} â€“ y_{A}) j + (z_{B} â€“ z_{A})k , here i, j and k denote the unit vectors along x, y and z axis respectively. **Some Key Points ** **â€¢ The magnitude of a vector is a scalar and scalars are denoted by normal letters. ** **â€¢ Vertical bars surrounding a boldface letter denote the magnitude of a vector. Since the magnitude is a scalar, it can also be denoted by a normal letter; |w| = w denotes the magnitude of a vector ** **â€¢ The vectors are denoted by either drawing a arrow above the letters or by boldfaced letters. ** **â€¢ Vectors can be multiplied by a scalar. The result is another vector. ** **Suppose c is a scalar and v = (a, b) is a vector, then the scalar multiplication is defined by cv= c (a,b)= (ca,cb). Hence each component of a vector is multiplied by the scalar. ** **â€¢ If two vectors are of the same dimension then they can be added or subtracted from each other. The result is gain a vector. ** **Then the sum of these two vectors is defined by** ** v + u = (a + e, b + f, c + g). ** **â€¢ We can also subtract two vectors of the same direction. The result is again a vector. As in the previous case subtracting vector u from v yields ** **v – u = (a – e, b – f, c – g). the difference of these vectors is actually the vector v – u = v + (-1)u. ** **Some Basic Rules of Vectors** If **u**, **v **and **w** are three vectors and c, d are scalars then the following hold true:

Â·u + v = v + u(the commutative law of addition)

Â·u + 0 = u

Â·u + (-u) = 0(existence of additive inverses)

Â· c(du) = (cd)u

Â·(c + d)u =cu +du

Â· c(u + v) =cu +cv

Â· 1u = u

Â·u + (v + w) = (u + v) + w(the associative law of addition)