10 vectors

By:-

DR. VIKRAM SINGHTANUSHREE SINGH

YEAR OF PUBLICATION-2010All rights reserved. No part of this publication may be reproduced,

stored in a retrieval system, transmitted in any form or by any means-

Electronic, Mechanical, Photocopying, Recording or otherwise, without

prior permission of the Authors and Publisher

SAVANT INSTITUTE

TM

CLASS XII

MATHEMATICS

Mathematics Vectors 1

SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi – 110065. Ph.: +91-11-26224417 www.savantgroup.org

10

VECTORSPre-requisites

§ Knowledge of basic arithmetic operations. § Knowledge of geometrical concepts regarding lines,

angles, triangle, etc. ______________________________________________ Concept Map

___________________ Slide 1 ____________________

History

§ Around 1636, Descartes and Fermat founded analytical geometry.

§ In 1804 Bolzano introduced certain operations on points, lives and plane which are predecessors of vectors.

___________________ Slide 2 ____________________

§ The foundation of the definition of vectors was Bellavitis notion of the bipoint.

§ In 1857, cay lay introduced the matrix notation. § In 1857 Gross man studied the barycentric calculus

initiated by Mobius.

§ Peano was the first to given the modern definition of vectors spaces and linear maps in 1888.

___________________ Slide 3 ____________________

Scalars and Vectors

A scalar is a quantity which has magnitude but does not posses direction. A vector is a quantity which has magnitude and is also related to a definite direction.

___________________ Slide 4 ____________________

For Example,

(i) A person covered a distance of 5 km is scalar as the distance covered is irrespective of direction which can be covered by person in number of ways as;

___________________ Slide 5 ____________________

(ii) A person covered 5 km due East is vector as fixed direction.

___________________ Slide 6 ____________________

Types of Vectors

1. Zero vector or Null vector: A Vector whose initial and terminal points are coincident is called the zero vector or the null vector. The magnitude of the zero vector is zero and it can have any arbitrary direction and any line as its line of support.

110 Vectors Mathematics


Slide 7

2. Unit Vector: A vector of unit magnitude is called a unit vector. Unit vectors are denoted by small letter with a cap on them. Thus, a is a unit vector of a. Where a 1,=

r i.e.,

If the vector a is divided by magnitude a ,r

then we

get a unit vector in the direction of a.r

Thus, a ˆ ˆ ˆa a a a, whereaa

= ⇔ =r r rr is the unit vector

in the direction of a.

___________________ Slide 8 ____________________

3. Parallel vectors: Two or more vectors are said to be parallel, if they have the same support or parallel support. Parallel vectors may have equal or unequal magnitudes and direction may be same or opposite. As shown in figure.

___________________ Slide 9 ____________________

4. Like and unlike vectors:

Two parallel vectors having the same direction are called like vectors. As shown in the figure.

__________________ Slide 10 ____________________

Two parallel vectors having opposite directions are called unlike vectors. As shown in the figure.

__________________ Slide 11 ____________________

5. Collinear vectors: Vectors which are parallel to the same vector and have either initial or terminal point in common are called collinear vectors. As shown in the figure. Here, OA,OB and OC

uuur uuur uuur are collinear vectors.

Slide 12

6. Co-initial vectors:

Vectors having same initial point are called co-initial vectors. As shown in the figure; Here, OA,OB,OCandOD

uuur uuur uuur uuur are co-initial vectors.

__________________ Slide 13 ____________________

7. Free vectors:

Vectors whose initial point is not specified are called free vectors. As shown in the figure.

8. Localised vectors:

A vector drawn parallel to a given vector, but through a specified point as initial point, is called localised vector.

__________________ Slide 14 ____________________

9. Equal vectors: Two vectors are said to be equal, if they have the same magnitude and same direction. Their lines of support may be parallel or coincident. As shown in the figure Here, AB CD=

uuur uuur

__________________ Slide 15 ____________________

A vector remains unaltered by one or more shifts parallel to itself. Thus, the vectors represented by the two opposite sides of a parallelogram are equal vectors, if their directions are same. As shown in the figure.



Slide 16

Algebra of Vectors

Addition of two vectors:

Let a a n d brr

be any two vectors. Such that;

OA a and OB b.= =uuur uuur rr

Then their sum of resultant, denoted by a b,+rr

is defined as vector OC

uuur given by the diagonal of the parallelogram

OACB as shown in the figure. i.e. OC OA OB a b= + = +

uuur uuur uuur rr

__________________ Slide 17 ____________________

Note.

(i) That this addition is commutative. OC OB BC b a= + = +uuur uuur uuur r r∵

Also from the ∆OAC, OC OA AC a b= + = +uuur uuur uuur rr

(ii) If OA and AC

uuur uuur are collinear, their sum is still OC

uuur

although in this case we do not have a triangle or a parallelogram in their usual sense. In general, known as (the polygon law of addition)

__________________ Slide 18 ____________________

(iii) As from the figure Which shows, 1 n1 2 n 1 nOA A A .... A A OA−+ + + =

uuur uuuuur uuuur uuur

By the polygon law of addition.

__________________ Slide 19 ____________________

(iv) Vector addition exhibits following properties 1. a b b a+ = +

r rr r (commutativity)

2. ( ) ( )a b c a b c+ + = + +r rr r r r

(associativity)

3. a 0 a+ =rr r

(additive identity) 4. ( )a a 0+ − =

rr r (additive inverse)

5. ( )1 2 1 2k k a k a k a+ = +r r r

(multiplication by scalars)

6. ( )k a b ka kb+ = +r rr r

(multiplication by scalars)

(v) Inequality law: 1. a b a b+ ≤ +

r rr r

2. a b a b− ≥ −r rr r

__________________ Slide 20 ____________________

Position Vector of a Point

If a point O is fixed in space as origin then for any point P, the vector OP a=

uuur r is called the position

vector (P.V.) of ‘P’ w.r.t. ‘O’. Let a and b

rr be the position vectors of the points A and B

respectively with reference to the origin. Then, OA a,OB b= =

uuur uuur rr

__________________ Slide 21 ____________________

By triangle law of addition of vectors, we have OA AB OB+ =

uuur uuur uuur

AB OB OA∴ = −uuur uuur uuur

AB b a= −uuur r r

__________________ Slide 22 ____________________

Solved Example

If a,b,crr r

be the vectors represented by the sides of a

triangle, taken in order, then prove that: a b c 0.+ + =r rr r

__________________ Slide 23 ____________________

Solution

Let ABC be a triangle such that;

BC a,CA bandAB c.

Then, a b c BC CA AB

BA AB AB AB

a b c 0

= = =

+ + = + +

= + = − +

+ + =

uuur uuur uuurrr ruuur uuur uuurrr r

uuur uuur uuur uuur

r rr r



Slide 24

Illustration

If c 3a 4b and 2c a 3b,= + = −r rr r r r

show that: (i) c and a

r r have the same direction and c a .>

r r

(ii) c and brr

have opposite direction and c b .>rr

__________________ Slide 25 ____________________

Solved Example

The position vectors of points A, B, C, D are a,b,2a 3b and a – 2b+

r r rr r r respectively. Show that

DB 3b–a and AC a 3b.= = +uuur uuurr rr r

__________________ Slide 26 ____________________

Solution

We have, DB =uuur

Position vector of B – Position vector of D

( )DB b – a – 2b 3 b – a⇒ = =uuur r r rr r

and, AC =uuur

Position vector of C – Position vector of A

( )AC 2a 3b – a a 3b.⇒ = + = +uuur r rr r r

__________________ Slide 27 ____________________

Illustration

Let ABCD be a parallelogram. If a,b,c,rr r

be the position vectors of A, B, C respectively with reference to the origin O, find the position vector of D with reference to O.

__________________ Slide 28 ____________________

Section Formula

(i) Internal Division:

Let A and B be two points with position vectors a and brr

respectively, and C be a point dividing AB internally in the ratio m : n. Then the position vector of C is given by:

mb naOC

m n+

=+

r ruuur

__________________ Slide 29 ____________________

(ii) External Division:

Let A and B be two points with position vectors a a n d brr

respectively and let C be a point dividing ABuuur

externally in

the ratio m : n. Then the position vector of Cur

is given by;

mb naOC .

m n−=−

r ruuur

__________________ Slide 30 ____________________

Illustration

Find the position vectors of the points which divide the join of the points 2a–3b and 3a–2b

r rr r internally and externally

in the ration 2 : 3.

__________________ Slide 31 ____________________

Illustration

Let a ,b , crr r

be the position vectors of three distinct points

A, B, C. If three exist scalars x, y, z (not all zero) such that xa yb zc 0+ + =

r rr r and x + y + z = 0, then show that A, B

and C lie of a line.

__________________ Slide 32 ____________________

Illustration

Prove using vectors: The diagonals of a quadrilateral bisect each other iff it is a parallelogram.

__________________ Slide 33 ____________________

Coplanar Vectors

Vector are said to be coplanar if all of them lie in the same plane. As shown in the figure.

Here, vectors a,b and c

rr r are coplanar because all of

them lie in the plane of the paper. Coplanar vectors may have any directions or magnitude.

__________________ Slide 34 ____________________

Components of a Vector in Two Dimensions

Let P (x, y) be a point in a plane with reference to OX and OY as the coordinate axes. As shown in the figure. Where OM = x and PM = y Let ˆ ˆi, j be unit vectors along OX and OY respectively.

Then, ˆ ˆOM xi andMP yj= =uuuur uuur



Slide 35

Vectors OM and MPuuuur uuur

are known as the components of

OPuuur

along x-axis and y-axis respectively.

__________________ Slide 36 ____________________

Then, ˆ ˆr xi yj= +r

(vector representation of a point) 2 2r x y ;⇒ = +

r (magnitude of a vector)

Thus, if a point P in a plane has co-ordinate (x, y), then (i) ˆ ÔP xi yj= +

uuur

(ii) 2 2OP x y= +uuur

__________________ Slide 37 ____________________

(iii) The component of OPuuur

along x-axis is a vector ˆxi, whose magnitude is |x| and whose direction is along OX or OX′ according as x is positive or negative.

(iv) The component of OPuuur

along y-axis is a vector yj, whose magnitude is |y| and whose direction is along OY or OY′ according as y is positive or negative.

__________________ Slide 38 ____________________

Components of a Vector ABuuur

in terms of Co-ordinates of A and B

( ) ( )2 2

2 1 2 1AB x x y y .= − + −uuur

__________________ Slide 39 ____________________

Addition, Subtraction, Multiplication of a Vector by a Scalar and Equality in Terms of Components

For any two vectors, 1 1 2 2ˆ ˆ ˆ â x i y j andb x i y j= + = +

rr

Then we define (i) ( ) ( )1 2 1 2

ˆ â b x x i y y j+ = + + +rr

(ii) ( ) ( )1 2 1 2ˆ â b x x i y y j− = − + −

rr

(iii) ( ) ( )1 1ˆ ˆma mx i my j,= +

r where m is scalar quantity

(iv) 1 1 2 2a b x y and x y .= ⇔ = =rr

__________________ Slide 40 ____________________

Solved Example

Find the values of x and y so that vectors ˆ ˆ ˆ ˆ2i 3 j and xi yj+ + are equal.

__________________ Slide 41 ____________________

Solution

1 1 2 2ˆ ˆ ˆ â i b j a i b j+ = +

1 2 1 2a a and b b⇔ = =

ˆ ˆ ˆ ˆ2i 3 j xi yi∴ + = +

⇒ x = 2 and y = 3.

__________________ Slide 42 ____________________

Illustration

Find the components along the coordinates of the position vector of each of the following points: (i) P(5, 4) (ii) S(−4, −5).

__________________ Slide 43 ____________________

Components of a Vector in Three Dimensions

Let P (x, y, z) be any point in space and ˆ ˆ î, j, k are the unit

vectors along the axes of co-ordinates. As shown in the figure. ˆ ˆ ˆ ˆ ˆ ÔP zk xi yj xi yj zk.= + + = + +

uuur

__________________ Slide 44 ____________________

Note: If P (x1, y1, z1) and Q (x2, y2, z2) are any two points in space, then ( ) ( ) ( )2 1 2 1 2 1

ˆ ˆ ˆPQ x x i y y j z z k= − + − + −uuur

Scalar components of PQ along the three axes are; (x2 – x1), (y2 – y1) and (z2 – z1) respectively and the corresponding vector components are ( ) ( ) ( )2 1 2 1 2 1

ˆ ˆ ˆx – x i, y – y j and z – z k.

In this case,

( ) ( ) ( )2 2 2

2 1 2 1 2 1r PQ x x y y z z .= = − + − + −uuurr



Slide 45

Solved Example

Find the sum of vectors ˆ ˆ ˆ ˆ ˆ ˆ â i – 2 j k , b –2i 4 j 5k= + = + +r

and

ˆ ˆ ˆc i –6 j–7k .=

__________________ Slide 46 ____________________

Solution

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

( ) ( ){ } ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ â b c i – 2 j k –2i 4 j 5k i – 6 j – 7 k⇒ + + = + + + + +rr r

( ) ( ) ( ){ } ( )ˆ ˆ ˆ ˆ ˆ â b c 1 – 2 i –2 4 j 1 5 k i – 6 j – 7 k⇒ + + = + + + + +rr r

( ) ( )ˆ ˆ ˆ ˆ ˆ â b c i 2 j 6k i – 6 j – 7 k⇒ + + = − + + +rr r

( ) ( ) ( )ˆ ˆ â b c –1 1 i 2 – 6 j 6 – 7 k⇒ + + = + + +rr r

ˆ ˆ â b c 0 i – 4 j – k .⇒ + + =rr r

__________________ Slide 47 ____________________

Solved Example

The position vectors of the points P, Q, R are ˆ ˆ ˆ ˆ ˆ ˆ ˆ î 2 j 3k, – 2 i 3 j 5k and 7 i – k+ + + + respectively. Prove that

P, Q and R are collinear

__________________ Slide 48 ____________________

Solution

We have, PQ =uuur

Position vector of Q − Position vector of P

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆPQ –2i 3j 5k – i 2 j 3k –3i j 2k and⇒ = + + + + = + +uuur

QR⇒ =uuur

Position vector of R − Position vector of Q

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆQR 7 i – k – –2i 3 j 5k 9 i – 3 j – 6 k⇒ = + + =uuur

Clearly, QR –3PQ.=uuur uuur

This shows that the vector PQ and QRuuur uuur

are collinear.

But, Q is common point between PQ and QR.uuur uuur

Therefore,

given points P, Q and R are collinear

__________________ Slide 49 ____________________

Illustration

If A, B, C have position vectors (2, 0, 0) (0, 1, 0), (0, 0, 2) show that ∆ABC is isosceles.

Slide 50

Collinearity of Vectors

If a and brr

are two collinear or parallel vectors, then there

exists a scalar λ such that a b o r b a= λ = λr rr r

__________________ Slide 51 ____________________

Note: (1) Two non-zero vectors a a n d b

rr are collinear if there

exists scalars x, y not both zero such that xa yb 0.+ =rr

(2) If a ,brr

are any two non-zero non-collinear vectors and

x, y are scalars, then xa yb 0 x y 0.+ = ⇒ = =rr

__________________ Slide 52 ____________________

Illustration

If a and brr

are non-collinear vectors, find the value of x

for which vectors: ( ) ( )x 2 a band 3 2x a 2bα = − + β = + −r rrr rr

are collinear.

__________________ Slide 53 ____________________

Collinearity of Points

Let A, B, C be three points. Then, each pair of the vector AB, BC; AB, AC; BC, ACuuur uuur uuur uuur uuur uuur

is a pair of collinear vectors.

Thus to check collinearity of three points, we can check the collinearity of any two vectors obtained with the help of three points.

__________________ Slide 54 ____________________

Note: Three points with position vectors a,b and crr

are

collinear if and only if there exists three scalars x, y, z not all zero simultaneously such that: xa yb zc 0+ + =

r rr r

together x + y + z = 0.

__________________ Slide 55 ____________________

Illustration

Show that the points A, B, C with position vectors 2a 3b 5c,a 2b 3c and 7a c− + + + + −

r rr r r r r r respectively, are

collinear.

__________________ Slide 56 ____________________

Product of Two Vectors

There are two ways in which the two vector quantities are multiplied. (a) Scalar or Dot product:

(Whose result is a number and does not involve direction)



(b) Vector or Cross product: (Whose result is associated with a definite direction and as such is a vector quantity). Now, we shall discuss these products.

__________________ Slide 57 ____________________

Scalar (or Dot) product of two vectors

The scalar product of a a n d brr

written as a.brr

is defined to

be the number a brr

cosθ, where θ is the angle between

the vectors a a n d brr

i.e., a.b a b cos= θr rr r

__________________ Slide 58 ____________________

Note: Let a a n d brr

be two given vectors forming an angle

θ, where 0 ≤ θ ≤ π. Then the scalar product of a and brr

(i) is positive, if θ is acute (ii) is negative, if θ is obtuse (iii) is zero, if θ is right angle (iv) a.b

rr is a pure number and not a vector

(v) The scalar product of a a n d brr

is sometimes denoted

by ( )a,b .rr

__________________ Slide 59 ____________________

Properties of Scalar Product

(i) 2 2a.a a a= =

rr r

ˆ ˆ ˆ ˆ ˆ î . i j . j k.k 1⇒ = = =

(ii) a.b b.a=r rr r

(i.e., commutative)

(iii) a.0 0=rr

(iv) ( ) ( )a.b acos b projectionof a onb b= θ =r rr r

also ( ) ( )a.b bcos a projectionofbona a= θ =r rr r

__________________ Slide 60 ____________________

(v) ( ) ( )a. b c a.b a.c distributive+ = +r rr r r r r

(vi) ( ) ( ) ( )l a . m a lm a.b=rr r r

(vii) a.b 0 a,b= ⇔r rr r

are perpendicular to each other

ˆ ˆ ˆ ˆ ˆ î . j j.k k . i 0⇒ = = =

(viii) ( ) ( ) ( )22 2a b a b . a b a b 2a.b± = ± ± = + ±

uur uurr r r rr r r r

__________________ Slide 61 ____________________

(ix) ( ) ( ) 22 2 2a b . a b a b a b+ − = − = −r r rr r r

If 1 2 3 1 2 3ˆ ˆ ˆ ˆ ˆ â a i a j a k and b b i b j b k= + + = + +

rr

Then 1 1 2 2 3 3a.b a b a b a b= + +rr

(x) If a,brr

are non-zero, then the angle between them is

given by a.bcosa b

θ =rrrr

or 1 1 2 2 3 3

2 2 2 2 2 21 2 3 1 2 3

a b a b a bcos

a a a b b b

+ +θ =

+ + + +

where 1 2 3 1 2 3ˆ ˆ ˆ ˆ ˆ â a i a j a k and b b i b j b k.= + + = + +

rr

__________________ Slide 62 ____________________

Solved Example

( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆFind a 3b . 2 a – b , i f a i j 2k and b 3 i 2j–k.+ = + + = +r rr r r

__________________ Slide 63 ____________________

Solution

We have, ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ â 3b i j 2k 3 3 i 2 j – k 10i 7 j – k+ = + + + + = +rr

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ând 2a–b 2 i j 2k – 3i 2 j – k – i 0 j 5k= + + + = + +rr

( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ â 3b . 2 a – b 10i 7 j – k . – i 0 j 5k∴ + = + + +r rr r

= (10) (−1) + (7) (0) + (−1) (5) = −10 0 − 5 = −15

__________________ Slide 64 ____________________

ALITER: We have a 1 1 4 6, b 9 4 1= + + = = + +rr

14anda.b 3 2 – 2 3.= = + =rr

( ) ( )a 3b . 2a–b∴ +r rr r

a.2a–a.b 3b.2a–3b.b= +r r r rr r r r

( ) 222 a –a.b 6 b.a – 3 b= +r r rr r r

( ) 222 a 5 a.b – 3 b= +r rr r

( ) ( ) ( )2 6 5 3 – 3 14 –15.= + =

__________________ Slide 65 ____________________

Solved Example

Find the value of λ so that the vectors ˆ ˆ â 2i j k= + λ +r

and

ˆ ˆ ˆb i – 2 j 3k= +r

are perpendicular to each other.



Slide 66

Solution

The vectors a a n d brr

are perpendicular to each other.

a.b 0∴ =rr

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ2i j k . i – 2 j 3k 0⇒ + λ + + =

⇒ (2) (1) + λ (−2) + (1) (3) = 0 ⇒ −2λ + 5 = 0 ⇒ λ = 5/2.

__________________ Slide 67 ____________________

Illustration

Find the value of p for which the vectors a 3 i 2 j 9k= + +r r rr

and b i p j 3k= + +r r r r

are

(i) perpendicular (ii) parallel.

__________________ Slide 68 ____________________

Components of a Vector br

along and Perpendicular to Vector a

r

Let a and brr

be two vectors represented by OA and OBuuur uuur

and let θ be the angle between a and b.rr

Draw BM ⊥ OA.

Shown as, b OM MB∴ = +

uuuur uuurr

or ( ) ( ) ( )ˆ ˆ ÔM OM a OBcos a b cos a= = θ = θuuuur r

__________________ Slide 69 ____________________

( ) ( )

2

a.b a.b aa.b a.bˆ ˆb a a aa a aa b a

= = = =

rr rr rr rr rr rr r r rr r

also b OM MB= +uuuur uuurr

2

a.bMB b OM b aa

∴ = − = −

rruuur uuuurr r rr

Thus, the components of br

along the perpendicular to ar

are 2 2

a.b a.ba and b aa a

−

r rr rrr rr r respectively.

Slide 70

Solved Example

Show that the projection vector of ( ) 2

a.ba o n b 0 is b.b

≠

rrr r rrr

__________________ Slide 71 ____________________

Solution

Let θ be the angle between a and br r

Shown as in the adjoining figure; Here OM is the length of projection of a on b

r r and

( ) ÔM OM b,=uuuur

i.e., OMuuuur

is the projection vector of a on brr

OMcos OM a cos

OA∴ θ = = = θ

r

__________________ Slide 72 ____________________

a.b a.bOM a

a b b

⇒ = =

r rr rr r rr (projection of a on brr

)

a.bOMb

⇒ =rr

r

Now, ( ) 2

a.b a.b b a.bˆ ÔM OM b b bb b b b

= = =

r r r rr r ruuuur rr r r r

∴ Projection vector of ( )2

a.ba o n b 0 is b.| b |

≠

rrr rr r

__________________ Slide 73 ____________________

Vector (or Cross) Product of two Vectors

The vector product of two vectors a a n d brr

denoted by

a b×rr

is defined as the vector â b sin ,θ ηrr

where θ is

the angle between the vectors â,b and ηrr

is the unit

vector perpendicular to the plane of a and brr

(i.e.,

perpendicular to both a and brr).

__________________ Slide 74 ____________________

The sense of η is obtained by the right hand thumb rule.

i.e., â,b and ηrr

form a right handed screw.



If we curl the figures of our right hand from a t o brr

through

the smaller angle (keeping the initial point of a a n d brr

same), the thumb points in the direction of ˆ.η

__________________ Slide 75 ____________________

In this case, ( )â,band or a b ,η ×r rr r

in that order are said to

form right handed system.

It is evident that a b absin× = θrr

__________________ Slide 76 ____________________

Properties:

(i) ( )a b b a× = − ×r rr r

(non-commutative)

a a 0× =r r

(ii) ˆ ˆ ˆ ˆ ˆ î i j j k k 0× = × = × =

__________________ Slide 77 ____________________

(iii) ( )a b c a b a c× + = × + ×r rr r r r r

(Distributive law)

(iv) ( ) ( ) ( )la mb lm a b× = ×r rr r

(v) a b 0 a a n d b× = ⇔r rr r

collinear (If none of a , o r brr

is a

zero vector)

__________________ Slide 78 ____________________

(vi) ˆ ˆ ˆ ˆ ˆ î j i j sin90 k k× = ° =

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ î j k, j k i,k i j× = × = × =

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆj i k, k j i, i k j× = − × = − × = −

Slide 79

(vii) If 1 2 3 1 2 3ˆ ˆ ˆ ˆ ˆ â a i a j a k and b b i b j b k= + + = + +

rr

then, 1 2 3

1 2 3

ˆ ˆ î j ka b a a a

b b b× =

rr

( ) ( ) ( )2 3 3 2 3 1 1 3 1 2 2 1ˆ ˆ â b – a b i a b – a b j a b – a b k= + +

__________________ Slide 80 ____________________

(viii) Any vector perpendicular to the plane of a and brr

is

( )a b ,λ ×rr

where λ is a real number. Unit vector

perpendicular to a ba a n d ba b

×= ±×

rrrr rr

__________________ Slide 81 ____________________

(ix) If a and brr

are adjacent sides of a triangle then;

Area of 1

a b2

∆ = ×rr

If a, brr

are adjacent sides of parallelogram,

Its Area a b= ×rr

__________________ Slide 82 ____________________

(x) If a,brr

are diagonals of parallelogram;

Its Area 1

| a b | .2

= ×rr

__________________ Slide 83 ____________________

Solved Example

ˆ ˆ ˆ ˆ ˆFind a b,i fa 2i k and b i j k.× = + = + +r rr r



Slide 84

Solution

ˆ ˆ î j kWe have, a b 2 0 1

1 1 1× =

rr

( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ â b 0 – 1 i – 2 – 1 j 2 – 0 k – i – j 2k.⇒ × = + = +r

__________________ Slide 85 ____________________

Solved Example

Find the area of the parallelogram determined by the vectors ˆ ˆ ˆ ˆ ˆ î 2 j 3k and 3 i – 2 j k.+ + +

__________________ Slide 86 ____________________

Solution

Let ˆ ˆ ˆ ˆ ˆ â 3i j – 2 k and b i – 3 j 4k.= + = +rr

Then,

ˆ ˆ î j ka b 3 1 –2

1 –3 4× =

rr

( ) ( ) ( )ˆ ˆ ˆ4 – 6 i – 12 2 j – 9 – 1 k= + + ˆ ˆ ˆ–2i–14j–10k=

( ) ( ) ( )2 2 2a b –2 –14 –10 300⇒ × = + + =rr

∴ Area of the parallelogram 1

a b2

= ×rr

1

300sq.units.2

=

__________________ Slide 87 ____________________

Illustration

Given a 10, b 2 and a.b 12, find a b .= = = ×r r rr r r

__________________ Slide 88 ____________________

Illustration

In a ∆ABC internal angle bisectors AI, BI and CI are produced to meet opposite sides in A′, B′ and C′ respectively. Prove that the maximum value of

AI.BI.CI 8is .

AA.BB.CC 27′ ′ ′

__________________ Slide 89 ____________________

Scalar Triple Product

It is defined for three vectors a ,b , crr r

in that order as the scalar

( )a b .c,×rr r

which can also be written simply as a b.c.×rr r

It denoted the volume of the parallelepiped formed by taking a,b,c

rr r as the coterminus edges.

i.e. V = magnitude of ( ) ( )a b .c a b . c× = ×r rr r r r

and is

represented by a b c . rr r

__________________ Slide 90 ____________________

Geometrical Interpretation of a Scalar Triple Product

Let a,b,crr r

be three vectors. Consider a parallelepiped

having coterminus edges OA,OB and OCuuur uuur uuur

such that

OA a,OB b and OC c.= = =uuur uuur uuurrr r

Then a b×rr

is a vector

perpendicular to the plane of a and b.rr

If η is a unit vector

along a b×rr

then φ is the angle between ˆ and c.ηr

__________________ Slide 91 ____________________

Now, ( )a b c = a b .c × r rr r r r

= (Area of parallelogram OADB) ˆ .cηr

= (Area of parallelogram OADB) ( )ˆ .cηr

= (Area of parallelogram OADB) ( )ˆc cosη φr

= (Area of parallelogram OADB) ( )c cosφr

= (Area of parallelogram OADB) (OL) = Area of base × height = volume of parallelepiped

__________________ Slide 92 ____________________

Properties:

(i) If a,b,crr r

are given by

1 2 3ˆ ˆ â a i a j a k,= + +

r

1 2 3ˆ ˆ ˆb b i b j b k= + +

r



1 2 3ˆ ˆ ˆc c i c j c k= + +

r

Then, ( )1 2 3

1 2 3

1 2 3

a a aa b .c b b b

c c c× =

rr r

__________________ Slide 93 ____________________

(ii) ( ) ( )a b .c a. b c× = ×r rr r r r

i.e., position of dot and cross can

be interchanged without altering the product. Hence it

is also represented by a b c .

rr r

(iii) a b c b c a c a b = = r r rr r r r r r

(iv) a b c b a c = − r rr r r r

(v) k a b c k a b c = r rr r r r

__________________ Slide 94 ____________________

(vi) a bcd acd bcd + = + r r r r rr r r r r

(vii) a,b,crr r

in that order form a right handed system if

a,b,c 0 > rr r

__________________ Slide 95 ____________________

(viii) The necessary and sufficient condition for three non-

zero, non-collinear vectors a,b,crr r

to be coplanar is

that abc 0 = rr r

i.e., a,b,crr r

are coplanar ⇔ abc 0 = rr r

(ix) Four points a ,b ,c ,dr rr r

are coplanar if,

dbc dca dab abc + + = r r r r r rr r r r r r

or abc a c d adb dbc + + = r r r r r rr r r r r r

(x) abc aab bba ccb 0 = = = = r r r r rr r r r r r r

i.e., if any two vectors are same then vectors are coplanar.

__________________ Slide 96 ____________________

Illustration

If a,b,c, l , mr rr r r

are vectors, prove that:

( )a b c

abc l m a.l b.l c.l

a.m b.m c.m

× =

rr rrr r r r rr r r r rrr r r r r

__________________ Slide 97 ____________________

Vector Triple Product

It is defined for three vector a,b,crr r

as the vector

( )a b c .× ×rr r

This vector being perpendicular to b c,×r r

is

coplanar with b a n d cr r

i.e., ( )a b c lb mc× × = +r rr r r

Take the scalar product of this equation with we get,

( ) ( )0 l a.b m a.c= +rr r r

( ) ( ) ( )a b c a.c b a.b c ⇒ × × = λ − r r rr r r r r r

__________________ Slide 98 ____________________

If we choose the co-ordinate axes in such a way that;

1 1 2 1 2 3ˆ ˆ ˆ ˆ ˆ â a i,b b i b j and c c i c j c k,= = + = + +

rr r it is easy to

show that λ = 1

Hence, ( ) ( ) ( )a b c a.c b a.b c× × = −r r rr r r r r r

In general, ( ) ( )a b c a b c× × ≠ × ×r rr r r r

( ) ( )a b c a b c× × = × ×r rr r r r

if some or all a,b,crr r

are zero

vectors or a a n d cr r

are collinear.

__________________ Slide 99 ____________________

Note: ( )a b c× ×rr r

is a linear combination of those two

vectors which are with in brackets.

If ( )r a b c , then r= × ×rr r r r

is perpendicular to ar

and lie in the

plane of bandc.r r

( ) ( ) ( )a b c a.c b a.b c× × = −r r rr r r r r r

( ) ( ) ( )a b c c.a b c.b a× × = −r r rr r r r r r

Aid to memory I × (II × III) = (I . III) II – (I . II) III.

__________________ Slide 100 ___________________

Illustration

If ( )ˆ ˆ ˆ ˆ â i j k , a.b 1 and a b j k,= + + = × = −r rr r r

Then br

is

(a) ˆ ˆ î j k− + (b) ˆ ˆ2 j k−

(c) i (d) ˆ2i.



Slide 101

Illustration

The unit vector which is orthogonal to the vector ˆ ˆ ˆ3i 2 j 6k+ + and is coplanar with the vectors ˆ ˆ ˆ2i j k+ + and

ˆ ˆ ˆi j k− + is

(a) ˆ ˆ ˆ2i 6 j k

41− +

(b) ˆ ˆ2i 3 j13−

(c) ˆ ˆ3 j k10−

(d) ˆ ˆ ˆ4i 3 j 3k .

34+ −

__________________ Slide 102 ___________________

Illustration

If A, B and Crr r

are vectors such that B C ,=rr

prove that

( ) ( ){ } ( ) ( )A B A C B C . B C 0.+ × + × × + =r r rr r r r r



CURRICULUM BASED WORKSHEET

Topics for Worksheet – I

§ Scalar and vector a quotation § Types of vectors. § Equality of two vectors § Addition of two vectors § Position vector of point § Section formula § Components of a vector § Addition, subtraction, multiplication of a vector by a

scalar and Equality in terms of compone nts § Components of a vector in three Dimensional.

Worksheet – I

1. Find the position vector of the midpoint of the vector joining the points ( )ˆ ˆ Â 2i 3 j 4k+ + and

( )ˆ ˆ ˆB 4 i j – 2 k .+

2. Points 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+ +

uuur uuur uuur is a vector parallel to

CK,uuur

where K divides AB in the ratio 1 : 3.

3. Let O be the origin and let P(−4, 3) be a point in the xy-plane. Express OP

uuur in terms of vector

ˆ î and j. Also find O P .uuur

4. Find the coordinates of the tip of the position vector which is equivalent to AB,

uuur where the

coordinates of A and B are (3, 1) and (5, 0) respectively.

5. Find the distance between the points A(2, 3, 1), B(−1, 2, −3), using vector method.

6. Prove that four points + +r rr r r r

2a 3b–c ,a–2b 3c,

3a 4b 2c and a – 6b 6c+ − +r rr r r r

are coplanar.

7. If $( ) $( )a i 2j 3k and b 2i 3 j k ,→ →

= + − = + +$ $ $ $ find a unit

vector in the direction of (a b).+r r

8. Find the components along the coordinates of the position vector of each of the following points: (i) Q(−4, 3) (ii) R(5, −7)

9. Show that the points with position vectors a – 2 b 3c,–2a 3 b – c a n d 4 a – 7 b 7c+ + +

r r rr r r r r r are

collinear.

Topics for Worksheet – II

§ Collinearity of vectors § Collinearity of points § Product of two vectors § Scalar and vector product § Products of vector product § Vector triple product

Worksheet – II

1. Let a a n d brr

be two given vectors such that

a 3, b 2 and a . b 6.= = =r rr r

Find the angle

between a and b.rr

2. Find the projection of ˆ ˆ â 2i j kon= − +r

ˆ ˆ ˆb i 2 j k.= − +r

3. If A(0, 1, 1) ,B(3, 1, 5) and C(0, 3, 3) be the vertices of a ∆ABC, show, using vectors, that ∆ABC is right angled at C.

4. Find a unit vector perpendicular to each one of the vectors ( ) ( )ˆ ˆ ˆ ˆ ˆ â 4 i – j 3k andb 2i 2 j – k .= + = +

rr

5. Using vector method, show that the points A(2, −1, 3). B(4, 3, 1) and C(3,1, 2) are collinear.

6. Find the angle between the vectors

( ) ( )ˆ ˆ ˆ ˆ ˆ â 3i 2 j k andb i 2 j 3k .= − + = − −rr

7. If ˆ ˆ ˆ ˆ ˆ â i j 2k andb 3i 2 j k,= + + = + −rr

find the value of

( ) ( )a 3b . 2 a – b .+r rr r

8. If ( ) ( )ˆ ˆ ˆ ˆ ˆ â i – 2 j 3k andb 2i 3 j – 5 k ,= + = +r r

then find

( )a b×rr

and verify that ( )a b×rr

is perpendicular to

each one of a and b.rr

9. If ( ) ( )ˆ ˆ â 2, b 7 and a b 3i 2 j 6k ,= = × = + +r r rr

find

the angle between aandb.rr

10. Find the area of the parallelogram whose adjacent sides are represented by the vectors

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ3i j –2k and i – 3 j 4k .+ +

CURRICULUM BASED CHAPTER ASSIGNMENT

1. Prove using vectors: Medians of a triangle are concurrent.

2. If D and E are the mid-points of sides AB and AC of a triangle ABC respectively. Show that

3BE DC BC.

2+ =

uuur uuur uuur



3. If the position vector ar

of a point (12, n) is such that a 13,=

r find the value of n.


of the point (5, n) is such that a 13,=


5. Find the scalar and vector components of the

vector with initial point A(2, 1) and terminal point

B(−5, 7). 6. If a

r is a position vector whose tip is (1, −3). Find

the coordinates of the point B such that AB a.=uuur r

If

A has coordinates (−1, 5). 7. Find a unit vector parallel to the vector ˆ ˆ3i 4j.− +

8. Show that the three points A(−2, 3, 5), B(1, 2, 3)

and C(7, 0, −1) are collinear . 9. If = + + = + −

rr ˆ ˆ ˆ ˆ ˆ â i 2 j 3kand b 2i 4 j 5k represent two

adjacent sides of a parallelogram, find unit vectors

parallel to the diagonals of the parallelogram.

10. Show that the points A, B and C with position vector = − = − +

rr ˆ ˆ ˆ ˆ â 3i 4j–4k, b 2i j k and ˆc i 3 j 5k= − −

r respectively. From the vertices of a

right angled triangle. 11. Find a.bwhen

rr

(i) = + − = − +rr ˆ ˆ ˆ ˆ ˆ â 2i 2 j k and b 6i 3 j 2k

(ii) ( ) ( )a 1, 1, 2 and b 3,2, 1= = −rr

12. Let = + − = − + = + −rr rˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ â 4 i 5 j k,b i 4 j 5kand c 3i j k.

Find a vector dr

which is perpendicular to both

aandb,rr

and satisfying =r rd.c 21.

13. Find the angle between the vectors + +ˆ ˆ ˆ5i 3 j 4k

and − −ˆ ˆ ˆ6 i 8 j k.

14. For any two vectors a a n d brr

prove that:

(i) 2 22a b a b 2a.b+ = + +

r r rr r r.

(ii) 2 22

a b a b 2a.b− = + −r r rr r r

(iii) ( )2 2 22a b a b 2 a b+ = − = +r r rr r r

Interpret the result geometrically.

(iv) 2

a b a b a b+ = − ⇔ ⊥r r rr r r

Interpret the result geometrically. (v) a b a b aisparallel t o b+ = + ⇔

r r rr r r

(vi) 2 22

a b a b a,bareorthogonal.+ = + ⇔r r rr r r

15. Find the components of a unit vector which is perpendicular to the vectors ˆ î 2j kand+ −

ˆ ˆ ˆ3i j 2k.− + 16. Find the magnitude of a

r given by

( ) ( )ˆ ˆ ˆ ˆ â i 3 j 2k i 3k= + − × − +r

.

17. Find a unit vector perpendicular to both the vectors − + + −ˆ ˆ ˆ ˆ î 2 j 3kand i 2 j k.

18. Find the area of the triangle whose vertices are A(3, −1, 2), B(1, −1, −3) and C(4, −3, 1).

19. If × = × × = ×rr r rr r r r

a b c d anda c b d, show that −rr

a d is

parallel to b c,−r r

where a dand b c.≠ ≠r rr r

20. If a ,b , crr r

are three vectors such that a b c 0,+ + =r rr r

then prove that a b b c c a× = × = ×r rr r r r

21. Prove that the normal to the plane containing

three points whose position vectors are rr r

a,b,c lies

in the direction b c c a a b.× + × + ×r rr r r r

22. For any two vectors rr

aand b, show that :

( ) ( ) ( ){ } ( )2 2221 a 1 b 1 a . b a b a b+ + = − + + + ×

r r r rr r r r

QUESTION BANK FOR COMPETITIONS

10. Find the position vector of the midpoint of the vector joining the points ( )ˆ ˆ Â 2i 3 j 4k+ + and

( )ˆ ˆ ˆB 4 i j – 2 k .+

(a) ( )ˆ ˆ ˆ3i 5 j k+ + (b) ( )ˆ ˆ ˆ3i 2 j k+ +

(c) ( )ˆ ˆ î 2 j k+ + (d) ( )ˆ ˆ ˆ9i 2 j k+ +

11. Points 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+ +

uuur uuur uuur is a vector parallel to

CK,uuur

where K divides AB in the ratio 1 : 3.

(a) 4 CK

10−

uuur (b)

10CK

4−

uuur

(c) 10

CK4

uuur (d)

4 CK10

uuur

12. Let O be the origin and let P(−4, 3) be a point in the xy-plane. Express OP

uuur in terms of vector

ˆ î and j. Also find O P .uuur

(a) 8 (b) 7

(c) 6 (d) 5



13. Find the coordinates of the tip of the position vector which is equivalent to AB,

uuur where the

coordinates of A and B are (3, 1) and (5, 0) respectively. (a) 1, –2 (b) –1, 2 (c) 2, –1 (d) –2, 1

14. Find the distance between the points A(2, 3, 1), B(−1, 2, −3), using vector method. (a) 26 (b) 40

(c) 29 (d) 29−

15. If $( ) $( )a i 2j 3k and b 2i 3 j k ,→ →

= + − = + +$ $ $ $ find a unit

vector in the direction of (a b).+r r

(a) 5 3 6ˆ ˆ î j k38 38 38

− −

(b) 3 5 2ˆ ˆ î j k38 38 38

− − +

(c) 3 5 2ˆ ˆ î j k38 38 38

+ −

(d) none of these 16. Let a a n d b

rr be two given vectors such that

a 3, b 2 and a . b 6.= = =r rr r

Find the angle

between a and b.rr

(a) 2π

(b) 4−π

(c) 4π

(d) none of these

17. Find the projection of ˆ ˆ â 2i j kon= − +r

ˆ ˆ ˆb i 2 j k.= − +r

(b) 5

6 (b)

56

(c) 56

(d) 5 56 6

18. If A(0, 1, 1) ,B(3, 1, 5) and C(0, 3, 3) be the vertices of a ∆ABC, show, using vectors, find the angle at C. (a) 45° (b) 60° (c) 30° (d) 90°

19. Find a unit vector perpendicular to each one of the vectors ( ) ( )ˆ ˆ ˆ ˆ ˆ â 4 i – j 3k andb 2i 2 j – k .= + = +

rr

(a) ( )1 ˆ ˆ î 2 j 9k3

+ − (b) ( )1 ˆ ˆ ˆ– i 2j 2k3

+ +

(c) ( )1 ˆ ˆ ˆ– i 2j 2k5

+ + (d) none of these

20. Find the angle between the vectors

( ) ( )ˆ ˆ ˆ ˆ ˆ â 3i 2 j k andb i 2 j 3k .= − + = − −rr

(a) 5

cos8

(b) 1 7cos

5− −

(c) 1 7cos

6− −

(d) 1 5cos

7−

21. If ˆ ˆ ˆ ˆ ˆ â i j 2k andb 3i 2 j k,= + + = + −rr

find the value of

( ) ( )a 3b . 2 a – b .+r rr r

(a) –21 (b) –15

(c) –18 (d) –24 22. If ( ) ( )ˆ ˆ ˆ ˆ ˆ â i – 2 j 3k andb 2i 3 j – 5 k ,= + = +

r r then find

( )a b×rr

(a) ( )ˆ ˆ ˆ9 i 6 j 3k+ + (b) ( )ˆ ˆ ˆ3 i 4 j 9k− −

(c) ( )ˆ ˆ î 11j 7k+ + (d) none of these

23. If ( ) ( )ˆ ˆ â 2, b 7 and a b 3i 2 j 6k ,= = × = + +r r rr

find

the angle between aandb.rr

(a) 6π

(b) 5π−

(c) 5π

(d) 65

−

24. Find the area of the parallelogram whose adjacent

sides are represented by the vectors

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ3i j –2k and i – 3 j 4k .+ +

(a) 15 3 (b) 10 2

(c) 5 3 (d) 10 3


of a point (12, n) is such that a 13,=


(a) 2± (b) 7±

(c) 3± (d) 5±


of the point (5, n) is such that a 13,=


(a) 12± (b) 11±

(c) 10± (d) none of these

27. If ar

is a position vector whose tip is (1, −3). Find

the coordinates of the point B such that AB a.=uuur r

If

A has coordinates (−1, 5).

(a) 2, 0 (b) –2, 0

(c) 2, –0 (d) 0, 2



28. Find a unit vector parallel to the vector ˆ ˆ3i 4j.− +

(a) 7 1ˆ î j5 5

− (b) 3 4ˆ î j5 5

− +

(c) 7 9ˆ î j5 5

+ (d) 3 4ˆ î j7 7

−

29. If = + + = + −rr ˆ ˆ ˆ ˆ ˆ â i 2 j 3kand b 2i 4 j 5k represent two

adjacent sides of a parallelogram, find unit vectors parallel to the diagonals of the parallelogram.

(a) ( )1 ˆ ˆ î 5 j 8k59

+ + (b) ( )5 ˆ ˆ î 2 j 8k79

− +

(c) ( )1 ˆ ˆ î 2 j 8k69

+ − (d) ( )1 ˆ ˆ î 4 j 5k91

+ −

30. Find a.bwhenrr

= + − = − +rr ˆ ˆ ˆ ˆ ˆ â 2i 2 j k and b 6i 3 j 2k

(a) 4 (b) 6 (c) 8 (d) 10

31. Let = + − = − + = + −rr rˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ â 4 i 5 j k,b i 4 j 5kand c 3i j k.

Find a vector dr

which is perpendicular to both

aandb,rr

and satisfying =r rd.c 21.

(a) ˆ ˆ ˆ5 i 5 j 5k− −r

(b) ˆ ˆ ˆ6 i 6 j 6k− −r

(c) ˆ ˆ ˆ7 i 7 j 7k− −r

(d) ˆ ˆ ˆ8 i 8 j 8k− −r

32. Find the angle between the vectors + +ˆ ˆ ˆ5i 3 j 4k

and − −ˆ ˆ ˆ6 i 8 j k.

(a) 22

cos5

(b) –1 5cos

5 101

−

(c) –1 5cos

5 11

−

(d) –1 2cos

5 101

33. Find the magnitude of ar

given by

( ) ( )ˆ ˆ ˆ ˆ â i 3 j 2k i 3k= + − × − +r

.

(a) 91 (b) 91

(c) 91− (d) 91− 34. Find a unit vector perpendicular to both the

vectors − + + −ˆ ˆ ˆ ˆ î 2 j 3kand i 2 j k.

(a) ( )1 ˆ ˆ î j k3

− + + (b) ( )1 ˆ ˆ î j k8

− − + +

(c) ( )1 ˆ ˆ î j k3

− − − (d) ( )1 ˆ ˆ î j k5

− − + +

35. Find the area of the triangle whose vertices are A(3, −1, 2), B(1, −1, −3) and C(4, −3, 1).

(a) 1

654

− (b) 5

9652

−

(c) 1

1652

(d) 1 2655

10 vectors

Documents