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**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

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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.

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**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.

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**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

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**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

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**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

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**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.

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**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

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**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.

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**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)

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**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:

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**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 ).

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**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

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**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.

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**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, ,

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**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 )

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**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.

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**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.

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**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:

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**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):

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**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)

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**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

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**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

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**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.

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**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).

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**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

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**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),

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**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

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**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)

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**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)

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**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),

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**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

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**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:

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**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.

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**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.

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**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.

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**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.

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**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)

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**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,

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**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 .

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**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.

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**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

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**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

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**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

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**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).

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**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)

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**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

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**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:

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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.

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**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.

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**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)

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**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.

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**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

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**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),

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**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),

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**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.

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**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

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**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)

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**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

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**Chapter Summary 13.2 Operations and Properties of Vectors**

1. Addition of Vectors 2. Subtraction of Vectors

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**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.

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**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

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**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.

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**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)

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**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.

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**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)

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**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

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**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.

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**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:

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**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):

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**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),

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**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.

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**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).

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**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

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**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),

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**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)

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**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)

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**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)

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**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):

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**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),

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**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:

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**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.

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**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.

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**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)

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**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)

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**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

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**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

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**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.

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**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)

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**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

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**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),

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**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),

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**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

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**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.

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Copyright © 2012, Elsevier Inc. All rights Reserved. 1 Chapter 7 Modeling Structure with Blocks.

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