# Vectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a.

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1 Vectors mesurement which onl descries the mgnitude (i.e. size) of the oject is clled sclr quntit, e.g. Glsgow is 11 miles from irdrie. vector is quntit with mgnitude nd direction, e.g. Glsgow is 11 miles from irdrie on ering of 270. The line joining the origin to the point (3, 4) cn e descried s: (3, 4) Directed line segment Position vector 4 Components Note tht if =, then = - The mgnitude of vector is its length, which cn e determined Pthgors Theorem. The mgnitude of is written s. Emple 1: Determine in the emple ove. If u = 2 2, then u = ddition of Vectors Two (or more) vectors cn e dded together to produce resultnt vector. For emple, we could 3 0 think of ove eing the resultnt vector of the ddition of = nd =. 0 4 In generl: c C C nd If u = nd v =, then u + v = d c d Emple 2: If p + q = r, find r when 2 7 p = nd q = Emple 3: Find vlues of nd such tht = 4-2 u u - v v Sutrction of Vectors Sutrction of vectors cn e considered s going long the second vector in the opposite direction. If u = c nd v =, then u - v = d c d I rson, mended M Dorn & E Mwell: irdrie cdem Mthemtics Deprtment, 2013/14 Pge 66 of 91

2 2 Multipliction Sclr Quntit If we go long twice, the resultnt vector is + = 2. s we hve not chnged direction, it follows tht 2 must e prllel to. If u = k then ku = k If v = ku, then u nd v re prllel Emple 4: If = nd c =, find: 5 ) 3 ) 2 + c c) c 2 1 unit vector is vector with mgnitude = 1. Unit Vectors Emple 5: Find the components of the unit vector prllel to h = 7-24 To find the components of unit vector: Find the mgnitude of the given vector Divide components the mgnitude Position Vectors Consider the vector in the digrm opposite. is the resultnt vector of going long in the opposite direction, followed in the correct direction. So, = - +, i.e.: = - Emple 6: is the point (4, 2), is the point (-5, -3). Find the components of. I rson, mended M Dorn & E Mwell: irdrie cdem Mthemtics Deprtment, 2013/14 Pge 67 of 91

3 Collinerit We hve seen collinerit in the Stright Line chpter; this cn lso e proven using vectors. Emple 7: Prove the points F (6, 1), G ( 4, 4), nd H (-2, 13) re colliner, nd find the rtio in which G divides FH. Section Formul P divides in the rtio 2:3. emining the digrm, we cn find formul for p (i.e. P ). P = + P 3 P 2 In generl, if P divides in the rtio m:n, then: P n p = 1 (n + m) n m m Emple 8: is the point (3, 2) nd is the point (7, 14). Find the coordintes of P such tht P divides in the rtio 1:3. I rson, mended M Dorn & E Mwell: irdrie cdem Mthemtics Deprtment, 2013/14 Pge 68 of 91

4 Three-Dimensionl Vectors The position of point in 3-D spce cn e descried if we dd third coordinte to indicte height. z D G C E M F (5, 4, 3) Emple 9: C DEFG is cuoid, where F is the point (5, 4, 3). Write down the coordintes of the points: ) ) D c) G d) M, the centre of fce FE ll rules for 2D vectors lso ppl to 3D vectors! z In the digrm opposite, = = = z Vectors in 3D cn lso e descried in terms of the three unit vectors i = which re prllel to the,, nd z es respectivel. Emple 10: v = 3i + 2j - 6k. ) Epress v in component form ) Find v 2 1 0, j = , nd k = 0 0 0, 1 The Sclr Product The sclr product is the result of tpe of multipliction of two vectors to give sclr quntit (i.e. numer with no directionl component). For vectors nd, the sclr product (or dot product) is given s: Note: nd point w from the verte. = cos I rson, mended M Dorn & E Mwell: irdrie cdem Mthemtics Deprtment, 2013/14 Pge 69 of 91

5 Emple 11: Find the sclr product in ech cse elow, where = 6 nd = 10. ) ) c) Component Form of the Sclr Product If = nd = 1 2 3, then We cn use this formul to find the sclr product when we hve een given the component forms of the two vectors ut not the ngle etween them.. = Emple 12: is the point (1, 2, 3), (6, 5, 4), nd C (-1, -2, -6). Evlute. C. Perpendiculr Vectors specil cse of the sclr product occurs when we hve perpendiculr vectors, i.e. when θ = 90. In the cse opposite,. = cos90 = 0 = 0 If. = 0, then nd re perpendiculr Emple 13: P, Q, nd R re the points (1, 1, 2), (-1, -1, 0) nd (3,-4, -1) respectivel. Find the components of QP nd QR, nd hence show tht the vectors re perpendiculr. I rson, mended M Dorn & E Mwell: irdrie cdem Mthemtics Deprtment, 2013/14 Pge 70 of 91

6 The ngle etween Two Vectors The first version of the sclr product includes the ngle etween the vectors, so we cn rerrnge this formul to otin: cos θ. Emple 14: = i + 2j + 2k, nd = 2i + 3j - 6k. Find the ngle etween nd. ther Uses of the Sclr Product For vectors,, nd c :. =..( + c ) =. +. c Emple 15: = 5 nd = 8. Find.( + ) 45 Pst Pper Emple 1: The digrm shows squre-sed prmid of height 8 units. Squre C hs side length of 6 units. The coordintes of nd D re (6, 0, 0) nd (3, 3, 8). C lies on the is. ) Write down the coordintes of. z D ) Determine the components of D nd D. C I rson, mended M Dorn & E Mwell: irdrie cdem Mthemtics Deprtment, 2013/14 Pge 71 of 91

7 c) Clculte the size of D. Pst Pper Emple 2: PQRSTU is regulr hegon of side 2 units. PQ, QR, nd RSrepresent the vectors,, nd c respectivel. Find the vlue of. ( c ) U T P S c Q R Vectors: Topic Checklist Topic Questions Done Help? Mgnitude of vector C Eercise 13N, p 247, Q 2 Y/N Y/N Resultnt vectors C Eercise 13N, p 248, Q 5, 6, 9, 11 Y/N Y/N Multipliction sclr C Eercise 13N, p 248, Q 10, 11 Y/N Y/N Unit Vectors C Eercise 13F, p 238, Q 1, 2 Y/N Y/N Position Vectors nd Eercise 13G, p 239, Q 1, 2 Y/N Y/N C Components Eercise 13N, p 249, Q 13, 14 Y/N Y/N Collinerit C Eercise 13N, p 249, Q 15 Y/N Y/N / Eercise 13N, p 249, Q 16 18, 23 Y/N Y/N Section Formul C Eercise 13N, p 249, Q 20 Y/N Y/N / Eercise 13N, p 249, Q 21, 22, 24 Y/N Y/N Sclr Product (ngle) C Eercise 13, p 250, Q 1 Y/N Y/N Sclr Product (component) C Eercise 13P, p 252, Q 1, 2 Y/N Y/N ngle etween vectors C Eercise 13Q, p 253, Q 1, 2 Y/N Y/N / Eercise 13S, p 255, Q 4 7 Y/N Y/N Perpendiculr Vectors C Eercise 13R, p 254, Q 1, 2, 3, 5 Y/N Y/N / Eercise 13R, p 254, Q 4, 6 8 Y/N Y/N Properties of Sclr Product C Eercise 13U, p 257, Q 1, 2, 4, 5 Y/N Y/N I rson, mended M Dorn & E Mwell: irdrie cdem Mthemtics Deprtment, 2013/14 Pge 72 of 91

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