1. 13 Vectors in Two-dimensional Space Case Study
13.1 Concepts of Vectors and Scalars
13.2 Operations and Properties of Vectors
13.3 Vectors in the Rectangular Coordinate System
13.4 Applications of Vectors
13.5 Scalar Products
13.6 Applications of Scalar Products
Chapter Summary

2. Case Study
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.
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
= ( ) km = 70 km
The displacement of the car, which is the distance between the initial position X and the final position Z of the car.
Displacement = XZ
Thus the displacement of the car is 50 km in the direction of around S53°E.

3. 13.1 Concepts of Vectors and Scalar
A. Definition of a Vector
Definition 13.1
A vector is a quantity which has both magnitude and direction.
A scalar is a quantity which has magnitude only.
For example, displacement, velocity and force are vectors.
For example, distance, temperature and area are scalars.

4. 13.1 Concepts of Vectors and Scalar
B. Representation of a Vector
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.
We can denote this vector by or XY.
The magnitude of the vector is specified by the length of XY and is denoted by
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

5. 13.1 Concepts of Vectors and Scalar
C. Different Types of Vectors
Definition 13.2
Two 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
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

6. 13.1 Concepts of Vectors and Scalar
C. Different Types of Vectors
When a vector has the same initial point and terminal
point, the magnitude of the vector is zero and it does
not have a specified direction.
Such a vector is called a zero vector and we denote
and
If the magnitude of a vector is 1 unit, then this vector is called a
unit vector and we denote the unit vector by

7. 13.2 Operations and Properties of Vectors
A. Addition of Vectors
Triangle Law of addition
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.
Step 2: Translate b in a parallel direction such that the
initial point of b coincides with the terminal
point of a.
Step 3: By the triangle law of addition, a third vector
a + b is obtained.

8. 13.2 Operations and Properties of Vectors
A. Addition of Vectors
Consider the parallelogram ABCD.
Since the opposite sides of a parallelogram are parallel and equal in length, we have
Hence we have the following:
Parallelogram law of addition
If ABCD is a parallelogram, then

9. 13.2 Operations and Properties of Vectors
B. Subtraction of Vectors
For two vectors and , their difference can be found by expressing it as
Negative vector
Triangle law of addition
Furthermore, for any vector a, we have a – a = 0.

10. Example 13.1T 13.2 Operations and Properties of Vectors Solution:
B. Subtraction of Vectors
Example 13.1T
Express the following vectors in terms of a, b, c and d.
(a) (b) (c)
Solution:
(a)
(b)
(c)

11. Example 13.2T 13.2 Operations and Properties of Vectors Solution:
B. Subtraction of Vectors
Example 13.2T
The figure shows two non-zero vectors a and b and an angle q. If OB // AC, express |b – a| in terms of |a|, |b| and q.
Solution:

12. 13.2 Operations and Properties of Vectors
C. Scalar 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 ).

13. 13.2 Operations and Properties of Vectors
D. Position Vectors
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
This 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
Let = x and = y.
Triangle law of addition

14. 13.2 Operations and Properties of Vectors
E. Rules 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.

15. 13.2 Operations and Properties of Vectors
E. Rules on Operations of Vectors
Proof of (b):
As shown in the figure, = a, = b and = c.
Since ,
Also, ,

16. 13.2 Operations and Properties of Vectors
E. Rules on Operations of Vectors
Proof of (g):
In the figure, = a, = b , and , where p > 0.
(common)
(ratio of 2 sides, inc. Ð )
(corr. sides, ~Ds )

17. 13.2 Operations and Properties of Vectors
E. Rules 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)b
Assume p – r  0, then
∵ is a scalar.
\ 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 ¹ 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.

18. Example 13.3T 13.2 Operations and Properties of Vectors Solution:
E. Rules on Operations of Vectors
Example 13.3T
In DABC, D and E are the mid-points of AB and AC respectively. Prove that BC // DE and BC = 2DE.
Solution:
\ BC // DE and BC = 2DE.

19. Example 13.4T 13.2 Operations and Properties of Vectors Solution:
E. Rules on Operations of Vectors
Example 13.4T
Given that Prove that
, where O is any reference point.
Solution:

20. Example 13.5T 13.2 Operations and Properties of Vectors Solution:
E. Rules on Operations of Vectors
Example 13.5T
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
Solution:
(a) (i)
(ii)
(1)  2 – (2):

21. Example 13.5T 13.2 Operations and Properties of Vectors Solution:
E. Rules on Operations of Vectors
Example 13.5T
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
Solution:
Substituting
into (1),
(a) (ii)
(b)

22. 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Vectors can be represented in the rectangular coordinate
plane.
Firstly, let i be the unit vector in the positive direction of
x-axis, and j be the unit vector in the positive direction of y-axis.
For any point P(x, y), the position vector can be expressed as
Note that and
where  is the angle between OP and the positive x-axis, measured in anticlockwise direction.
By Pythagoras’ theorem
If we consider OMP in the figure, we have
and

23. 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate 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.
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.
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
Hence we have
and

24. Example 13.6T 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Example 13.6T
Given two points X(–2, 1) and Y(4, –1), find the magnitude and direction of
Solution:
Let  be the angle between and the positive x-axis.
(cor. to 3 sig. fig.)
makes an angle of 342 with the positive x-axis.

25. Example 13.7T 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Example 13.7T
If the coordinates of Y are (0, –5) and , find the coordinates of X .
Solution:
\ The coordinates of X are (1, 5).

26. Example 13.8T 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Example 13.8T
Given two points X(0, 0) and Y(–1, 1). Find the unit vector in the direction of
Solution:
\ Unit vector

27. Example 13.9T 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Example 13.9T
If x = –2i + 3j and y = 4i – j, express 5j in terms of x and y.
Solution:
Let
From (1), m = 2n
Substituting m = 2n into (2),

28. 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
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.
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)
rz + sz = sx + ry
(r + s)z = sx + ry \
If Z is the mid-point of XY, i.e., XZ : ZY = r : s = 1 : 1, then

29. Example 13.10T 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
Example 13.10T
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)
Solution:
(a)
(b)

30. Example 13.11T 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
Example 13.11T
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.
Solution:
(a)
(b)

31. Example 13.11T 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
Example 13.11T
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.
Solution:
(c) From the results of (a) and (b),
From (1),
Substituting r = 1 into (2),

32. 13.4 Applications of Vectors
A. Parallelism
If two vectors are in the same or opposite directions, then they are parallel.
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.
The converse of the above fact is also true.
If two non-zero vectors u and v are parallel, then u = kv, where k is a non-zero real number.
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.,
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

33. Example 13.12T 13.4 Applications of Vectors Solution: A. Parallelism
If the vectors a = 2ci + 8j and b = 2i + (c + 2)j are parallel but in the opposite directions, find the value of c.
Solution:

34. Example 13.13T 13.4 Applications of Vectors Solution: A. Parallelism
Given a parallelogram ABCD. M and N are points on the
diagonal AC such that AM = NC. Using the vector method,
prove that MBND is a parallelogram.
Solution:
\ BN // MD and BN = MD.
\ MBND is a parallelogram.

35. 13.4 Applications of Vectors
B. Prove Three Points are Collinear by Vectors
Suppose there are three distinct points A, B and C on
the same plane.
If , where k is a non-zero real constant, then
the line segments AB and AC must be parallel.
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.
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.

36. Example 13.14T 13.4 Applications of Vectors Solution:
B. Prove Three Points are Collinear by Vectors
Example 13.14T
In DABC, 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.
Solution:
(a)
(b)
(c)
\ and are parallel.
\ A, E and F are collinear.

37. 13.4 Applications of Vectors
C. Find the Ratio of Line Segments on a
Straight Line by Vectors
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.

38. Example 13.15T 13.4 Applications of Vectors Solution:
C. Find the Ratio of Line Segments on a
Straight Line by Vectors
Example 13.15T
In DABC, 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.
Solution:
(a)
(b)

39. Example 13.15T 13.4 Applications of Vectors Solution:
C. Find the Ratio of Line Segments on a
Straight Line by Vectors
Example 13.15T
In DABC, 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.
Solution:
(c) Since B, F and D are collinear,

40. 13.5 Scalar Products A. Definition
Suppose a and b are two vectors on the same plane.
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 .

41. 13.5 Scalar Products A. Definition
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:
a  b = |a||b|cos 
Note:
The scalar product must be written as a  b. It cannot be written as ab or a × b.
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
a  a = |a|2.
For the unit vectors i and j, we have
i  i = j  j = 1.

42. 13.5 Scalar Products A. Definition
If a and b are perpendicular to each other, then a  b = |a||b| cos 90° = 0
Conversely, if a and b are two non-zero vectors such that a  b = 0, then |a||b| cos  = 0
cos  = 0
 = 90°
 a and b are orthogonal, i.e. perpendicular to each other.
So we can conclude that:
For two non-zero vectors a and b, they are orthogonal if and only if a  b = 0.
Since the unit vectors i and j are orthogonal,
i  i = j  j = 0

43. 13.5 Scalar Products B. Properties of Scalar Product
If a, b and c are vectors and k is a real number, then
(a) a  b = b  a
a  a = 0 if and only if a = 0
a  (b + c) = a  b + a  c
(ka)  b = k(a  b) = a  kb
|a||b|  |a  b|
|a – b|2 = |a|2 + |b|2 – 2(a  b)
Proof of (a):
a  b = |a||b| cos 
b  a = |b||a| cos 
\ a  b = b  a

44. 13.5 Scalar Products B. Properties of Scalar Product Proof of (c):
Let = a, = b and = c.
a  (b + c) = |a||b + c| cos ÐAOC
= (OA)(OC)cos ÐAOC
= (OA)(OF)
= (OA)(OE + EF)
= (OA)(OE) + (OA)(EF)
= (OA)(OB cos ÐAOB ) + (OA)(BC cos ÐDBC )
= |a||b| cos ÐAOB + |a||c| cos ÐDBC
= a  b + a  c
\ a  (b + c) = a  b + a  c

45. 13.5 Scalar Products B. Properties 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| (–cos )
= k(a  b)
Combining all the results above, we have (ka)  b = k(a  b).
Similarly, we can prove that a  kb = k(a  b).

46. 13.5 Scalar Products B. Properties 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)

47. Example 13.16T 13.5 Scalar Products Solution:
B. Properties of Scalar Product
Example 13.16T
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
Solution:
(a) y  x = |y||x| cos 
(c) |x – y|2
= (1)(1) cos 135°
= (x – y)  (x – y)
= (x – y)  (x – y)
= x  x – x  y – y  x – y  y
= |x|2 + |y|2 – 2x  y
(b) (x + 3y)  (2y + 5x)
= 2x  y + 5x  x + 6y  y + 15y  x
= 17x  y + 5|x|2 + 6|y|2

48. Example 13.17T 13.5 Scalar Products Solution:
B. Properties of Scalar Product
Example 13.17T
If x, y and z are unit vectors such that 3x – 2y – z = 0, find the value of x  z.
Solution:

49. 13.5 Scalar Products
C. Calculation of Scalar Product in the Rectangular
Coordinate 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.

50. Example 13.18T 13.5 Scalar Products Solution:
C. Calculation of Scalar Product in the Rectangular
Coordinate System
Example 13.18T
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.
Solution:
(a) r  s
= (2i + 5j)  (i – 2j)
= (2)(1) + (5)(–2)
(b) Let  be the angle between r and s.
(cor. to the nearest degree)
\ The angle between r and s is 132.

51. Example 13.19T 13.5 Scalar Products Solution:
C. Calculation of Scalar Product in the Rectangular
Coordinate System
Example 13.19T
Given four points W(2, 2), X(–2, 1), Y(–1, –3) and Z(3, –2). Prove that
(a)
(b)
Solution:
(a)
(b)

52. 13.6 Applications of Scalar Products
A. Projection 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
As is in the same direction as a, we can find by multiplying its length with the unit vector of a, that is,
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.

53. Example 13.20T 13.6 Applications of Scalar Products Solution:
A. Projection of a Vector onto Another Vector
Example 13.20T
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.
Solution:
Consider a  b = (9i + 4j)  (2i – 11j)
= (9)(2) + (4)(–11)
= –26
|a|2 = = 97
|b|2 = 22 + (–11)2 = 125
(a) Projection of a on b
(b) Projection of b on a

54. Example 13.21T 13.6 Applications of Scalar Products Solution:
A. Projection of a Vector onto Another Vector
Example 13.21T
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.
Solution:
Let r = mi + nj, where m and n are non-zero constants.
Since r is a unit vector,
|r|2 = m2 + n2 = 1…………(1)
Projection of p on r
Projection of q on r
Combining (2) and (3),

55. Example 13.21T 13.6 Applications of Scalar Products Solution:
A. Projection of a Vector onto Another Vector
Example 13.21T
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.
Solution:
Substituting (4) into (1),

56. 13.6 Applications of Scalar Products
B. Determination 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.

57. Example 13.22T 13.6 Applications of Scalar Products Solution:
B. Determination of Orthogonality by Vectors
Example 13.22T
In the figure, O is a centre of the circle. M is the mid-point of the chord AB. Show that OM  AB.
Solution:
Let and

58. Example 13.23T 13.6 Applications of Scalar Products Solution:
B. Determination of Orthogonality by Vectors
Example 13.23T
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.
Solution:
(a)

59. Example 13.23T 13.6 Applications of Scalar Products Solution:
B. Determination of Orthogonality by Vectors
Example 13.23T
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.
Solution:
(b) If OC is the shortest distance from O to AB, OC  AB.
The shortest distance

60. Chapter Summary 13.2 Operations and Properties of Vectors
1. Addition of Vectors
2. Subtraction of Vectors

61. Chapter Summary 13.2 Operations and Properties 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.

62. Chapter Summary 13.3 Vectors in the Rectangular Coordinate System
If P(x, y) is a point on the rectangular coordinate system, then
(a)
(b)
(c)
2. If C is a point on AB such that AC : BC = m : n, then

63. Chapter Summary 13.4 Applications of Vectors
1. For two non-zero vectors u and v and a scalar k given, if u = kv, then u and v are parallel.
2. If , then A, B and C are collinear.

64. Chapter Summary 13.5 Scalar Products
If a = x1i + y1j and b = x2i + y2j are non-zero vectors,
and q is the angle between them, then
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)

65. Chapter Summary 13.6 Applications of Scalar Products
1. For two non-zero vectors a and b, the projection of b on a is
given by
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.

66. Follow-up 13.1 13.2 Operations and Properties of Vectors Solution:
B. Subtraction of Vectors
Follow-up 13.1
Express the following vectors in terms of a, b, c and d.
(a) (b) (c)
Solution:
(a)
(b)
(c)

67. Follow-up 13.2 13.2 Operations and Properties of Vectors Solution:
B. Subtraction of Vectors
Follow-up 13.2
The figure shows two non-zero vectors a and b. If |a| = 2, |b|= 3 and the angle between them is 60°, find the value of |a – b|.
Solution:
a  b

68. Follow-up 13.3 13.2 Operations and Properties of Vectors Solution:
E. Rules on Operations of Vectors
Follow-up 13.3
ABCD is a quadrilateral. If Q is a mid-point of BD and AC, prove that ABCD is a parallelogram.
Solution:
\ AD and BC are in the same direction and their magnitude are the same.
\ ABCD is a parallelogram.

69. Follow-up 13.4 13.2 Operations and Properties of Vectors Solution:
E. Rules on Operations of Vectors
Follow-up 13.4
Given that Prove that
Solution:

70. Follow-up 13.5 13.2 Operations and Properties of Vectors Solution:
E. Rules on Operations of Vectors
Follow-up 13.5
ABCD is a parallelogram. E and F are the mid-points of AB and BC respectively. AF and DE intersect at G. Let = p and = q.
(a) (i) Express and in terms of p and q.
(ii) Express p and q in terms of and
(b) Prove that
Solution:
(a) (i)
(ii)
(1)  2 + (2):
(2)  2 – (1):

71. Follow-up 13.5 13.2 Operations and Properties of Vectors Solution:
E. Rules on Operations of Vectors
Follow-up 13.5
ABCD is a parallelogram. E and F are the mid-points of AB and BC respectively. AF and DE intersect at G. Let = p and = q.
(a) (i) Express and in terms of p and q.
(ii) Express p and q in terms of and
(b) Prove that
Solution:
(b) Let and ,
where a and b are constants.
From (1),
Substituting into (2),

72. Follow-up 13.6 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Follow-up 13.6
Given the points P(0, 3) and Q(−1, −2), find the
magnitude and direction of
Solution:
Let  be the angle between and the positive x-axis.
(cor. to 3 sig. fig.)
makes an angle of 259 with the positive x-axis.

73. Follow-up 13.7 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Follow-up 13.7
If the coordinates of P are (–2, 4) and = –2i – 5j, find the coordinates of Q.
Solution:
\ The coordinates of Q are (4, 1).

74. Follow-up 13.8 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Follow-up 13.8
Given two points P(0, –2) and Q(5, 10). Find the unit
vector in the direction of
Solution:
\ Unit vector

75. Follow-up 13.9 13.3 Vectors in the Rectangular Coordinate System
A. Representation of Vectors in the Rectangular
Coordinate System
Follow-up 13.9
If p = i – 2j and q = –3i + j, express –10i – j in terms of p and q.
Solution:
Let
Substituting into (3),
From (2),
Substituting (3) into (1),

76. Follow-up 13.10 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
Follow-up 13.10
Given that OABC is a parallelogram with = a and
= c. OD : DA = 1 : 3 and AE : EC = 2 : 1. Express the following vectors in terms of a and c.
(a) (b)
Solution:
(a)
(b)

77. Follow-up 13.11 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
Follow-up 13.11
In the parallelogram ABCD, E is the mid-point of BC. AC intersects BD at F and AE intersects FB at G such that FG : GB = 1 : r and
AG : GE = 1 : s. Let = a and = b.
(a) By considering DCBF, express in terms of r, a and b.
(b) By considering DCAE, express in terms of s, a and b.
(c) Hence find the values of r and s.
Solution:
(a)

78. Follow-up 13.11 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
Follow-up 13.11
In the parallelogram ABCD, E is the mid-point of BC. AC intersects BD at F and AE intersects FB at G such that FG : GB = 1 : r and
AG : GE = 1 : s. Let = a and = b.
(a) By considering DCBF, express in terms of r, a and b.
(b) By considering DCAE, express in terms of s, a and b.
(c) Hence find the values of r and s.
Solution:
(b)

79. Follow-up 13.11 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
Follow-up 13.11
In the parallelogram ABCD, E is the mid-point of BC. AC intersects BD at F and AE intersects FB at G such that FG : GB = 1 : r and
AG : GE = 1 : s. Let = a and = b.
(a) By considering DCBF, express in terms of r, a and b.
(b) By considering DCAE, express in terms of s, a and b.
(c) Hence find the values of r and s.
Solution:
(c) From the results of (a) and (b),
(2)  (1):

80. Follow-up 13.11 13.3 Vectors in the Rectangular Coordinate System
B. Point of Division
Follow-up 13.11
In the parallelogram ABCD, E is the mid-point of BC. AC intersects BD at F and AE intersects FB at G such that FG : GB = 1 : r and
AG : GE = 1 : s. Let = a and = b.
(a) By considering DCBF, express in terms of r, a and b.
(b) By considering DCAE, express in terms of s, a and b.
(c) Hence find the values of r and s.
Solution:
(c)
Substituting (3) into (1),
Substituting r = 2 into (3),

81. Follow-up 13.12 13.4 Applications of Vectors Solution: A. Parallelism
If the vectors p = 2i – bj and q = (b + 1)i – 6j are parallel but in opposite directions, find the value of b.
Solution:

82. Follow-up 13.13 13.4 Applications of Vectors Solution: A. Parallelism
The figure shows a triangle ABC. D, E and F are the mid-points of AB, BC and CA respectively. It is given that = b and = c. Using the vector method, prove that BEFD is a parallelogram.
Solution:
Hence and
\ BEFD is a parallelogram.

83. Follow-up 13.14 13.4 Applications of Vectors Solution:
B. Prove Three Points are Collinear by Vectors
Follow-up 13.14
In the square PQRS, T is a point on QR such that
QT : TR = 3 : 1. SR is produced to a point U such that
SR : RU = 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 the points P, T and U are collinear.
Solution:
(a)
(b)
(c)
\ and are parallel.
\ P, T and U are collinear.

84. Follow-up 13.15 13.4 Applications of Vectors Solution:
C. Find the Ratio of Line Segments on a
Straight Line by Vectors
Follow-up 13.15
In DABC, 2AD = 3DB and BE = 4CE. DE produced meets AC produced at F such that DE : EF = 1 : k. Let = a, = b and = a
(a) By considering DABC, express in terms of a and b only.
(b) By considering DADF, express in terms of a, b, k and a.
(c) Hence find the values of k and a.
Solution:
(a)
(b)

85. Follow-up 13.15 13.4 Applications of Vectors Solution:
C. Find the Ratio of Line Segments on a
Straight Line by Vectors
Follow-up 13.15
In DABC, 2AD = 3DB and BE = 4CE. DE produced meets AC produced at F such that DE : EF = 1 : k. Let = a, = b and = a
(a) By considering DABC, express in terms of a and b only.
(b) By considering DADF, express in terms of a, b, k and a.
(c) Hence find the values of k and a.
Solution:
(c) By (a) and (b),
Substituting into (2),
From (1), 15k = 5(1 + k)

86. Follow-up 13.16 13.5 Scalar Products Solution:
B. Properties of Scalar Product
Follow-up 13.16
Suppose |p|= 2, |q| = 3 and the angle between p and q is 120°, find the values of each of the following expressions.
(a) p  q
(b) (p + q)  (p – q)
(c) |p – q|
Solution:
(a) p  q = |p||q|cos 
(c) |p – q|
= (2)(3)cos 120°
(b) (p + q)  (p – q)
= p  p – p  q + q  p – q  q
= |p|2 – |q|2
= 22 – 32

87. Follow-up 13.17 13.5 Scalar Products Solution:
B. Properties of Scalar Product
Follow-up 13.17
If u, v and w are unit vectors such that u + 2v + 3w = 0, find the values of
(a) |2v + 3w|, (b) u  w.
Solution:
(a) |2v + 3w|
= |–u|
(Note that –u is a unit vector)
(b) u + 2w + 3w = 0
u + 3w = –2v

88. Follow-up 13.18 13.5 Scalar Products Solution:
C. Calculation of Scalar Product in the Rectangular
Coordinate System
Follow-up 13.18
Given two vectors m = i + 4j and n = 7i + 3j.
(a) Find the value of m  n.
(b) Hence find the angle between m and n, correct to the nearest degree.
Solution:
(a)
(b) Let  be the angle between m and n.
q = 53° (cor. to the nearest degree)
\ The angle between m and n is 53.

89. Follow-up 13.19 13.5 Scalar Products Solution:
C. Calculation of Scalar Product in the Rectangular
Coordinate System
Follow-up 13.19
Given four points P(–1, –1), Q(3, –4), R(6, 0) and S(2, 3).
Prove that
(a) (b)
Solution:
(a)
(b)

90. Follow-up 13.20 13.6 Applications of Scalar Products Solution:
A. Projection of a Vector onto Another Vector
Follow-up 13.20
Two vectors u = i – 4j and v = 3i + 7j are given. Find
(a) the projection of u on v, and
(b) the projection of v on u.
Solution:
(a) Projection of u on v
(b) Projection of v on u

91. Follow-up 13.21 13.6 Applications of Scalar Products Solution:
A. Projection of a Vector onto Another Vector
Follow-up 13.21
Let m = –5i + 2j and n = 3i – 8j. Find a unit vector v such
that the projection of m on v is equal to the projection of n on v.
Solution:
Let v = ai + bj, where a and b are non-zero constants.
Since v is a unit vector,
Projection of m on v
Projection of n on v
Combining (2) and (3),

92. Follow-up 13.21 13.6 Applications of Scalar Products Solution:
A. Projection of a Vector onto Another Vector
Follow-up 13.21
Let m = –5i + 2j and n = 3i – 8j. Find a unit vector v such
that the projection of m on v is equal to the projection of n on v.
Solution:
Substituting (4) into (1),

93. Follow-up 13.22 13.6 Applications of Scalar Products Solution:
B. Determination of Orthogonality by Vectors
Follow-up 13.22
In the figure, ABCD is a rhombus. Show that AC  BD.
Solution:
Let and

94. Follow-up 13.23 13.6 Applications of Scalar Products Solution:
B. Determination of Orthogonality by Vectors
Follow-up 13.23
Given two points P(3, 4) and Q(7, 2). R is a point on PQ that divides PQ in the ratio 1 : r. Find the shortest distance from O to PQ.
Solution:
\ The shortest distance
If OR is the shortest distance from O to PQ, then OR  PQ.