13.1concepts of vectors and scalars 13.2operations and properties of vectors 13.3vectors in the...

13.1 Concepts of Vectors and Scalars13.2 Operations and Properties of Vectors13.3 Vectors in the Rectangular Coordinate System

Chapter Summary

Case Study

Vectors in Two-dimensional Space13

13.4 Applications of Vectors13.5 Scalar Products13.6 Applications of Scalar Products

P. 2

Last Sunday, Mr. Chan drove his car from his home to the cinema in Town A which is 30 km due south of his home.

Case StudyCase Study

Then he drove 40 km east to visit his grandmother. During the whole trip, the total distance and displacement of the car is said to be 70 km and 50 km in the direction of around S53°E respectively. The distance travelled by the car refers to how long the car has travelled, that is the total distance of XY and YZ. Total distance = XY + YZ The displacement of the car, which is the distance between the initial position X and the final position Z of the car. Displacement = XZ

= (30 + 40) km = 70 km

km 50km 4030 22 .1.5330

40tan and 1

Thus the displacement of the car is 50 km in the direction of around S53°E.

P. 3

For example, displacement, velocity and force are vectors.

1133 .1 .1 Concepts of Vectors and ScalarConcepts of Vectors and Scalar

A. A. Definition of a VectorDefinition of a Vector

Definition 13.1A vector is a quantity which has both magnitude and direction.A scalar is a quantity which has magnitude only.

For example, distance, temperature and area are scalars.

P. 4

The directed line segment from point X to point Y in the direction of XY is called a vector from X to Y.

X is called the initial point and Y is called the terminal point.

BB. . Representation of a VectorRepresentation of a Vector

We can denote this vector by or XY. XY

The magnitude of the vector is specified by the length of XY and is denoted by . XY

Note: 1. The notation represents the fact that the vector is pointing from X to Y. 2. If the initial and terminal points of the vector are not specified, it can be denoted by a single lowercase letter such as , a or a and the magnitude of the vector can be denoted by , or .

XY

aa aa


P. 5

CC. . Different Types of VectorsDifferent Types of Vectors

Definition 13.2Two vectors are equal if they have the same magnitude and direction.

If two vectors and are equal, then they are said to be equal vectors and we denote them by .

AB CDCDAB

From the definition above, equal vectors are not required to have the same initial points and terminal points. Therefore, vectors defined in this way are called free vectors.

If two vectors have the same magnitude but are in opposite directions, then one of the vectors is called the negative vector of the other. The negative vector of is denoted by .AB BAAB


P. 6

CC. . Different Types of VectorsDifferent Types of Vectors

When a vector has the same initial point and terminalpoint, the magnitude of the vector is zero and it doesnot have a specified direction.

00AASuch a vector is called a zero vector and we denote and .0AA


If the magnitude of a vector is 1 unit, then this vector is called aunit vector and we denote the unit vector by .

AB^

AB

P. 7

In general, for any two vectors a and b, we can find the addition of these two vectors in the following way:

Step 1: Given a and b are two vectors on the same plane.

1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors

Step 2: Translate b in a parallel direction such that theinitial point of b coincides with the terminalpoint of a.

AA. . Addition of VectorsAddition of Vectors

Triangle Law of addition ACBCAB

Step 3: By the triangle law of addition, a third vectora + b is obtained.

P. 8

Consider the parallelogram ABCD. Since the opposite sides of a parallelogram are parallel and equal in length, we have


AA. . Addition of VectorsAddition of Vectors

Parallelogram law of addition If ABCD is a parallelogram, then .ACBCAB

ADABBCABAC

Hence we have the following:

P. 9


BB. . SubtractionSubtraction of Vectors of Vectors

)( OBOA

OBOA

Furthermore, for any vector a, we have a – a = 0.

For two vectors and , their difference can be found by expressing it as .

OBOA OBOA )( OBOA

BA

OABO

Negative vector

Triangle law of addition

BOOA

P. 10

Example 13.1T

Solution:


Express the following vectors in terms of a, b, c and d. (a) (b) (c) EABEAC

(a) AC BCAB CBAB

ba

(b) BE DECDBC DECDCB

dcb

(c) EA BACBDCED ABDBCDDE

abcd


P. 11

Example 13.2T

Solution:



The figure shows two non-zero vectors a and b and an angle . If OB // AC, express |b – a| in terms of |a|, |b| and .

AB ab

AB ab

)cos(222 baba

cos222 baba

P. 12


CC. . Scalar MultiplicationScalar Multiplication

When a vector a is multiplied by a scalar k, their product, which is denoted by ka, is a vector that is defined according to the following conditions:

1. If k = 0, ka = 0.2. If k > 0, the magnitude of ka is k|a| and the direction of

ka is the same as a.3. If k < 0, the magnitude of ka is |k||a| and the direction of

ka is opposite to that of a..

Using the definitions above, for any non-zero vector a, the unit vector

is denoted by (or ).

a

a

aa ˆ a

a

1

P. 13


For a given point X on a plane, if we choose a point O on the same plane as the reference point, then the position of X can be determined by the vector .OXThis vector is called the position vector of the point X with respect to the point O. Similarly, the point Y can be determined by the position vector OY

Let = x and = y.

OX OY

OYXOXY

xy OYOX )(

Triangle law of addition

DD. . Position VectorsPosition Vectors

P. 14


EE. . Rules on Operations of VectorsRules on Operations of Vectors

Property 13.1 Rules of operations of vectors Given that a, b and c are vectors and p, q, r and s are real numbers. Then (a) a + b = b + a(b) a + (b + c) = (a + b) + c(c) a + 0 = 0 + a = a(d) 0a = 0(e) p(qa) = (pq)a(f) (p + q)a = pa + qa(g) p(a + b) = pa + qb(h) If pa + qb = ra + sb, where a and b are non-zero and not

parallel to each other, then p = r and q = s.

P. 15



Proof of (b): As shown in the figure, = a, = b and = c. AB BC CD

Since , b a BCABAC

CDACAD cba )(

Also, , c b CDBCBD

BDABAD )( cba

cbacba )()(

P. 16



Proof of (g): In the figure, = a, = b , and , where p > 0.

DA AE apDApBA bpAEpAC

EADCABpAE

AC

DA

BA and (common)

ADEABC ~ (ratio of 2 sides, inc.)

pDE

BC (corr. sides, s)

DEpBC )( AEDApACBA

)( baba ppp

P. 17



Proof of (h): We prove this rule by contradiction. Suppose pa + qb = ra + sb, where a and b are non-zero and not parallel to each other, and p, q, r and s are constants. Then we have

(p – r)a = (s – q)bAssume p – r 0, then

barp

qs

∵ is a scalar.rp

qs

This indicates that a and b are parallel. However, this contradicts to the assumption that a and b are non-zero and not parallel. Therefore, the assumption that p – r 0 is incorrect. p – r = 0 and hence s – q = 0, i.e., p = r and q = s.

The proofs of other rules can be done by using the basic definition of vectors. These are left to the students as exercise.

P. 18

Example 13.3T

Solution:

InABC, D and E are the mid-points of AB and AC respectively. Prove that BC // DE and BC = 2DE.



ABACBC ADAEDE

ABAC2

1

2

1

)(2

1ABAC

BC2

1

BC // DE and BC = 2DE.

P. 19

Example 13.4T

Solution:

Given that . Prove that , where O is any reference point.OAOCOB 34

CABCAB 53



CABCAB 53

OCOAOB 862 OAOCOB 34

)(5)(3)( OCOAOBOCOAOB

P. 20

Example 13.5T

Solution:



ABCD is a square. X and Y are the mid-points of BC and CD respectively. It is given that and .(a) (i) Express p and q in terms of and . (ii) Express and in terms of p and q.

(b) Prove that .

BCABBCAB

pAX qAY

qp3

2

3

2 AC

(a) (i) BXAB p

BCAB2

1

DYAD q

ABBC2

1

ABBC2

1

(ii)

)2(2

1

)1(2

1

............ q

............ p

ABBC

BCAB

(1) 2 – (2):

AB2

32 qp

qp3

2

3

4 AB

P. 21



Substituting into (1),

qp3

2

3

4AB

BC2

1

3

2

3

4 qpp

qp3

2

3

1

2

1BC

qp3

4

3

2 BC

BCAB

AC

(b)

qpqp

3

4

3

2

3

2

3

4

qp3

2

3

2

(a) (ii)

Example 13.5T

Solution:

ABCD is a square. X and Y are the mid-points of BC and CD respectively. It is given that and .(a) (i) Express p and q in terms of and . (ii) Express and in terms of p and q.

(b) Prove that .

BCABBCAB

pAX qAY

qp3

2

3

2 AC

P. 22

Firstly, let i be the unit vector in the positive direction ofx-axis, and j be the unit vector in the positive direction of y-axis.

1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System

Vectors can be represented in the rectangular coordinateplane.

AA. . Representation of Vectors in the Rectangular Representation of Vectors in the Rectangular Coordinate SystemCoordinate System

Note that and

where is the angle between OP and the positive x-axis, measured in anticlockwise direction.

j.i yxOP 22 yxOP

,tanx

y

By Pythagoras’ theorem

If we consider OMP in the figure, we have

and 22

cosyx

x

OP

x

22sin

yx

y

OP

y

For any point P(x, y), the position vector can be expressed asOP

P. 23

Once we represent the vectors in terms of i and j, they can be added or subtracted by adding or subtracting their i and j components.


Note: As all the vectors in the rectangular coordinate system can be expressed in terms of i and j, the unit vectors i and j are called unit base vectors.


For example, as shown in the figure, X(x1, y1) and Y(x2, y2) are two points on the coordinate plane. Then the vector can be expressed as XY

Hence we have

OXOYXY )()( 1122 jiji yxyx

)j)i 1212 (( yyxx

and

)j)i 1212 (( yyxxXY 2

122

12 (( )) yyxxXY 12

12tanxx

yy

P. 24

Example 13.6T

Solution:



Given two points X(–2, 1) and Y(4, –1), find the magnitude and direction of . XY

ji 2OX

ji 4OYOXOYXY

)2()4( jiji ji 26

22 )2(6 XY 102Let be the angle between and the positive x-axis. XY

3

1

6

2tan

342 (cor. to 3 sig. fig.)

makes an angle of 342 with the positive x-axis. XY

P. 25

Example 13.7T

Solution:



If the coordinates of Y are (0, –5) and , find the coordinates of X .

iXY

j5OYOXOYXY OX ji 5

ji 5OX

The coordinates of X are (1, 5).

P. 26

Example 13.8T

Solution:



Given two points X(0, 0) and Y(–1, 1). Find the unit vector in the direction of . XY

0OX

ji OY

ji OXOYXY

21)1( 22

XY

Unit vector 2

ji

XY

ji2

1

2

1

P. 27

Example 13.9T

Solution:If x = –2i + 3j and y = 4i – j, express 5j in terms of x and y.



Let . yxj nm 5)4()32(5 jijij nmji )3()24( nmmn

)2(............ 53)1(............ 024

nmmn

From (1), m = 2n Substituting m = 2n into (2),

5)2(3 nn

155

nn

2)1(2 myxj 25

P. 28

In junior forms, we learnt how to find the coordinates of the point that divide a line segment in a particular ratio. We can apply the same concepts in vectors.


BB. . Point of DivisionPoint of Division

As shown in the figure, the point Z divides the line segment XY in the ratio r : s, i.e., XZ : ZY = r : s. Let x, y and z be the position vectors of X, Y and Z with respect to the reference point O respectively. Then we have

XZ : ZY = r : s

r(z – y) = s(x – z) XZsZYr

rz + sz = sx + ry (r + s)z = sx + ry

sr

rs

yx

z

If Z is the mid-point of XY, i.e., XZ : ZY = r : s = 1 : 1, then

2

yxz

P. 29

Example 13.10T

Solution:



Given that OABC is a square with = a and = c. D is the mid-point of OC. E is a point on AC such that AE : EC = 3 : 1. Express the following vectors in terms of a and c. (a) (b) DEAD

OCOA

(a) OAOCAC ac

2

ACAOAD

acaca

2

1

2

)(

OCDC2

1(b) c

2

1

ca2

1DA

13

13

DADC

DE4

2

1

2

13

cac

ca4

1

4

1

P. 30

Example 13.11T

Solution:



The figure shows that the trapezium ABCD with AB // DC. E is the mid-point of AB such that CE // DA. BD intersects CE at F such that

and . and .

(a) Express in terms of r, a and b.(b) Express in terms of s, a and b.(c) Hence find the values of r and s.

rDF

BF sCF

EF aAE bAD

AFAF

r

ADrABAF

1

1(a)

r

r

1

2 baba

r

r

r

11

2

(b) DCADAC ab

s

ACsAEAF

1

1s

s

1

)( ababa

s

s

1

P. 31

Example 13.11T

Solution:


(c) From the results of (a) and (b),

babas

s

r

r

r

111

2

)2(11

)1(11

2

..........

.................

s

s

r

rr

From (1), 112

rr

Substituting r = 1 into (2),

s

s

111

1

ss 21 1s


The figure shows that the trapezium ABCD with AB // DC. E is the mid-point of AB such that CE // DA. BD intersects CE at F such that

and . and .

(a) Express in terms of r, a and b.(b) Express in terms of s, a and b.(c) Hence find the values of r and s.

rDF

BF sCF

EF aAE bAD

AFAF

P. 32

If two vectors are in the same or opposite directions, then they are parallel.

1133 ..44 Applications of VectorsApplications of Vectors

AA. . ParallelismParallelism

The converse of the above fact is also true.

In particular, if two non-zero parallel vectors u and v can be expressed as the scalar sum of two non-parallel vectors a and b, i.e.,

For two non-zero vectors u and v, if u = kv, where k is a real number, then u and v are parallel.When k > 0, u and v are in the same direction.When k < 0, u and v are in the opposite directions.

If two non-zero vectors u and v are parallel, then u = kv, where k is a non-zero real number.

u = m1a + n1b and v = m2a + n2b where m1, m2, n1 and n2 are real numbers and m2 and n2 are non-zero, we can conclude that

.2

1

2

1

n

n

m

m

P. 33

Example 13.12T

Solution:

If the vectors a = 2ci + 8j and b = 2i + (c + 2)j are parallel but in the opposite directions, find the value of c.



ba //

2

8

2

2

c

c

16)2(2 cc0822 cc0)2)(4( cc

(rejected) 2or 4c

P. 34

Example 13.13T

Solution:

Given a parallelogram ABCD. M and N are points on thediagonal AC such that AM = NC. Using the vector method,prove that MBND is a parallelogram.



CNBCBN MAAD

MD

BN // MD and BN = MD. MBND is a parallelogram.

P. 35

Suppose there are three distinct points A, B and C onthe same plane.


BB. . Prove Three Points are Collinear by VectorsProve Three Points are Collinear by Vectors

As A is a common point for AB and AC, we can conclude that A, B and C lie on the same straight line. In this case, we say that A, B and C are collinear.

If , where k is a non-zero real constant, thenthe line segments AB and AC must be parallel.

ACkAB

Similarly, if either or , where m and n are non-zero real constants, we can also conclude that A, B and C are collinear using the above argument.

BCmAB BCnAC

P. 36

Example 13.14T

Solution:


BB. . Prove Three Points are Collinear by VectorsProve Three Points are Collinear by Vectors

In ABC, E is the mid-point of BC. D is a point on AB such that AD : DB = 1 : 2. CF : FD = 3 : 1. Let = a and = b. (a) Express in terms of a and b. (b) Express in terms of a and b. (c) Hence determine whether A, E and F are collinear.

AEAF

ACAD

13

13

ACAD

AF(a)4

3 ba ba4

1

4

3

a3AB(b)

2

ACABAE

2

3 ba ba2

1

2

3

(c) ba2

1

2

3AE

ba

4

1

4

32 AF2

and are parallel. AE AFA, E and F are collinear.

P. 37

Using the knowledge about the division of line segments and collinearity of three points, we can determine the ratio of line segments on a straight line.


CC. . Find the Ratio of Line Segments on a Find the Ratio of Line Segments on a Straight Line by VectorsStraight Line by Vectors

P. 38

Example 13.15T

Solution:

In ABC, D and E are points on AC and AB respectively. CD : DA = 1 : 3 and AE : EB = 2 : 1. Let = a, = b and CF : FE = 1 : k. (a) Express in terms of a, b and k. (b) Express in terms of a and b. (c) Hence find BF : FD.

BDBF

BCBA

(a)

(b)



a3

1BE

k

BCkBEBF

1

1

k

k

13

1ba

bak

k

k

1)1(3

1

31

31

BCBA

BD baba

4

3

4

1

4

3

P. 39



(c) Since B, F and D are collinear, . BDBF //

4

31

4

1)1(3

1

k

kk

)1(3

4

)1(3

4

k

k

k

144

kk

ba11

1

)11(3

1

BF ba

2

1

6

1

BFBDFD

baba

2

1

6

1

4

3

4

1

ba4

1

12

1

ba

2

1

6

1

2

1BF

2

1

1:2: FDBF

Example 13.15T

Solution:

In ABC, D and E are points on AC and AB respectively. CD : DA = 1 : 3 and AE : EB = 2 : 1. Let = a, = b and CF : FE = 1 : k. (a) Express in terms of a, b and k. (b) Express in terms of a and b. (c) Hence find BF : FD.

BDBF

BCBA

P. 40

Suppose a and b are two vectors on the same plane.

1133 ..55 Scalar ProductsScalar Products

AA. . DefinitionDefinition

Case 1: a and b have the same initial point but different terminal points. Then the included angle is the angle between a and b.

Case 2: The terminal point of a coincides with the initial point of b.Translate a along its direction such that the initial points of both vectors coincide with each other.Then 180° is the angle between a and b.

Case 3: The terminal points of both vectors coincide with each other.Translate a and b along their direction such that their initial points coincide with each other.Hence the angle between a and b is .

P. 41

As shown in the figure, a and b are two non-zero vectors and (where 0 180) is the angle between them.The scalar product (or dot product) of a and b, denoted by a b, is defined as:



The scalar product of two vectors is a number, which may be positive, negative or zero depending on whether is acute, obtuse or a right angle. In particular, if b = a, then we have

For the unit vectors i and j, we have

a b = |a||b|cos

a a = |a|2.

i i = j j = 1.

Note: The scalar product must be written as a b. It cannot be written as ab or a × b.

P. 42

If a and b are perpendicular to each other, then a b = |a||b| cos = 0



a and b are orthogonal, i.e. perpendicular to each other.

Since the unit vectors i and j are orthogonal,

For two non-zero vectors a and b, they are orthogonal if and only if a b = 0.

i i = j j = 0

Conversely, if a and b are two non-zero vectors such that a b = 0, then |a||b| cos = 0

cos = 0 = 90°

So we can conclude that:

P. 43


BB. . Properties of Scalar ProductProperties of Scalar Product

Proof of (a): a b = |a||b| cos

Properties of Scalar ProductIf a, b and c are vectors and k is a real number, then(a) a b = b a (b) a a = 0 if and only if a = 0(c) a (b + c) = a b + a c (d) (ka) b = k(a b) = a kb (e) |a||b| |a b|(f) |a – b|2 = |a|2 + |b|2 – 2(a b)

b a = |b||a| cos a b = b a

P. 44



Proof of (c): Let = a, = b and = c.

a (b + c) = |a||b + c| cos AOC BCOBOA

= (OA)(OC)cos AOC = (OA)(OF) = (OA)(OE + EF) = (OA)(OE) + (OA)(EF) = (OA)(OB cos AOB ) + (OA)(BC cos DBC ) = |a||b| cos AOB + |a||c| cos DBC

= a b + a c a (b + c) = a b + a c

P. 45



Proof of (d): If k = 0, then it is obvious that (ka) b = k(a b) = 0.If k > 0, then ka and a are in the same direction.

(ka) b = |ka||b|cos = |k||a||b|cos = k(a b)

If k < 0, then ka and a are in opposite directions.

(ka) b = |ka||b|cos (180° – )

= |k||a||b| (–cos )

= k(a b)

= –k|a||b| (–cos )

Combining all the results above, we have (ka) b = k(a b).Similarly, we can prove that a kb = k(a b).

P. 46



Proof of (f): |a – b|2 = (a – b) (a – b)

= a a – a b – b a + b b

= |a|2 – a b – b a + |b|2

= |a|2 – 2(a b) + |b|2

|a – b|2 = |a|2 + |b|2 – 2(a b)

P. 47

Example 13.16T

Solution:

If |x|= 1, |y| = 1 and the angle between x and y is 135°, find the values of the following. (a) y x(b) (x + 3y) (2y + 5x)(c) |x – y|2



(a) y x = |y||x| cos = (1)(1) cos 135°

2

2

(b) (x + 3y) (2y + 5x)= 2x y + 5x x + 6y y + 15y x

= 17x y + 5|x|2 + 6|y|2

22 )1(6)1(52

217

2

21711

(c) |x – y|2

= (x – y) (x – y) = (x – y) (x – y)

= x x – x y – y x – y y= |x|2 + |y|2 – 2x y

2

2211 22

22

P. 48

Example 13.17T

Solution:

If x, y and z are unit vectors such that 3x – 2y – z = 0, find the value of x z.



0zyx 23yzx 23 yzx 23

2)3()3( zxzx469 22 zxzx

4619 zx66 zx1zx

P. 49


CC. . Calculation of Scalar Product in the Rectangular Calculation of Scalar Product in the Rectangular Coordinate SystemCoordinate System

If a = x1i + y1j and b = x2i + y2j are two non-zero vectors, then

a b = x1x2+ y1y2

where is the angle between a and b.

22

22

21

21

2121cosyxyx

yyxx

P. 50

Example 13.18T

Solution:

Two vectors r = 2i + 5j and s = i – 2j are given. (a) Find the value of r s. (b) Hence find the angle between r and s, correct to the nearest degree.



(a) r s= (2i + 5j) (i – 2j)= (2)(1) + (5)(–2)

8

(b) Let be the angle between r and s.

sr

srcos2222 )2(152

8

145

8

132 (cor. to the nearest degree) The angle between r and s is 132.

P. 51

Example 13.19T

Solution:



(a)

Given four points W(2, 2), X(–2, 1), Y(–1, –3) and Z(3, –2). Prove that (a) (b)

XZWY WZXY //

OWOYWY )22()3( jiji ji 53 OXOZXZ )2()23( jiji ji 35

)35()53( jiji XZWY )3)(5()5)(3( 0XZWY

(b) OXOYXY )2()3( jiji ji 4OWOZWZ )22()23( jiji ji 4 XY

WZXY //

P. 52

1133 ..66 Applications of Scalar ProductsApplications of Scalar Products

AA. . Projection of a Vector onto Another VectorProjection of a Vector onto Another Vector

In the figure, a and b are two vectors and is the angle between them.

Suppose C is the foot of perpendicular from B to OA. Then we call the projection of b on a, and the

length of OC is given by .

OC

a

ba

aa

ba

a

a

a

ba

2OC

Suppose the angle between a and b is obtuse.Since cos is negative, the projection of b on a is also negative, which means is in the opposite direction to a.

As is in the same direction as a, we can find by multiplying its length with the unit vector of a, that is,

OC OC

P. 53

Example 13.20T

Solution:Consider a b = (9i + 4j) (2i – 11j)

Two vectors a = 9i + 4j and b = 2i – 11j are given. Find (a) the projection of a on b, and (b) the projection of b on a.

(a) Projection of a on b



= (9)(2) + (4)(–11) = –26

|a|2 = 92 + 42 = 97

|b|2 = 22 + (–11)2 = 125

)112(125

26ji

ji

125

286

125

52

(b) Projection of b on a )49(97

26ji

ji

97

104

97

234

P. 54

Example 13.21T

Solution:Let r = mi + nj, where m and n are non-zero constants.

Let p = 6i – 4j and q = 7i + 3j. Find a unit vector r such that the projection of p on r is equal to the projection of q on r.

Projection of p on r



Since r is a unit vector, |r|2 = m2 + n2 = 1…………(1)

)()()46(

22ji

jijinm

nm

nm

)2(............ )46()46( ji nnmmnm

Projection of q on r )()()37(

22ji

jijinm

nm

nm

)3(............ )37()37( ji nnmmnm

Combining (2) and (3), mnmmnm )37()46(

)4(............ 77

nmnm

P. 55

Example 13.21T

Substituting (4) into (1),

Let p = 6i – 4j and q = 7i + 3j. Find a unit vector r such that the projection of p on r is equal to the projection of q on r.



1)7( 22 nn

150 2 n

50

12 n

25

1n

25

7m

jir

25

1

25

7

ji

25

1

25

7

Solution:

P. 56


BB. . Determination of Orthogonality by VectorsDetermination of Orthogonality by Vectors

In the last section, we learnt that the scalar product of two non-zero vectors is zero if and only if they are perpendicular to each other or orthogonal. Thus we can use this property to test whether two given vectors are perpendicular or not.

P. 57

Example 13.22T

Solution:


In the figure, O is a centre of the circle. M is the mid-point of the chord AB. Show that OM AB.

Let and aOA bOB

ab OAOBAB

)(2

1

)(2

1

ba

OBOAOM

)()(2

1abba

ABOM

)(2

1abbbaaba

0)(2

1 22 ab

ABOM


P. 58

Example 13.23T



(a) OAOBAB

jijiji

42)32()4(

k

OAkOBOC

1

1

k

k

1

)32()4( jiji

jik

k

k

k

1

13

1

)2(2

Given two points A(2, 3) and B(4, 1). C is a point on AB such that it divides AB in the ratio 1 : k. (a) Express and in terms of k, i and j. (b) Find the shortest distance from O to AB.

OCAB

Solution:

P. 59

Example 13.23T

Solution:



(b) If OC is the shortest distance from O to AB, OC AB.

0 ABOC

0)42(1

13

1

)2(2

jijik

k

k

k

01

)13(4

1

)2(4

k

k

k

k

2

3041284

k

kk

ji

2

31

12

33

2

31

22

32

OC

ji5

7

5

14

The shortest distance OC

22

5

7

5

14

5

7

Given two points A(2, 3) and B(4, 1). C is a point on AB such that it divides AB in the ratio 1 : k. (a) Express and in terms of k, i and j. (b) Find the shortest distance from O to AB.

OCAB

P. 60

13.2 Operations and Properties of Vectors

Chapter Chapter SummarySummary

1. Addition of Vectors

BCABAC 2. Subtraction of Vectors

OAOBAB

P. 61

13.2 Operations and Properties of Vectors


Given that a, b and c are vectors and p, q, r and s arereal numbers. Then (a) a + b = b + a(b) a + (b + c) = (a + b) + c(c) a + 0 = 0 + a = a(d) 0a = 0(e) p(qa) = (pq)a(f) (p + q)a = pa + qa(g) p(a + b) = pa + qb(h) If pa + qb = ra + sb, where a and b are non-zero and not parallel to each other, then p = r and q = s.

P. 62

13.3 Vectors in the Rectangular Coordinate System


1. If P(x, y) is a point on the rectangular coordinate system, then

(a)

(b)

(c)

,ji yxOP

,22 yxOP

.tanx

y

2. If C is a point on AB such that AC : BC = m : n, then

.nm

OBmOAnOC

P. 63

1. For two non-zero vectors u and v and a scalar k given, if u = kv, then u and v are parallel.

13.4 Applications of Vectors


2. If , then A, B and C are collinear.ACAB //

P. 64

13.5 Scalar Products


If a = x1i + y1j and b = x2i + y2j are non-zero vectors, and is the angle between them, then

2121

cosyyxx

baba

22

22

21

21

2121cosyxyx

yyxx

If a, b and c are vectors and k is a real number, then(a) a b = b a (b) a a = 0 if and only if a = 0(c) a (b + c) = a b + a c (d) (ka) b = k(a b) = a kb (e) |a||b| |a b|(f) |a – b|2 = |a|2 + |b|2 – 2(a b)

P. 65

13.6 Applications of Scalar Products


1. For two non-zero vectors a and b, the projection of b on a is

given by .aa

ba

2

2. For two non-zero vectors a and b, a b = 0 if and only if a and b are orthogonal, i.e. they are perpendicular to each other.

13.1concepts of vectors and scalars 13.2operations and properties of vectors 13.3vectors in the...

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