B.Tech 1st Year 1st SemesterMathematics(M101)Teacher Name: Kakali Ghosh

LECTURE-- 1

Vector algebra and vector calculus

Introduction : Vectors are frequently used in many branches of pure and applied mathematics and in physical and engineering science.

Scalar: A scalar is a physical quantity which has magnitude only but no definite direction in space. For example density, volume , temperature , work , speed, heat etc.

Vectors: A vector is a physical quantity which has magnitude and is related to a definite direction in space. For example Velocity, Acceleration, Force etc.

A vector is a directed segment of straight line on which there are distinct initial and terminal points. The arrows indicate the direction of vectors. The length of the line segment is the magnitude of the vector. For example, is a vector directed from P to Q.

P Q

Thus = .

Unit vector: A vector whose magnitude is unity is called unit vector and is denoted by .

Null Vector : A vector whose magnitude is 0 is called Null vector, denoted by .

Equal vector : If two vectors a ( ) and b ( ) are said to be equal if they have equal magnitudes

and same direction and denoted by .

Addition of two vectors : Let a and b be any two given vectors.

+ B

O A

If three points O , A , B are taken such that OA��������������

= , = , then the vector is called vector

sum or the resultant of the given vectors and and and write as = + .

Subtraction of two vectors: We define the difference of two vectors and to be the sum

of the vectors and - , i.e. = ( )a b

Multiplication of a vector by a real number: Let be scalar. Then is a vector whose magnitude is | | times that of and direction is the same as that of or opposite of , according as is positive or negative.

Collinear vectors: Two vectors and are said to be Collinear or parallel if = where is a scalar. A system of vectors is said to be collinear if they are parallel to the same straight line.

Coplanar vectors: A system of vectors is said to be Coplanar if they are parallel to the same plane.

Linearly dependent and Linearly independent vectors: A set of vectors is

said to be linearly dependent , if there exist a set of scalars x,y,z,……not all zero, such that x + y + z +………..= 0.

Otherwise they form a linearly independent set of vectors. Thus for a set of linearly independent

vectors if x + y + z +………..= 0 , then we have x = y = z = ……..= 0.

Position vector of a point: The position vector (p.v ) of a point P with respect to a fixed origin O in space is the vector . If = a

, we write P ( ) as the position vector of P is .

OIf a and b are position vectors of P and Q respectively , then = - = p.v. Q – p.v. of P.

The position vector of the point P whose Cartesian coordinates are (x,y,z) is given by

. Obviously | | = where direction cosines of =

( ).

Let OP makes with the rectangular axes at O ( The figure above). Then cos , cos , cos are called the direction cosines of OP and we can write

x = | |cos , y = | |cos , z = | |cos

The unit vector in the direction of is given by

,

where are unit vectors along the coordinate axes and (x,y,z) is position of P w.r.t O.

b

P Q -

Y

ZP

X

O

Projection or component of a vector on an axis : Let be a vector and OX be an axis. A plane passing through A which cuts OX perpendicularly at P. Then P is the point of projection of A on OX.

Similarly , we take point of projection Q of B on OX. Then PQ is called projection or component of the vector on the axis OX.

A

P QO

B

X

If makes an angle with OX , then component of on OX = | |Cos .

Illustrative examples:

1) Show that the vectors i – 3j + 5k , 3i – 2j+k , 2i + j - 4k form a right angle triangle.

Soln: Let a = i – 3j + 5k, b = 3i – 2j+k and c = 2i + j - 4k .

We see that a + c = b, Therefore a,b,c form a triangle in a plane. Now

|a| = =

|b| = =

|c| = =

. Therefore the given vectors form a right angle triangle.

2.a) Show that the vectors (2,4,10) and (3,6,15) are linearly dependent.

2.b) Show that the vectors (1,2,3) and (4,-2,7) are linearly independent.

Soln 2a) Let a = (2,4,10) and b = (3,6,15)

Let x and y be two scalars , such that x a + y b = 0

or, x(2,4,10) + y(3,6,15) = 0or, (2x + 3y, 4x+6y,10x+15y) = (0,0,0)

Equating from both sides, we get

2x + 3y = 0

4x+6y =0

10x+15y =0

Solving these , we get x= 3, y = -2, which are not all zero. Hence 3a– 2b = 0 Therefore the vectors a , b are linearly dependent.

Soln 2b) Let a = (1,2,3) and b = (4,-2,7)

Let x and y be two scalars , such that

x a + y b = 0Therefore x(1,2,3) + y(4,-2,7) = 0

Equating both sides, we get x + 4y =0

2x -2y =0

3x +7y =0

Solving we get , x = y = 0

Therefore, x a + y b = 0 , only if x = y=0.Thus the given vectors are linearly independent.

3) Show that the following vectors are coplanar:

3a – 7b -4c , 3a -2b + c , a + b +2c where a , b ,c are any three non coplanar vectors

Soln: If the given vectors be coplanar , then it will be possible to express one of them as a linear combination of the other two.

Let 3a – 7b -4c = x (3a -2b + c) + y (a + b +2c) , x and y are scalars.

Comparing the coefficients of a,b,c from both sides , we get,

3x + y =3 , -2x + y = -7 , x + 2y = -4

Solving the first two equations we get , x= 2 and y = -3 .

These values of x and y satisfy the 3rd equation. Thus

3a – 7b -4c = 2 (3a -2b + c) + (-3) (a + b +2c)

Therefore the 1st vector can be expressed as linear combination of the other two.

Hence , the three given vectors are coplanar.

Assignment:

1) If a = i -2j+2k then show that |a| =3 and direction cosines are 1/3 , (-2/3), 2/3

2) Prove that the vectors (2,3,-6) , (6,-2,3) and (4,-5,9) form the sides of an isosceles triangle.

3) Show that the vectors a = (1,2,3) , b = (2,-1,4) and c = (-1,8,1) are linearly dependent and also show that the vectors a = (1,-3,2) , b = (2,-4,-1) and c = (3,2,-1) are linearly independent.

4) Determine the values of and for which the vectors (-3i + 4j + k) and ( i + 8j + 6k) are collinear.5) Find the constant m such that the vectors

are coplanar

Multiple Choice Questions

1)The unit vector along the vector is

(a) (b) (c) (d) none

2) If then and are (a) Coplanar (b) independent (c) collinear (d) none

3)If for two vectors and

| | = | | , then and are (a) Parallel (b) orthogonal (c) collinear (d) none

LECTURE- 2

PRODUCT OF VECTORS :

DOT PRODUCT

Objective: Definition of Dot product

Angle between two vectors

Projection

Proof of Cosine formula.

Definition:

The dot product of a and b, written as a.b, is defined by a.b = a b cos where a and b are the magnitudes of a and b and is the angle between the two vectors.

The dot product is distributive: a.(b + c) = a.b + a.c and commutative: a.b = b.a Knowing that the angles between each of the i, j, and k vectors is /2 radians (90 degrees) and cos /2 = 0, we can derive a handy alternative definition: Let, u = ai + bj + ck v = xi + yj + zk then, u.v = (ai + bj + ck).( xi + yj + zk)

=>u.v = (ai + bj + ck). xi + (ai + bj + ck).yj + (ai + bj + ck).zk The angle between any nonzero vector and iteself is 0, and cos 0 = 1, so i.i = 1 etc., Hence,

u.v = a x + b y + c z

This means that for any vector, a,a. a = a2

Finding the angle between two vectors

We can now, given the coordinates of any two nonzero vectors u and v find the angle between them: u = ai + bj + ck v = xi + yj + zk u.v = u v cos u.v = a x + b y + c z

=> u v cos = a x + b y + c z

=> = cos-1 o (a x + b y + c z) / ( u v ) p To get used to this method check out this applet What would happen if one of the vectors was the null vector 0, from (0,0,0) to (0,0,0). This is the only vector without a direction and it isn't meaningful to ask the angle between this vector and another vector. How does our method fail if we try? One of the main uses of the dot product is to determine whether two vectors, a and b, are othogonal (perpendicular). If a . b = 0, then either, a is orthogonal to b, or a = 0, or b = 0.

Projection

It will often be useful to find the component of one vector in the direction of another:

We have a given vector a, and we want to see how far it extends in a direction given by the unit vector n. The distance is d, which, from simple trigonometry

we can calculate as, d = a cos => d = n a cos => d = a . n

Proof of the cosine formula

You have two sides of a triangle, a and b, and the angle in between, C, - the problem is to find the remaining side c. You kill the problem by recalling the cosine formula: c2 = a2 + b2 - 2 a b cos C but have you ever seen a proof? The proof by geometry isn't very friendly but with vectors it takes all of 3 lines (using the second triangle above): c.c = (b - a).(b - a) => c2 = b.b + a.a - 2a.b => c2 = a2 + b2 - 2ab cos C

Finding the distance between two places, along the surface of the Earth

From the latitudes and longitudes of two places on the Earth together with the radius of the Earth we can determine the position vectors of the two places with the origin at the centre of the Earth. If you have two points on the circumference of a circle then the radius of the circle times the angle (in radians) subtended by the two points at the centre of the circle gives the arc distance between the two points. Using the dot product we can find the angle subtended by our two position vectors, multiply by the radius of the Earth, and hey presto we have the great circle distance.

WORKED OUT EXAMPLE:

1) Find the angle between and .

Solution: We know that a.b = a b cos where a and b where a and b are the magnitudes of a and b and is the angle between the two vectors.

Now a = =3 , b = =7

a.b =2.6+2.(-3) + (-1).2 =12 – 6 -2 =4

Then = and = approximately.

2) If | | = 3 and | | = 4, then find the values of the scalar for

which the vectors and will be perpendicular to one another.

Solution: The vectors and will be perpendicular to one another if

( ) . ( ) = 0 i.e =0 i.e | |2 - =0 i.e 32 - 42 = 0

i.e

3 ) Given two vectors ; Express in the form where is parallel to

and is perpendicular to .

Soln: The vector is parallel to for all values of the scalar , since

( ) = ( )=

Therefore, = = ( )

Let =

Since is perpendicular to , therefore, . =0 3x-y = 0 or, y=3x

Therefore =

Now

Therefore, = ( )+( )=(3 +x) +(3x- ) +z

Equating coefficients of , , from both sides , we get ,

3 +x = 2, 3x- = 1, z=-3

Solving first two equations we get, x=1/2,

= 1/2

Therefore y=3/2

Therefore, = ½( ) and =

A ssignment:

(1) Find a vector of magnitude 11 perpendicular to the plane of vectors and

.

(2) If be unit vectors satisfying the condition , then show that

(3) If and , then find the vector which satisfies

and

(4) If , then shopw the projection of in the

direction of is .

Multiple choice question: (1) The angle between the vectors the vectors and is

(a) , (b) , (c) , (d) none.

(2) The value of p for which the vectors and are perpendicular is (a) 5 ,(b) 0 , (c) -5 , (d) 3 . (3) The component of vector on the vector is (a) 19 , (b) 19/9 , (c) 9/19 , (d) 9 . (4) If =0 and =3 , =5 , =7 then angle between and is

(a) ,(b) , (c) , (d) .

LECTURE-3

Cross Product:Objective: Definition Finding normal vectors Some properties of cross product .

Definition

The cross product of a and b, written a x b, is defined by: a x b = n a b sin where a and b are the magnitude of vectors a and b; is the angle between the vectors, and n is the unit vector (vector with magnitude = 1) that is perpendicular (at 90 degrees to/ orthogonal to/ normal to) both a and b. But there are two vectors that this could be - one on either side of the plane formed by the two vectors), so we choose n to be the one which makes (a, b, n) a right handed triad.

Like in the definition of the dot product where we pulled out of a hat and said it was the angle between the two vectors without any way of finding it, so we need a way of finding n for out definition of the cross product to be any use. Again the i, j, k vectors come to our rescue, giving us an equivalent definition: let, a = a1 i + a2 j + a3 k b = b1 i + b2 j + b3 k then, a x b = ( a1 i + a2 j + a3 k) x (b1 i + b2 j + b3 k) The cross product of any two parallel vectors is the null vector since sin 0 = 0, and also i x j = k j x k = i k x i = j and j x i = -k k x j = -i i x k = -j Using these, we can eventually find: a x b = (a2b3 - a3b2)i + (a3b1 - a1b3)j + (a1b2 - a2b1)k That's our equivalent definition. If you're familiar with determinants you may see this can be written more conveniently as,

| i j k| | a1 a2 a3| | b1 b2 b3|

Finding the normal vectors

Using the equivalence of our two definitions, a x b = (a2b3 - a3b2)i + (a3b1 - a1b3)j + (a1b2 - a2b1)k = n a b sin you can now find n To get used to the cross product try out this cross product applet

Some Properties of the Cross Product

(1)The cross product is anti commutative: a x b = - b x a (2)The cross product of parallel vectors is the null vector, in particular: a x a = 0 (3)Also | a x b | is the area of the parallelogram formed by a and b,

a = a b sin

Worked out examples:

(1)Find the unit vector perpendicular to both 1 ˆˆ ˆ( 3 5 11 )

| | 155

a bi j k

a b

and

. Find also the angle between them.

Solution:Let = and =

=(1-4) -(-2-3) +(8+3)

=-3 +5 +11

Hence the required unit vector is

Now | | = = , =

So the required angle between the given vectors is

=

(2) Show that .

Solution: We have ,

= . =

=

(3) Prove that

Solution: L.H.S =

Assignment:

(1) Find a vector of magnitude 9 which is perpendicular to both the vector

and .

(2) If are unit vectors and is the angle between them , show that

(3) Find a unit vector perpendicular to each of the vector and and obtain the sine of the angle between them.

Multiple Choice Questions:

(1) If , and then is

(a) , (b) (c) , (d) none

(2) If = 0 , then the vectors are

(a) coplanar , (b) independent , (c) collinear ,(d) none