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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
1133 .1 .1 Concepts of Vectors and ScalarConcepts of Vectors and Scalar
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
1133 .1 .1 Concepts of Vectors and ScalarConcepts of Vectors and Scalar
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
1133 .1 .1 Concepts of Vectors and ScalarConcepts of Vectors and Scalar
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
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:
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
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
BB. . SubtractionSubtraction of Vectors of Vectors
P. 11
Example 13.2T
Solution:
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
BB. . SubtractionSubtraction of Vectors of Vectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
EE. . Rules on Operations of VectorsRules on Operations of Vectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
EE. . Rules on Operations of VectorsRules on Operations of Vectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
EE. . Rules on Operations of VectorsRules on Operations of Vectors
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.
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
EE. . Rules on Operations of VectorsRules on Operations of Vectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
EE. . Rules on Operations of VectorsRules on Operations of Vectors
CABCAB 53
OCOAOB 862 OAOCOB 34
)(5)(3)( OCOAOBOCOAOB
P. 20
Example 13.5T
Solution:
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
EE. . Rules on Operations of VectorsRules on Operations of Vectors
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
1133 ..22 Operations and Properties of Operations and Properties of VectorsVectors
EE. . Rules on Operations of VectorsRules on Operations of Vectors
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.
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
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.
AA. . Representation of Vectors in the Rectangular Representation of Vectors in the Rectangular Coordinate SystemCoordinate System
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:
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
AA. . Representation of Vectors in the Rectangular Representation of Vectors in the Rectangular Coordinate SystemCoordinate System
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:
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
AA. . Representation of Vectors in the Rectangular Representation of Vectors in the Rectangular Coordinate SystemCoordinate System
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:
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
AA. . Representation of Vectors in the Rectangular Representation of Vectors in the Rectangular Coordinate SystemCoordinate System
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.
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
AA. . Representation of Vectors in the Rectangular Representation of Vectors in the Rectangular Coordinate SystemCoordinate System
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.
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
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:
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
BB. . Point of DivisionPoint of Division
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:
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
BB. . Point of DivisionPoint of Division
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:
BB. . Point of DivisionPoint of Division
(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
1133 ..33 Vectors in the Rectangular Vectors in the Rectangular Coordinate SystemCoordinate System
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.
1133 ..44 Applications of VectorsApplications of Vectors
AA. . ParallelismParallelism
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.
1133 ..44 Applications of VectorsApplications of Vectors
AA. . ParallelismParallelism
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.
1133 ..44 Applications of VectorsApplications of Vectors
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:
1133 ..44 Applications of VectorsApplications of Vectors
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.
1133 ..44 Applications of VectorsApplications of Vectors
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)
1133 ..44 Applications of VectorsApplications of Vectors
CC. . Find the Ratio of Line Segments on a Find the Ratio of Line Segments on a Straight Line by VectorsStraight Line by Vectors
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
1133 ..44 Applications of VectorsApplications of Vectors
CC. . Find the Ratio of Line Segments on a Find the Ratio of Line Segments on a Straight Line by VectorsStraight Line by Vectors
(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:
1133 ..55 Scalar ProductsScalar Products
AA. . DefinitionDefinition
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
1133 ..55 Scalar ProductsScalar Products
AA. . DefinitionDefinition
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
1133 ..55 Scalar ProductsScalar Products
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
1133 ..55 Scalar ProductsScalar Products
BB. . Properties of Scalar ProductProperties of Scalar Product
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
1133 ..55 Scalar ProductsScalar Products
BB. . Properties of Scalar ProductProperties of Scalar Product
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
1133 ..55 Scalar ProductsScalar Products
BB. . Properties of Scalar ProductProperties of Scalar Product
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
1133 ..55 Scalar ProductsScalar Products
BB. . Properties of Scalar ProductProperties of Scalar Product
(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.
1133 ..55 Scalar ProductsScalar Products
BB. . Properties of Scalar ProductProperties of Scalar Product
0zyx 23yzx 23 yzx 23
2)3()3( zxzx469 22 zxzx
4619 zx66 zx1zx
P. 49
1133 ..55 Scalar ProductsScalar Products
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.
1133 ..55 Scalar ProductsScalar Products
CC. . Calculation of Scalar Product in the Rectangular Calculation of Scalar Product in the Rectangular Coordinate SystemCoordinate System
(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:
1133 ..55 Scalar ProductsScalar Products
CC. . Calculation of Scalar Product in the Rectangular Calculation of Scalar Product in the Rectangular Coordinate SystemCoordinate System
(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
1133 ..66 Applications of Scalar ProductsApplications of Scalar Products
AA. . Projection of a Vector onto Another VectorProjection of a Vector onto Another Vector
= (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
1133 ..66 Applications of Scalar ProductsApplications of Scalar Products
AA. . Projection of a Vector onto Another VectorProjection of a Vector onto Another Vector
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.
1133 ..66 Applications of Scalar ProductsApplications of Scalar Products
AA. . Projection of a Vector onto Another VectorProjection of a Vector onto Another Vector
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
1133 ..66 Applications of Scalar ProductsApplications of Scalar Products
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:
1133 ..66 Applications of Scalar ProductsApplications of Scalar Products
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
BB. . Determination of Orthogonality by VectorsDetermination of Orthogonality by Vectors
P. 58
Example 13.23T
1133 ..66 Applications of Scalar ProductsApplications of Scalar Products
BB. . Determination of Orthogonality by VectorsDetermination of Orthogonality by Vectors
(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:
1133 ..66 Applications of Scalar ProductsApplications of Scalar Products
BB. . Determination of Orthogonality by VectorsDetermination of Orthogonality by Vectors
(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
Chapter Chapter SummarySummary
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
Chapter Chapter SummarySummary
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
Chapter Chapter SummarySummary
2. If , then A, B and C are collinear.ACAB //
P. 64
13.5 Scalar Products
Chapter Chapter SummarySummary
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
Chapter Chapter SummarySummary
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.