JEE (Ad.) Roll No.2003111, CBSE Roll No.9111240 OP … Of Formulae By OP Gupta Page - [1] u Standard Formulae Of Integrals: A. 1, 1 1 n xdx k nn x n B. 1 dx x klog x

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LIST OF

FORMULAE [ For Class XII ]

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u

Standard Formulae Of Integrals:

A. 1

, 11

n

n xx dx k n

n

B. 1

logdx x kx

C. 1

logx xa dx a k

a

D. 1ax axe dx e ka

E. 1

sin cosax dx ax ka

F. 1

cos sinax dx ax ka

G. tan log secxdx x k

OR log cos x k

H. cot log sinxdx x k

OR log cosec x k

I. sec log sec tanxdx x x k

OR log tan4 2

xk

J. cosec log cosec cot xdx x x k

OR log tan2

xk

K. 2sec tanxdx x k

L. 2cosec cotxdx x k

M. sec . tan sec x xdx x c

N. cosec .cot cosec x xdx x k

O. 2

11sec

1dx x k

x x

P. 2 2

11 1tan

xdx k

a x a a

Q. 2 2

1 1log

2

a xdx k

a x a a x

R. 2 2

1 1log

2

x adx k

x a a x a

S. 2 2

2 2

1logdx x x a k

x a

T. 2 2

2 2

1logdx x x a k

x a

U. 2 2

11sin

xdx k

aa x

V. 2

2 2 2 2 2 2log2 2

x a

x a dx x a x x a k

W. 2

2 2 2 2 2 2log2 2

x ax a dx x a x x a k

X. 2

2 2 2 2 1sin2 2

x a xa x dx a x k

a

Y. 1 1

log dx ax b k

ax b a, where ‘a’ is any constant (and obviously, k is the integral constant)

Z. dx x k , where is a constant (and, k is the integral constant).

Integral Calculus Formulae For By OP Gupta [Indira Award Winner, +91-9650 350 480]

List Of Formulae for Class XII By OP Gupta (Electronics & Communications Engineering)

Page - [2] List Of Formulae By OP Gupta

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Methods of Integration:

Though there is no general method for finding the integral of a function, yet here we have considered the following methods based on observations for evaluating the integral of a function:

a) Integration by Substitution Method–

In this method we change the integral f x dx ,

where independent variable is x, to another integral in which independent variable is t (say) different

from x such that x and t are related by x g t .

Let u f x dx then, du

f xdx

.

Again as x g t so, we have dx

g tdt

.

Now . . du du dx

f x g tdt dx dt

On integrating both sides w.r.t. t, we get

dudt f x g t dt

dt

u f g t g t dt

i.e., f x dx f g t g t dt where x g t .

So, it is clear that substituting x g t in

f x dx will give us the same result as obtained

by putting g t in place of x and g t dt in place

of dx .

b) Integration by Partial Fractions–

Consider

f x

g x defines a rational polynomial

function.

If the degree of numerator i.e. f x is

greater than or equal to the degree of

denominator i.e. g x then, this type of rational

function is called an improper rational

function. And if degree of f x is smaller than

the degree of denominator i.e. g x then, this

type of rational function is called a proper rational function.

In rational polynomial functions if the degree (i.e. highest power of the variable) of numerator (Nr) is greater than or equal to the degree of denominator (Dr), then (without any doubt) always perform the division i.e., divide the Nr by Dr before doing anything and thereafter use the following:

Numerator RemainderQuotient

Denominator Denominator .

On doing this, the rational function is resolved into partial fractions. The table shown below lists the types of simpler partial fractions that are to be associated with various kinds of rational functions which will be dealt in our current study:

TABLE DEMONSTRATING PARTIAL FRACTIONS OF VARIOUS FORMS

Form of the Rational Function Form of the Partial Fraction

px q

x a x b, a b

A B

x a x b

2

px q

x a

2

A B

x a x a

2

px qx r

x a x b x c

A B C

x a x b x c

2

2

px qx r

x a x b

2

A B C

x a x bx a

2

2

px qx r

x a x bx c

where 2 x bx c can’t be factorized further.

2

A Bx C

x a x bx c

MATHEMATICS – List Of Formulae for Class XII By OP Gupta (+91-9650350480)



c) Integral By Parts–

If and V be two functions of x then,

I II.

dV dx Vdx Vdx dx

dx

In finding integrals by this method, proper choice of functions and V is crucial. Though there is no

fixed rule for taking and V (their choice is possible by practice) yet, following rule is found to be quite

helpful in deciding the functions and V :

If and V are of different types, take that function as which comes first in the word ILATE .

Here I stands for Inverse trigonometrical function, L stands for Logarithmic function, A stands for Algebraic function, T stands for Trigonometrical function and E stands for the Exponential function.

If both the functions are trigonometrical, take that function as V whose integral is easier.

If both the functions are algebraic, take that function as whose differentiation is easier.

Some integrands are such that they are not product of two functions. Their integrals may be found by integrals by parts taking 1 as the second function. Logarithmic and inverse trigonometric functions are examples of such functions.

The result of integral ( ) ( ) ( ) x xe f x f x dx e f x k can be directly applied in case of the

objective type questions. Making the Perfect Square:

STEP1– Consider the expression 2 ax bx c .

STEP2– Make the coefficient of 2x as unity i.e., 1 by taking a common, after doing so the original

expression will look like, 2

b ca x x

a a.

STEP3– Add and subtract

2

2

b

a to the expression obtained in STEP2 as depicted here i.e.,

2 2

2

2 2

b c b ba x x

a a a a.

STEP4– The perfect square of 2 ax bx c will be

2 2

2 2

b c ba x

a a a.

Various Integral forms:

Integrals of the form 2

px q

dxax bx c

, 2

px qdx

ax bx c, 2 px q ax bx c dx : Express the

numerator px q as shown here, i.e., 2 d

px q A ax bx c Bdx

. Then on, obtain the values of A and

B by equating the coefficients of like powers of x and constants terms on both the sides. Then, integrate it

after replacing px q by 2 d

A ax bx c Bdx

using the values of A and B.

Integrals of the form sin cos

sin cos

a x b x

dxc x d x

: Express Numerator Denominator Denominator d

A Bdx

.

Then obtain the values of A and B by equating the coefficients of sin x and cos x on both the sides and proceed.


sin cos

a x b x cdx

p x q x r

: Note that the previous integral form can be considered as a

special case of this form. Express Numerator Denominator Denominator d

A B Cdx

. Then obtain the




values of unknowns i.e., A, B and C by equating the coefficients of sin x , cos x and the constant terms on both the sides and hence proceed.

Integrals of the form 2 2sin cos

dx

a x b x,

2sindx

a b x,

2cosdx

a b x,

2

sin cos

dx

a x b x and

2 2sin cos dx

a b x c x: Divide the Nr and Dr both by 2cos x . Replace 2sec x , if any, in Dr by 21 tan x

and then put tan tx and proceed.


dx

a x b x,

sindx

a b x,

cosdx

a b x and

sin cosdx

a x b x:

Use 2

2 tan2sin

1 tan2

x

xx

and/ or

2

2

1 tan2cos

1 tan2

x

xx

. Replace 21 tan2

x

in the Nr by 2sec2

x and then put

tan t2

x

and then after proceed.

Integrals of the form 1

M N dx where M and N are linear or quadratic expressions in x :

M N Substitutions

Linear Linear 2t N

Quadratic Linear 2t N

Linear Quadratic 1t

M

Quadratic Quadratic 2 N 1t or t

M

x

A Few Useful Quickies:

a)

1

, 11

n

n f xf x f x dx k n

n b)

log

f x

dx f x kf x

c)

1

1

n

n

f xf xdx k

nf x d)

1

1

1

nn ax b

ax b dx ka n

.

Formulae & Properties Of Definite Integrals:

P.01 F F F b

b

aa

f x dx x b a

P.02 b a

a b

f x dx f x dx

P.03 b b

a a

f x dx f t dt

P.04 , b m b

a a m

f x dx f x dx f x dx a m b




P.05 b b

a a

f x dx f a b x dx

P.06

0

2 , if is even function . .,

0 , if is odd function . .,

a

aa f x dx f x i e f x f x

f x dx

f x i e f x f x

P.07 0 0

2 2

a a a

a

f x dx f x dx f x dx

0 0

2a a

f x dx f a x dx

P.08

0

/2

0

2 , if

0, if

mm f x dx f m x f x

f x dx

f m x f x

Proof Of A Few Important Properties:

P.04 , b m b

a a m

f x dx f x dx f x dx a m b .

PROOF We know, F F F b

b

aa

f x dx x b a …(i)

Consider F F F m

m

aa

f x dx x m a …(ii)

And F F F b

b

mm

f x dx x b m …(iii)

Adding the equations (ii) and (iii), we get

F F m b

a m

f x dx f x dx b a

b

a

f x dx . [ By (i)

Hence, , b m b

a a m

f x dx f x dx f x dx a m b . [H.P.]

P.05 b b

a a

f x dx f a b x dx .

PROOF Consider b

a

f x dx .

Let x a b t dx dt . Also when x a t b and, when x b t a .

So, b a

a b

f x dx f a b t dt

b

a

f a b t dt [By using P.02

b

a

f a b t dt




b a

a b

f x dx f a b x dx [Replacing t by x, P.03

Hence, b b

a a

f x dx f a b x dx . [H.P.]

SPECIAL CASE OF P.05 Take 0a and b a .

Then, 0 0

a a

f x dx f a x dx .

The proof for the special case is same as is for the P.05, so it has been left as an exercise for you!

P.06

0

2 , . .,

0, . .,

a

a

a

f x dx if f x is even function i e f x f xf x dx

if f x is odd function i e f x f x

.

PROOF We know that 0

0

a a

a a

f x dx f x dx f x dx …(i) [By using P.04

Consider 0

a

f x dx .

Let x t dx dt . Also when x a t a and when 0 0 x t .

So, 0 0

a a

f x dx f t dt

0

a

f t dt [By using P.02

0

a

f t dt

0

0

a

a

f x dx f x dt [Replacing t by x, P.03

Therefore equation (i) becomes,

0 0

a a a

a


0

a a

a

f x dx f x f x dx

0

2 ,

0,

a

a

a

f x dx if f x f xf x dx

if f x f x

i.e.,.

0

2 , is even function

0, is odd function

a

a

a

f x dx if f xf x dx

if f x

[H.P.]

P.07 0 0

2 2

a a a

a


0 0

2a a

f x dx f a x dx .




PROOF We know 0 0

2 2

a a a

a

f x dx f x dx f x dx …(i) [By using P.04

Consider 2

a

a

f x dx .

Let 2 x a t dx dt . Also when x a t a and when 2 0 x a t .

So, 2 0

2 a

a a

f x dx f a t dt

0

2

a

f a t dt [By using P.02

0

2 a

f a t dt

2

0

2 a a

a

f x dx f a x dx [Replacing t by x, P.03

So equation (i) becomes,

0 0

2

0

2 a a a

f x dx f x dx f a x dx . [H.P]

P.08

0

0

2 2 , 2

0, 2

aa f x dx if f a x f x

f x dx

if f a x f x

.

PROOF We know 0 0

2 2

a a a

a

f x dx f x dx f x dx …(i)

Consider 2

a

a

f x dx .

Let 2 x a t dx dt . Also when x a t a and when 2 0 x a t .

So, 2 0

2 a

a a

f x dx f a t dt

0

2

a

f a t dt [By using P.02

0

2 a

f a t dt

2

0

2 a a

a

f x dx f a x dx [Replacing t by x, P.03

So equation (i) becomes,

0 0

2

0

2 a a a

f x dx f x dx f a x dx

0

2 a

f x f a x dx




Hence,

0

0

2 2 , 2

0, 2

aa f x dx if f a x f x

f x dx

if f a x f x

. [H.P.]

Definite integral as the Limit Of A Sum (First Principle Of Integrals):

Take that function whose integral value is to be calculated as f x and then use the given relation,

0

lim 2 ... ( 1)b

ha

f x dx h f a f a h f a h f a n h

or, 0

lim 2 3 ...b

ha

f x dx h f a h f a h f a h f a nh

i.e., 1

0

limb n

nra

f x dx h f a rh

, such that as ,n 0h and nh b a

or, 1

limb n

nra

f x dx h f a rh

, such that as ,n 0h and nh b a .




u

Important Terms, Definitions & Formulae

01. Differential Equation - A basic introduction: It is an equation consisting of an independent variable, dependent variable and differential coefficients of dependent variable with respect to the independent variable.

02. Order and Degree of a differential equation:

a) Order of a differential equation: It is the order of the highest order derivative appearing in the differential equation.

b) Degree of a differential equation: It is the degree (power) of the highest order derivative, when the differential coefficients are made free from the radicals and the fractions.

03. Formation of the differential equations: If the equation of the family of curves is given then its differential equation is obtained by eliminating arbitrary constants occurring in its equation with the help of equation of the curve and the equations formed by differentiation of equation of the curve.

ALGORITHM FOR THE FORMATION OF DIFFERENTIAL EQUATION

STEP1- Write down the given equation of the curve. STEP2- Differentiate the given equation with respect to the independent variable as many times as the number of arbitrary constants. STEP3- Eliminate the arbitrary constants by using given equation and the equations obtained by the differentiation in STEP2.

04. Solutions of differential equations:

a) General solution: The solution which contains as many as arbitrary constants as the order of the

differential equation. e.g. cos sin y x x is the general solution of 2

20

d yy

dx.

b) Particular solution: Solution obtained by giving particular values to the arbitrary constants in the general solution of a differential equation is called a particular solution. e.g. 3cos 2sin y x x is a

particular solution of the differential equation 2

20

d yy

dx.

c) Solution of differential equation by Variable Separable Method: A variable separable form of the

differential equation is the one which can be expressed in the form of f x dx g y dy . The solution is

given by f x dx g y dy k where k is the constant of integration.

d) Homogeneous Differential Equations and their solution:

First of all we shall learn how to identify a homogeneous differential equation.

Identifying a Homogeneous differential equation:

STEP1- Write down the given differential equation in the form ,dy

f x ydx

.

STEP2- If , , nf kx ky k f x y then, the given differential equation is homogeneous of degree ‘ n ’.

Solving a homogeneous differential equation:

If ,dy

f x ydx

If ,dx

f x ydy

Put y vx Put x vy

dy dv

v xdx dx

dx dv

v ydy dy

Then after, separate the variables to get the required solution.

Differential Equations Formulae For By OP Gupta [Indira Award Winner, +91-9650 350 480]




e) Solution of Linear differential equations:

Linear differential equation in y : It is of the form P( ) Q( ) dy

x y xdx

where P( )x and Q( )x are

functions of x only. SOLVING OF LINEAR DIFFERENTIAL EQUATION IN y

STEP1- Write the given differential equation in the form P( ) Q( ) dy

x y xdx

.

STEP2- Find the Integration Factor P( )

(I.F.) x dx

e .

STEP3- The solution is given by, .(I.F.) Q( ).(I.F.) y x dx k where k is the constant of integration.

Linear differential equation in x : It is of the form P( ) Q( ) dx

y x ydy

where P( )y and Q( )y are

functions of y only. P( )

(I.F.) x dx

e

SOLVING OF LINEAR DIFFERENTIAL EQUATION IN x

STEP1- Write the given differential equation in the form P( ) Q( ) dx

y x ydy

.

STEP2- Find the Integration Factor P( )

(I.F.) y dy

e .

STEP3- The solution is given by, . (I.F.) Q( ).(I.F.) x y dy k where k is the constant of integration.




u


01. Vector - Basic Introduction: A quantity having magnitude as well as the direction is called vector. It is

denoted as AB

or a . Its magnitude (or modulus) is AB

or a otherwise, simply AB or a .

Vectors are denoted by symbols such as a or a or a.

[Pictorial representation of vector]

02. Initial and Terminal points: The initial and terminal points mean that point from which the vector originates and terminates respectively.

03. Position Vector: The position vector of a point say P( , , )x y z is OP ˆˆ ˆ

r x y zi j k and the

magnitude is 2 2 2 r x y z . The vector OP ˆˆ ˆ

r x y zi j k is said to be in its component form.

Here , ,x y z are called the scalar components or rectangular components of r

and ˆˆ ˆ, ,x y zi j k are the

vector components of r

along x-, y-, z- axes respectively.

Also, AB Position Vector of B Position Vector of A

. For example, let

1 1 1A( , )x y ,z and 2 2 2B( , )x y ,z . Then, 2 2 2 1 1 1ˆ ˆˆ ˆ ˆ ÂB ( ) ( )

x i y j z k x i y j z k .

Here ,ˆ î j and k are the unit vectors along the axes OX, OY and OZ respectively. (The

discussion about unit vectors is given later in the point 05(e).)

04. Direction ratios and direction cosines: If ˆˆ ˆ ,r x y zi j k

then coefficients of ˆˆ ˆ, ,i j k in r

i.e., , ,x y z

are called the direction ratios (abbreviated as d.r.’s) of vector r

. These are denoted by , ,a b c (i.e.

, , a x b y c z ; in a manner we can say that scalar components of vector r

and its d.r.’s both are the

same).

Also, the coefficients of ˆˆ ˆ, ,i j k in r (which is the unit vector of r

) i.e., 2 2 2

x

x y z,

2 2 2

y

x y z,

2 2 2

z

x y z are called direction cosines (which is abbreviated as d.c.’s) of vector r

.

These direction cosines are denoted by l, m, n such that cos , cos , cos l m n and

2 2 2 2 2 21 cos cos cos 1 l m n .

It can be easily concluded that cos , cos , x y

l mr r

cos z

nr

.

Therefore, cos cosˆ ˆˆ ˆ ˆ ˆ(cos ) r mr nrlri j k r i j k . (Here

r r .)

[See the OAP in Fig.1] Angles , , are made by the vector r

with the positive directions of x, y, z-axes

respectively and these angles are known as the direction angles of vector r

).

For a better understanding, you can visualize the Fig.1.

05. TYPES OF VECTORS a) Zero or Null vector: Its that vector whose initial and terminal points are coincident. It is denoted

by 0

. Of course its magnitude is 0 (zero). Any non-zero vector is called a proper vector. b) Co-initial vectors: Those vectors (two or more) having the same initial point are called the co- initial vectors.

Vector Algebra Formulae For By OP Gupta [Indira Award Winner, +91-9650 350 480]

BASIC ALGEBRA OF VECTORS




c) Co-terminous vectors: Those vectors (two or more) having the same terminal point are called the co-terminous vectors. d) Negative of a vector: The vector which has the same magnitude as the r

but opposite direction. It

is denoted by r

. Hence if, AB BA

r r . That is AB BA, PQ QP

etc.

e) Unit vector: It is a vector with the unit magnitude. The unit vector in the direction of vector r

is

given by ˆ rr

r such that 1r

. So, if ˆˆ ˆr x y zi j k

then its unit vector is:

2 2 2 2 2 2 2 2 2

ˆˆ ˆˆ x y zi j k

x y z x y z x y zr

.

Unit vector perpendicular to the plane of a and

b is:

a b

a b.

f) Reciprocal of a vector: It is a vector which has the same direction as the vector r

but magnitude

equal to the reciprocal of the magnitude of r

. It is denoted as 1r . Hence 1 1r

r

.

g) Equal vectors: Two vectors are said to be equal if they have the same magnitude as well as direction, regardless of the positions of their initial points.

Thus

a b

a b

and have same direction

a b .

Also, if 1 2 3 1 2 3 1 1 2 2 3 3ˆ ˆˆ ˆ ˆ ˆ , ,

a b a i a j a k b i b j b k a b a b a b .

h) Collinear or Parallel vector: Two vectors a and

b are collinear or parallel if there exists a non-

zero scalar such that a b

.

It is important to note that the respective coefficients of ˆˆ ˆ, ,i j k in a and

b are proportional

provide they are parallel or collinear to each other.

Consider 1 2 3 1 2 3ˆ ˆˆ ˆ ˆ ˆ,

a a i a j a k b b i b j b k , then by using a b

, we can conclude

that: 31 2

1 2 3

aa a

b b b .

The d.r.’s of parallel vectors are same (or are in proportion).

The vectors a and

b will have same or opposite direction as is positive or negative.

The vectors a and

b are collinear if 0

a b .

i) Free vectors: The vectors which can undergo parallel displacement without changing its magnitude and direction are called free vectors.

06. ADDITION OF VECTORS a) Triangular law: If two adjacent sides (say sides AB and BC) of a triangle ABC are represented by

a and

b taken in same order, then the third side of the triangle taken in the reverse order gives the

sum of vectors a and

b i.e., AC AB BC

AC

a b . See Fig.2.

Also since AC CA AB BC CA AA 0

.

And AB BC AC

AB BC AC 0 AB BC CA 0

.

b) Parallelogram law: If two vectors a and

b are represented in magnitude and the direction by the

two adjacent sides (say AB and AD) of a parallelogram ABCD, then their sum is given by that

diagonal of parallelogram which is co-initial with a and

b i.e., OC OA OB

. For the

illustration, see Fig.3.

Multiplication of a vector by a scalar Let a be any vector and k be any scalar. Then the product

ka is defined as a vector whose

magnitude is k times that of a and the direction is

(i) same as that of a if k is positive, and (ii) opposite to that of

a if k is negative.




07. PROPERTIES OF VECTOR ADDITION

• Commutative property: a b b a

.

Consider 1 2 3ˆˆ â a i a j a k

and 1 2 3

ˆˆ ˆb b i b j b k

be any two given vectors.

Then 1 1 2 2 3 3ˆˆ â b a b i a b j a b k b a

.

• Associative property: a b c a b c

• Additive identity property: 0 0a a a

.

• Additive inverse property: ( ) 0 ( )

a a a a .

08. Section formula: The position vector of a point say P dividing a line segment joining the points A and B

whose position vectors are a

and b

respectively, in the ratio m:n

(a) internally, is OP

mb na

m n (b) externally, is OP

mb na

m n.

Also if point P is the mid-point of line segment AB then, OP2

a b.


01. PRODUCT OF TWO VECTORS

a) Scalar product or Dot product: The dot product of two vectors a

and b

is defined by,

. cosa b a b

where θ is the angle between a

and b

, 0 . See Fig.4.

Consider 1 2 3 1 2 3ˆ ˆˆ ˆ ˆ ˆ,a a i a j a k b b i b j b k

. Then 1 1 2 2 3 3.a b a b a b a b

.

Properties / Observations of Dot product

• ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ. cos0 1 . 1 . .i i i i i i j j k k .

• ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ. cos 0 . 0 . .2

i j i j i j j k k i

.

• . R,

a b where R is real number i.e., any scalar.

• . .a b b a

(Commutative property of dot product).

• . 0a b a b

.

• If θ 0= then, .

a b a b . Also 2 2.

a a a a ; as θ in this case is 0 .

Moreover if θ = then, .

a b a b .

• . . .a b c a b a c

(Distributive property of dot product).

• . . .a b a b a b

.

• Angle between two vectors a

and b

can be found by the expression given below:

.

cos

a b

a b or, 1 .

cos

a b

a b

.

• Projection of a vector a

on the other vector say b

is given as .a b

b

i.e., ˆ.a b

.

PRODUCT OF VECTORS

DOT PRODUCT & CROSS PRODUCT




This is also known as Scalar projection or Component of a

along b

.

• Projection vector of a

on the other vector say b

is given as . ˆ.

a bb

b

.

This is also known as the Vector projection.

• Work done W in moving an object from point A to the point B by applying a force F

is

given as W F.AB

.

b) Vector product or Cross product: The cross product of two vectors a

and b

is defined by,

ˆsina b a b

, where is the angle between the vectors a

and b

, 0 and is a

unit vector perpendicular to both a

and b

. For better illustration, see Fig.5.

Consider 1 2 3 1 2 3ˆ ˆˆ ˆ ˆ ˆ,a a i a j a k b b i b j b k

.

Then, 1 2 3 2 3 3 2 1 3 3 1 1 2 2 1

1 2 3

ˆˆ ˆ

ˆˆ ˆ

i j k

a b a a a a b a b i a b a b j a b a b k

b b b

.

Properties / Observations of Cross product

• ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆsin 0. 0 0i i i i j i i j j k k

.

• ˆ ˆˆ ˆ ˆ ˆ sin .2

i j i j k k

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ, ,i j k j k i k i j .

• Fig.6 at the end of chapter can be considered for memorizing the vector product of ˆˆ ˆ, ,i j k .

• a b

is a vector c

(say) and this vector c

is perpendicular to both the vectors a

and b

.

• ˆsin sin

a b a b a b i.e., sin

a b ab .

• 0a b

//a b

or, 0, 0

a b .

• 0a a

.

•

a b b a (Commutative property does not hold for cross product).

• ;a b c a b a c b c a b a c a

(Distributive property of the vector product or cross product).

• Angle between two vectors a

and b

in terms of Cross-product can be found by the

expression given here: sin

a b

a b

or, 1sin

a b

a b

.

• If a

and b

represent the adjacent sides of a triangle, then the area of triangle can be

obtained by evaluating 1

2

a b .

• If a

and b

represent the adjacent sides of a parallelogram, then the area of parallelogram

can be obtained by evaluating

a b .

• If p

and q

represent the two diagonals of a parallelogram, then the area of

parallelogram can be obtained by evaluating 1

2 p q .

If a and

b represent the adjacent sides of a parallelogram, then the

diagonals 1

d and 2

d of the parallelogram are given as:

1 d a b , 2

d b a .




02. Relationship between Vector product and Scalar product [Lagrange’s Identity]

Consider two vectors a

and b

. We also know that ˆsina b a b

.

Now, ˆsin

a b a b

sin

a b a b

2 22 2sin

a b a b

2 22 2

2 2 22 2 2

1 cos

cos

a b a b

a b a b a b

22 22

2 2 22

cos

.

a b a b a b

a b a b a b

or, 2 22 2

.a b a b a b

.

Note that 2 . .

. .

a a a ba b

a b b b

. Here the RHS represents a determinant of order 2.

03. Cauchy- Schwartz inequality:

For any two vectors a

and b

, we always have .a b a b

.

Proof: The given inequality holds trivially when either 0

a or 0 b i.e., in such a case . 0a b a b

.

So, let us check it for 0

a b .

As we know, . cosa b a b

2 22 2. cosa b a b

Also we know 2cos 1 for all the values of .

2 22 22cosa b a b

2 22

.a b a b

.a b a b

. [H.P.]

04. Triangle inequality:

For any two vectors a

and b

, we always have

a b a b .

Proof: The given inequality holds trivially when either 0

a or 0 b i.e., in such a case

0

a b a b . So, let us check it for 0

a b .

Then consider 2 22

2 .

a b a b a b

2 22

2 cos

a b a b a b

For cos 1 , we have: 2 cos 2a b a b

2 22 2

2 cos 2a b a b a b a b

22

a b a b

a b a b . [H.P.]





01. SCALAR TRIPLE PRODUCT:

If a

,b

and c are any three vectors, then the scalar product of

a b with

c is called scalar triple

product of a

,b

and c .

Thus, ( ).

a b c is called the scalar triple product of a

,b

and c .

Notation for scalar triple product: The scalar triple product of a

,b

and c is denoted

by [ ]

a b c . That is, ( ). [ ]

a b c a b c .

Scalar triple product is also known as mixed product because in scalar triple product, both the signs of dot and cross are used.

Consider 1 2 3 1 2 3 1 2 3ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ, ,

a a i a j a k b b i b j b k c c i c j c k .

Then,

1 2 3

1 2 3

1 2 3

[ ]

a a a

a b c b b b

c c c

.

Properties / Observations of Scalar Triple Product

• ( ). .( )

a b c a b c . That is, the position of dot and cross can be interchanged without

change in the value of the scalar triple product (provided their cyclic order remains the same).

• [ ] [ ] [ ]

a b c b c a c a b . That is, the value of scalar triple product doesn’t change

when cyclic order of the vectors is maintained.

Also [ ] [ ]; [ ] [ ]

a b c b a c b c a b a c . That is, the value of scalar triple

product remains the same in magnitude but changes the sign when cyclic order of the vectors is altered.

• For any three vectors a

, b

, c and scalar , we have [ ] [ ]

a b c a b c .

• The value of scalar triple product is zero if any two of the three vectors are identical.

That is, [ ] 0 [ ] [ ]

a a c a b b a b a etc.

• Value of scalar triple product is zero if any two of the three vectors are parallel or collinear.

• Scalar triple product of ˆ ˆ,i j and k is 1 (unity) i.e., ˆˆ ˆ[ ] 1i j k .

• If [ ] 0

a b c then, the non-parallel and non-zero vectors a

, b

and c are coplanar.

Volume Of Parallelopiped

If a

, b

and c represent the three co-terminus edges of a parallelopiped, then its

volume can be obtained by: [ ] ( ).

a b c a b c . That is,

( ). Base area of Parallelopiped Height of Parallelopiped on this basea b c

.

If for any three vectors a

, b

and c , we have [ ] 0

a b c , then volume of

parallelepiped with the co-terminus edges as a

, b

and c , is zero. This is possible

only if the vectors a

, b

and c are co-planar.

SCALAR TRIPLE PRODUCT OF VECTORS




u


01. Distance formula:

The distance between two points 1 1 1A , ,x y z and 2 2 2B , ,x y z is given by the expression

2 2 2

2 1 2 1 2 1AB unitsx x y y z z .

02. Section formula:

The coordinates of a point Q which divides the line joining the points 1 1 1A , ,x y z and

2 2 2B , ,x y z in the ratio :m n

a) internally, are 2 1 2 1 2 1, ,

mx nx my ny mz nz

m n m n m n

b) externally i.e., internally in the ratio ( ) : ( )m n , are 2 1 2 1 2 1, ,

mx nx my ny mz nz

m n m n m n.

03. Direction Cosines of a Line:

If A and B are two points on a given line L then, the direction cosines of vectors AB

and BA

are the direction cosines (d.c.’s) of line L. Thus if α, β, γ are the direction-angles which the line L

makes with the positive direction of x, y, z- axes respectively then, its d.c.’s are cosα, cosβ, cosγ

(See Fig.1). If direction of line L is reversed, the direction angles are replaced by their supplements i.e., α, β, γ and so are the d.c.’s i.e., the direction cosines become

cosα, cosβ, cos γ . So, a line in space has two sets of d.c.’s viz. cosα, cosβ, cos γ .

The d.c.’s are generally denoted by , ,l m n . Also 2 2 2 1 l m n and so, we can deduce that 2 2 2cos α cos β cos γ 1 . Also 2 2 2sin α sin β sin γ 2 .

The d.c.’s of a line joining the points 1 1 1A , ,x y z and 2 2 2B , ,x y z are

2 1 2 1 2 1, ,AB AB AB

x x y y z z ; where AB is the distance between points A and B i.e.,

2 2 2

2 1 2 1 2 1AB x x y y z z .

In order to obtain the d.c.’s of a line which does not pass through the origin, we draw a line through the origin and parallel to the given line. As parallel lines have same set of the d.c.’s, so the d.c.’s of given line can be obtained by taking the d.c.’s of the parallel line through origin.

In a unit vector, the coefficient of ˆˆ ˆ, ,i j k are called d.c.’s. For example in ˆˆ â li mj nk The

d.c.’s are , ,l m n . 04. Direction Ratios of a Line:

Any three numbers , ,a b c (say) which are proportional to d.c.’s i.e., , ,l m n of a line are called the

direction ratios (d.r.’s) of the line. Thus, , , a l b m c n for any R 0 .

Consider, 1

(say)

l m n

a b c

3 Dimensional Geometry Formulae For By OP Gupta [Indira Award Winner, +91-9650 350 480]

RECAPITULATION, DIRECTION COSINES,

DIRECTION RATIOS & EQUATION OF LINES




, ,

a b c

l m n

2 2 2

1

a b c 2 2 2Using 1 l m n

2 2 2 a b c

Therefore, 2 2 2 2 2 2 2 2 2

, ,

a b cl m n

a b c a b c a b c.

The d.c.’s of a line joining the points 1 1 1A , ,x y z and 2 2 2B , ,x y z are

2 1 2 1 2 1, ,x x y y z z or 1 2 1 2 1 2, ,x x y y z z .

Direction ratios are sometimes called as Direction Numbers as well.

For a line if , ,a b c are its d.r.’s then, , ,ka kb kc ; 0k is also a set of its d.r.’s. So, for a line there are infinitely many sets of the direction ratios.

05. Relation between the direction cosines of a line:

Consider a line L with d.c’s , ,l m n . Draw a line passing through the origin and P , ,x y z and

parallel to the given line L. From P draw a perpendicular PA on the X-axis. Suppose OP r . (See Fig.2)

Now in OAP we have, OA

cosαOP

x

x lrr

.

Similarly we can obtain y mr and z nr .

Therefore, 2 2 2 2 2 2 2 x y z r l m n .

But we know that 2 2 2 2 x y z r .

Hence , 2 2 2 1 l m n . 06. Equation of a line in space passing through a given point and parallel to a given vector:

Consider the line L is passing through the given point 1 1 1A , ,x y z with the position vector a

, b

is the given vector with d.r.’s , ,a b c and r

is the position vector of any arbitrary point P , ,x y z

on the line. See Fig.3.

Thus 1 1 1ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ÔA , OP ,a x i y j z k r xi yj zk b ai bj ck

.

a) Vector equation of a line: As the line L is parallel to given vector b

and points A and P are

lying on the line so, AP

is parallel to the b

.

AP ,b

where R , set of real nos.i.e.

r a b

r a b

. This is the vector equation of line.

b) Parametric equations: If d.r.’s of the line are , ,a b c then by using r a b

we get,

1 1 1ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆx y z x y z a b ci j k i j k i j k .

Now, as we equate the coefficients of ˆˆ ˆ, ,i j k we get the Parametric equations of line given as,

1 1 1, ,x x a y y b z z c .

Coordinates of any point on the line considered here are 1 1 1, ,x a y b z c .

c) Cartesian equation of a line: If we eliminate the parameter from the Parametric equations of a line, we get the Cartesian equation of line as




1 1 1x x y y z z

a b c

.

If , ,l m n are the d.c.’s of the line then, Cartesian equation of line becomes

1 1 1x x y y z z

l m n

.

The Cartesian equation of line is also called the symmetrical equation or one point form of line. In the symmetrical form the coefficient of x, y, z are unity i.e., 1.

Note that b

is parallel to the line L. So they both have the same d.r.’s. 07. Equation of a Line passing through two given points:

Consider the two given points as 1 1 1A , ,x y z and 2 2 2B , ,x y z with position vectors a

and b

respectively. Also assume r

as the position vector of any arbitrary point P , ,x y z on the line L

passing through A and B. See Fig.4.

Thus 1 1 1 2 2 2ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ÔA , OB , OPa x i y j z k b x i y j z k r xi yj zk

.

a) Vector equation of a line: Since the points A, B and P all lie on the same line which means they are all collinear points.

Further it means, AP r a

and AB b a

are collinear vectors, i.e.,

AP AB

r a b a

,r a b a

where R .

This is the vector equation of line.

b) Cartesian equation of a line: By using the vector equation of the line r a b a

we get,

1 1 1 2 1 2 1 2 1ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆx y z x y z x x y y z zi j k i j k i j k

On equating the coefficients of ˆˆ ˆ, ,i j k we get,

1 2 1 1 2 1 1 2 1, ,x x x x y y y y z z z z ....( )i

On eliminating we have,

1 1 1

2 1 2 1 2 1

x x y y z z

x x y y z z

.

This is the Cartesian equation of line.

c) Parametric equations: By using (i), we get

1 2 1 1 2 1 1 2 1, ,x x x x y y y y z z z z .

These are called the Parametric equations of line. 08. Angle between two Lines:

a) When d.r.’s or d.c.’s of the two lines are given: Consider two lines 1L and 2L with d.r.’s as 1 1 1, ,a b c and 2 2 2, ,a b c ; d.c.’s as 1 1 1, ,l m n and

2 2 2, ,l m n . Consider 1 1 1 1ˆˆ ˆb a i b j c k

and 2 2 2 2

ˆˆ ˆb a i b j c k

. These vectors 1b

and 2b

are

parallel to the given lines 1L and 2L . So in order to find the angle between the Lines 1L and

2L , we need to get the angle between the vectors 1b

and 2b

. Consider the Fig.5.

So the acute angle θ between the vectors 1b

and 2b

(and hence lines 1L and 2L ) can be

obtained as,

1 2 1 2. cosθb b b b




Thus, 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

cosθa a b b c c

a b c a b c

.

1 2 1 2 1 2Also, in terms of d.c.’s : cosθ l l m m n n .

2 2 2

1 2 2 1 1 2 2 1 1 2 2 1

2 2 2 2 2 21 1 1 2 2 2

Sine of angle is given as: sinθa b a b b c b c c a c a

a b c a b c

.

b) When vector equations of two lines are given:

Consider vector equations of lines 1L and 2L as 1 1 1r a b

and 2 2 2r a b

respectively.

Then, the acute angle θ between the two lines is given by the relation

1 2

1 2

.cosθ

b b

b b

.

c) When Cartesian equations of two lines are given: Consider the lines 1L and 2L in Cartesian form as,

1 1 1 2 2 21 2

1 1 1 2 2 2

L : , L :x x y y z z x x y y z z

a b c a b c

.

Then the acute angle θ between the lines 1L and 2L can be obtained by,

1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

cosθa a b b c c

a b c a b c

.

For two perpendicular lines: 1 2 1 2 1 2 0a a b b c c ; 1 2 1 2 1 2 0l l m m n n .

For two parallel lines: 1 1 1

2 2 2

a b c

a b c ; 1 1 1

2 2 2

l m n

l m n .

09. Shortest Distance between two Lines:

If two lines are in the same plane i.e. they are coplanar, they will intersect each other if they are non-parallel. Hence shortest distance between them is zero. If the lines are parallel then the shortest distance between them will be the perpendicular distance between the lines i.e. the length of the perpendicular drawn from a point on one line onto the other line. Adding to this discussion, in space, there are lines which are neither intersecting nor parallel. In fact, such pair of lines are non-coplanar and are called the skew lines.

Skew lines: Two straight lines in space which are neither parallel nor intersecting are known as the skew lines. They lie in different planes and are non-coplanar.

Shortest Distance: There exists unique line perpendicular to each of the skew lines 1L and

2L , this line is known as line of shortest distance (S.D.).

a) Shortest distance between two Skew lines: When lines are in vector form:

Consider the two skew lines 1 1 1r a b

and 2 2 2r a b

. Assume that A and B are two

points on the lines 1L and 2L with position vectors 1a

and 2a

respectively. Let us assume

that the Shortest Distance between the two lines is PQ d . See Fig.6.




Now PQ is perpendicular to both the lines 1L and 2L . That means PQ

is perpendicular to

both 1b

and 2b

. But we know that 1 2b b

is perpendicular to both 1b

and 2b

. So we can

deduce that PQ

is in the direction of 1 2b b

.

Let the unit vector in the direction of 1 2b b

is n .

So, 1 2

1 2

ˆb b

nb b

ˆPQ PQ. n

1 2

1 2

PQ PQb b

b b

The shortest distance PQ is basically the projection of AB

on PQ

.

ˆ. . PQ AB.i e , n

Then the S.D. between them is given as follow,

1 2 2 1

1 2

.PQ

b b a ad

b b

.

When the lines are in Cartesian form: Consider the two skew lines as,

1 1 1 2 2 21 2

1 1 1 2 2 2

L : , L :x x y y z z x x y y z z

a b c a b c

.

Then the S.D. between them is given as follow

2 1 2 1 2 1

1 1 1

2 2 2

2 2 2

1 2 2 1 1 2 2 1 1 2 2 1

x x y y z z

a b c

a b cd

a b a b b c b c c a c a

.

b) Shortest distance between two parallel lines:

If the lines are parallel then they are coplanar i.e. they lie in the same plane.

Consider the two parallel lines as 1 1L : r a b

and 2 2L : r a b

. Assume that A and B

are two points on the lines 1L and 2L with position vectors 1a

and 2a

respectively. Also

assume that the lines are parallel to b

. Let AB

makes angle θ with the line 1L . So the angle

between the AB

and b

will be θ . See Fig.7. Draw 1BP L . Now BP represents the perpendicular distance between 1L and 2L .

In APB , we have BP = ABsin θ ...( )i

Now consider ÂB AB sin( θ)b b n

ÂB AB sin θ [ 1b b n

AB AB sinθb b

AB BP [ ( )b b By using i




AB

BPb

b

.

Assume that the Shortest Distance between the two lines is BP d .

Then the S.D. between them is given as follow,

2 1b a ad

b

.

Note that the S.D. between two parallel lines in the Cartesian form can be obtained by simply

replacing 2 1 2 1 2 1 2 1ˆˆ ˆ( ) ( ) ( )a a x x i y y j z z k

and ˆˆ ˆb ai bj ck

in the

expression obtained above for the Vector form.


01. Plane and its equation: A plane is a surface such that if any two points are taken on it, the line segment joining them lies completely on the surface. Plane is symbolized by the Greek letter .

a) Equation of plane in Normal unit vector form:

Consider a plane at distance d from the origin such that ON

is the normal from the origin to

plane and n is a unit vector along ON

. Then ÔN dn

if ON d . Consider r

be the

position vector of any arbitrary point P , ,x y z on the plane. See Fig.8.

Vector form of the equation of plane: Since P lies on the plane so NP

is perpendicular to

the vector ON

. That implies NP.ON 0

ˆ ˆ. 0r dn dn

ˆ ˆ. 0 [ 0r dn n d

ˆ ˆ ˆ. . 0r n dn n

ˆ.r n d

ˆ ˆ[ . 1n n

This is the vector equation of the plane.

Cartesian form of the equation of plane: If , ,l m n are d.c.’s of the normal n to the given

plane. Then by using ˆ.r n d

we get,

.ˆ ˆˆ ˆ ˆ ˆx y z l m ni j k i j k d

lx my nz d .

This is the Cartesian equation of the plane.

Also if , ,a b c are the d.r.’s of the normal n to the plane then, the Cartesian equation of

plane becomes ax by cz d .

b) Equation of plane Perpendicular to a given vector and passing through a given point:

Assume that the plane passes through a point 1 1 1A , ,x y z with the position vector a

and is

perpendicular to the vector m

with d.r.’s as A,B,C ˆˆ Â B Cm i j k

.

Also consider P , ,x y z as any arbitrary point on the plane with position vector as r

.

Consider the Fig.9.

PLANE & ITS EQUATION IN VARIOUS FORMS




Vector form of the equation of plane: As AP

lies in the plane and m

is perpendicular to

the plane. So AP

is perpendicular to m

.

AP. 0m

. 0r a m

.

This is the Vector equation of the plane.

The above obtained equation of plane can also be expressed as . .r m a m

.

Cartesian form of the equation of plane: As 1 1 1ˆˆ ÂP x x i y y j z z k

, so by

using . 0r a m

we get,

1 1 1ˆ ˆˆ ˆ ˆ ˆ. A B C 0x x i y y j z z k i j k

1 1 1A B C 0x x y y z z .


c) Equation of plane passing through three non- collinear points:

Assume that the plane contains three non- collinear points 1 1 1R , ,x y z , 2 2 2S , ,x y z and

3 3 3T , ,x y z with the position vectors as a

, b

and c

respectively. Let P , ,x y z be any

arbitrary point in the plane whose position vector is r

.

Vector form of the equation of plane: As RS

and RT

are in the plane, so

RS RT m

( )say will be perpendicular to the plane containing the points R, S and T. Also

since r

is position vector of P which lies in the plane, therefore RP m

. See Fig.10.

RP. 0m

. RS RT 0 [ RS RTr a m

. 0r a b a c a

.

This is the Vector equation of the plane.

Cartesian form of the equation of plane:

The position vector of RP

, RS

and RT

is given as,

1 1 1 1 1 1

2 2 2 1 1 1 2 1 2 1 2 1

3 3 3 1 1 1 3 1 3 1 3 1

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆRP

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆRS

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆRT

xi yj zk x i y j z k x x i y y j z z k

x i y j z k x i y j z k x x i y y j z z k

x i y j z k x i y j z k x x i y y j z z k

Substituting these in the above obtained vector equation of plane, we get

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

0

x x y y z z

x x y y z z

x x y y z z

.


d) Intercept form of the equation of plane: Consider the equation of plane A B C D 0,x y z D 0 and the plane makes intercepts

, ,a b c on , ,x y z -axes respectively. This implies that the plane meets , ,x y z -axes at ,0,0a ,

0, ,0b , 0,0,c respectively. See Fig.11.

Therefore, D

A. B.0 C.0 D 0 A ,a + + + = =a




D

A.0 B. C.0 D 0 B+ b + + = =b

and

D

A.0 B.0 C. D 0 C+ + c + = =c

.

Substituting these values in A B C D 0,x y z , we get 1x y z

a b c .

This is the equation of plane in intercept form.

NOTE

Equation of XY-plane : 0,

Equation of YZ-plane : 0,

Equation of ZX-plane : 0.

z

x

y

02. Equation of plane passing through the intersection of two given planes:

The intersection of two planes say 1 and 2 is always a straight line. For instance, we can

visualize the intersection of xy - plane and xz - plane to form x- axis.

a) Vector equation of the plane: Consider two planes 1 1 1: .r m d

and 2 2 2: .r m d

. So if

h

is the position vector of any arbitrary point on the line of intersection of 1 and 2 then, it

must satisfy both the equations of planes i.e.,

1 1.h m d

and 2 2.h m d

1 1. 0h m d

and 2 2. 0h m d

.

Therefore for all R (set of all real nos.) , we get

1 1 2 2. . 0h m d h m d

1 2 1 2.h m m d d

As the obtained equation is of the form .r m d

(Note that in .r m d

, d is not the

perpendicular distance of plane from the origin. Rather d is perpendicular distance from the origin in ˆ.r n d

.)

So it represents a plane 3 (say).

Hence, the required plane is: 1 2 1 2.r m m d d

.

This is the Vector equation of plane.

b) Cartesian equation of the plane:

Assume 1 1 1 1ˆˆ Â B Cm i j k

, 2 2 2 2

ˆˆ Â B Cm i j k

and ˆˆ ˆr xi yj zk

.

Then by using 1 2 1 2.r m m d d

, we get

1 2 1 2 1 2 1 2A A B B C Cx y z d d

i.e., 1 1 1 1 2 2 2 2A B C A B C 0x y z d x y z d .

This is the Cartesian equation of plane. (You can visualize the situation discussed here in the Fig.12)

03. Co- planarity of two Lines:

Assume the given lines are 1 1 1L : r a b

and 2 2 2L : r a b

such that 1L passes through

1 1 1A , ,x y z with position vector 1a

and is parallel to 1b

with d.r.’s 1 1 1, ,a b c . Also 2L passes

through 2 2 2B , ,x y z with position vector 2a

and is parallel to 2b

with the d.r.’s 2 2 2, ,a b c .

a) Vector form of co- planarity of lines:




We know 2 1AB a a

. Now the lines 1L and 2L are coplanar iff AB

is perpendicular to

1 2b b

. That implies, 1 2AB. 0b b

2 1 1 2. 0a a b b

.

b) Cartesian form of co- planarity of lines:

We know that 2 1 2 1 2 1ˆˆ ÂB x x i y y j z z k

, 1 1 1 1

ˆˆ ˆb a i b j c k

and

2 2 2 2ˆˆ ˆb a i b j c k

. So by using 2 1 1 2( ).( ) 0a a b b

, we get

2 1 2 1 2 1

1 1 1

2 2 2

0

x x y y z z

a b c

a b c

.

Note that only coplanar lines can intersect each other in the plane they exist.

04. Angle between two planes:

The angle between two planes is the angle between their normals 1m

and 2m

(say). Therefore if θ

is the angle between the planes 1 and 2 then o180 θ is also the angle between the two planes.

Though we shall be taking acute angle θ only as the angle between two planes. Observe the Fig.13.

a) Vector form for the angle between two planes:

Consider the planes 1 1 1: .r m d

and 2 2 2: .r m d

. If θ is the angle between the normals

to the plane drawn from some common point. Then,

1 2

1 2

.cosθ

m m

m m

[Using dot product of vectors

1 1 2

1 2

.θ cos

m m

m m

.

b) Cartesian form for the angle between two planes: Assume the planes, 1 1 1 1A B C D 0x y z and 2 2 2 2A B C D 0x y z where

1 1 1A , B , C and 2 2 2A , B , C are the d.r.’s of normals (to the planes) 1m

and 2m

respectively.

Then, 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

A A B B C Ccosθ

A B C A B C

+ +

+ + + +

1 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

A A B B C Cθ cos

A B C A B C

+ +

+ + + +

.

For the parallel planes, we have: 1 1 1

2 2 2

A B C

A B C .

For the perpendicular planes, we have: 1 2 1 2 1 2 0A A B B C C .

05. Distance of a point from a plane:

a) Vector form for the distance of a point from a plane:

Let : .r m d

be the plane and 1 1 1P , ,x y z be the point with position vector a

. Let PA be

the length of perpendicular on the plane. See Fig.14. Since line PA passes through P a

and

is parallel to the m

which is normal to the plane. So the vector equation of the line PA is ...( )r a m i




Since A is point of intersection of line (i) and the given plane. So we have,

.a m m d

2

.d a m

m

Putting the value of in (i), we get the position vector of A given as follow,

2

.d a mr a m

m

Since PA OA OP

PA r a

2

.PA

d a ma m a

m

2

.PA

d a mm

m

2

.PA PA

d a mm

m

2

.PA

d a mm

m

.. ., PA

d a mi e

m

Hence, length of the perpendicular PA ( )p say from a point having position vector a

to the

plane .r m d

is given by

.d a m

pm

.

b) Cartesian form for the distance of a point from a plane: Let A, B, C be the d.r.’s of the

normal m

to the given plane. So by using the relation .d a m

pm

we can obtain,

1 1 1

2 2 2

A B C

A B C

x y z dp

.

If d is the distance from the origin and l, m, n are the d.c.’s of the normal vector to the

plane through origin, then the coordinates of the foot of perpendicular is , ,ld md nd .

c) Distance between two parallel palnes: Assume the two planes as,

1 1. i.e., A B C D 0r m d x y z

and 2 2. i.e., A B C D 0r m d x y z

.

Then the distance p (say) between them is given as

(i) Vector form: 1 2d dp

m

.

(ii) Cartesian form: 1 2

2 2 2

D D

A B C

p .

06. Angle between a line and a plane:




The angle between a line and a plane is complementary to the angle between the line and normal to

the plane. Let θ is the angle between b

(which is parallel to the line) and normal m

of the plane.

This implies that 90 θ is the angle between the line r a b

and plane .r m d

.

Consider Fig.15.

Now the angle between b

and m

, .

cosθb m

b m

[By using dot product of vectors

So the angle ( )say between the line and plane is given as 90 θ i.e.,

sin sin 90 θ cosθ

i.e., .

sinb m

b m

1 .

sinb m

b m

.

This is the angle between line and a plane.




u


01. Basics Of Probability: Let S and E be the sample space and an event in an experiment respectively.

Then, ENumber of favourable events

Probability =Total number of elementary events S

n

n .

Also as E S

E Sn n n

or 0 E Sn n

E S0

S S S

n n

n n n

0 P E 1 .

Hence, if P E denotes the probability of occurrence of an event E then, 0 P E 1 and

P E = 1 P E such that P E denotes the probability of non-occurrence of the event E.

Note that P E can also be represented as P E .

02. Mutually Exclusive or Disjoint Events: Two events A and B are said to be mutually exclusive if occurrence of one prevents the occurrence of the other.

Consider an example of throwing a die. We have the sample spaces as, S 1,2,3,4,5,6 .

Suppose A 5,6the event of occurence of a number greater than 4 ,

B 1,3,5the event of occurence of an odd number and

C 2,4,6the event of occurence of an even number .

In these events, the events B and C are mutually exclusive events but A and B are not mutually exclusive events because they can occur together (when the number 5 comes up). Similarly A and C are not mutually exclusive events as they can also occur together (when the number 6 comes up).

If A and B are mutually exhaustive events then we always have,

P A B 0 As A Bn

P A B P A P B .

If A, B and C are mutually exhaustive events then we always have,

P A B C P A P B P C .

03. Independent Events: Two events are independent if the occurrence of one does not affect the occurrence of the other. Consider an example of drawing two balls one by one with replacement from a bag containing 3 red and 2 black balls. Suppose A the event of occurence of a red ball in first draw ,

B the event of occurence of a black ball in the second draw .

Then 3

P A5

, 2

P B5

.

Here probability of occurrence of event B is not affected by occurrence or non-occurrence of the event A. Hence event A and B are independent events.

Probability Formulae For By OP Gupta [Indira Award Winner, +91-9650 350 480]

RECAPITULATION & CONDITIONAL PROBABILITY




But if the two balls would have been drawn one by one without replacement, then the probability of

occurrence of a black ball in second draw when a red ball has been drawn in first draw 2

P B4

.

Also if a red ball is not drawn in the first draw, then the probability of occurrence of a black ball in the

second draw 1

P B4

(After a black ball is drawn there are only 4 balls left in the bag). In this case

the event of drawing a red ball in the first draw and the event of drawing a black ball in the second draw are not independent.

For independent events A and B, we always have P A B P A .P B .

For independent events A, B and C, we always have P A B C P A .P B .P C .

Also we have for independent events A and B, P A B 1 P A .P B .

Since for independent events A and B we have P A B P A .P B , so the conditional

probability (discussed later in this chapter) of event A when B has already occurred is given as,

P A B

P A|BP B

P A .P B

i.e., P A|BP B

P A|B P A .

04. Exhaustive Events: Two or more events say A, B and C of an experiment are said to be exhaustive events if, a) their union is the total sample space i.e. A B C S b) the events A, B and C are disjoint in pairs i.e. A B , B C and C A .

c) P A P B P C 1 .

Consider an example of throwing a die. We have S 1,2,3,4,5,6 .

Suppose A 2,4,6the event of occurence of an even number ,

B 1,3,5the event of occurence of an odd number and

C 3,6the event of getting a number multiple of 3 .

In these events, the events A and B are exhaustive events as A B S but the events A and C or the events B and C are not exhaustive events as A C S and similarly B C S .

05. Conditional Probability: By the conditional probability we mean the probability of occurrence of event A when B has already occurred. You can note that in case of occurrence of event A when B has already occurred, the event B acts as the sample space and A B acts as the favourable event. The ‘conditional probability of occurrence of event A when B has already occurred’ is sometimes also called as probability of occurrence of event A w.r.t. B.

P A BP A | B , B . . P B 0

P Bi e

P A B

P B | A , A . . P A 0P A

i e

P A BP A | B , P B 0

P B

P A B

P A | B , P B 0P B

P A BP A | B , P B 0

P B

P A | B P A | B 1, B .

06. Useful formulae:

a) P A B P A P B P A B . . P A or B P A P B P A and Bi e

b) P(A B C) P(A) P(B) P(C) P(A B) P(B C) P(C A) P(A B C)

c) P A B P only B P B A P B but not A P B P A B




d) P A B P onlyA P A B P A but not B P A P A B

e) P A B P neither A nor B 1 P A B

Pictorial Description Of The Playing Cards:

!C , C C such that

! !n n n

r r n r

nn r

r n r

! .( 1).( 2)...5.4.3.2.1n n n n . Also, 0! 1 .

07. Events and Symbolic representations:

Verbal description of the event Equivalent set notation Event A A

Not A A or A

A or B (occurrence of atleast one of A and B ) A B or A B

A and B (simultaneous occurrence of both A and B ) A B or A B

A but not B ( A occurs but B does not) A B or A B Neither A nor B A B At least one of A, B or C A B C All the three of A, B and C A B C


01. Bayes’ Theorem: If 1 2 3E , E , E ,..., En are n non- empty events constituting a partition of sample

spaces S i.e., 1 2 3E , E , E ,..., En are pair wise disjoint and 1 2 3E E E ... E Sn and A is any

event of non-zero probability then,

1

P E .P A | EP E | A , 1,2,3,...,

P E .P A | E

i ii n

j jj

i n

For example,

1 1

1

1 1 2 2 3 3

P E .P A | EP E |A

P E .P A | E P E .P A | E P E .P A | E

.

Bayes’ Theorem is also known as the formula for the probability of causes.

If 1 2 3E , E , E ,..., En form a partition of S and A be any event then,

1 1 2 2P A P E .P A | E P E .P A | E ... P E .P A | En n

P(E A) P(E ).P(A|E )i i i

SET OF 52 PLAYING CARDS Face Cards

Suit Clubs (Black card)

Diamonds (Red card)

Hearts (Red card)

Spades (Black card)

TOTAL PROBABILITY & BAYES' THEOREM




The probabilities 1 2P E , P E ,..., P En which are known before the experiment takes

place are called priori probabilities and P A | En are called posteriori probabilities.


01. Bernoulli Trials: Trials of a random experiment are called Bernoulli trials, if they satisfy the following four conditions:

a) The trials should be finite in numbers. b) The trials should be independent of each other. c) Each of the trial yields exactly two outcomes i.e. success or failure. d) The probability of success or the failure remains the same in each of the trial.

02. Binomial Distribution: Let E be an event. Let p probability of success in one trial (i.e.,

occurrence of event E in one trial) and, 1q p probability of failure in one trial (i.e.,non-

occurrence of event E in one trial). Let X = number of successes (i.e., number of times event E occurs in n trials). Then, Probability of X successes in n trials is given by the relation,

P(X ) P( ) Cn r n rrr r p q

where 0,1,2,3,...,r n ; p probability of success in one trial and 1q p probability of failure

in one trial.

P(X )r or P( )r is also called probability of occurrence of event E exactly r times in n

trials.

Here !

C! ( )!

nr

n

r n r

.

Note that Cn r n rr p q

is the ( 1)thr term in the binomial expansion of n

q p .

0

.P( )n

r

Mean r r np

22

0

. P( ) Meann

r

Variance r r npq

Standard Deviation npq

, P( 1) P( )1

n r pRecurrence formula x r + r

r q

.

A Binomial Distribution with n Bernoulli trials and probability of success in each trial as p is

denoted by B ,n p . Here n and p are known as the parameters of binomial distribution.

The expression P x r or P r is called the probability function of the binomial

distribution.

If an experiment is repeated n times under the similar conditions, we say that n trials of the experiment have been made.

The result P(X ) P( ) Cn r n rrr r p q can be used only when:

(i) the probability of success in each trial is the same. (ii) each trial must surely result in either a success or a failure.

BERNOULLI TRIALS & BINOMIAL DISTRIBUTION




Important Terms, Definitions & Formulae 01. Random Variable: A random variable is a real valued function defined over the sample space of an experiment. In other words, a random variable is a real-valued function whose domain is the sample space of a random experiment. A random variable is usually denoted by upper case letters X, Y, Z etc.

Discrete random variable: It is a random variable which can take only finite or countably infinite number of values. Continuous random variable: It is a random variable which can take any value between two given limits is called a continuous random variable.

02. Probability Distribution Of A Random Variable: If the values of a random variable together with the corresponding probabilities are given, then this description is called a probability distribution of the random variable.

Mean or Expectation of a random variable 1

Pn

i ii

X x

Variance

2 2 2

1

( ) Pn

i ii

x

Standard Deviation Variance .

Hii, All!

I hope this texture may have proved beneficial for you. While going through this material, if you noticed any error(s) or, something which doesn’t make sense to you, please bring it in my notice through SMS or Call at +91-9650 350 480 or Email at [email protected].

With lots of Love & Blessings!

- OP Gupta Electronics & Communications Engineering, Indira Award Winner www.theOPGupta.WordPress.com

Click on the following link to go for a pleasant surprise: http://theopgupta.wordpress.com/maths-rockers/

PROBABILITY DISTRIBUTION

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JEE (Ad.) Roll No.2003111, CBSE Roll No.9111240 OP … Of Formulae By OP Gupta Page - [1] u Standard Formulae Of Integrals: A. 1, 1 1 n xdx k nn x n B. 1 dx x klog x