B.Tech 1st Year 1st Semester Mathematics(M101)

Teacher Name: Kakali Ghosh, Rahuldeb Das, Amalendu Singha Mahapatra, Raicharan Denra

LECTURE-- 1

Vector algebra and vector calculus

Objective : Vectors are frequently used in many branches of pure and applied mathematics and in physical and engineering science. Objective of Vector algebra is to learn a set of rules which are gainfully employed in combining a vector with another vector or a vector with a scalar.

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

,

b

P Q -

Y

ZP

X

O

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

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

P QO

B

X

|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 = 0

Therefore 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