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1 3 .1 Concepts of Vectors and Scalars

1 3. Vectors in Two-dimensional Space. Case Study. 1 3 .1 Concepts of Vectors and Scalars. 1 3 .2 Operations and Properties of Vectors. 1 3 .3 Vectors in the Rectangular Coordinate System. 1 3 .4 Applications of Vectors. 1 3 .5 Scalar Products. 1 3 . 6 Applications of Scalar Products.

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1 3 .1 Concepts of Vectors and Scalars

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  1. 13 Vectors in Two-dimensional Space Case Study 13.1Concepts 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 13.6Applications 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 = (30 + 40) 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. 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 . 13.1 Concepts of Vectors and Scalar B. Representation of a Vector Thedirected 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.

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

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

  7. Triangle Law of addition 13.2Operations and Properties of Vectors A. Addition of Vectors 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. Parallelogram law of addition If ABCD is a parallelogram, then . 13.2Operations 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:

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

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

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

  12. Using the definitions above, for any non-zero vector a, the unit vector is denoted by (or ). 13.2Operations 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..

  13. For a given point X on a plane, if we choose a point O on the same plane as the reference point, then the position of X can be determined by the vector 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 Triangle law of addition Let = x and = y. 13.2Operations and Properties of Vectors D. Position Vectors

  14. 13.2Operations 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. Proof of (b): As shown in the figure, = a, = b and = c. Since , Also,, 13.2Operations and Properties of Vectors E. Rules on Operations of Vectors

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

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

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

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

  22. 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 and 13.3Vectors in the Rectangular Coordinate System A. Representation of Vectors in the Rectangular Coordinate System Vectors can be represented in the rectangular coordinate plane. Firstly, leti be the unit vector in the positive direction of x-axis, and j be the unit vector in the positive direction of y-axis. If we consider OMP in the figure, we have

  23. Then the vector can be expressed as and 13.3Vectors 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. Hence we have

  24. Given two points X(–2, 1) andY(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. 13.3Vectors in the Rectangular Coordinate System A. Representation of Vectors in the Rectangular Coordinate System Example 13.6T Solution:

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

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

  27. Let . 13.3Vectors 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: From (1), m = 2n Substituting m = 2n into (2),

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

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

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

  32. 13.4Applications 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. 13.4Applications of Vectors A. Parallelism Example 13.12T If the vectorsa = 2ci + 8j and b = 2i + (c + 2)j are parallel but in the opposite directions, find the value of c. Solution:

  34. 13.4Applications of Vectors A. Parallelism Example 13.13T 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 // MDand BN = MD. \ MBND is a parallelogram.

  35. 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. 13.4Applications of Vectors B. Prove Three Points are Collinear by Vectors Suppose there are three distinct points A, B and C on the same plane. 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.

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

  37. 13.4Applications 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. In DABC, D and Eare points onACand AB respectively.CD : DA= 1 : 3 and AE : EB= 2 : 1. Let = a, = b andCF : FE = 1 : k. (a)Express in terms of a,band k. (b)Express in terms of a and b. (c)Hence find BF : FD. 13.4Applications of Vectors C. Find the Ratio of Line Segments on a Straight Line by Vectors Example 13.15T Solution: (a) (b)

  39. In DABC, D and Eare points onACand AB respectively.CD : DA= 1 : 3 and AE : EB= 2 : 1. Let = a, = b andCF : FE = 1 : k. (a)Express in terms of a,band k. (b)Express in terms of a and b. (c)Hence find BF : FD. (c) Since B, F and D are collinear, . 13.4Applications of Vectors C. Find the Ratio of Line Segments on a Straight Line by Vectors Example 13.15T Solution:

  40. 13.5Scalar 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. 13.5Scalar 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. 13.5Scalar 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. 13.5Scalar Products B. Properties of Scalar Product • 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 ifa = 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. Proof of (c): Let = a, = b and = c. a  (b+ c) = |a||b+ c|cos ÐAOC 13.5Scalar Products B. Properties of Scalar Product = (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. 13.5Scalar 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. 13.5Scalar 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. 13.5Scalar Products 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 = (x – y)  (x – y) = (1)(1) cos 135° = (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. 13.5Scalar Products B. Properties of Scalar Product Example 13.17T If x, y and zare unit vectors such that 3x – 2y – z = 0, find the valueof x  z. Solution:

  49. a  b= x1x2+ y1y2 where  is the angle between a and b. 13.5Scalar 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

  50. 13.5Scalar Products 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.

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