# Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

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1 hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude Equl Vectors Addition nd Subtrction of Vectors 13 6 Multipliction by Sclr Position Vectors Bsis Vectors Collinerity Dividing Lines in Rtio The Sclr Product The Angle Between Vectors Perpendiculr Vectors Properties of the Sclr Product 148 HSN3100 This document ws produced specilly for the HSN.uk.net website, nd we require tht ny copies or derivtive works ttribute the work to Higher Still Notes. For more detils bout the copyright on these notes, plese see

2 OUTCOME 1 Vectors 1 Vectors nd Sclrs A sclr is quntity with mgnitude (size) only for exmple, n mount of money or length of time. Sometimes size lone is not enough to describe quntity for exmple, directions to the nerest shop. For this we need to know mgnitude (i.e. how fr), nd direction. Quntities with both mgnitude nd direction re clled vectors. A vector is nmed either by using the points t the end of directed line segment (e.g. AB represents the vector strting t point A nd ending t point B) or by using bold letter (e.g. u). You will see bold letters used in printed text, but in hndwriting you should just underline the letter (e.g. ). B Throughout these notes, we will show vectors bold nd underlined (e.g. u ). Components A A vector my be represented by its components, which we write in column. For exmple, 3 is vector in two dimensions. In this cse the first component is nd 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 strts t the origin, it will look like: y 3 u O x Pge 18 HSN3100

3 Note tht we write the components in column to void confusing them with coordintes. The following digrm lso shows the vector 3, but in this cse it does not strt t the origin. y ( 1, ) ( 1, 1 ) O x Vectors in Three Dimensions In vector with three components, the first two tell us how mny units to move in the x- nd y-directions, s before. The third component specifies how fr to move in the z-direction. When looking t pir of ( x, y ) -xes, the z-xis points out of the pge from the origin. z A set of 3D xes cn be drwn on pge s shown to the right. z O y For exmple, is vector in three dimensions. This vector is shown in the digrm, strting from the origin. z O 4 1 y 3 x x Zero Vectors Any vector with ll its components zero is clled zero vector nd cn be 0 written s 0, e.g. 0 = 0. 0 Pge 19 HSN3100

4 3 Mgnitude The mgnitude (or length) of vector u is written s u. It cn be clculted s follows. EXAMPLES 1. Given = If PQ = b then PQ = + b. If PQ = b then PQ = + b + c. c u 5 = 1, find u. u = 5 + ( 1) 169 = 13 units.. Find the length of 5 = 6. 3 ( ) = + + = 50 = 5 units. Unit Vectors Any vector with mgnitude of one is clled unit vector. For exmple: if 1 = 0 3 u then 3 ( ) ( ) So u is unit vector. u = = 4 4 = 1 unit. Pge 130 HSN3100

5 Distnce in Three Dimensions The distnce between the points A nd B is d AB = AB units. So given 1 AB = 5, we find d ( ) AB = = 30. In fct, there is three dimensionl version of the distnce formul. The distnce d between the points ( x, y, z ) nd (,, ) EXAMPLE ( ) ( ) ( ) x y z is units. d = x x + y y + z z Find the distnce between the points ( 1,4,1 ) nd ( 0,5, 7) The distnce is 4 Equl Vectors ( x x ) + ( y y ) + ( z z ) = ( 0 ( 1) ) + ( 5 4) + ( 7 1) = ( 8) = = 66 units.. Vectors with the sme mgnitude nd direction re sid to be equl. For exmple, ll the vectors shown to the right re equl. If vectors re equl to ech other, then ll of their components re equl, i.e. p r q s t d if b = e then = d, b = e nd c = f. c f Conversely, two vectors re only equl if ll of their components re equl. Pge 131 HSN3100

6 5 Addition nd Subtrction of Vectors Consider the following vectors: b c Addition We cn construct + b s follows: b + b + b mens followed by b. Similrly, we cn construct + b + c s follows: + b + c mens followed by b followed by c. To dd vectors, we position them nose-to-til. Then the sum of the vectors is the vector between the first til nd the lst nose. Subtrction b + b + c c Now consider b. This cn be written s + ( b ), so if we first find b we cn use vector ddition to obtin b. b b is just b but in the opposite direction. b b nd b hve the sme mgnitude, i.e. b = b. Therefore we cn construct b b s follows: b b mens followed by b. Pge 13 HSN3100

7 Using Components If we hve the components of vectors, then things become much simpler. The following rules cn be used for ddition nd subtrction. d + d b + e = b + e c f c + f dd the components d d b e = b e c f c f subtrct the components EXAMPLES 1. Given 1 u = 5 nd 1 1 u + v = Given 0 = 7 4 p = 3 nd p q = = v =, clculte u + v nd u v = 5 0 = 3. u v. 1 q = 3, clculte p q nd q + p q + p = =. 9 5 Pge 133 HSN3100

8 6 Multipliction by Sclr A vector u which is multiplied by sclr k > 0 will give the result ku. This vector will be k times s long, i.e. its mgnitude will bek u. Note tht if k < 0 this mens tht the vector ku will be in the opposite direction to u. For exmple: u 3u u 1 u k If u = b then ku = kb. c kc Ech component is multiplied by the sclr. EXAMPLES 1. Given 1 v = 5, find 3v v = 3 5 = Given 6 r = 3, find 4r r = 4 3 = Pge 134 HSN3100

9 Negtive Vectors The negtive of vector is the vector multiplied by 1. If we write vector s directed line segment AB, then AB = BA : AB A 7 Position Vectors B OA is clled the position vector of point A reltive to the origin O, nd is written s. A B AB = BA OB is clled the position vector of point B, written b. Given P ( x, y, z ), the position vector OP or p hs components x y. z z O y x P y O A b B x To move from point A to point B we cn move bck long the vector to the origin, nd long vector b to point B, i.e. AB = AO + OB = OA + OB = + b = b. For the vector joining ny two points P nd Q, PQ = q p. Pge 135 HSN3100

10 EXAMPLE R is the point (,, 3) From the coordintes, RS = s r 4 = = 8. 4 nd S is the point ( 4, 6, 1) r = nd 3 4 s = Find RS. Note You don t need to write this line down in the exm. 8 Bsis Vectors A vector my lso be defined in terms of the bsis vectors i, j nd k. These re three mutully perpendiculr unit vectors (i.e. they re perpendiculr to ech other). These bsis vectors cn be written in component form s 1 i = 0, 0 0 j = 1 nd 0 0 k = 0. 1 Any vector cn be written in bsis form using i, j nd k. For exmple: k i j v = 3 = = i 3j + 6 k There is no need for the working bove if the following is used: i + bj + ck = b. c Pge 136 HSN3100

11 9 Collinerity In Stright Lines (Unit 1 Outcome 1), we lerned tht points re colliner if they lie on the sme stright line. The points A, B nd C in 3D spce re colliner if AB is prllel to BC, with B common point. Note tht we cnnot find grdients in three dimensions insted we use the following. Non-zero vectors re prllel if they re sclr multiples of the sme vector. For exmple: 6 u = 1, v = 3 = 31 = 3 u So u nd v re prllel p = 9 = 3 3, q = 1 = So p nd q re prllel. EXAMPLE A is the point ( 1,, 5), B( 8, 5, 9) nd C(, 11,17 ). Show tht A, B nd C re colliner. AB = b BC = c b = 5 = = 3 = 6 = BC = AB, so AB nd BC re prllel, nd since B is common point, A, B nd C re colliner. Pge 137 HSN3100

12 10 Dividing Lines in Rtio There is simple process for finding the coordintes of point which divides line segment in given rtio. EXAMPLE 1. P is the point (, 4, 1) nd R is the point ( 8, 1,19 ). The point T divides PR in the rtio : 3. Find the coordintes of T. Step 1 Mke sketch of the line, showing the rtio in which the point divides the line segment. Step Using the sketch, equte the rtio of the two line segments with the given rtio. Step 3 Cross multiply, then chnge directed line segments to position vectors. Step 4 Rerrnge to give the position vector of the unknown point. Step 5 From the position vector, stte the coordintes 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 Pge 138 HSN3100

13 Using the Section Formul The previous method cn be condensed into formul s shown below. If the point P divides the line AB in the rtio m : n, then n + mb p =, n + m where, b nd p re the position vectors of A, B nd P respectively. This is referred to s the section formul. It is not necessry to know this, since the pproch explined bove will lwys work. EXAMPLE. P is the point (, 4, 1) nd R is the point ( 8, 1,19 ). The point T divides PR in the rtio : 3. Find the coordintes of T. The rtio is : 3, so m = nd n = 3. Hence: np + mr t = n + m 3p + r = ( 3( ) + ( 8) ) 1 = 5 ( 3( 4) + ( 1) ) 1 5 ( 3( 1) + ( 19) ) =. 7 So T is the point (,, 7 ). Note If you re confident with rithmetic, this step cn be done mentlly. Pge 139 HSN3100

14 Further Exmples EXAMPLES 3. The cuboid OABCDEFG is shown in the digrm below. E F H D A G B O C The point A hs coordintes ( 0,0,5 ), C( 8,0,0 ) nd G( 8,1,0 ). The point H divides BF in the rtio 4 :1. Find the coordintes of H. From the digrm: OH = OA + AB BF = OA + OC CG ( g c) h = + c = + c g 5c = c + 5 g = = 5. 5 So H hs coordintes 48 ( ) 8, 5,5. 4. The points P( 6,1, 3), Q ( 8, 3,1) nd R ( 9, 5,3) re colliner. Find the rtio in which Q divides PR. Since the points re colliner PQ = kqr for some k. Working with the first components: 8 6 = k ( 9 8) k =. Therefore PQ = QR so Q divides PR in the rtio :1. Note BH 4 =, so BH= 4 5 BF. BF 5 Note The rtio is : 1 since PQ =. QR 1 Pge 140 HSN3100

15 5. The points A ( 7, 4, 4), B( 13,5, 7) nd C re colliner. Given tht B divides AC in the rtio 3 :, find the coordintes of C. 3 AB = 5 AC b = 3 5 ( c ) b = 3 3 5c c = b 5 A c = 5 3b = = So C hs coordintes ( 17,11, 9). Note A sketch my help you to see this: B C 11 The Sclr Product So fr we hve dded nd subtrcted vectors nd multiplied vector by sclr. Now we will consider the sclr product, which is form of vector multipliction. The sclr product is denoted by b. (sometimes it is clled the dot product) nd cn be clculted using the formul: b. = b cos θ, where θ is the ngle between the two vectors nd b. This is given in the exm. Pge 141 HSN3100

16 The definition bove ssumes tht the vectors nd b re positioned so tht they both point wy from the ngle, or both point into the ngle. θ b θ b However, if one vector is pointing wy from the ngle, while the other points into the ngle, θ b θ b we find tht b. = b cosθ. EXAMPLES 1. Two vectors, nd b hve mgnitudes 7 nd 3 units respectively nd re t n ngle of 60 to ech other s shown below. Wht is the vlue of b.? b. = b cosθ = 7 3 cos 60 = 1 1 = 1. b 60. The vector u hs mgnitude k nd v is twice s long s u. The ngle between u nd v is 30, s shown below. v 30 u Find n expression for uv. in terms of k. uv. = u v cosθ = k k cos30 = k = 3 k. 3 Remember When one vector points in nd one points out, uv. = u v cosθ. Pge 14 HSN3100

17 The Component Form of the Sclr Product The sclr product cn lso be clculted s follows: b. = b + b + b where This is given in the exm. nd 1 = 3 b b. 1 = b b 3 EXAMPLES 3. Find pq,. given tht pq. = p q + p q + p q p = nd 3 ( ) = ( 1 ) + ( ) + ( 3) 3 = = If A is the point ( ) AB.AC. C( 1, 3, 6) B( 1, 4, ) A (, 3, 9) q =. 3, 3, 9, B( 1, 4, ) nd C( 1, 3, 6), clculte We need to use the position vectors of the points: AB = b AC = c 1 = = 1 11 AB.AC = (( 1) ( 3) ) + ( 1 0) + ( 11) ( 15) = = 168. ( ) 1 = = Pge 143 HSN3100

18 1 The Angle Between Vectors The formule for the sclr product cn be rerrnged to give the following equtions, both of which cn be used to clculte θ, the ngle between two vectors.. cosθ = b b or θ b + b + b b cos =. Look bck to the formule for finding the sclr product, given on the previous pges. Notice tht the first eqution is simply rerrnged form of the one which cn be used to find the sclr product. Also notice tht the second simply substitutes b. for the component form of the sclr product. These formule re not given in the exm but cn both be esily derived from the formule on the previous pges (which re given in the exm). EXAMPLES 1. Clculte the ngle θ between vectors p = 3i + 4j k nd q = 4i + j + 3k. 3 p = 4 nd 4 q = 1 3 p q + p q + p q cosθ = p q = = ( 3 4) + ( 4 1) + (( ) 3) ( ) θ = cos 9 6 = (to 1 d. p.) (or rdins (to 3 d. p.)) Pge 144 HSN3100

19 . K is the point ( 1, 7, ) Strt with sketch: L( 3, 3, 4) θ, L( 3, 3, 4) nd ( ) M(, 5,1) M, 5,1. Find ɵ KLM. K ( 1, 7, ) Now find the vectors pointing wy from the ngle: LK = k l = 7 3 = 10, LM = m l = 5 3 = Use the sclr product to find the ngle: ɵ LK.LM cosklm = LK LM = = ( 4 5) + ( 10 ) + ( ( 3) ) 4 + ( 10) + ( ) ( 3) ɵ 1 6 KLM = cos = (to 1 d. p.) (or rdins (to 3 d. p.)) Pge 145 HSN3100

20 3. The digrm below shows the cube OPQRSTUV. The point R hs coordintes ( 4,0,0 ). () Write down the coordintes of T nd U. (b) Find the components of RT nd RU. O (c) Clculte the size of ngle TRU. () From the digrm, T( 0,4,4 ) nd U( 4,4,4 ). z S P T V R U x Q y (b) (c) RT = t r = 4 0 = , RU = u r = 4 0 = RT.RU cos TRU = RT RU = ( 4 0) + ( 4 4) + ( 4 4) ( ) = = 6 1 TRU = cos 6 = (to 1 d. p.) (or rdins (to 3 d. p.)) Pge 146 HSN3100

21 13 Perpendiculr Vectors If nd b re perpendiculr then b. = 0. This is becuse b. = b cosθ = b cos90 ( θ = 90 since perpendiculr) = 0 (since cos90 = 0). Conversely, if b. = 0 then nd b re perpendiculr. EXAMPLES 1. Two vectors re defined s = 4i + j 5k nd b = i + j + k. Show tht nd b re perpendiculr. b. = b + b + b ( 4 ) ( 1) (( 5) ) = + + = = 0. Since b. = 0, nd b re perpendiculr.. 4 PQ = 7 nd RS = 3 nd RS where is constnt. Given tht PQ Since PQ nd RS re perpendiculr, PQ.RS = ( 3) + 7 = = = 0 =. re perpendiculr, find the vlue of. Pge 147 HSN3100

22 14 Properties of the Sclr Product Some useful properties of the sclr product re s follows: b. = b. ( ). b + c = b. + c. (Expnding brckets). =. Note tht these re not given in the exm, so you need to remember them. EXAMPLES 1. In the digrm, p = 3, r = 4 nd q =. Clculte p. ( q + r ). p q r ( ) p. q + r = pq. + pr. = p q cosθ + = = p r cosθ 1 = 3 cos cos 45. In the digrm below, = c = nd b = 3. c 30 b 30 b c. Clculte.( + + ) ( + + ). b c =. + b. + c. = + b cosθ c cosθ 3 = = = + 3 cos30 cos10 = 1. Remember. c= c cosθ since points into θ nd c points wy. Pge 148 HSN3100

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