Vector Theory E

MATHS

Vector Vectors and their representation :

Vector quantities are specified by definite magnitude and definite direction. A vector is generally represented by a directed line segment, say AB . A is called the initial point and B is called the terminal point. The magnitude of vector AB is expressed by  AB . Zero vector: A vector of zero magnitude i.e. which has the same initial and terminal point, is called a zero vector. It is denoted by O. The direction of zero vector is indeterminate. Unit vector:   A vector of unit magnitude in the direction of a vector a is called unit vector along a and is denoted by  ˆa , symbolically aˆ  a . |a| Example # 1 : Find unit vector of ˆi  2ˆj  3kˆ Solution :



 a = ˆi  2ˆj  3kˆ if

 a = a x ˆi + a y ˆj + a zkˆ

then

 |a| =

 |a| =

14

 a aˆ = | a | =

1

ˆ – 14 i

2

2

ax  ay  az

2 14

3

ˆj +

14

2

kˆ

Equal vectors: Two vectors are said to be equal if they have the same magnitude, direction and represent the same physical quantity. Collinear vectors: Two vectors are said to be collinear if their directed line segments are parallel irrespective of their directions. Collinear vectors are also called parallel vectors. If they have the same direction they are named as like vectors otherwise unlike vectors.     Symbolically, two non-zero vectors a and b are collinear if and only if, a  b , where   R   a1 = b1, a2 = b2, a3 = b3 a  b  a1ˆi  a 2 ˆj  a 3 kˆ =  b1ˆi  b 2 ˆj  b 3kˆ 









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MATHS 

a a1 a = 2 = 3 ( =) b1 b2 b3

 a3 a1 a2  Vectors a = a1 ˆi + a 2 ˆj + a 3kˆ and b = b1 ˆi + b 2 ˆj + b 3kˆ are collinear if = = b1 b2 b3

Example # 2 : Find values of x & y for which the vectors  a = (x + 2) ˆi – (x – y) ˆj + kˆ  b = (x – 1) ˆi + (2x + y) ˆj + 2 kˆ are parallel. Solution :

  1 yx x2 = = a and b are parallel if 2 2 x  y x 1

x = – 5, y = – 20 Coplanar vectors: A given number of vectors are called coplanar if their line segments are all parallel to the same plane. Note that “two vectors are always coplanar”.

Multiplication of a vector by a scalar :    If a is a vector and m is a scalar, then m a is a vector parallel to a whose magnitude is m times that    of a . This multiplication is called scalar multiplication. If a and b are vectors and m, n are scalars, then :       , m (a )  (a ) m  m a m (na )  n(m a )  (mn )a        , (m  n ) a  m a  n a m (a  b )  m a  m b

Self Practice Problems : (1)

(2)

(3)

Given a regular hexagon ABCDEF with centre O, show that (i)

OB – OA = OC – OD

(iii)

AD + EB + PC = 4 AB

(ii)

OD + OA = 2 OB + OF

 The vector  ˆi  ˆj  kˆ bisects the angle between the vectors c and 3 ˆi  4ˆj . Determine the unit  vector along c . The sum of the two unit vectors is a unit vector. Show that the magnitude of the their difference is 3 .

Answers :

(2)



1 ˆ 2 ˆ 14 ˆ i j k 3 15 15

Addition of vectors : (i)

    If two vectors a and b are represented by OA and OB , then their sum a  b is a vector

represented by OC , where OC is the diagonal of the parallelogram OACB.

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MATHS

(iv)

    a  b  b  a (commutative)      a0  a  0a

(vi)

    |ab||a| |b|

(ii)

(v)

      (a  b)  c  a  ( b  c) (associative)      a  ( a )  0  (  a )  a

(vii)

    |ab|  ||a|  |b||

(iii)

  Example # 3 : If a  ˆi  2ˆj  3kˆ and b  2ˆi  4ˆj  5kˆ represent two adjacent sides of a parallelogram, find unit

Solution :

vectors parallel to the diagonals of the parallelogram.   Let ABCD be a parallelogram such that AB = a and BC = b . Then,



AB + BC = AC   AC = a  b = 3ˆi  6ˆj  2kˆ

| AC | =

9  36  4 = 7

AB + BD = AD



  BD = AD  AB = b  a = ˆi  2ˆj  8kˆ | BD | =

1  4  64 =

69





AC 1 Unit vector along AC = = 3ˆi  6ˆj  2kˆ 7 | AC |

and

Unit vector along BD =

BD | BD |

1 =

69



ˆi  2ˆj  8kˆ 

Example # 4 : ABCDE is a pentagon. Prove that the resultant of the forces AB , AE , BC , DC , ED and AC

Solution :

is 3 AC .  Let R be the resultant force   R = AB + AE + BC + DC + ED + AC   R = ( AB + BC ) + ( AE + ED + DC ) + AC   R = AC + AC + AC   R = 3 AC . Hence proved.

Position vector of a point:   Let O be a fixed origin, then the position vector of a point P is the vector OP . If a and b are position vectors of two points A and B, then   AB = b  a = position vector (p.v.) of B  position vector (p.v.) .) of A.

DISTANCE FORMULA

    Distance between the two points A (a) and B (b) is AB = a  b SECTION FORMULA   If a and b are the position vectors of two points A and B, then the p.v. of

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MATHS    na  m b a point which divides AB in the ratio m: n is given by r  . mn   ab Note : Position vector of mid point of AB = . 2 Example # 5 : ABCD is a parallelogram. If L, M be the middle point of BC and CD, express AL and AM in

3 AC . 2   Let the position vectors of points B and D be respectively b and d referred to A as origin of reference. terms of AB and AD . Also show that AL + AM =

Solution :

Then AC = AD + DC = AD + AB [ DC = AB ]       AB = b , AD = d AC = d + b   i.e. position vector of C referred to A is d + b 

AL = p.v. of L, the mid point of BC .

=





1 1    1 [p.v. of B + p.v. of C] = b  d  b = AB + 2 2 2 AD 1 AM = 2

Similarly

d  d  b  = AD + 21

AB

  1  1  b AL + AM = b + d + d+ 2 2



=

 3  3  3  3 b + d = (b + d ) = AC . 2 2 2 2

Example # 6 : If ABCD is a parallelogram and E is the mid point of AB. Show by vector method that DE trisect AC and is trisected by AC.   Solution : Let AB = a and AD = b    Then BC = AD = b and AC = AB + AD = a + b Also let K be a point on AC, such that AK : AC = 1 : 3 

AK =



AK =

1 AC 3  1  (a + b ) 3

.........(i)

Again E being the mid point of AB, we have



AE =

1  a 2

Let M be the point on DE such that DM : ME = 2 : 1   ba AD  2AE AM = = ..........(ii) 3 1 2 From (i) and (ii) we find that AK =

 1  ( a + b ) = AM , and so we conclude that K and M coincide. i.e. DE trisect AC and is 3

trisected by AC. Hence proved.

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MATHS Self Practice Problems (4)

Express vectors BC , CA and AB in terms of the vectors OA , OB and OC

(5)

    If a, b are position vectors of the points (1, –1), (–2, m), find the value of m for which a and b are collinear.

(6)

The position vectors of the points A, B, C, D are ˆi  ˆj  kˆ , 2ˆi  5ˆj , 3ˆi  2ˆj  3kˆ , ˆi  6ˆj  kˆ respectively. Show that the lines AB and CD are parallel and find the ratio of their lengths.

(7)

(8)

   The vertices P, Q and S of a PQS have position vectors p, q and s respectively..   (i) If M is the mid point of PQ, then find position vector of M in terms of p and q    (ii) Find t , the position vector of T on SM such that ST : TM = 2 : 1, in terms of p, q  and s .  (iii) If the parallelogram PQRS is now completed. Express r , the position vector of the    point R in terms of p, q and s D, E, F are the mid-points of the sides BC, CA, AB respectively of a triangle.

Show FE = (9)

1 BC and that the sum of the vectors AD , BE , CF is zero. 2

The median AD of a ABC is bisected at E and BE is produced to meet the side AC in F. Show that AF =

(10)

1 1 AC and EF = BF.. 3 4

Point L, M, N divide the sides BC, CA, AB of ABC in the ratios 1 : 4, 3 : 2, 3 : 7 respectively. Prove that AL + BM + CN is a vector parallel to CK , when K divides AB in the ratio 1 : 3. Answers :

(4) (6)

BC  OC  OB , CA  OA  OC , AB  OB  OA 1:2

(7)

 1   (p  q) , m = 2

(5)

m=2

  1    1    (p  q  s ) , r = (q  p  s) t = 2 2

Angle between two vectors : It is the smaller angle formed when the initial points or the terminal points of the two vectors are brought together. Note that 0º    180º .

Vector equation of a line :

      Parametric vector equation of a line passing through two point A (a ) and B(b) is given by r = a  t(b  a) ,   where 't' is a parameter. If the line passes through the point A (a ) and is parallel to the vector b , then     its equation is r  a  t b . Note : r is the p.v. of the point on the line.

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MATHS   A vector in the direction of the bisector of the angle between the two vectors a and b is





  a b    . a b

  Hence bisector of the angle between the two vectors a and b is  a  b , where  R+. Bisector





  of the exterior angle between a and b is  a  b ,  R+.













Note that the equations of the bisectors of the angles between the lines r = a +  b and r = a +  c are :

 

 

    r = a + t b  c and r = a + p c  b .

Scalar product (Dot Product) of two vectors : Geometrical interpretation of scalar product :   Let a and b be vectors represented by OA and OB respectively. Let  be the angle between OA and

OB . Draw BL  OA and AM  OB. From OBL and OAM, we have OL = OB cos  and OM = OA cos . (b)   Here OL and OM are known as projections of b on a   and a on b respectively..        Now, = | a | | b | cos  = | a |(| b | cos  ) a.b (a)   = | a | (OB cos  ) = | a | (OL)    = (Magnitude of a ) (Projection of b on a ) ........(i)       Again a . b = | a | | b | cos  = | b | (| a | cos  )   = | b | (OA cos ) = | b | (OM)    = (magnitude of b ) (Projection of a on b ) ........(ii) Thus geometrically interpreted, the scalar product of two vectors is the product of modulus of either vector and the projection of the other in its direction. (i) (ii) (iii)

ˆi . ˆi = ˆj . ˆj = kˆ . kˆ = 1; ˆi . ˆj = ˆj . kˆ = kˆ . ˆi = 0   a . b  Projection of a on b   |b|     If a = a1 ˆi + a2 ˆj + a3 kˆ and b = b1 ˆi + b2 ˆj + b3 kˆ , then a . b = a1b1 + a2b2 + a3b3

 a  (iv)

(v)

2

2

a1  a 2  a 3

2

,

 b 

2

2

b1  b 2  b 3

2

    a.b  , 0  The angle  between a and b is given by cos    |a| |b|

    a . b  a b cos , (0    )     note that if  is acute, then a . b > 0 and if  is obtuse, then a . b < 0

(vi) (vii)

  a  b =

    | a |2  | b |2  2 | a || b | cos  , where  is the angle between the vectors

  2  a . a  a  a2

(viii)

    a . b  b . a (commutative)

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MATHS (ix) (x) (xi) Note: (a) (b) (c)

       a . (b  c )  a . b  a . c (distributive)       a.b  0  a  b (a  0 , b  0 )       (m a ) . b = a . (mb ) = m (a . b) (associative), where m is a scalar..

    a . b is  a  b      Minimum value of a . b is –  a  b       Any vector a can be written as a = a . i  i  a . j j  a . k  k . Maximum value of

  Example # 7 : Find the value of p for which the vectors a  3 ˆi  2ˆj  9kˆ and b  ˆi  pˆj  3kˆ are (i) Solution :

(i)

perpendicular   ab 

(ii)

(ii)

 3 + 2p + 27 = 0   vectors a = 3ˆi  2ˆj  9kˆ and b = ˆi  pˆj  3kˆ are parallel iff

  a.b = 0

parallel

3ˆi  2ˆj  9kˆ  . ˆi  pˆj  3kˆ  = 0



2 9 3 = = p 3 1



3=

2 p





p=

p = – 15

2 3

         Example # 8 : If a + b + c = 0 , | a | = 3, | b | = 5 and | c | = 7, find the angle between a and b .     Solution : We have, a  b  c  0            a  b . a  b =  c  .  c  ab = –c





  ab



 a



2

2

 = | c |2

 + b

2

 a



 +2 a

 b

 

2

 + b

 cos  = c

9 + 25 + 2 (3) (5) cos  = 49





2

   + 2a . b = c

2

2

cos  =

1 2



=

 . 3

  Example # 9 : Find the values of x for which the angle between the vectors a = 2x 2 ˆi + 4x ˆj + kˆ and b = 7 ˆi

– 2 ˆj + x kˆ is obtuse. Solution :

  a.b   |a||b|   a.b  