st
st
B.Tech 1 Year 1 Semester Mathematics(M101) Teacher Name: Kakali Ghosh
LECTURE-- 1 Vector algebra Objective : Vectors are frequently used in many branches o f pure and applied mathematics and in physical and engineering science. Objective Object ive of Vector algebra is to learn a set of rules which are gainfully employed in combining a vecto r 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 o f straight line on which there are distinct initial and t erminal points. The arrows indicate indicate the direction of vectors. The length length of the line segment is the magnitude magnitude uuur of the vector. For example, PQ is a vector directed from from P to Q. PpQ r a
uuur r Thus PQ = a .
r
Unit vector: A vector a whose magnitude is unity is called unit vecto r and is denoted by aÖ .
r
r
Null Vector : A vector a whose magnitude is 0 is called Null vector, denoted by O .
r
r
Equal vector : If two vectors a ( a ) and b ( b ) are said to be equal if they have equal magnitudes
r
r
and same direction and denoted by a ! b . Addition
of two vectors : Let a and b be any two given vectors.
r
r
a + b
B
r
r
a
O
b
A
If three points O , A , B are taken such that
uuur
A
r
r uuur r uuur AB = b , then the vector O B is called vector r uuur r r
= a,
sum or the resultant of the given g iven vectors and a and b and write as
O B
= a + b.
r r r r Subtraction of two vectors: We define the difference a b of two vectors a and b to be the sum r r r r r r of the vectors a and - b , i.e. a b = a ( b ) r
Multiplication of a vector by a real number: Let P be scalar. Then P a is a vector whose r r r magnitude is | P | times that of a and direction is the same as that of a or opposite of a , according as P is positive or negative.
r
r
r
r
Co llinear or parallel if a = P b where P Collinear vectors: Two vectors a and b are said to be Collinear is a scalar. A system of vectors is said to be collinear if they are parallel to the t he same straight line. Cop lanar if they are parallel are parallel to the same plane. p lane. Coplanar vectors: A system of vectors is said to be Coplanar Linearly dependent and Linearly in dependent vectors:
A
x y z set of vectors s uuur , uuur , uuur is OP
OP
OP
r
said to be linearly linearly dependent , if there exist a set of scalars scalars x,y,z,««not all zero, such that xa +
r
r
y b + z c +«««..= 0. Otherwise they form a linearly independent set of vectors. Thus for a set of linearly independent
r r r r r r vectors a, b , c , .... if x a + y b + z c +«««..= 0 , then we have x = y = z = ««..= 0.
vector of a point: The position vector ( vector (p.v ) of a point point P with respect to a fixed fixed origin O uuur uuur r r r in space is the vector O P . If O P = a , we write P ( a ) as the position vector of P is a . Position
P
r
r
Q
b - a
r
a ar
r
b O
r r uuur If a and b are position vectors vectors of P and Q respectively respectively , then PQ = b - a = p.v. Q ± p.v. of P. The position vector of the po int P whose Cartesian coordinates are (x,y,z) is given by uuur r r 2 2 2 Ö Ö Ö r ! xi yj zk . Obviously | r | = x y z where direction cosines of O P = x y z ( uuur , uuur , uuur ). | O P | | O P | | O P |
Z
P
Y
X
EF K with the rectangular axes at O ( The figure above). Then cos E , cos F , cos K Let OP makes EF are called the direction cosines of OP and we can write uuur uuur uuur x = | O P |cos E , y = | O P |cos F , z = | O P |cos K uuur The unit vector in the direction of O P is given by uuur 1 O P uuur ! uuur ( xÖ i Öyj zÖk) ! cos Ö i cos Ö j cos K Ök, O P O P where iÖ, Öj , k Ö are unit vectors along the coordinate coo rdinate axes and (x,y,z) is position of P w.r.t O.
uuur or component of a vector on an axis : Let A 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. Projection
B A Q
P
O
X
Similarly , we take point point of projection Q of B on OX. Then PQ is called projection or component of uuur the vector A on the axis OX. uuur uuur uuur If A makes an angle U with OX , then component of A B on OX = | A B |Cos U .
Illustrative
examples:
1) Show that the vectors i ± 3j + 5k , 3i ± 2j+k , 2i + j - 4k form a right angle triangle. t riangle.
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 2
2
2
|a| =
1 ( 3) 5
=
35
|b| =
3 ( 2) 1 =
14
2
2
2
|c| =
@
b
2 (1) ( 4) = 2
2
c
2
2
2
21
! 1 4 2 1 ! 35 !
a
2
. Therefore the given vecto rs form a right angle triangle.
2.a) Show that the vectors (2,4,10) an d (3,6,15) are linearly dependent. 2.b) Show that the vectors (1,2,3) an d (4,-2,7) are linearly in dependent.
(3,6,15) Soln 2a) Let a = (2,4,10) and b = (3,6,15) Let x and y be two scalars , such that xa+yb=0 x(2,4,10) + y(3 y(3,6,15) ,6,15) = 0 or, or, (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. 1,2,3) and b = (4,-2,7) Soln 2b) Let a = (1,2,3
Let x and y be two scalars , such that xa+yb=0 Therefore x( x(1,2,3 1,2,3) + y( 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 co planar , then it will be possible to express one o f 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 =3 , -2x + y = -7 , x + 2y = -4 Solving the first two equations we get , x= 2 and y = -3 -3 . These values of x and y satisfy the 3rd equation. Thus 3a ± 7b -4c = 2 (3a -2b + c) + (-3) (a + b +2c) st
Therefore the 1 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
2,3,-6) , (6,-2,3 6,-2,3) and (4,-5,9) form the sides of an isosceles triangle. 2) Prove that the vectors (2,3
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 P and Q for which the vectors (-3i + 4j + P k ) and ( Q i + 8j + 6k ) are collinear. 5) Find the constant constant m such that the vectors
r
a
) r ) ) ) ) ) r ) ) ) ! 2 i j k , b ! i 2 j 3k , c ! 3i m j 5k are coplanar
Multiple Choice Questions ) ) ) 1)The unit vector along the vector 2i j k is ) ) ) ) ) ) ) ) ) 1 1 (a) 2i j 5k (b) 2i j 5k (c) 2i j 5k 30 30 r r r r r r r r r r 2) If a ! 2 i 3 j k , b ! 6 i 9 j 3 k then a and b are (a) Coplanar (b) in dependent (c) collinear ( d) none
(d) none
r r 3)If for two vectors a and b r )r ) r ) ) ) ) r ) ) ) r r r r r | a b m j 5 ka 2 i j k, b i 2 j 3k , c ! 3i | = | a b | , then a and b are (a) Parallel (b) orthogonal (c) collinear ( d) none
Product
of vectors
Scalar product or dot product or inner product: The scalar product of two vectors a and is the the scal scalar ar qua quanti ntity ty,, | a | | b | cosU and is denoted b, whose directions are inclined at an angle U , is by a. b.
B
b U
O
a
N
A
Thus a.b = |a||b| cos U , where |a| and |b| represent the magnitudes of the vectors a and b respectively. The scalar product is positive or negative according as U is acute or obtuse.
Properties
on scalar pro duct:
1) Scalar product of two vectors is commutative i.e. a.b = b. a 2) If U is the angle between a and b, then r r ¨ a .b ¸ U ! cos 1 © r r ¹ ª | a || b | º 3) If two vectors a and b are perpendicular to each other, then U ! 90o and cos U ! 0 . Therefore a.b = 0. This is the condition of orthogonali orthogo nality ty of o f two non zero vectors a and b r 4) From definition of dot product we get a.a = |a||a|cos0 = | a |2 , write as | a |2 = a 2
5) For the unit vectors iÖ, iÖ. Öj ! Öj.kÖ ! kÖ.iÖ ! 0 r r r r r r r 6) a.(b c ) ! a.b a.c
Ö =1 and Öj , k Ö , we have iÖ 2 ! iÖ.iÖ = 1 . Similarly Öj2 ! Öj. Öj = 1 and kÖ 2 ! kÖ.k
(Distributive law.)
r r r r 7) P ( a b ) ! P a P b
8) Component of
r
a
along
r
b
r r = | a | cos U ! r |b | r
a.b
