Vector Theory E 705g42

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. 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 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 of p and q    (ii) Find t , the position vector of T on SM such that ST : TM = 2 : 1, in of p, q  and s .  (iii) If the parallelogram PQRS is now completed. Express r , the position vector of the    point R in 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 ing through two point A (a ) and B(b) is given by r = a  t(b  a) ,   where 't' is a parameter. If the line es 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   <0 |a||b|

  The angle  between vectors a and b is given by cos  =

Now,  is obtuse





  a.b < 0



14x 2 – 8x + x < 0

cos  < 0



  [ | a |, | b |  0 ] 

7x (2x – 1) < 0

1 2 Hence, the angle between the given vectors is obtuse if x  (0, 1/2) 

x(2x – 1) < 0



0<x<

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MATHS Example # 10 : D is the mid point of the side BC of a ABC, show that AB2 + AC2 = 2 (AD2 + BD2) Solution :

We have AB = AD + DB



AB2 = ( AD  DB )2



AB2 = AD2 + DB2 + 2 AD . DB

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

Also we have AC = AD + DC 

2 AC2 = ( AD  DC)



AC2 = AD2 + DC2 + 2 AD . DC

........(ii)

Adding (i) and (ii), we get AB2 + AC2 = 2AD2 + 2BD2 + 2 AD . (DB  DC )   AB2 + AC2 = 2(AD2 + BD2)  DB + DC = 0   Example # 11 : If a = ˆi + ˆj + kˆ and b = 2 ˆi – ˆj + 3 kˆ , then find      (i) Component of b along a . (ii) Component of b in plane of a & b  but  to a .    a.b     Solution : (i) Component of b along a is   2  a | a |    Here a . b = 2 – 1 + 3 = 4  | a |2 = 3

   a.b   4  4   Hence   2  a = a = ( ˆi + ˆj + kˆ ) 3 3 | a |  (ii)

    a.b    1     Component of b in plane of a & b but  to a is b –   2  a . = 2ˆi  7 ˆj  5kˆ 3 | a | 





Self Practice Problems : (11) (12)

(13)

(14)

  |ab|    If a and b are unit vectors and  is angle between them, prove that tan =   . |ab| 2   Find the values of x and y if the vectors a = 3ˆi  xˆj  kˆ and b = 2ˆi  ˆj  ykˆ are mutually perpendicular vectors of equal magnitude.

   Let a = x 2 ˆi  2ˆj  2kˆ , b = ˆi  ˆj  kˆ and c = x 2 ˆi  5 ˆj  4kˆ be three vectors. Find the values of x     for which the angle between a and b is acute and the angle between b and c is obtuse.

      The points O, A, B, C, D are such that OA  a , OB  b , OC  2a  3b , OD  a  2b . Given that the length of OA is three times the length of OB . Show that BD and AC are perpendicular.

(15)

ABCD is a tetrahedron and G is the centroid of the base BCD. Prove that AB2 + AC2 + AD2 = GB2 + GC2 + GD2 + 3GA2

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MATHS Answers :

(12)

x=–

31 41 , y= 12 12

(13)

(– 3, –2)  (2, 3)

Vector product (Cross Product) of two vectors: (i)

     If a , b are two vectors and  is the angle between them, then a x b  a

 b sin  nˆ , where nˆ is the

    unit vector perpendicular to both a and b such that a, b and nˆ forms a right handed screw system. (ii)

Geometrically

  axb

= area of the parallelogram whose two adjacent sides are represented by

  a and b .

(iii)

 ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0 ; ˆi  ˆj  kˆ, ˆj  kˆ  ˆi, kˆ  ˆi  ˆj

(iv)

ˆi ˆj     If a = a1 ˆi +a2 ˆj + a3 kˆ and b = b1 ˆi + b2 ˆj + b3 kˆ , then a  b  a1 a 2 b1 b 2

(v) (vi) (vii) (viii)

kˆ a3 b

3     (not commutative) axbbxa     (m a )  b = a  m b = m a  b (associative), where m is a scalar..        a x ( b  c )  (a x b)  (a x c ) (distributive)          a  b  0  a and b are parallel (collinear) (a  0 , b  0 ) i.e. a  K b , where K is a scalar..

( )

(

)

(ix)

    ab Unit vector perpendicular to the plane of a and b is nˆ =    | ab |

(x)

    r (a  b ) A vector of magnitude ‘r’ and perpendicular to the plane of a and b is    | ab |

(xi)

(xii)

  axb   If  is the angle between a and b, then sin     a b

   If a, b and c are the position vectors of 3 points A, B and C respectively, then the vector area of ABC =

(xiii)

(xiv)

    1          a x b  bx c  cx a . The points A, B and C are collinear if a x b  b x c  c x a  0 2





   1  d1 x d2 Area of any quadrilateral whose diagonal vectors are d1 and d2 is given by 2

   2   Lagrange's Identity : For any two vectors a and b ; (a x b)  a

2

 b

2

    a.a  (a . b ) 2    a.b

  a.b   b.b

Example # 12 : Find a vector of magnitude 9, which is perpendicular to both the vectors 4 ˆi – ˆj  3kˆ and  2ˆi  ˆj  2kˆ .

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9

MATHS Solution :

  Let a = 4ˆi  ˆj  3kˆ and b =  2ˆi  ˆj  2kˆ . Then

  ab =

ˆj

4

1

3

2

1

2

  | ab | =

 

ˆi

kˆ = (2 – 3) ˆi – (–8 + 6) ˆj + (4 – 2) kˆ =  ˆi  2ˆj  2kˆ

( 1)2  2 2  2 2 = 3

   ab  9 Required vector =  9     =  ( ˆi  2ˆj  2kˆ ) = ± (  3 ˆi  6 ˆj  6kˆ ) 3 | a  b |  

   Example # 13 : For any three vectors a, b, c , show that        Solution : We have, a × (b  c ) + b × (c  a) + c             = ab  ac  bc  ba  c a  c b             = ab  ac  bc  ab  ac bc =

 Example # 14 : For any vector a , prove that Solution :

          a  (b  c )  b  ( c  a )  c  ( a  b )  0 .   × (a  b )  0

[Using distributive law]     [ b  a  – a  b etc]

    | a  ˆi |2 + | a  ˆj |2 + | a  kˆ |2 = 2 | a |2

 Let a = a1ˆi  a 2 ˆj  a 3 kˆ . Then  ˆ ˆ a  ˆi = (a1ˆi  a 2 ˆj  a 3kˆ ) × ˆi = a1 ( ˆi  ˆi ) + a2 ( ˆj  ˆi ) + a3 (kˆ  ˆi ) = –a2 k  a 3 j   | a  ˆi |2 = a22 + a32

 a  ˆj = (a1ˆi  a 2 ˆj  a 3kˆ ) × ˆj = a1kˆ  a 3 ˆi 

 | a  ˆj |2 = a21 + a32



  a  kˆ  (a i ˆi  a 2 ˆj  a 3 kˆ ) × k = – a i ˆj  a 2 ˆi  | a  kˆ |2 = a12 + a22



   | a  ˆi |2 + | a  ˆj |2 + | a  kˆ |2 = a22 + a33 + a12 + a32 + a12 + a22  = 2 (a12 + a22 + a32) = 2 | a |2

    Example # 15 : Let OA = a , OB = 10 a + 2b and OC = b where O is origin. Let p denote the area of the quadrilateral OABC and q denote the area of the parallelogram with OA and OC as adjacent sides. Prove that p = 6q. Solution : We have, p = Area of the quadrilateral OABC



p=

1 1 | OB  AC | = | OB  (OC  OA ) | 2 2



p=

       1 1      | (10a  2b)  (b  a) | = | 10(a  b)  10(a  a)  2(b  b)  2(b  a) | 2 2

  1     | 10(a  b)  0  0  2(a  b) | = 6 | a  b | .....(i) 2 and q = Area of the parallelogram with OA and OC as adjacent sides   q = | OA  OC | = | a  b | ........(ii) 

p=

From (i) and (ii), we get

p = 6q

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MATHS Self Practice Problems : (16)

  If p and q are unit vectors forming an angle of 30º. Find the area of the parallelogram having       a  p  2q and b  2p  q as its diagonals.

(17)

Prove that the normal to the plane containing the three points whose position vectors are          a, b, c lies in the direction b  c  c  a  a  b

(18)

ABC is a triangle and EF is any straight line parallel to BC meeting AC, AB in E, F respectively. If BR and CQ be drawn parallel to AC, AB respectively to meet EF in R and Q respectively, prove that  ARB = ACQ. Answers :

(16)

3/4 sq. units

Shortest distance between two lines : If two lines in space intersect at a point, then obviously the shortest distance between them is zero. Lines which do not intersect and are also not parallel are called skew line. For Skew lines the direction of the shortest distance would be perpendicular to both the lines. Let LM be the shortest distance vector between the lines L1 and L2 . Then LM is perpendicular to both     p and q i.e. LM is parallel to p × q . Therefore the magnitude of the shortest distance vector (i.e. | LM |) would be equal to that of the projection of AB along the direction of the line of shortest distance. 

(i) i.e. (ii)

  | LM |  Pr ojection of AB on LM = Pr ojection of AB on p x q       AB . (p x q) (b  a) . (p x q)  =     pxq pxq   The two lines directed along p and q will intersect only if shortest distance = 0             (b  a) . (p x q)  0 i.e. b  a lies in the plane containing p and q .  b  a · p  q  0 .       If two parallel lines are given by r1  a1  Kb and r2  a 2  Kb , then distance (d) between them is    b x (a 2  a1 ) given by d   b









Scalar triple product (Box Product) (S.T.P.) : (i)

         The scalar triple product of three vectors a, b and c is defined as: a x b . c  a b c sin  . cos  ,

       ) and  is the angle between a x b and c where  is the angle between a , b (i.e. a b θ





          (i.e. ( a  b) c = ) . It is (i.e. a  b . c ) also written as a b c and spelled as box product. (ii)

Scalar triple product geometrically represents the volume of the parallelopiped whose three coterminous edges are       represented by a, b and c i.e. V = [ a b c ]

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MATHS (iii)

(iv)

(v)

In a scalar triple product the position of dot and cross can be interchanged i.e.          a . (b x c )  (a x b) . c  [ a b c]  [ b c a ]  [ c a b ]         a . (b x c )   a . (c x b ) i.e. [ a b c ]   [ a c b ]

a1 a 2 a3       ˆ ˆ ˆ ˆ ˆ ˆ If a = a1 ˆi + a2 j + a3 k ; b = b1 ˆi +b2 j +b3 k and c = c 1 ˆi + c 2 j + c 3 k , then [ a b c ]  b1 b 2 b 3 . c1 c 2 c 3             In general, if a  a1  a 2 m  a 3n ; b  b 1   b 2 m  b 3 n and c  c 1   c 2 m  c 3n a1 a 2    then a b c  b1 b 2





c1 c 2 (vi) (vii) (viii)

a3





     b 3  m n , where , m and n are non-coplanar vectors. c3

      If a, b and c are coplanar  [ a b c ]  0 .  Scalar product of three vectors, two of which are equal or parallel is 0  [ a         If a , b , c are non-coplanar, then [a b c]  0 for right handed system and [a b

  b c]0 ,  c]  0 for left handed

system. (ix)

[ ˆi ˆj kˆ ] = 1

(xii)

a  b



(xiii)



      (x) [Ka b c ]  K [ a b c ]

        b  c c  a = 0 and a  b b  c



  a.a      2 a b c = b.a   c .a





  a.b   b .b   c .b

          (xi) [(a  b) c d ]  [ a c d ]  [ b c d ]      ca = 2 a b c .







  a.c   b.c   c .c

Tetrahedron and its properties : (a)

The volume of the tetrahedron OABC with O as origin and the position vectors of A, B and C being       a, b and c respectively is given by V  1 a b c 6     If the position vectors of the vertices of tetrahedron are a, b, c and d , then the position vector of its



(b)



1     (a  b  c  d) . 4 note that this is also the point of concurrency of the lines ing the vertices to the centroids of the opposite faces and is also called the centre of the tetrahedron. In case the tetrahedron is regular it is equidistant from the vertices and the four faces of the tetrahedron. centroid is given by

Example # 16 : Find the volume of a parallelopiped whose sides are given by  3ˆi  7 ˆj  5kˆ ,  5ˆi  7 ˆj  3kˆ and 7ˆi  5ˆj  3kˆ Solution :

   Let a  3ˆi  7ˆj  5kˆ , b  5 ˆi  7ˆj – 3kˆ and c  7ˆi  5 ˆj  3kˆ .    We know that the volume of a parallelopiped whose three adjacent edges are a, b, c is    [a b c ] .

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MATHS

Now

3 7 5  5 7 3 = –3 (–21 – 15) – 7 (15 + 21) + 5 (25 – 49) [a b c ] = 7 5 3

= 108 – 252 – 120 = –264    So required volume of the parallelopiped = [ a b c ] = | – 264 | = 264 cubic units

  Example # 17 : Simplify [ a  b Solution :

  bc

  c  a]

We have :             [By definition] [ a  b b  c c  a ] = {(a  b)  (b  c )} . (c  a)           = (a  b  a  c  b  b  b  c ) . ( c  a ) [By distribution law]            = (a  b  c  a  b  c ) . ( c  a ) [  bb  0 ]                   = (a  b) . c – (a  b) . a + (c  a) . c – (c  a) . a + (b  c ) . c – (b  c ) . a [By distribution law]                   = [a b c ] – [a b a] + [ c a c ] – [ c a a ] + [ b c c ] – [ b c a ]       = [a b c ] – [ b c a ] [ When any two vectors are equal, scalar triple product is zero ]             = [a b c ] – [a b c ] = 0 [ [ b c a ] = [a b c ] ]

    Example # 18 : Find the volume of the tetrahedron whose four vertices have position vectors a , b , c and d .     Solution : Let four vertices be A, B, C, D with position vectors a , b , c and d respectively..    = (a – d) DA   = (b – d) DB   = (c – d) DC

Hence volume V =

 1  [a – d 6

  b – d

  c – d]

=

    1   ( a – d ) . [( b – d ) × ( c – d )] 6

=

     1    (a – d ) . [b × c – b × d + c × d ] 6

=

   1          {[ a b c ] – [ a b d ] + [ a c d ] – [ d b c ]} 6

=

   1          {[ a b c ] – [ a b d ] + [ a c d ] – [ b c d ]} 6

   Example # 19 : Show that the vectors a  2 ˆi  4 ˆj  2 kˆ , b  4 ˆi  2 ˆj  2 kˆ and c  2 ˆi  2 ˆj  4 kˆ are coplanar..

Solution :

   [a b c ] =

2

4

2

 2  2 = – 2(–8 – 4) – 4(16 – 4) – 2(–8 – 4) 2 2 4 4

= 24 – 48 + 24 = 0    So vectors a , b , c are coplanar

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13

MATHS Self Practice Problems :





(19)

         Show that {( a + b + c ) × ( c – b )} . a = 2 a b c .

(20)

      Show that a . (b  c )  (a  b  c )  0

(21)

One vertex of a parallelopiped is at the point A (1, –1, –2) in the rectangular cartesian co-ordinate. If three adjacent vertices are at B(–1, 0, 2), C(2, –2, 3) and D(4, 2, 1), then find the volume of the parallelopiped.

(22)

Find the value of m such that the vectors 2ˆi  ˆj  kˆ , ˆi  2ˆj  3kˆ and 3ˆi  mˆj  5kˆ are coplanar..

(23)

         Show that the vector a, b, c are coplanar if and only if b  c , c  a , a  b are coplanar.. Answers : (21) 72 (22) –4

Vector triple product :

      Let a, b and c be any three vectors, then the expression a x (b x c) is a vector & is called a vector triple product.    Geometrical interpretation of a x (b x c)







Consider the expression a x ( b x c ) which itself is a vector, since it is a cross product of two vectors            a and (b x c) . Now a x ( b x c ) is a vector perpendicular to the plane containing a and (b x c) but b x c      is a vector perpendicular to the plane containing b and c , therefore a x ( b x c ) is a vector which lies         in the plane of b and c and perpendicular to a . Hence we can express a x (b x c) in of b and c      i.e. a x (b x c) = xb  yc , where x , y are scalars.   

         a x ( b x c ) = (a . c) b  (a . b) c          (a x b) x c = (a . c) b  (b . c) a       In general (a x b) x c  a x (b x c)

      Example # 20 : For any vector a , prove that ˆi  (a  ˆi )  ˆj  (a  ˆj )  kˆ  (a  k ) = 2a Solution :

Let Then

 a  a1ˆi  a 2 ˆj  a 3 kˆ . ˆi  (a  ˆi )  ˆj  (aˆ  ˆj )  kˆ  (a  kˆ )       = { ( ˆi . ˆi ) a  ( ˆi . a) ˆi } + {( ˆj . ˆj ) a  ( ˆj . a) ˆj} + {(kˆ . kˆ ) a  (kˆ . a) kˆ }       = {(a  ( ˆi . a) ˆi }  {a  ( ˆj . a) ˆj} + {a  (kˆ . a) kˆ }         = 3a  {( ˆi . a) ˆi  ( ˆj . a) ˆj  (kˆ . a) kˆ = 3a  (a1ˆi  a 2 ˆj  a 3kˆ ) = 3a  a  2a

Example # 21 :

            Prove that a  {b  (c  d)} = (b . d)(a  c ) – (b. c ) (a  d)            Solution : We have, a  {b  (c  d)} = a  {(b . d) c  (b . c ) d}         = a  {(b . d) c }  a  {(b . c ) d} [by dist. law]         = (b . d) (a  c )  (b . c ) (a  d)

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MATHS    Example # 22 : Let a =  ˆi + 2ˆj – 3kˆ , b = ˆi + 2ˆj – 2kˆ and c = 2ˆi – ˆj + kˆ . Find the value(s) of , if any,,        such that ×  ca  = 0. ab  bc             Solution : ×  ca  = a b c b ×  ca  ab  bc          = a b c a.b c  b.c a          which vanishes if (i) a . b c = b . c a (ii) a b c = 0       (i) a . b c = b . c a leads to the equation 2 3 + 10  + 12 = 0, 2 + 6 = 0 and 6 – 6 = 0,

                           

(ii)





which do not have a common solution.    a b c =0





 1

3 2

2 2

2 



=0

3 = 2



1

=

2 3

     ab  a         Example # 23 : If A  B = a , A . a = 1 and A  B = b , then prove that A = and | a |2





    b  a  a | a |2 1   . B = | a |2 Solution :    

  

   Given A  B  a      a. A B = a.a       a . A  a .B = a . a  2   1 + a.B = |a|    a . B = | a |2 – 1    Given A  B  b      a A B = a × b         a .B A – a . A B  ab      | a |2 1 A  B = a  b





.....(i)



...........(ii)



     

...........(iii)

[Using equation (ii)]

solving equation (i) and (iii) simultaneously, we get        ab  a   b  a  a | a |2 1   and B = A = | a |2 | a |2





        Example # 24 : Solve for r satisfying the simultaneous equations r  b  c  b , r . a  0 provided a is not  perpendicular to b .        (r  c) × b = 0 Solution :  r  c and b are collinear     r  c  kb     ........(i) r = c  kb    r .a = 0

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MATHS  

   (c  kb ) . a = 0   a.c k=–   putting in (i) we get a.b

    a.c  r c   b a.b

       Example # 25 : If x  a  k x  b , where k is a scalar and a, b are any two vectors, then determine x in

Solution :



  of a, b and k.     x  a  k x  b ..........(i)

 Premultiply the given equation vectorially by a        a  ( x  a ) + k (a  x ) = a  b           (a . a) x  (a . x ) a  k(a  x )  a  b ..........(ii)  Premultiply (i) scalarly by a        [a x a] + k ( a . x ) = a . b     k(a . x )  a . b .......(iii)     Substituting x  a from (i) and a . x from (iii) in (ii) we get       (a . b)   1  x = a 2  k 2 kb  (a  b)  k a  

Self Practice Problems : (24)

         Prove that a  (b  c )  b  (c  a)  c  (a  b)  0 .

(25)

Find the unit vector coplanar with ˆi + ˆj + 2 kˆ and ˆi + 2 ˆj + kˆ and perpendicular to ˆi + ˆj + kˆ .

(26)

        Prove that a  {a  (a  b)}  (a . a) (b  a) .

(27)

  1     1      Given that x   (p . x ) p  q , show that p . x  p . q and find x in of p and q . 2 2 p

(28)

          If x . a = 0, x . b = 0 and x . c = 0 for some non-zero vector x , then show that [a b c ] = 0

            ( r . a ) (b  c ) ( r . b ) (c  a ) ( r . c ) (a  b )           (29) Prove that r = + + [a b c ] [a b c ] [a b c ]    where a, b, c are three non-coplanar vectors    p.q   1   ˆ ˆ Answers : (25) ± ( – j + k ) and x = q –   2  p 2 2|p| 

Linear combinations :   









Given a finite set of vectors a , b , c ,...... , then the vector r  xa  yb  zc  ........ is called a linear

  

(a)

combination of a , b , c ,...... for any x, y, z..... R. We have the following results:       If a , b are non zero, noncollinear vectors, then xa  yb  x ' a  y ' b  x  x ' , y  y '

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MATHS (b)

 



Fundamental Theorem in plane : Let a , b be non zero, non collinear vectors, then any vector r coplanar

    with a , b can be expressed uniquely as a linear combination of a and b i.e. (c)



 

there exist some unique x, y  R such that xa  yb  r .

  

If a , b , c are nonzero, noncoplanar vectors, then

      xa  yb  zc  x' a  y' b  z' c  x  x' , y  y' , z  z' (d)

  

Fundamental theorem in space: Let a , b , c be nonzero, noncoplanar vectors in space.

  



Then any vector r can be uniquely expressed as a linear combination of a , b , c

i.e. there exist

    some unique x,y, z  R such that xa  yb  zc  r . (e)

(f)

   If x 1,x 2 ,......, x n are n non zero vectors and k 1 ,k 2 ,.....,k n are n scalars and if the linear     com bination k 1x 1  k 2 x 2  .......  k n x n  0  k 1  0, k 2  0 ,....., k n  0 , then we say that vectors    x1, x 2, ......, x n are linearly independent vectors.    If x 1,x 2 ,......, x n are not linearly independent then they are said to be linearly dependent vectors.       i.e. if k 1x 1  k 2 x 2  k 3 x 3 ......  k r x r  ......  k n x n  0 and if there exists at least one k r  0, then    x 1, x 2, ......, x n are said to be linearly dependent vectors.

      Note 1: If k r  0; k 1x 1  k 2 x 2  k 3 x 3 ......  k r x r  ......  k n x n  0 

       k r x r  k 1x1  k 2 x 2  .......  k r 1 . x r 1  k r 1 . x r 1 ...... k n x n



k r

1  1  1  1  1  x r  k1 x1  k 2 x 2  ..... k r 1 . x r 1 ..... k n xn kr kr kr kr kr

      x r  c 1x 1  c 2 x 2  ......  c r –1x r –1 + c r 1x r 1  .......  c n x n       i.e. x r is expressed as a linear combination of vectors x1, x 2 ,........, x r 1, x r 1, .......... , x n       Hence x r with x1, x 2 , ........, x r 1, x r 1, ......., x n forms a linearly dependent set of vectors.



Note 2: 

  If a = 3 ˆi + 2 ˆj + 5 kˆ then a is expressed as a Linear Combination of vectors ˆi , ˆj , kˆ . Also

 a , ˆi , ˆj ,

kˆ form a linearly dependent set of vectors. In general, in 3 dimensional space every set of four vectors is a linearly dependent system. 





ˆi , ˆj , kˆ are Linearly Independent set of vectors. For  K1 ˆi + K2 ˆj + K3 kˆ = 0  K1= K2= K3 = 0         Two vectors a and b are linearly dependent  a is parallel to b i.e. a  b  0 linear dependence of a       and b . Conversely if a  b  0 then a and b are linearly independent.       If three vectors a, b, c are linearly dependent, then they are coplanar i.e. [a b c ] = 0. Conversely if    [a b c ]  0 then the vectors are linearly independent.

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MATHS Example # 26 : Given that position vectors of points A, B, C are respectively         a – 2 b + 3 c , 2 a + 3 b – 4 c , – 7 b + 10 c then prove that vectors AB and AC are linearly dependent. Solution : Let A, B, C be the given points and O be the point of reference then         and OA = a – 2 b + 3 c , OB = 2 a + 3 b – 4 c OC = – 7 b + 10 c Now AB = p.v. of B – p.v. of A    = OB – OA = ( a + 5 b – 7 c ) and



AC = p.v. of C – p.v of A    = OC  OA = – (a  5b  7c ) = – AB

AC =  AB where  = – 1. Hence AB and AC are linearly dependent

         Example # 27 : Prove that the vectors 5 a + 6 b + 7 c , 7 a – 8 b + 9 c and 3 a + 20 b + 5 c are linearly    dependent, where a , b , c being linearly independent vectors. Solution : We know that if these vectors are linearly dependent , then we can express one of them as a linear combination of the other two. Now let us assume that the given vector are coplanar, then we can write          5 a + 6 b + 7 c = ( 7 a – 8 b + 9 c ) + m (3 a + 20 b + 5 c ) where , m are scalars    Comparing the coefficients of a , b and c on both sides of the equation 5 = 7 + 3m ..........(i) 6 = – 8 + 20 m ..........(ii) 7 = 9 + 5m ..........(iii) From (i) and (iii) we get

1 = m which evidently satisfies (ii) equation too. 2 Hence the given vectors are linearly dependent . 4 = 8



=

Self Practice Problems : (30)

(31)

(32)

(33)

     Does there exist scalars u, v, w such that ue1  ve 2  we 3  ˆi where e1  kˆ , e 2  ˆj  kˆ ,  e 3   ˆj  2kˆ ?       Consider a base a, b, c and a vector  2a  3b  c . Compute the co-ordinates of this vector               relatively to the base p , q, r where p  2a  3b , q  a  2b  c , r  3a  b  2c .

      If a and b are non-collinear vectors and A =(x + 4y) a + (2x + y + 1) b and B = (y – 2x + 2)     a + (2x – 3y – 1) b , find x and y such that 3 A  2B .   If vectors a, b,c be linearly independent, then show that         (i) a  2b  3c ,  2a  3b  4c ,  b  2c are linearly dependent         (ii) a  3b  2c ,  2a  4b  c , 3a  2b  c are linearly independent.

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MATHS (34)

Given that ˆi  ˆj , ˆi  2ˆj are two vectors. Find a unit vector coplanar with these vectors and perpendicular to the first vector ˆi  ˆj . Find also the unit vector which is perpendicular to the plane of the two given vectors.

(35)

(36)

If with reference to a right handed system of mutually perpendicular unit vectors ˆi, ˆj, kˆ ,           3 ˆi  ˆj ,   2ˆi  ˆj  3kˆ . Express  in the form   1   2 where 1 is parallel to  and   2 is perpendicular to  .  Prove that a vector r in space can be expressed linearly in of three non-coplanar,,              [ r b c] a  [ r c a ] b  [ r a b ] c    non-zero vectors a, b, c in the form r     [ a b c]

Answers :

(30)

No

(34)

±

(31)

1 2

( ˆi  ˆj ) ; kˆ

(0, – 7/5, 1/5) (32) (35)

x = 2, y = –1

  3 1 1 3 1  ˆi  ˆj ,  2  ˆi  ˆj  3kˆ 2 2 2 2

Reciprocal system of vectors :            If a, b, c & a' , b' , c' are two sets of non-coplanar vectors such that a.a' = b. b' = c. c' = 1 , then the two systems are called Reciprocal System of vectors.          bxc cxa axb Note : a =    , b     and c     [a b c ] [a b c ] [a b c ]

      Example # 28 : If a , b , c and a, b, c  be the reciprocal system of vectors, prove that              (i) (ii) a . a  b . b   c . c   3 a  a   b  b  c  c   0       Solution : (i) We have : a . a  = b . b = c . c  = 1       a . a  + b . b + c . c  = 1 + 1 + 1 = 3 (ii)

and 

   1       We have : a =  (b  c ) , b =  (c  a) and c  =  (a  b) , where  =    [a b c ]               a  a  a  (b  c )   {a  (b  c )}   {(a . c ) b  (a . b) c }               b  b  b  (c  a)   {b  (c  a )}   {(b . a) c  (b . c ) a}               c  c   c  (a  b)   {c  (a  b )}   {(c . b) a  (c . a ) b}       a  a  b  b  c  c                    =  {(a . c ) b  (a . b) c }   {(b . a) c  (b . c ) a}   {(c . b) a  (c . a) b}                   =  [(a . c ) b  (a . b) c  (b . a ) c  (b . c ) a  (c . b ) a  (c . a) b]                     =  [(a . c ) b  (a . b) c  (a . b ) c  (b . c ) a  (b . c ) a  (a . c ) b] = 0  0

Equation of a plane : (i)

   The equation ( r  r0 ) . n  0 represents a plane containing the point with position vector   r0 , where n is a vector normal to the plane.

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MATHS     The above equation can also be written as r . n  d , where d = r0 . n

(ii)

(iii)

(iv)

Angle between two planes is the angle between two normals drawn to the planes and the angle between a line and a plane is the compliment of the angle between the line and the normal to the plane.    The length of perpendicular (p) from a point having position vector a to the plane r . n  d is   |a.nd|  given by p = |n|       If ( r  a ) . n1  0 and ( r  a ) . n2  0 are the equations of two planes, then the equation of line     of intersection of these planes is given by r  a   (n1  n 2 ) .

Test of collinearity :

  

Three points A,B,C with position vectors a, b, c respectively are collinear, if & only if there exist

    scalars x, y, z not all zero simultaneously such that xa  yb  zc = 0 , where x + y + z = 0.

Test of coplanarity :

   

Four points A, B, C, D with position vectors a, b, c, d respectively are coplanar if and only if there      exist scalars x, y, z, w not all zero simultaneously such that xa + yb + zc + wd = 0 , where x + y + z + w = 0.          Example # 29 : Show that the vectors 2a  b  3c , a  b  2c and a  b  3c are non-coplanar vectors.

Solution :

Let, the given vectors be coplanar. Then one of the given vectors is expressible in of the other two.          Let 2a  b  3c = x a  b  2c + y a  b  3c , for some scalars x and y..       2a  b  3c = (x + y) a + (x + y) b + (–2x – 3y) c   2 = x + y, –1 = x + y and 3 = – 2x – 3y. Solving first and third of these equations, we get x = 9 and y = –7. Clearly these values do not satisfy the second equation. Hence the given vectors are not coplanar.









            Example # 30 : Prove that four points 2a  3b  c , a  2b  3c , 3a  4b  2c and a  6b  6c are coplanar..

Solution :

Let the given four points be P, Q, R and S respectively. These points are coplanar if the vectors

PQ , PR and PS are coplanar. These vectors are coplanar iff one of them can be expressed as a linear combination of other two. So let PQ = x PR + y PS            a  5b  4c = x a  b  c + y  a  9b  7c         a  5b  4c = (x – y) a + (x – 9y) b + (–x + 7y) c     x – y = –1, x – 9y = –5, –x + 7y = 4 [Equating coeff. of a, b, c on both sides]









1 1 ,y= . 2 2 These values also satisfy the third equation. Hence the given four points are coplanar. Solving the first two equations of these three equations, we get x = –

Self Practice Problems : (37)

    If a, b, c, d are any four vectors in 3-dimensional space with the same initial point and such

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MATHS      that 3a  2b  c  2d  0 , show that the terminal A, B, C, D of these vectors are coplanar. Find the point (P) at which AC and BD meet. Also find the ratio in which P divides AC and BD.

(38)

           Show that the vector a  b  c , b  c  a and 2a  3b  4c are non-coplanar, where a, b, c are any non-coplanar vectors.

(39)

Find the value of  for which the four points with position vectors  ˆj  kˆ , 4ˆi  5ˆj  kˆ , 3ˆi  9ˆj  4kˆ and  4ˆi  4ˆj  4kˆ are coplanar..    3a  c p Answers : (37) P divides AC in 1 : 3 and BD in 1 : 1 ratio (39) 4

=1

Application of vectors : (a) (b)

(c)



Work done against a constant force F over a displacement

   s is defined as W  F . s

     The tangential velocity V of a body moving in a circle is given by V   x r , where r is the position vector of the point P.

     The moment of F about ’O’ is defined as M  r x F , where r is the position vector of P w.r.t. ’O’. The

 



direction of M is along the normal to the plane OPN such that r , F

(d)

 & M form a right handed system.

     Moment of the couple = ( r1  r2 ) x F , where r1 and r2 are position vectors of the point of the application   of the forces F and  F.

Example # 31 : Forces of magnitudes 5 and 3 units acting in the directions 6ˆi  2ˆj  3kˆ and 3ˆi – 2ˆj  6kˆ

Solution :

respectively act on a particle which is displaced from the point (2, 2, –1) to (4, 3, 1). Find the work done by the forces.   Let F be the resultant force and d be the displacement vector. Then,

1 (6ˆi  2ˆj  3kˆ ) (3ˆi – 2ˆj  6kˆ )  +3 = (39 ˆi  4ˆj  33kˆ ) F = 5 7 36  4  9 9  4  36 and 

 d = ( 4ˆi  3ˆj  kˆ ) – (2ˆi  2ˆj  kˆ ) = 2ˆi  ˆj  2kˆ

  1 (39 ˆi  4ˆj  33kˆ ) . (2ˆi  ˆj  2kˆ ) Total work done = F . d = 7 =

1 148 (78 + 4 + 66) = units. 7 7

Self Practice Problems :

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MATHS (40)

A point describes a circle uniformly in the ˆi , ˆj plane taking 12 seconds to complete one revolution. If its initial position vector relative to the centre is ˆi and the rotation is from ˆi to ˆj , find the position vector at the end of 7 seconds. Also find the velocity vector.

(41)

The force represented by 3ˆi  2kˆ is acting through the point 5ˆi  4ˆj  3kˆ . Find its moment about the point ˆi  3ˆj  kˆ .

(42)

Find the moment of the couple formed by the forces 5ˆi  kˆ and  5 ˆi  kˆ acting at the points (9, –1, 2) and (3, –2, 1) respectively

Answers :

(40)

–

 3 ˆi  ˆj , 12 ˆi 

1 2

3 ˆj



(41)

2ˆi  20 ˆj  3kˆ

(42)

ˆi  ˆj  5kˆ

Miscellaneous solved examples Example # 32 : Show that the points A, B, C with position vectors 2ˆi  ˆj  kˆ , ˆi  3ˆj  5kˆ and 3ˆi  4ˆj  4kˆ

Solution :

respectively are the vertices of a right angled triangle. Also find the remaining angles of the triangle. We have, AB

= Position vector of B – Position vector of A = ( ˆi  3ˆj  5kˆ ) – (2ˆi  ˆj  kˆ ) =  ˆi  2ˆj  6kˆ

BC

= Position vector of C – Position vector of B = (3 ˆi  4 ˆj  4kˆ ) – ( ˆi  3ˆj  5kˆ ) = 2ˆi  ˆj  kˆ

and,

= Position vector of A – Position vector of C

CA

= (2ˆi  ˆj  kˆ ) – (3 ˆi  4 ˆj  4kˆ ) =  ˆi  3 ˆj  5kˆ  Since AB + BC + CA = ( ˆi  2ˆj  6kˆ ) + (2ˆi  ˆj  kˆ ) + ( ˆi  3 ˆj  5kˆ ) = 0 So A, B and C are the vertices of a triangle.

Now, 

BC . CA = (2ˆi  ˆj  kˆ ) . ( ˆi  3 ˆj  5kˆ ) = –2 – 3 + 5 = 0 

BC  CA

BCA =

 2

Hence ABC is a right angled triangle. Since A is the angle between the vectors AB and AC . Therefore

( ˆi  2ˆj  6kˆ ) . ( ˆi  3ˆj  5kˆ )

AB . AC cos A =

| AB | | AC |

=

( 1)2  ( 2) 2  ( 6)2

1  6  30 =



| BA | | BC |

41 35

=

35 41

35 41

=

12  2 2  6 2 2 2  ( 1)2  (1)2

226 cos B =

=

( ˆi  2ˆj  6kˆ ) . (2ˆi  ˆj  kˆ )

BA .BC cos B =

35

1  4  36 1  9  25 A = cos –1

12  ( 3)2  ( 5)2

41 6

=

6  41

B = cos –1

6 41

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MATHS       Example # 33 : If a, b, c are three mutually perpendicular vectors of equal magnitude, prove that a  b  c is    equally inclined with vectors a, b and c .       Solution : Let | a | = | b | = | c | =  (say). Since a, b, c are mutually       perpendicular vectors, therefore a . b = b . c = c . a = 0 ..............(i)  2               abc Now, = a . a + b . b + c . c + 2a . b + 2b . c + 2c . a  2   = | a | | + | b |2 + | c |2 [Using (i) ]    = 32 [ | a | = | b | = | c | = ]     ..............(ii) | a  b  c | = 3       Suppose a  b  c makes angles 1, 2, 3 with a, b and c respectively. Then,           a . (a  b  c ) a.aa.ba.c     cos1 =     = |a||a bc | |a||a bc |   |a| | a |2   =    = =   |abc | |a||a bc |

 3

1 =

3

[Using (ii)]

 1   1 = cos –1    3



 1   and  = cos –1 Similarly, 2 = cos –1   3  3

 1       3



 1 =  2 =  3.       Hence, a  b  c is equally inclineded with a, b and c Example # 34 : Prove using vectors : If two medians of a triangle are equal, then it is isosceles. Solution :

Let ABC be a triangle and let BE and CF be two equal medians. Taking A as the origin, let the   position vectors of B and C be b and c respectively. Then, P.V. of E =



1  1  c and P.V. of F = 2 b 2

 1  BE = 2 (c  2b) 1   CF = 2 (b  2c ) 

Now,

BE = CF



| BE | = | CF | 

   

2

2

| BE | = | CF |

 1  (c  2b) 2

2

=

 1  (b  2c ) 2

2

   1  1     | c  2b |2 = | b  2c |2  | c  2b |2 = | b  2c |2 4 4         (c  2b) . (c  2b) = (b  2c ) . (b  2c )             c . c – 4b . c + 4b . b = b . b – 4b . c + 4c . c         | c |2 – 4b . c + 4 | b |2 = | b |2 – 4b . c + 4 | c |2

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MATHS 

  3 | b |2 = 3 | c |2



AB = AC

  | b |2 = | c |2



Hence triangle ABC is an isosceles triangle.

Example # 35 : Using vectors : Prove that cos (A + B) = cos A cos B – sin A sin B Solution :

Let OX and OY be the coordinate axes and let ˆi and ˆj be unit vectors along OX and OY respectively. Let XOP = A and XOQ = B. Drawn PL  OX and QM  OX. Clearly angle between OP and OQ is A + B In OLP, OL = OP cos A and LP = OP sin A. Therefore OL = (OP cos A) ˆi and

 

LP = (OP sin A)  ˆj

Now,

OL + LP = OP

OP = OP [(cos A ) ˆi – (sin A) ˆj ]  ......(i) In OMQ, OM = OQ cos B and MQ = OQ sin B. Therefore, OM = (OQ cos B) ˆi , MQ = (OQ sin B) ˆj Now,

OM + MQ = OQ

OQ = OQ [(cos B) ˆi  (sin B)ˆj]  From (i) and (ii), we get

......(ii)

OP . OQ = OP [(cos A) ˆi – (sin A) ˆj ] . OQ [(cos B) ˆi + (sin B) ˆj ] = OP . OQ [cos A cos B – sin A sin B] But, OP . OQ = | OP | | OQ | cos (A + B) = OP . OQ cos (A + B)  OP . OQ cos (A + B) = OP . OQ [cos A cos B – sin A sin B]  cos (A + B) = cos A cos B – sin A sin B Example # 36 : Prove that in any triangle ABC (i) c 2 = a2 + b2 – 2ab cos C Solution : 

(i)

(ii)

c = bcosA + acosB.

In ABC, AB + BC + CA = 0

BC + CA = – AB Squaring both sides

......(i)

( BC )2 + ( CA )2 + 2 ( BC ). CA = ( AB )2  

a2 + b2 + 2 ( BC . CA ) = c 2 c 2 = a2 + b2 – 2ab cosC

(ii)

( BC + CA ). AB = – AB . AB



c 2 = a2 + b2 + 2 ab cos ( – C)

BC . AB + CA . AB = – c2 – ac cosB – bc cos A = – c 2 acosB + bcosA = c. Example # 37 : If D, E, F are the mid-points of the sides of a triangle ABC, prove by vector method that area of DEF = Solution :

1 (area of ABC) 4

  Taking A as the origin, let the position vectors of B and C be b and c respectively. Then the

position vectors of D, E and F are

1   1  1  (b  c ) , c and 2 b respectively.. 2 2

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MATHS Now,

and



 1  1   b (b  c ) = DE = c – 2 2 2    1  1 c = ( (b  c ) = b – 2 2 2

DF

Vector area of DEF =

=

=

=

=

Hence area of DEF =

(vector area of ABC)

area of ABC.

Example # 38 : P, Q are the mid-points of the non-parallel sides BC and AD of a trapezium ABCD. Show that APD = CQB. Solution :

Let

=

and

=

Now DC is parallel to AB  there exists a scalar t such that 

=

+

2

=

×

2

and

respectively..

×

=

Also

=t

=

The position vectors of P and Q are Now

=t

=

=

×

(1 + t)

=

=

×

=

=

=

=

Hence Prove.

Example # 39 : Let

and

are unit vectors and

the value of Solution :

=

and

then find

.

Given

=



×



is a vector such that

×

and

=

(as



(using



) .

= 1 and

= 0, since unit vector)





=0

(as

 0)

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

Now   

(given

+ u)

 (as

from (i))

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MATHS 

–



–0

. (as

1

= 0)

(as



=

= 1)

=1

Example # 40 : In any triangle, show that the perpendicular bisectors of the sides are concurrent. Solution :

Let ABC be the triangle and D, E and F are respectively middle points of sides BC, CA and AB. Let the perpendicular bisectors of BC and CA meet at O. OF. We are required to prove that OF is  to AB. Let the position vectors of A, B, C with O as origin of reference be

,

and



respectively..

=

Also = Since OD  BC 

(

+

).(

( +

),

,

=

–

–

= –

(

+

and

=

(

+

).(

(

+

–

).(

(

+

)

–

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

)=0

 a2 = c 2 from (i) and (ii) we have a2 – b2 = 0 

=

)=0

 b2 = c 2 Similarly OE  CA 

) and

–

............(ii)



)=0

(

+

).(

)=0 

–

Example # 41 : A, B, C, D are four points in space. using vector methods, prove that AC2 + BD2 + AD2 + BC2  AB2 + CD2 what is the implication of the sign of equality. Solution :

Let the position vector of A, B, C, D be AC2 + BD2 + AD2 + BC2 = =

+

–2

+

=

+

–2

+

. +

.

+

= AB2 + CD2 +

+

+

–

. –2

+ –

+

+ –

+ +

. –

+ –

+ .

2

for the sign of equality to hold,



+

+

 AB + CD AC2 + BD2 + AD2 + BC2  AB2 + CD2 2



respectively then .

–2

+

.

+ –2

+

+2 =

and

and

=0

are collinear, the four points A, B, C, D are collinear

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