St Andrew s Academy Mathematics Department Higher Mathematics

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1 St Andrew s Academy Mathematics Department Higher Mathematics VECTORS

2 hsn.uk.net Higher Mathematics Vectors Contents Vectors 1 1 Vectors and Scalars EF 1 Components EF 1 3 Magnitude EF 3 4 Equal Vectors EF 4 5 Addition and Subtraction of Vectors EF 5 6 Multiplication by a Scalar EF 7 7 Position Vectors EF 8 8 Basis Vectors EF 9 9 Collinearity EF Dividing Lines in a Ratio EF The Scalar Product EF 14 1 The Angle Between Vectors EF Perpendicular Vectors EF 0 14 Properties of the Scalar Product EF 1 CfE Edition This document was produced specially for the HSN.uk.net website, and we require that any copies or derivative works attribute the work to Higher Still Notes. For more details about the copyright on these notes, please see 1

3 Higher Mathematics Vectors Vectors 1 Vectors and Scalars EF A scalar is a quantity with magnitude (size) only for example, an amount of money or a length of time. Sometimes size alone is not enough to describe a quantity for example, directions to the nearest shop. For this we need to know a magnitude (i.e. how far), and a direction. Quantities with magnitude and direction are called vectors. A vector is named either by using the letters at the end of a directed line segment (e.g. AB represents a vector starting at point A and ending at point B) or by using a bold letter (e.g. u). You will see bold letters used in printed text, but in handwriting you should just underline the letter. B A Throughout these notes, we will show vectors in bold as well us underlining them (e.g. u). Components EF A vector may be represented by its components, which we write in a column. For example, 3 is a vector in two dimensions. In this case, the first component is and this tells us to move units in the x-direction. The second component tells us to move 3 units in the y-direction. So if the vector starts at the origin, it will look like: y 3 u O x hsn.uk.net Page 1 CfE Edition

4 Higher Mathematics Vectors Note that we write the components in a column to avoid confusing them with coordinates. The following diagram also shows the vector 3, but in this case it does not start at the origin. y 1, 1, 1 O x Vectors in Three Dimensions In a vector with three components, the first two tell us ho many units to move in the x- and y-directions, as before. The third component specifies how far to move in the z-direction. z When looking at a pair of x, y -axes, the z-axis points out of the page from the origin. A set of 3D axes can be drawn on a page as shown to the right. z y O For example, is a vector in three dimensions. This vector is shown in the diagram, starting from the origin. z O 4 y 1 3 x x Zero Vectors Any vector with all components zero is called a zero vector and can be written 0 as 0, e.g hsn.uk.net Page CfE Edition 3

5 Higher Mathematics Vectors 3 Magnitude EF The magnitude (or length) of a vector u is written as u. It can be calculated as follows. EXAMPLES 1. Given u If If 5 u 1, find u units.. Find the length of a units. a u b then u a b a u b then u a b c c 5 a 6. 3 Unit Vectors Any vector with a magnitude of one is called a unit vector. For example: if u then u 1 0 So u is a unit vector unit. hsn.uk.net Page 3 CfE Edition 4

6 Higher Mathematics Vectors Distance in Three Dimensions The distance between the points A and B is d AB AB units. For example, given 1 AB 5, we find d AB In fact, there is a three-dimensional version of the distance formula. The distance d between the points EXAMPLE x y z and,, 1, 1, x y z is d x x y y z z units. Find the distance between the points 1, 4,1 and 0,5, The distance is x x y y z z units. 4 Equal Vectors EF Vectors with the same magnitude and direction are equal. For example, all the vectors shown to the right are equal. If vectors are equal to each other, then all of their components are equal, i.e. p r q s t a d if b e then a d, b e and c f. c f Conversely, two vectors are only equal if all of their components are equal. hsn.uk.net Page 4 CfE Edition 5

7 Higher Mathematics Vectors 5 Addition and Subtraction of Vectors EF Consider the following vectors: a b c Addition We can construct a b as follows: b a a b a b means a followed by b. Similarly, we can construct a b c as follows: a a b c means a followed by b followed by c. To add vectors, we position them nose-to-tail. Then the sum of the vectors is the vector between the first tail and the last nose. Subtraction Now consider we can use vector addition to obtain b a b c c a b. This can be written as a b. a b, so if we first find b, b b is just b but in the opposite direction. b b and b have the same magnitude, i.e. b b. Therefore we can construct b a b as follows: a b a a b means a followed by b. hsn.uk.net Page 5 CfE Edition 6

8 Higher Mathematics Vectors Using Components If we have the components of vectors, then things become much simpler. The following rules can be used for addition and subtraction. a d a d b e b e c f c f add the components a d a d b e b e c f c f subtract the components EXAMPLES 1. Given 1 u 5 and 1 1 u v Given 4 3 p and p q v, calculate u v and u v u v q 3 6, calculate p q and q p q p hsn.uk.net Page 6 CfE Edition 7

9 Higher Mathematics Vectors 6 Multiplication by a Scalar EF A vector u which is multiplied by a scalar k 0 will give the result ku. This vector will be k times as long, i.e. the magnitude will be k u. Note that if k 0 this means that the vector ku will be in the opposite direction to u. For example: u 3u u 1 u a ka If u b then ku kb. c kc Each component is multiplied by the scalar. EXAMPLES 1. Given 1 v 5, find 3v v Given 6 r 3, find 4r r hsn.uk.net Page 7 CfE Edition 8

10 Higher Mathematics Vectors Negative Vectors The negative of a vector is the vector multiplied by 1. If we write a vector as a directed line segment AB, then AB BA: B B AB AB BA A A 7 Position Vectors EF OA is called a position vector of point A relative to the origin O, and is written as a. OB is called the position vector of point B, written b. z y Given P x, yz,, the position vector OP or P x p has components y. z O x y O a A b B x To move from point A to point B we can move back along the vector a to the origin, and along vector b to point B. AB AO OB OA OB a b b a For the vector joining any two points P and Q, PQ q p. hsn.uk.net Page 8 CfE Edition 9

11 Higher Mathematics Vectors EXAMPLE R is the point,, 3 From the coordinates, RS s r and S is the point 4, 6, 1 r and 3 4 s Find RS. 8 Basis Vectors EF A vector may also be defined in terms of the basis vectors i, j and k. These are three mutually perpendicular unit vectors (i.e. they are perpendicular to each other). These basis vectors can be written in component form as 1 i 0, 0 0 j 1 and 0 0 k 0. 1 Any vector can be written in basis form using i, j and k. For example: i 3j 6k There is no need for the working above if the following is used: a ai bj ck b. c k i j hsn.uk.net Page 9 CfE Edition 10

12 Higher Mathematics Vectors 9 Collinearity EF In Straight Lines, we learned that points are collinear if they lie on the same straight line. The points A, B and C in 3D space are collinear if AB is parallel to BC, with B a common point. Note that we cannot find gradients in three dimensions instead we use the following. Non-zero vectors are parallel if they are scalar multiples of the same vector. For example: If If EXAMPLE u 1 and p and 6 6 v 3 313u then u and v are parallel q then p and q are parallel. 8 A is the point 1,, 5, B8, 5,9 and C, 11,17. Show that A, B and C are collinear. AB b a BC c b BC AB, so AB and BC are parallel and since B is a common point, A, B and C are collinear. hsn.uk.net Page 10 CfE Edition 11

13 Higher Mathematics Vectors 10 Dividing Lines in a Ratio EF There is a simple process for finding the coordinates of a point which divides a line segment in a given ratio. EXAMPLE 1. P is the point, 4, 1 and R is the point 8, 1,19. The point T divides PR in the ratio :3. Find the coordinates of T. Step 1 Make a sketch of the line, showing the ratio in which the point divides the line segment. Step Using the sketch, equate the ratio of the two lines with the given ratio. Step 3 Cross multiply, then change directed line segments to position vectors. Step 4 Rearrange to give the position vector of the unknown point. Step 5 From the position vector, state the coordinates of the unknown point. R 3 PT TR 3 3PT TR t p r t 3 T 3t 3p r t 3t t r 3p 8 5t t t t 7 So T is the point,, 7. P hsn.uk.net Page 11 CfE Edition 1

14 Higher Mathematics Vectors Using the Section Formula The previous method can be condensed into a formula as shown below. If the point P divides the line AB in the ratio m: n, then: na mb p, n m where is a, b and p are the position vectors of A, B and P respectively. It is not necessary to know this, since the approach explained above will always work. EXAMPLE. P is the point, 4, 1 and R is the point 8, 1,19. The point T divides PR in the ratio :3. Find the coordinates of T. The ratio is :3, so let m and n 3, then: np mr t n m 3p r So T is the point,, 7. Note If you are confident with arithmetic, this step can be done mentally. hsn.uk.net Page 1 CfE Edition 13

15 Higher Mathematics Vectors Further Examples EXAMPLES 3. The cuboid OABCDEFG is shown in the diagram. E F H D A G B O C The point A has coordinates 0,0,5, C8,0,0 and G8,1,0. The point H divides BF in the ratio 4:1. Find the coordinates of H. From the diagram: OH OA AB 4 5 BF OA OC 4 5 CG h a c 4 5 g c a c g 5c a 1 4 5c 5 g So H has coordinates 8, 48 5,5. 4. The points P6,1, 3, Q8, 3,1 and R9, 5,3 are collinear. Find the ratio in which Q divides PR. Since the points are collinear PQ kqr for some k. Working with the first components: 86k 98 k. Therefore PQ QR so Q divides PR in the ratio :1. Note BH 4 BF 5, so BH 4 5 BF. Note The ratio is :1 since PQ QR 1. hsn.uk.net Page 13 CfE Edition 14

16 Higher Mathematics Vectors 5. The points A 7, 4, 4, B13,5, 7 and C are collinear. Given that B divides AC in the ratio 3:, find the coordinates of C. 3 AB 5 AC 3 b a 5 c a to see this: 3 3 b a 5c 5a 3 3 5c b 5a A 5 c 3b 3a So C has coordinates 17,11, 9. Note A sketch may help you B C 11 The Scalar Product EF So far we have added and subtracted vectors and multiplied a vector by a scalar. Now we will consider the scalar product, which is a form of vector multiplication. The scalar product is denoted by ab. (sometimes it is called the dot product) and can be calculated using the formula: ab. a b cosθ, where θ is the angle between the two vectors a and b. This formula is given in the exam. hsn.uk.net Page 14 CfE Edition 15

17 Higher Mathematics Vectors The definition above assumes that the vectors a and b are positioned so that they both point away from the angle, or both point into the angle. a θ b a θ b However, if one vector is pointing away from the angle, while the other points into the angle, a θ b a θ b we find that ab. a b cosθ. EXAMPLES 1. Two vectors, a and b have magnitudes 7 and 3 units respectively and are at an angle of 60 to each other as shown below. b 60 What is the value of ab?. a ab. a bcosθ 73cos The vector u has magnitude k and v is twice as long as u. The angle between u and v is 30, as shown below. v Find an expression for uv. in terms of k. uv. u vcosθ kkcos30 k 3 3 k. 30 u Remember When one vector points in and one points out, uv. u v cosθ. hsn.uk.net Page 15 CfE Edition 16

18 Higher Mathematics Vectors The Component Form of the Scalar Product The scalar product can also be calculated as follows: ab. ab ab ab where This is given in the exam a1 a a and a 3 b1 b b b 3 EXAMPLES 3. Find pq,. given that pq. pq pq pq p and If A is the point AB.AC. C1,3, 6 B1,4, q. 3, 3, 9, B1,4, and C 1,3, 6, calculate We need to use the position vectors of the points: AB b a AC c a A, 3, AB.AC hsn.uk.net Page 16 CfE Edition 17

19 Higher Mathematics Vectors 1 The Angle Between Vectors EF The formulae for the scalar product can be rearranged to give the following equations, both of which can be used to calculate θ, the angle between two vectors.. cosθ ab ab ab ab ab cosθ ab or Look back to the formulae for finding the scalar product, given on the previous pages. Notice that the first equation is simply a rearranged form of the one which can be used to find the scalar product. Also notice that the second simply replaces ab. with the component form of the scalar product. These formulae are not given in the exam but can both be easily derived from the formulae on the previous pages (which are given in the exam). EXAMPLES 1. Calculate the angle θ between vectors p 3i 4j k and q 4i j 3k. 3 p 4 and q pq pq pq cosθ pq θ cos (to 1 d.p.) (or radians (to 3 d.p.)) hsn.uk.net Page 17 CfE Edition 18

20 Higher Mathematics Vectors. K is the point 1, 7, Start with a sketch: L 3,3,4 θ, L 3,3,4 and M,5,1 M,5,1. Find KLM. K1, 7, Now find the vectors pointing away from the angle: LK k l , LM ml Use the scalar product to find the angle: LK.LM cosθ LK LM θ cos (to 1 d.p.) (or radians (to 3 d.p.)) hsn.uk.net Page 18 CfE Edition 19

21 Higher Mathematics Vectors 3. The diagram below shows the cube OPQRSTUV. T z U y S V P O Q The point R has coordinates 4,0,0. (a) Write down the coordinates of T and U. (b) Find the components of RT and RU. (c) Calculate the size of angle TRU. (a) From the diagram, T0,4,4 and U4,4,4. R x (b) RT t r 40 4, RU u r RT.RU (c) cos TRU RT RU cos TRU (to 1 d.p.) (or radians (to 3 d.p.)) hsn.uk.net Page 19 CfE Edition 0

22 Higher Mathematics Vectors 13 Perpendicular Vectors EF If a and b are perpendicular then ab. 0. This is because ab. a bcosθ abcos90 ( θ 90 since perpendicular) 0 (since cos900) Conversely, if ab. 0 then a and b are perpendicular. EXAMPLES 1. Two vectors are defined as a 4i j 5k and b i j k. Show that a and b are perpendicular. ab. ab 1 1ab ab Since ab. 0, a and b are perpendicular.. 4 PQ a 7 and RS 3 a and RS where a is a constant. Given that PQ Since PQ and RS are perpendicular, PQ.RS 0 4 3a 7a 0 83a7a 0 84a 0 a. are perpendicular, find the value of a. hsn.uk.net Page 0 CfE Edition 1

23 Higher Mathematics Vectors 14 Properties of the Scalar Product EF Some properties of the scalar product are as follows: ab. ba. a. b c ab. ac. (Expanding brackets) aa. a Note that these are not given in the exam. EXAMPLES 1. In the diagram, 3 Calculate p. q r. p, r 4 and q. p q r p. q r p. q p. r pqcosθ1 prcosθ 3cos6034cos In the diagram below a c and b 3. a c 30 a a b c. Calculate. a. a b c aa. ab. ac. a a b cosθ1 a c cosθ 3 cos30 cos b 30 Remember ac. a c cosθ since a points to θ and c points away. hsn.uk.net Page 1 CfE Edition

24 Exercise 1 : Exercise : 3

25 Exercise 3 : Exercise 4 : 4

26 Exercise 5 : 5

27 6

28 7

29 Exercise 6 : 8

30 Exercise 7 : 9

31 30

32 ANSWERS 31

33 3

34 33

35 Vectors 008 P1 Ans C 008 P1 Ans B 34

36 008 P1 Ans C 008 P1 Ans C 35

37 008 P 5 Ans (a) P(8,0,4), Q(0,4,3) (b) 8 0 PQ 4 PA (c) P1 4 Ans 007 P 1 5 Ans 36

38 006 P Ans 006 P Ans 37

39 005 P1 4 1 Ans 005 P 5 Ans 38

40 005 P 4 Ans 004 P1 3 Ans 004 P 5 Ans 003 P1 Ans Vectors are perpendicular. 003 P1 3 Ans 39

41 003 P 4 Ans 00W P1 Ans 00W P1 3 Ans 0 00W P 4 40

42 Ans 00 P1 3 Ans 00 P 1 4 Ans 41

43 3 001 P1 3 Ans 4

44 001 P 7 Ans 000 P1 3 Ans 000 P Ans t = 4 43

45 000 P 1 6 Ans Specimen P1 4 Ans 44

46 Specimen P 5 Ans 45

47 Specimen 1 P1 5 6 Ans Specimen 1 P

48 Ans 47