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13.1Concepts of Vectors and Scalars 13.2Operations and Properties of Vectors 13.3Vectors in the Rectangular Coordinate System Chapter Summary Case Study.

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Presentation on theme: "13.1Concepts of Vectors and Scalars 13.2Operations and Properties of Vectors 13.3Vectors in the Rectangular Coordinate System Chapter Summary Case Study."— Presentation transcript:

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

2 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 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 = ( ) km = 70 km Thus the displacement of the car is 50 km in the direction of around S53°E.

3 P. 3 For example, displacement, velocity and force are vectors 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, distance, temperature and area are scalars.

4 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. B. Representation of a Vector 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 Concepts of Vectors and Scalar

5 P. 5 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 Concepts of Vectors and Scalar

6 P. 6 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 Concepts of Vectors and Scalar 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 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 Operations and Properties of Vectors Vectors Step 2:Translate b in a parallel direction such that the initial point of b coincides with the terminal point of a. A. Addition of Vectors Triangle Law of addition Step 3:By the triangle law of addition, a third vector a + b is obtained.

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

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

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

11 P. 11 Example 13.2T Solution: 13.2 Operations and Properties of Vectors Vectors B. Subtraction 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 .

12 P Operations and Properties of Vectors 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 P Operations and Properties of Vectors 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 D. Position Vectors

14 P Operations and Properties of Vectors Vectors E. Rules on Operations of Vectors Property 13.1Rules 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 P Operations and Properties of Vectors Vectors E. Rules on Operations of Vectors Proof of (b): As shown in the figure, = a, = b and = c. Since, Also,,

16 P Operations and Properties of Vectors 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.  ) (corr. sides,  s  )

17 P Operations and Properties of Vectors 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  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 P. 18 Example 13.3T Solution: In  ABC, D and E are the mid-points of AB and AC respectively. Prove that BC // DE and BC = 2DE Operations and Properties of Vectors Vectors E. Rules on Operations of Vectors  BC // DE and BC = 2DE.

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

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

21 P Operations and Properties of Vectors Vectors E. Rules on Operations of Vectors Substituting into (1), (b) (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.

22 P. 22 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 Vectors in the Rectangular Coordinate System Coordinate System Vectors can be represented in the rectangular coordinate plane. A. Representation of Vectors in the Rectangular Coordinate System Coordinate System 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 For any point P(x, y), the position vector can be expressed as

23 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 Vectors in the Rectangular Coordinate System 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. A. Representation of Vectors in the Rectangular Coordinate System Coordinate System For example, as shown in the figure, X(x 1, y 1 ) and Y(x 2, y 2 ) are two points on the coordinate plane. Then the vector can be expressed as Hence we have and

24 P. 24 Example 13.6T Solution: 13.3 Vectors in the Rectangular Coordinate System Coordinate System A. Representation of Vectors in the Rectangular Coordinate System Coordinate System Given two points X(–2, 1) and Y(4, –1), find the magnitude and direction of. 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 P. 25 Example 13.7T Solution: 13.3 Vectors in the Rectangular Coordinate System Coordinate System A. Representation of Vectors in the Rectangular Coordinate System Coordinate System If the coordinates of Y are (0, –5) and, find the coordinates of X.  The coordinates of X are (  1,  5).

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

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

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

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

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

32 P. 32 If two vectors are in the same or opposite directions, then they are parallel Applications of Vectors A. Parallelism 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 = m 1 a + n 1 b andv = m 2 a + n 2 b where m 1, m 2, n 1 and n 2 are real numbers and m 2 and n 2 are non-zero, we can conclude that

33 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 Applications of Vectors A. Parallelism

34 P. 34 Example 13.13T Solution: 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 Applications of Vectors A. Parallelism  BN // MD and BN = MD.  MBND is a parallelogram.

35 P. 35 Suppose there are three distinct points A, B and C on the same plane Applications of Vectors B. Prove 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, then the line segments AB and AC must be parallel. 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 P. 36 Example 13.14T Solution: 13.4 Applications of Vectors B. Prove 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. (a) (b)(b) (c)(c)  and are parallel.  A, E and F are collinear.

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

38 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. (a) (b)(b) 13.4 Applications of Vectors C. Find the Ratio of Line Segments on a Straight Line by Vectors Straight Line by Vectors

39 P Applications of Vectors C. Find the Ratio of Line Segments on a Straight Line by Vectors Straight Line by Vectors (c) Since B, F and D are collinear,. 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.

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

42 P. 42 If a and b are perpendicular to each other, then a  b = |a||b| cos  = Scalar Products A. Definition  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:

43 P Scalar Products B. Properties of Scalar Product Proof of (a): a  b = |a||b| cos  Properties of Scalar Product 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) b  a = |b||a| cos   a  b = b  a

44 P Scalar Products B. Properties of Scalar Product Proof of (c): Let = a, = b and = c. a  (b + c) = |a||b + c| cos  AOC = (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

45 P 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) = –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).

46 P 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 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| Scalar Products B. Properties of Scalar Product (a) y  x = |y||x| cos  = (1)(1) cos 135° (b) (x + 3y)  (2y + 5x) = 2x  y + 5x  x + 6y  y + 15y  x = 17x  y + 5|x| 2 + 6|y| 2 (c) |x – y| 2 = (x – y)  (x – y) = x  x – x  y – y  x – y  y = |x| 2 + |y| 2 – 2x  y

48 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 Scalar Products B. Properties of Scalar Product

49 P Scalar Products C. Calculation of Scalar Product in the Rectangular Coordinate System Coordinate System If a = x 1 i + y 1 j and b = x 2 i + y 2 j are two non-zero vectors, then a  b = x 1 x 2 + y 1 y 2 where  is the angle between a and b.

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

52 P 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. 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,

53 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 13.6 Applications of Scalar Products A. Projection of a Vector onto Another Vector = (9)(2) + (4)(–11) = –26 |a| 2 = = 97 |b| 2 = (–11) 2 = 125 (b)Projection of b on a

54 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 13.6 Applications of Scalar Products A. Projection of a Vector onto Another Vector Since r is a unit vector, |r| 2 = m 2 + n 2 = 1…………(1) Projection of q on r Combining (2) and (3),

55 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 Applications of Scalar Products A. Projection of a Vector onto Another Vector Solution:

56 P 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 P. 57 Example 13.22T Solution: 13.6 Applications 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 B. Determination of Orthogonality by Vectors

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

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

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

61 P Operations and Properties of Vectors Chapter Summary 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 P Vectors in the Rectangular Coordinate System Chapter Summary 1.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 P For two non-zero vectors u and v and a scalar k given, if u = kv, then u and v are parallel Applications of Vectors Chapter Summary 2. If, then A, B and C are collinear.

64 P Scalar Products Chapter Summary If a = x 1 i + y 1 j and b = x 2 i + y 2 j are non-zero vectors, and  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 P Applications of Scalar Products Chapter Summary 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 Solution: 13.2 Operations and Properties of Vectors Vectors B. Subtraction of Vectors Express the following vectors in terms of a, b, c and d. (a)(b)(c) (a) (b) (c)

67 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: 13.2 Operations and Properties of Vectors Vectors B. Subtraction of Vectors a  ba  b

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

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

70 Follow-up Operations and Properties of Vectors Vectors E. Rules on Operations of Vectors (a) (i) (ii) (1)  2 + (2): (2)  2 – (1): Solution: 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.

71 Follow-up 13.5 Solution: 13.2 Operations and Properties of Vectors Vectors E. Rules on Operations of Vectors (b) Let and, where a and b are constants. From (1),Substituting into (2), 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.

72 Solution: Follow-up Vectors in the Rectangular Coordinate System Coordinate System A. Representation of Vectors in the Rectangular Coordinate System Coordinate System Given the points P(0, 3) and Q(−1, −2), find the magnitude and direction of. 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 Solution: 13.3 Vectors in the Rectangular Coordinate System Coordinate System A. Representation of Vectors in the Rectangular Coordinate System Coordinate System If the coordinates of P are (–2, 4) and = –2i – 5j, find the coordinates of Q.  The coordinates of Q are (  4,  1).

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

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

76 Follow-up Solution: 13.3 Vectors in the Rectangular Coordinate System Coordinate System B. Point of Division 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) (a) (b)

77 Follow-up Solution: 13.3 Vectors in the Rectangular Coordinate System Coordinate System B. Point of Division 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  CBF, express in terms of r, a and b. (b) By considering  CAE, express in terms of s, a and b. (c) Hence find the values of r and s. (a)

78 Follow-up Solution: 13.3 Vectors in the Rectangular Coordinate System Coordinate System B. Point of Division (b)(b) 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  CBF, express in terms of r, a and b. (b) By considering  CAE, express in terms of s, a and b. (c) Hence find the values of r and s.

79 Follow-up Solution: 13.3 Vectors in the Rectangular Coordinate System Coordinate System B. Point of Division (c) From the results of (a) and (b), (2)  (1): 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  CBF, express in terms of r, a and b. (b) By considering  CAE, express in terms of s, a and b. (c) Hence find the values of r and s.

80 Follow-up Vectors in the Rectangular Coordinate System Coordinate System B. Point of Division Substituting (3) into (1), Substituting r = 2 into (3), Solution: 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  CBF, express in terms of r, a and b. (b) By considering  CAE, express in terms of s, a and b. (c) Hence find the values of r and s. (c)(c)

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

82 Follow-up Solution: 13.4 Applications of Vectors 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. Hence and.  BEFD is a parallelogram.

83 Follow-up Solution: 13.4 Applications of Vectors B. Prove Three Points are Collinear by Vectors 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. (a) (b)(b) (c)(c)  and are parallel.  P, T and U are collinear.

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

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

86 Follow-up Solution: 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| 13.5 Scalar Products B. Properties of Scalar Product (a) p  q = |p||q|cos  = (2)(3)cos 120° (b) (p + q)  (p – q) = p  p – p  q + q  p – q  q = |p| 2 – |q| 2 (c) |p – q| = 2 2 – 3 2

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

88 Follow-up Solution: 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 Scalar Products C. Calculation of Scalar Product in the Rectangular Coordinate System Coordinate System (a) (b)Let  be the angle between m and n.  = 53° (cor. to the nearest degree)  The angle between m and n is 53 .

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

90 Follow-up Solution: 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 Applications of Scalar Products A. Projection of a Vector onto Another Vector (a)Projection of u on v (b)Projection of v on u

91 Follow-up Solution: 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 Applications of Scalar Products A. Projection of a Vector onto Another Vector 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 Solution: 13.6 Applications of Scalar Products A. Projection of a Vector onto Another Vector Substituting (4) into (1), 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.

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

94 Follow-up Solution: 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 Applications of Scalar Products B. Determination of Orthogonality by Vectors If OR is the shortest distance from O to PQ, then OR  PQ.  The shortest distance


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