1. Vectors & Their Representation:
Vector quantities are specified by definite magnitude and definite directions. A vector is generally represented by a directed line segment, say AB. A is called the initial point & B is called the terminal point. The magnitude of vector AB is expressed by AB.
Zero Vector:
A vector of zero magnitude is a zero vector. i.e. which has the same initial & 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 aˆ symbolically, ,
| a | a aˆ . Equal Vectors:
Two vectors are said to be equal if they have the same magnitude, direction & 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 vectorsa andb are collinear if and only if,,a Kb, where K R
Vectors a = a1iˆ + a2jˆ + a3kˆ and b = b1iˆ + b2jˆ + b3kˆ are collinear
if 1 1 b a = 2 2 b a = 3 3 b a 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”.
1.Example
Find values of x & y for which the vectors a = (x + 2)iˆ – (x – y)jˆ + kˆ
b= (x – 1)iˆ + (2x + y)jˆ + 2kˆ are parallel.
Solution
a and b are parallel if x 1 2 x = y x 2 x y = 2 1 x = – 5, y = – 20
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. It should be noted that 0º 180º .
3. Addition Of Vectors:
If two vectors a & b are represented by OA &OB , then their sum a isb a vector represented byOC , where OC is the diagonal of the parallelogram OACB.
ab b a (commutative) (ab) c a( bc) (associativity) a 0 a 0a a ( a) 0 ( a) a
|ab||a||b| |ab|||a||b||
ab = |a|2 |b|2 2|a||b|cos where is the angle between the vectors
A vector in the direction of the bisector of the angle between the two
vectors a&b is a a b b
. Hence bisector of the angle between the two
vectorsaand b is
a b
, where R+. Bisector of the exterior angle between
a&b is
a b
, R+.4. Multiplication Of A Vector By A Scalar:
Ifa is a vector & m is a scalar, then m a is a vector parallel to a whose modulus is m times that of a. This multiplication is called SCALAR MULTIPLICATION. If a and bare vectors & m, n are scalars, then:
a m m ) a ( ) a ( m m(na)n(ma) (mn)a a n a m a ) n m ( m(ab) ma mb
2. Example
I f aiˆ2jˆ3kˆ a n d b 2iˆ4jˆ5kˆ r e p re s e n t t w o a d ja c e n t s i de s of a parall elogram, fi nd uni t vectors paral l el to the diagonals of the parallelogram.
Solution.
Let ABCD be a parallelogram such that AB = a and BC = b. Then,AB + BC = AC AC = ab = 3iˆ6jˆ2kˆ and AB + BD = AD AD + AD = AB BD = AD AB = ba Now, AC = 3iˆ6jˆ2kˆ |AC| = 9364 = 7 and, BD = iˆ2jˆ8kˆ |BD| = 1464 = 69
Unit vector along AC = | AC | AC = 7 1
3iˆ6jˆ2kˆ
and, Unit vector along BD = | BD | BD = 69 1
iˆ2jˆ8kˆ
3. ExampleABCDE is a pentagon. Prove that the resultant of the forces AB, AE, BC, DC, ED and AC is 3AC.
Solution.
Let R be the resultant force
R = AB + AE + BC + DC + ED + AC R = (AB + BC) + (AE + ED + DC) + AC = AC + AC + AC
= 3AC. Hence proved.
5. 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,
DISTANCE FORMULA
Distance between the two points A(a) and B(b) is AB = ab
SECTION FORMULA
If a and b are the position vectors of two points A & B then the p.v. of a point which divides AB in the ratio m: n is given by:
n m b m a n r .
Note p.v. of mid point of AB = 2
b a .
4. Example
ABCD is a parallelogram. If L, M be the middle point of BC and CD, express AL and AM in terms of AB and AD, if AL + AM = k AC. find k
Solution.
Let the position vectors of points B and D be respectively b and a referred to A as origin of reference.
Then AC = AD + DC = AD + AB [ DC = AB] = d + b AB = b, AD = d
i.e. position vector of C referred to A is d + b AL = p.v. of L, the mid point of BC.
= 2 1 [p.v. of D + p.v. of C] = 2 1
b d b
= AB + 2 1 AD AM = 2 1
ddb
= AD + 2 1 AB AL + AM = b + 2 1 d + d+ 2 1 b = 2 3 b + 2 3 d = 2 3 (b + d) = 2 3 AC.6. Scalar Product Of Two Vectors:
Geometrical interpretation of Scalar Product
Let a and b be vectors represented by y OA and OB respectively. Let be the angle between OA and OB. Draw BL OA and AM OB.
From s OBL and OAM, we have OL = OB cos and OM = OA cos . Here OL and OM are known as projections of b on a and a on b respectively.ely. Now, a.b = |a| | b| cos
= |a| (OB cos ) = |a| (OL)
Againa.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.
1. i.i = j.j = k.k = 1; i.j = j.k = k.i = 0
projection of | b | b . a b on a 2. ifa = a1i + a2j + a3k &b = b1i + b2j + b3k then a b. = a1b1 + a2b2 + a3b3 a a12a22a32 , b b12b22b32
3. the angle betweena & b is given by |a| |b| b . a cos 0 4. a.b a b cos(0) ,
note that if is acute then a b . > 0 & if is obtuse then a b. < 0 5. a.a a2a2,a.bb.a (commutative) a.(bc)a.ba.c (distributive)
6. a.b 0 a b (a0 b0)
7.
(
m a)
.b =a.(
m b)
= m(
a b .)
(associative) where m is scalar..Note :
(i) Maximum value ofa . b is a b (ii) Minimum value ofa .b is – a b
(iii) Any vectora can be written as,a =
a i i.
a. j j
a k k.
.5. Example
If a + b + c = 0, |a| = 3, |b| = 5 and |c| = 7, find the angle between a and b. Solution. We have, abc0 ab = –c
ab
.
ab
=
c .
c ab 2 = 2 | c | a 2 + b 2 + 2a.b = c 2 a 2 + 2 b + 2 a b cos = c 2 9 + 25 + 2 (C) (5) cos = 49 cos = 2 1 = 3 .6. Example
D is the mid point of the side BC of a triangle ABC, if AB2 + AC2 = k (AD2
+ BD2) find k Solution. We have AB = AD + DB AB2 = 2 ) DB AD ( = AD2 + DB + AD 2 . DB ...(i) Also we have AC = AD + DC AC2 = (ADDC)2 = AD2 + DC2 + AD 2 . DC ...(ii)
Adding (i) and (ii), we get
AB2 + AC2 = 2AD2 + 2BD2 + 2
AD . (DBDC) = 2(DA2 + DB2), for
DB + DC = 0
Solved Example
If a = iˆ + jˆ + kˆ and b = 2iˆ – jˆ + 3kˆ, then find (i) Component of b along a.
(ii) Component of b perpendicular to along a.
Solution. (i) Component of b along a is 2 | a | b . a a Here a . b = 2 – 1 + 3 = 4 2 | a | = 3 Hence 2 | a | b . a a = 3 4 a = 3 4 (iˆ + jˆ + kˆ) (ii) Component of b perpendicular to along a is b – 2 | a | b . a a. = 3 1
2iˆ7jˆ5kˆ
7. Vector Product Of Two Vectors:
1. Ifa&b are two vectors & is the angle between them thena x b a b sin n, wheren is the unit vector perpendicular to botha&b such that a b, & n forms a right handed screw system.
2. Geometricallya x b = area of the parallelogram whose two adjacent sides are represented bya & b.
3. iˆiˆ jˆjˆkˆkˆ0 ; iˆjˆkˆ,jˆkˆ iˆ,kˆiˆ jˆ
4. Ifa = a1iˆ +a2jˆ + a3kˆ & b = b1iˆ + b2jˆ + b3kˆ then 3 2 1 3 2 1 b b b a a a kˆ jˆ iˆ b a 5. a x b b x a (not commutative)
6.
(
m a)
b =a (
m b)
= m(
a b)
(associative) where m is a scalar.. 7. a x b c( ) (a x b ) (a x c ) (distributive)8. a x b0 a& b are parallel (collinear) (a0 , b0 i.e.) aK b , where K is a scalar.
9. Unit vector perpendicular to the plane of
a b is n a x b
a x b &
A vector of magnitude ‘r’ & perpendicular to the palne of
a b is r a x b
a x b &
If is the angle between
a b then a x b
a b & sin
If a b, & are the pv’s of 3 points A, B & C then the vector area of trianglec
AB C = 1
2 a x b b x c cx a . T h e p oi n t s A , B & C a r e c ol l i n e a r i f 0 a x c c x b b x a Area of any quadrilateral whose diagonal vectors ared1 & d2 is given by
1
2 1 2
d x d
Lagrange's Identity: for any two vectors
a b a x b a b a b a a a b a b b b & ; ( ) ( . ) . . . . 2 2 2 2 7. Example
Find a vector of magnitude 9, which is perpendicular to both the vectors kˆ 3 jˆ iˆ 4 and 2iˆ jˆ2kˆ. Solution.
b a = 2 1 2 3 1 4 kˆ jˆ iˆ = (2 – 3) iˆ – (–8 + 6) jˆ + (4 – 2) kˆ = iˆ2jˆ2kˆ |ab| = (1)22222 = 3 Required vector = 9 | b a | b a = 3 9 ) kˆ 2 jˆ 2 iˆ ( = 3iˆ6jˆ6kˆ 8. Example
For any vector a, if | a iˆ|2 + | a jˆ|2 + | a kˆ|2 = k |a|2 find k
Solution.
Let a = a1iˆa2jˆa3kˆ. Then iˆ
a = (a1iˆa2jˆa3kˆ) × iˆ = a1( iˆ iˆ)+ a2( jˆ iˆ) + a3 ( kˆ iˆ) = –a2kˆa3iˆ
| a iˆ|2 = a22 + a32 jˆ a = (a1iˆa2jˆa3kˆ) × jˆ = a1kˆa3iˆ | a jˆ|2 = a21 + a32 | a kˆ|2 = a 1 2 + a 2 2 | a iˆ|2 + | a jˆ|2 + | a kˆ|2 = a22 + a33 + a12 + a32 + a12 + a22 2 (d12 + a 2 2 + a 3 2) = 2 |a|2 9. Example
Let OA = a, OB = 10a + 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. find k if p = kq.
Solution.
We have,
p = Area of the quadrilateral OABC
= 2 1 | AC OB | = 2 1 | ) OA OC ( OB | = 2 1 | ) a b ( ) b 2 a 10 ( | = 2 1 | ) a b ( 2 ) b b ( 2 ) a a ( 10 b a ( 10 | = 2 1 | ) b a ( 2 0 0 ) b a ( 10 |
and, q = Area of the parallelogram with OA and OC as adjacent sides = |OAOC| = |a b|
...(ii)
8. Scalar Triple Product:
The scal ar tri pl e product of three vectors a b, & c i s defi ned as:
a x b c. a b c sin cos where is the angle betweena & b & is the angle
between a x b& c. It is also written as[ a b c]and spelled as box product.
Scalar triple product geometrically represents the volume of the parallelopiped whose three coterminous edges are represented by
a b, &c i e V. . [a b c]
In a scalar triple product the position of dot & cross can be interchanged i.e. a. (b x c ) ( a x b c ). OR [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 ]
If a= a1i+a2j+a3k;b = b 1i+b2j+b3k &c = c1i+c2j+c3k then[a b c ]
a a a b b b c c c 1 2 3 1 2 3 1 2 3 . In general, ifaa l1 a m2 a n3;b b l1 b m2 b n3 & c c l1 c m2 c n3 then
a b c
a a a b b b c c c l m n 1 2 3 1 2 3 1 2 3; where,m &n are non coplanar vectors.
Ifa b c , , are coplanar[a b c ]0.
Scalar product of three vectors, two of which are equal or parallel is 0 i.e.[ a b c 0] ,
If a b c, , are non coplanar then[a b c 0 ] for right handed system & [ a b c 0] for left handed system.
[i j k] = 1 [K a b c ]K a b c[ ] [ (a b c d ) ][a c d ][b c d ]
The volume of the tetrahedron OABC with O as origin & the pv’s of A, B
and C beinga b , &c respectively is given by V 1 a b c
6 [ ]
The positon vector of the centroid of a tetrahedron if the pv’s of its
vertices area b c , , &d are given by 1
4 [ ]
a b cd .
Note that this is also the point of concurrency of the lines joining 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.
10. Example
Find the volume of a parallelopiped whose sides are given by 3iˆ7jˆ5kˆ, kˆ 3 jˆ 7 iˆ 5 and 7iˆ5jˆ3kˆ Solution.
Let a3iˆ7jˆ5kˆ, b5iˆ7jˆ3kˆ and c7iˆ5jˆ3kˆ.
We know that the volume of a parallelopiped whose three adjacent edges are a,b,c is [a,b,c] . Now, [abc] = 3 5 7 3 7 5 5 7 3 = –3 (–21 – 15) – 7 (15 + 21) + 5 (25 – 49) = 108 – 252 – 120 = –264
So, required volume of the parallelopiped = [a,b,c] = | – 264 | = 264 cubic units
9. Vector Triple Product:
Leta b c , , be any three vectors, then the expressiona x b x c( ) is a vector & is called a vector triple product.
Geometrical Interpretation of a x b x c ( )
Consider the expressiona x b x c( ) which itself is a vector, since it is a
cross product of two vectorsa& ( b x c). Nowa x b x c ( ) is a vector perpendicular
to the plane containinga& ( b x c) butb x c is a vector perpendicular to the
plane containingb & c, thereforea x b x c ( ) is a vector which lies in the
plane ofb& c and perpendicular toa. Hence we can expressa x b x c ( ) in
terms ofb&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
(a x b x c) a x b x c ( ), in general
11. Example
For any vector a, P.T. iˆ(aiˆ)jˆ(ajˆ)kˆ(ak) = 2a
Solution.
Let aa1iˆa2jˆa3kˆ. Then, iˆ(aiˆ)jˆ(aˆjˆ)kˆ(akˆ) = {(iˆ.iˆ)a(iˆ.a)iˆ} + {(jˆ. jˆ)a(jˆ.a)jˆ} + {(kˆ.kˆ)a(kˆ.a)kˆ} = {(a(iˆ.a)iˆ}{a(jˆ.a)jˆ} + {a(kˆ.a)kˆ}
= 3a{(iˆ.a)iˆ(jˆ.a)jˆ(kˆ.a)kˆ = 3a(a1iˆa2jˆa3kˆ)
12. Example
Let a = iˆ + 2jˆ – 3kˆ, b = iˆ + 2jˆ – 2kˆ and c = 2iˆ – jˆ + kˆ. Find the value(s) of , if any, such that
ab
bc
×
ca
= 0. Find the vector product when = 0. Solution.
ab bc
×
ca
=
a b c
b ×
ca
=
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, which do not have a common solution.(ii)
a b c
= 0 1 2 2 2 1 3 2 = 0 3 = 2 = 3 2 when = 0,
a b c
= – 10, a.b = 6, b.c = 0 and the vector product is – 60
2 iˆ kˆ
.13. Example
If A B = a, A . a = 1 and AB = b, then find A and B.
Solution. Given A B a ...(i) a.
AB
= a.a a.A a . B = a.a 1 + a.B = |a|2 a.B = |a|2 – 1 Given ABb a
AB
= a × b
a.B
A –
a.A
B ab
|a|2 1
A B = ab [using equation (B)] solving equation (1) and (5), simultaneously, we getA = 2 | a | a b a and B =
2
2 | a | 1 | a | a a b 14. Example
Solve for r , the simultaneous equations rbcb, r.a 0 provided a is not perpnedicular to b.
Solution
) c r
( × b = 0 r c and b are collinear rc kb r = ckb ...(i) a . r = 0 (ckb) . a = 0 k = – b . a c . a
putting in (i) we get
b . a c . a c r b
10. 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. Note:
c b a b x a c c b a a x c b c b a c x b = a 11. Linear Combinations:Given a finite set of vectorsa b c , , ,... then the vector rxa yb zc ...is called a linear combination of a b c, , ,... for any x, y, z... R. We have the following results:
(a) If a b, are non zero, noncollinear vectors then xayb x a'y b' xx y' ; y' (b) FundamentalTheorem: Let a b, be non zero, non collinear vectors. Then
any vectorr coplanar with a b, can be expressed uniquely as a linear combination of a b,
i.e. There exist some uniquly x, y R such thatxayb r. (c) If a b c, , are nonzero, noncoplanar vectors then:
xa yb zc x a y b z c ' ' ' xx' , yy' ,zz'
(d) Fundamental Theorem In Space: Leta b c , , be nonzero, noncoplanar vectors in space. Then any vectorr, can be uniquly expressed as a linear com bi nat i on of a b c, , i .e. There ex i st som e un i que x,y R su ch thatxa yb zc r.
(e) Ifx x 1, 2,...xn are n non zero vectors, & k1, k2,...kn are n scalars & if the linear combination k x1 1 k x2 2 ...k xnn 0k1 0,k2 0...kn 0 then we say that vectorsx x 1, 2,...xn are LINEARLY INDEPENDENT VECTORS.
(f) Ifx x 1, 2,...xn are not LINEARLY INDEPENDENT then they are said to be LINEARLY DEPENDENT vectors. i.e. ifk x1 1 k x22 ... k xnn 0 & if there exists at least one kr 0 thenx x 1, 2,...xn are said to be
L
INEARLYD
EPENDENT. iˆ,jˆ,kˆ are Linearly Independent set of vectors. For K1iˆ+ K2jˆ + K3kˆ= 0 K1= K2= K3 = 0
Two vectorsa & b are linearly dependent a is parallel to
b i.e.a x b 0 linear dependence ofa & b. Conversely ifa x b 0 thena & 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.
15. Example
Given A that the points a – 2b + 3c, 2a + 3b – 4c, – 7b + 10c, A, B, C have position vector 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 OA = a – 2b + 3c, OB = 2a + 3b – 4c and OC= – 7b + 10c
Now AB = p.v. of B – p.v. of A = OB – OA = (a + 5b – 7c) = – AB
AC = AB where = – 1. Hence AB and AC are linearly dependent
16. Example
Prove that the vectors 5a + 6b + 7c, 7a – 8b + 9c and 3a + 20b + 5c are linearly dependent 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 5a + 6b + 7c = ( 7a – 8b + 9c) + m (3a + 20b + 5c)
where , m are scalars
Comparing the coefficients of a,b and c on both sides of the equation
5 = 7 + 3 ...(i)
6 = – 8 + 20 m ...(ii)
7 = 9 + 5m ...(iii)
From (i) and (iii) we get
4 = 8 = 2 1
= m which evidently satisfies (ii) equation too. Hence the given vectors are linearly dependent .
Not e: 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;xayb zc 0, where x + y + z = 0.
Note: 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 thatxa + yb + zc + wd = 0 where, x + y + z + w = 0.
17. Example
Show that the vectors 2ab3c, ab2c and ab3c are non-coplanar vectors.
Solution.
Let, the given vectors be coplanar.
Then one of the given vectors is expressible in terms of the other two. Let 2ab3c = x
ab2c
+ y
ab3c
, for some scalars x and y.. 2ab3c = (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 thrid equation. Hence, the given vectors are not coplanar.
18. Example
Prove that four points 2a 3b c , a2b3c , 3a4b2c and a6b6c 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
a5b4c = x
abc
+ y
a9b7c
a5b4c = (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]
Solving the first of these three equations, we get x = – 2 1 , y = 2 1 . These values also satisfy the third equation. Hence, the given four points are coplanar.