Vectors

May 2, 2009

Abstract

Vectors are those mathematical structures that have magnitude and

direction associated with them. To master vectors is to think them as

coordinates in 3D geometry.

1

CONTENTS CONTENTS

Contents

I Theory, Discussion & Formulae 4

1 Vectors & Scalars 4

2 Types of Vectors 4

3 Length of a vector 4

4 Equality of vectors 4

5 Collinear vectors 4

6 Coplanar vectors 5

7 Addition of two vectors 5

8 Linear combination of vectors 5

9 Linear Independence & Dependence. 6

10 Points in Vector space 7

11 Product in Vectors 7

11.1 Scalar Multiplication of a vector . . . . . . . . . . . . . 7

11.2 Scalar Product or Dot product . . . . . . . . . . . . . . 8

11.3 Vector Product or Cross product . . . . . . . . . . . . . 8

11.3.1 In cartesian coordinate system . . . . . . . . . . 8

11.3.2 Generalised properties . . . . . . . . . . . . . . . 8

11.4 Scalar Triple product or Box product . . . . . . . . . . . 9

11.5 Vector Triple product . . . . . . . . . . . . . . . . . . . 9

11.5.1 How to Derive this? . . . . . . . . . . . . . . . . 9

12 Projection & Component of Vectors 10

13 Solving Problems-Part I 10

13.1 Solving Vectors Equations . . . . . . . . . . . . . . . . . 10

14 Useful Formulae 11

15 Area & Volume of geometrical structures 12

16 Vector Geometry 12

16.1 Equation of a line in Vector form . . . . . . . . . . . . . 12

16.2 Skew lines . . . . . . . . . . . . . . . . . . . . . . . . . . 12

16.3 Equation of angle bisector . . . . . . . . . . . . . . . . . 13

16.4 Vector equation of a plane . . . . . . . . . . . . . . . . . 13

16.4.1 Normal form . . . . . . . . . . . . . . . . . . . . 13

16.4.2 Point-two vector form . . . . . . . . . . . . . . . 13

16.4.3 Three point form . . . . . . . . . . . . . . . . . . 13

2

CONTENTS CONTENTS

16.5 Section made by a plane . . . . . . . . . . . . . . . . . . 14

17 Reciprocal system of Vectors 14

II Problems 15

18 Level I Problems 15

19 Level 2 problems 15

3

5 COLLINEAR VECTORS

Part I

Theory, Discussion & Formulae

1 Vectors & Scalars

A vector is a dened as that mathematical structure that has magnitude and

direction associated with it. Scalars are those mathematical structures that

have only magnitude no direction.

2 Types of Vectors

Any vector

AB is a vector in the direction joining the points A & B inthe vector space, which can be expressed in form of basic vectors called as

position vectors.

Position vectors specify the position of a point with respect to origin.

Any vector joining two points can be written in terms of position vectors

of those two points. Like

AB = b aNotation : I prefer to denote any vector i+ j + k as (1, 1, 1) and assumeto have the i,j & k. It saves lot of time and things are clear with this way of

writing vectors.

3 Length of a vector

Length of a vector or Modulus of a vector say |a| which is the position vectoris distance of the point from origin. Or if for any vector

AB, |AB| is distancebetween the points A & B.

Given position vector a = a1i+ a2j + a3k then |a| =a21 + a

22 + a

33

Given any vector joining A & B points,|AB| = |b a|

=

(b1 a1)2 + (b2 a2)2 + (b3 a3)2

4 Equality of vectors

Two vectors are equal if have same direction and magnitude.

5 Collinear vectors

Two vectors are collinear if they have the same direction.

If they turn to have the same magnitude as well then they become equal

vectors.

4

8 LINEAR COMBINATION OF VECTORS

6 Coplanar vectors

Two vectors are coplanar if they lie in the same plane. Any two vectors are

always coplanar if they have the same starting point.

7 Addition of two vectors

Addition of two vectors is given by Triangle law or Parellogram law.

Vector addition is associative a+ (b+ c) = (a+ b) + c Vector addition is Cummutative a+ b = b+ a

8 Linear combination of vectors

If vectors a & b are two position vectors of A & B points. What will bec = a+ b?

c above is produced by linear combination of two vectors. Now there aretwo cases to this problem

If a & b are non-collinear vectors then the linear combination will beguided by the parellogram law.

If a & b are collinear vectors. i.e. they are having the same direction,then any linear combination is going to give another vector of dierent

length in the same direction.

This analysis above shows that linear combination has the power toproduce new vectors. If both the vectors are in line means collinear

then the linear combination will produce the new vector in the same

line. But here two vectors are not really required. So

to produce a new vector in the same line linear combination is a

to produce a new vector in the plane containing two vectors. If

these two vectors lie in the same line means are collinear then the

linear combination will produce another vector in the same line.

But we want to produce another vector in the plane. So we need

two non-collinear vectors. Hence a + b will produce the newvector in the plane. And for dierent values of & any vectorin that plane containing a & can be produced.

to produce a new vector in the 3D space containing three vectors.

If these three vectors are coplanar then the linear combination will

produce a new vector in the plane containing these three vectors.

Hence a+ b+ c will produce the new vector in the 3D spaceif all three vectors are non-coplanar.

5

9 LINEAR INDEPENDENCE & DEPENDENCE.

Linear combination is the method by which we produce new

vectors.

9 Linear Independence & Dependence.

The idea of linearly independent vectors is generalized way of speaking about

collinearity or coplanarity of two or three vectors respectively.

In 1 dimensional space (i.e. line), how many vectors are sucient toproduce a new vector using linear combination?

Only one vector is enough to produce any other vector in 1D space

(i.e. line) using linear combination a

In 2 dimensional space (i.e. plane), how many vectors are sucientto produce a new vector using linear combination & what condition

should they satisfy?

Only two vectors are sucient to produce a new vector in 2D space

provided these two are non-collinear. Since if they are collinear

they will again produce a new vector in 1D space. i.e. line con-

taining both the lines.

In 3 dimensional space (i.e. space), how many vectors are sucientto produce a new vector using linear combination & what condition

should they satisfy?

Only 3 vectors are sucient to produce a new vector in 3D space

provided these vectors are non-coplanar. Since if they are copla-

nar then their linear combination will produce a new vector in the

plane and not in 3D space.

NOTE: Number of vectors under some condition required for producing new

vectors is always equal to the dimension of the space.

Now we come to the point of denining what is linear independence?

Linear-Independence - Given vectors are linearly independent if one can-

not be produced from the other.

Linear-Dependence - Given Vectors are said to be linearly dependent if

one can be produced from the other.

So this denition is nothing but in

1 dimensional space (i.e. line) we require only one linearly independentvector to produce a new vector.

6

11 PRODUCT IN VECTORS

2 dimensional space (i.e plane) we require two linearly independent vec-tors to produce a new vector. So two vectors are linearly independent

in 2D space means they are non-collinear

3 dimensional space (i.e. space) we require three linearly independentvectors to produce a new vector. So three vectors are linearly indepen-

dent in 3D space means they are non-coplanar.

So now you should know why the idea of linear independence is constructed

? or why we study that ?

Purely since we want to know what is that condition that is required to

produce new vectors in n-dimensional space! And how we can extend the

idea of non-collinear & non-coplanar into higher dimension. So ultimately is

nothing but non-collinearity & non-coplanarity.

10 Points in Vector space

Any point in vector space can be uniquely represented by its position vector.

1. Three points B,C & D are collinear i

(a) there exists x,y,z not all zero such that

i. xu+ yv + zw = 0ii. x+ y + z = 0 ( this is directly evident from section formula)

(b)

BC =

CD

2. Four points A,B,C & D are coplanar i

(a) there exists x,y,z & w not all zero such that

i. xa+ yb+ zc+ wd = 0ii. x+ y + z + w = 0

(b) There exists , not both zero such that AB +

AC =

AD istrue.

11 Product in Vectors

11.1 Scalar Multiplication of a vector

Any vector multiplied by a scalar (a real number) is dened as scalar multi-

plied to a vector.

Geometrically, the length of the vector is increased scalar times. i.e.

a = b tells |a| = |b| and another crucial info of they both pointing inthe same direction. Whenver we have vectors thinking of them in terms of

lines containing those vectors helps. So two vectors are collinear means lines

containing these two vectors is parellel.

7

11.2 Scalar Product or Dot product 11 PRODUCT IN VECTORS

11.2 Scalar Product or Dot product

Dened as a b = |a||b| cos

So a b = 0 |a| = 0 or |b| = 0 or cos = 0

a b is a scalar quantity

aa = a2 = |a|2; this is a very important identity that allows movementfrom vectors to real numbers.

a b = b a; dot product is commutative

We cant talk about associativity since three vectors having dot productdoesnt make sense. Why??

11.3 Vector Product or Cross product

Dened as a b = |a||b| sin n, where n: is perpendicular both a & b

11.3.1 In cartesian coordinate system

1. i j = k, i k = j Note: if in alphabetic (cyclic) order then positiveelse negative.

2. i i = j j = k k = 0

3. When a & b when represented in terms of i, j, k then a b =i j ka1 a2 a3b1 b2 b3

11.3.2 Generalised properties

a b is perpendicular to both a & b

a & b are parallel to each other a b = 0

a b = b a

|a b|2 + (a b)2 = |a|2|b|2 Lagrange's identity. Relation between crossand dot product.

Area of a parellogram = |a b|

Area of a Triangle = |ab|2

8

11 PRODUCT IN VECTORS11.4 Scalar Triple product or Box product

11.4 Scalar Triple product or Box product

Dened as a (b c) a (b c) = a b c - bracket doesn't makes sense since the output isa scalar (hence called scalar triple product).

If a, b, c are represented using i, j, k then a b c =a1 a2 a3b1 b2 b3c1 c2 c3

a b c = a b c -interchange the & values remains same. [a b c] = a b c [a b c] = [b c a] = [c a b] = [a c b] change them cyclically and theyall have same value, the order is changed and the sign changes.

Geometrically [a b c] represents volume of a parellopiped with edgesa, b & c

So 0 = [a b a] represents area of a parellopiped with edges which arecomplanar making volume zero.

For any four coplanar points (note this are points and not vectors)given by position vector a,b, c & d we have [abc] = [dab] + [dbc] + [dca](can be proved by the fact that vectors

AB,

AC, &

AD are coplanar.

11.5 Vector Triple product

Dened as a (b c) = (a c)b (a b)cAlso (a b) c = (a c)b (b c)aWay to remember : a (b c) lies in the plane of vectorsin the bracket.

11.5.1 How to Derive this?

Let r = a (b c). r a & r b ci.e. r lies in the plane of b & ci.e. r = b+ ci.e. a r = 0 = (a b) + (a c) ac = ab = k (assume) r = k(a c)b+ k(a b)c = k((a c)b (a b)c)Value of k can be found out by trying values of a = i, b = j & c = ir = a (b c) = i (j i) = i (k) = jr = k((i i)j (i j)i) = k(j 0) = kjEquating r we get k = 1

9

13 SOLVING PROBLEMS-PART I

12 Projection & Component of Vectors

We have two vectors a & b, each non-zero then

Projection of a along b = ab|b|

Projection of a b = |ab||b|

Component of a along b = (ab)b|b|2

Component of a b = b(ab)|b|2 = a(ab)b|b|2

13 Solving Problems-Part I

13.1 Solving Vectors Equations

1. (IITJEE-2009) Angle between the vectors a & b where a, b and c areunit vectors satisfying a+ b+

3c = 0

Solution : We need to nd the angle between a & b, So we need toarrange the given info such that we are able to produce just that

and rest others is given to us.

a+ b =

3c

|a+ b| =|

3c||a+ b|2 =|

3c|2

(a+ b) (a+ b) =3c c|a|2 + |b|2 + 2a b =3|c|2

2 + 2a b =3|a||b| cos =1

2 =

pi

3

Alternate-Solution : Another way to look this problem is by sym-

metry. In a+ b+

3c = 0 terms a & b are very symmetric. Given|a| = |b| = 1 So the problem is problem of statics. The forcesare all balanced hence along one component. Therefore we get,

|a| cos + |b| cos = 3|c| 2 cos = 3 = pi6 . Thereforeangle between a& b is 2 = pi3

10

14 USEFUL FORMULAE

14 Useful Formulae

1. |a b|2 + (a b)2 = |a|2|b|2

LHS = |ab|2+(ab)2 = |a|2|b|2 sin2 +|a|2|b|2 cos2 = |a|2|b|2 =RHS

2. |a b| |a||b|

Solution : |a b| = |a||b| | sin ||n| |a||b|

3. |a b| |a||b|

Solution : |a b| = |a||b|| cos | |a||b|

4. (a b) (c d) = a c a db d b d

Solution :Let n = c d

LHS = a b n= a b n= a b (c d)= a {(b d)c (b c)d}= (b d)(a c) (b c)(a d)= RHS

5.

(a b) (c d) = [a b d]c [a b c]d= [a c d]b [b c d]a

Solution :Let n = a bLHS = n (c d)

= (n d)c (n c)d= [a b d]c [a b c]d6. Let r be any vector & a, b & c be non-zero, non-coplanar vectors then

r = [r b c][a b c]

a+ [r c a][a b c]

b+ [r a b][a b c]

c

(a) Solution: Let r = 1a + 2b + 3c as a, b & c are non-coplanar.Now take dot product with b c and nd 1similarly nd theother constants by taking dot product with other symmetric cross

products.

7. [a b b c c a] = [abc]2

11

16 VECTOR GEOMETRY

15 Area & Volume of geometrical structures

Area of a parellogram with adjacent sides given by a & b = |a b|

with diagonals d1 & d2 = 12 |d1 d2|

Area of a Triangle :

a & b are the edges of the triangle then Area = 12 |a b| If a, b & c are the position vectors of the vertices A,B & C thenArea =

12 |a b+ b c+ c a|

Volume of a parellopiped : [a b c] Volume of a tetrahedron : 16 [a b c] Volume of a triangular prism : 12 [a b c]

16 Vector Geometry

16.1 Equation of a line in Vector form

Two point form : Equation of line passing through the pointsA(a) &B(b)is derived as

r = a+ AB = a+ (b a) = (1 )a+ b

Slope point form : Equation of a line passing through the point A(a)and parellel to the vector b

r = a+ b

16.2 Skew lines

Any two lines which are neither intersecting nor parellel are called skew lines.

This will happen in 3D space.

Note: Two lines that are parallel never intersect and if they are

not parallel then they always intersect- this happens only in a

2D space. If we expand this property to hold true in 3D space it

is not true in form of skew lines. But if u take this analogy to 3D

space you have to increase the property by one dimension. Like

- Two planes if they are parallel then they never interesect and

if they are not parallel then they always intersect in a line.

So if line L1 : r = a + c & L2 : r = b + d. If L1, L2 are skew linesthen shortest distance between the lines is p = [ab c d]|cd|

12

16 VECTOR GEOMETRY 16.3 Equation of angle bisector

16.3 Equation of angle bisector

Equation of angle bisector of angle shown by sides parallel to b & c andpassing through point A(a)

r = a+ (b

|b| c

|c|)

to specify the internal and external bisector of the angle.

16.4 Vector equation of a plane

16.4.1 Normal form

1

Vector equation of a plane at a distance of p units from origin and n beingnormal to it is given as

r n = p

16.4.2 Point-two vector form

Equation of plane passing through point A(a) and containing b & c on it is

(r a) b c = 0

or

[r b c] = [a b c]

16.4.3 Three point form

Equation of a plane containing 3 points A(a), B(b) & C(c) is

[rab] + [rbc] + [rca] = [abc]

or

r (a b+ b c+ c a) = a b c1

Any equation of the form r n = p (Note here p is not the distance between theplane from origin since n is not the unit normal) can be converted to have unit normal bydividing by |n| on both sides to get the vector equation. So p|n| represent the distance ofthis plane from origin.

13

16.5 Section made by a plane17 RECIPROCAL SYSTEM OF VECTORS

16.5 Section made by a plane

Let A(a) & B(b) are two points outside a plane r n = p. We will use basicproportionality theorem used in schools to nd the ratio in which plane

divides line joining AB.

Plane parallel to r n = p and passing through point A & B are givenby r n = a n & r n = b n. Let us see the transverse section ofthese three planes passing through the point of section and points A & B.

So here the section made by the plane r n = p is AZZB . Using schoolgeometry we know using basic proportionality theorem,

AZ

ZB=AY

AX=p1 p0p0 p2 =

a n p0b n p0

17 Reciprocal system of Vectors

p, q & r are reciprocal system of three non-coplanar vectors a, b & c if

p =b c[abc]

, q =c a[abc]

& r =a b[abc]

Properties we get out of these system are

p a = q b = r c = 1

p b = p c = q a = q c = r a = r b = 0

p a+ q b+ r c = 3

[abc][pqr] = 1

Any vector v = (v a)p+ (v b)q+ (v c)r = (v p)a+ (v q)b+ (v r)c

14

19 LEVEL 2 PROBLEMS

Part II

Problems

18 Level I Problems

1. Which of the following makes sense?

(a) (a b) c(b) a b c(c) a (b c)(d) (a b)c(e) (a b)c(f) (a b) (c d) e2. Given r = a b which of the statements makes sense(a) r a & r b(b) r a or r b(c) r & a b are collinear vectors(d) none of these

3. Maximum value of [a b c] where a, b, c are non-coplanar vectors is

(a) 1

(b) abc

(c) 0

(d) none of these

19 Level 2 problems

1. Derive following in triangles using vectors

(a) a cosB + b cosA = c(b) a2 = b2 + c2 2bc cosA(c)

asinA =

bsinB =

csinC

(d) = 12bc sinA

2. Prove pythogoras theorem

(a) i.e. prove if angle A is

pi2 then b

2 + c2 = a2

(b) and also prove the converse of the theorem.

15