Topic 3 - Plane Vectors - Topic3:PlaneVecctors Scalar&Ve...

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Unformatted text preview: Topic 3:: Plane Vecctors Scalar & Ve ector quanttities Properties o P of Vector q quantities Graphical re G epresentattion of vecctors Unit vector U s , , Sum and diffference of vector qu uantities Dot & cross D s product & Vector Q Quantities: Scalar & The mathematical quantities that are used to deescribe thee motion of objects ccan ded into tw wo categories. The qu uantity is eeither a vecctor or a sccalar. Thesse be divid two categories can be distin nguished frrom one annother by ttheir distin nct definitiions: Scalars are quantitiess that are fully described by a m magnitude (or numerrical value) alone e. Vectors are V e quantities that are ffully descrribed by bo oth a magn nitude and a direction. Vectors have magnitude and direction, scalars only have magnitude. The fact that magnitude occurs for both scalars and vectors can lead to some confusion. There are some quantities, like speed, which have very special definitions for scientists. By definition, speed is the scalar magnitude of a velocity vector. A car going down the road has a speed of 50 mph. Its velocity is 50 mph in the northeast direction. It can get very confusing when the terms are used interchangeably! Another example is mass and weight. Weight is a force which is a vector and has a magnitude and direction. Mass is a scalar. Weight and mass are related to one another, but they are not the same quantity. Mass and energy are scalar quantities, while momentum is a vector quantity. Scalar Quantities A scalar quantity is defined as a quantity that has magnitude only. Typical examples of scalar quantities are time, speed, temperature, and volume. A scalar quantity or parameter has no directional component, only magnitude. For example, the units for time (minutes, days, hours, etc.) represent an amount of time only and tell nothing of direction. Additional examples of scalar quantities are density, mass, and energy. Vector Quantities A vector quantity is defined as a quantity that has both magnitude and direction. To work with vector quantities, one must know the method for representing these quantities. Magnitude, or "size" of a vector, is also referred to as the vector's "displacement." It can be thought of as the scalar portion of the vector and is represented by the length of the vector. By definition, a vector has both magnitude and direction. Direction indicates how the vector is oriented relative to some reference axis. Using north/south and east/west reference axes, vector "A" is oriented in the NE quadrant with a direction of 45°. Giving direction to scalar "A" makes it a vector. The length of "A" is representative of its magnitude or displacement. The vector A can be represented by a single letter in bold type or with an arrow above the letter, such as or Propertties of Vector quantiities: Unit Ve ector: A unit vvector is a vector pointing in a ggiven direcction with a magnitude of one.. Essentiaally, it merrely indicattes directio on. In a Carrtesian sysstem the th hree unit vectors are called i, j, and k can also be represennted as iˆ, ˆj and kˆ . In choosingg i, j, once i and j are chose en, k must point to aa particularr direction,, so that a and k, o commo on convention called "right‐han nd rule" hoolds. Matheematically,, this can b be compacctly expressed as, kˆ iˆ ˆj Unit vecctors are ggenerally ch hosen to b be orthogo nal. That iss, each unit vector iss perpend dicular to e each of the e others. Vector C Componen nts: Every ve ector may be expressed as the sum of itss n unit vecctors. A a x iˆ a y ˆj a z kˆ The quaantities a x , a y and ponents of vector A. a z are called the ve ctor comp Sometim mes they aare represe ented simp ply as an orrdered trip plet. e.g. a x , a y , a z . Vector A Algebra: Negatio on ering a vector represe ented grap phically by an arrow, the negattive of a veector Conside would b be represented by a vvector of the same leength but o opposite d direction. Vector n negation o of A is presented as A (a x iˆ a y ˆj a z kˆ) a x iˆ a y ˆj a z kˆ of a vector:: Norm o The disttance betw ween the in nitial pointt and the teerminal po oint of a veector v is called the norm or the magn nitude of vv and is dennoted by v . n 2‐space is given by The norrm of the vvector v v1 , v 2 in v v12 v22 And the e norm of aa vector v v1 , v2 , v3 in 3‐space is giveen by v v12 v 22 v32 Normallizing a vecctor: We can find a unitt vector that has th he same diirection ass some giveen nonzero o vector and is givven by u v 1 v v v Example e 1: Find th he unit vecctor that has the sam me directio on as v 2i 2 j k Solution n: The vecttor has le ength v 22 22 (1) 2 3 So the u unit vectorr in the same directtion as iss u 1 1 2 2 v i j k 3 3 3 3 Example e 2: Given v 3,1,2 find a unit vecttor that, (a a) Points in n the same e direction as (b b) Points in n the oppo osite directtion as Solution n: (a) The vector h has length v 32 (1) 2 (2) 2 14 The unit vector th hat will point in the saame directtion as iss given by v 3 1 2 , , u v 14 14 14 (b) Sincce u is a un nit vector tthat pointss in the sam me directio on as theen its negative will be aa unit vector that points in the opposite ddirections as . So th he unit vecctor which p points in th he opposite e direction n of is givven by v 3 1 2 , , u v 14 14 14 Multiplication Scalar M kA ka x iˆ ka y ˆj ka z kˆ negation iss merely m multiplicatio on by a scaalar, wheree that scalaar is Note that vector n nted graphically woulld point in the same direction aas ‐1. A scaaled vector represen the origginal vector but have its magnittude scaledd by a facttor of k. Addition of Vecto ors: Illustrattion of head‐to‐tail addition A B (a x iˆ a y ˆj a z kˆ) (b x iˆ b y ˆj bz kˆ) (a x bx )iˆ (a y b y ) ˆj (a z bz ) kˆ Two vecctors can b be added ggraphically by placingg the tail off the secon nd vector (here, B B) coincidental with tthe tip of the first vecctor (A). Th he resultan nt vector A A + B is the ve ector draw wn from the e tail of A tto the tip o of B. Any num mber of ve ectors can be added iin this mannner. Vector aaddition is commutative: A B C C B A Also Vector additiion is assocciative: ( A B) C A ( B C ) oduct: Dot Pro When w we multiplyy two vectors, we can either appply a multtiplication rule that produce es a scalar as the end d result, orr one that pproduces aa vector ass the end result. TThe first on ne that pro oduces a sccalar is callled dot pro oduct. In m mathematiical texts, th his is often n called inn ner productt, and som me older texxts will reffer to this aas scalar p product); th hey are all the same. Dot produuct has all the usual propertiess of productts, such as associativity, commutativity, aand the disstributive p property. Geomettrically, do ot product is defined as: If u and v are two vectors in n 2‐space o or 3‐space and is tthe angle b between them th hen the dot product, denoted b by ∙ is defined ass, ∙ ‖ ‖‖ ‖ ccos e it as, We can also write coss ∙ Wheree 0 ‖ ‖‖ ‖ ndard unit vectors arre Note that the stan 0,0,1 0 1,,0,0 , 0,1,0 and iˆ iˆ ˆj ˆj kˆ kˆ 1 iˆ ˆj ˆj kˆ kˆ iˆ 0 his, we can n define do ot product in terms o of componeent vectors as follow ws: Using th A B ( Ax iˆ Ay ˆj Az kˆ) ( Bx iˆ B y ˆj Bz kˆ ) Ax Bx Ay B y Az Bz Note that the dot product iss sometime es called thhe scalar p product or the Euclid dean inner prroduct. Example 1: Compute u v and also the e angle which is fo ormed by tthe two vectors u & v: (a) 3, 1,6 4,2,0 Solution n: u v (3)( ) 4) ( 1)((2) (6)(0) 10 u (3) 2 (1) 2 (6) 2 46 v (4) (2) (0) 20 2 Now, cos (b) uv u v 9, 2 2 2 10 0.3297 3 cos 1 (0.33297) 700.75 46 20 4,18 Solution: u v (9)(4) ( 2)(18) 0 u (9) 2 (2) 2 85 v (4) (18) 340 2 Now, cos 2 0 u v 0 cos 1 (0) 90 u v 85 340 Note: If the angle between two vectors is 90 , then such perpendicular vectors are called orthogonal. 1. Two non‐zero vectors u & v are orthogonal if and only if u v 0 . 2. If u v 0 , then the angle between the two vectors is Acute. 3. If u v 0 , then the angle between the two vectors is Obtuse. The orthogonal projection of u on a(vector component of u along a) is denoted by proja u and is given by projau ua a 2 a And the vector component of u orthogonal to a is given by u projau u ua a 2 a Example 2: Compute the orthogonal projection of u on a and the vector component of u orthogonal to a for the following. 3,1 & 7,2 u a (3)(7) (1)(2) (19) Solution n: a 7 2 53 2 2 2 Hence, the orthoggonal proje ection of u on a is proj a u ua a 2 a 19 133 38 7,2 , 53 53 53 ector component of u orthogonal to a is & the ve u proj a u u ua a 2 a 3,1 133 38 26 91 , , 553 553 53 53 Cross Product: Cross prroduct is also known n as outer p product orr vector pro oduct. Thee product ccan be defin ned with th he two rule es, first spe ecifying thhe product vector’s direction an nd the seco ond specifyying its maagnitude. 1. A B is perpendiculaar to A and d B (that iss, perpend dicular to the plane defined by these two vectors). TThis leavess two possible directions along the line perpen ndicular to the plane. One of the two direections is called by a d rule": Ho old out inde ex finger, m middle fingger, and th he thumb sso "right‐hand th hat they arre all perpe endicular tto each othher. Let the index fin nger point to owards dirrection of A , and the middle ffinger towaards B . Th hen the thu umb points towa p ards the direction of A B . Thee orderingg is importaant here (n note exchangingg A and B m makes the tthumb point in the o opposite direction). & v are two non‐zero o vectors aand is th he angle beetween thee 2. Suppose u & tw wo vectorss then, u v u v sinn Note: U Using the sstandard unit vectorss in , wee have iˆ ˆj ˆj iˆ kˆ ˆj kˆ kˆ ˆj iˆ kˆ iˆ iˆ kˆ ˆj iˆ iˆ ˆj ˆj kˆ kˆ 0 omponents, we have e And in tterms of co A B ( Ay Bz B y Az )iˆ ( Az Bx Bz Ax ) ˆj ( Ax B y Bx Ay )kˆ It turns out we can write thiis complicaated relatio onship as aa determin nant of a 3 x 3 matrix: . Some properties o of cross product, succh as A B B A and A A 0 can be erty of the e determinant of the matrix. derived as a prope Example 1: Find u v for 4, 9,1 9 3, 2,7 Solution n: j k i uv 4 9 1 3 2 7 i (63 6 2) j (28 3) k (8 27) (61)i 25 j 19k u v 61,25,19 Example 2: If 1, 1,0 and 0, 2,0 the angle between the two given vectors. then find u v and also find Solution: i j k u v 1 1 0 0 2 0 i (0) j (0) k (2 0) 0i 0 j 2k 0,0,2 u Also, v 12 (1)2 (0)2 2 02 (2)2 (0)2 2 u v (0) 2 (0) 2 (2) 2 2 sin uv 2 1 u v 2 2 2 1 sin 1 45 2 Properties of Cross Product: 1. u v (v u ) Example: Let 3, 1,4 Prove that u v (v u ) Solution: We have and 3, 1,4 2,0,1 and . Find u v and v u . Also 2,0,1 . i j k u v 3 1 4 2 0 1 i (1 0) j (3 8) k (0 (2)) i 5 j 2k 1,5,2 Now, i j k vu 2 0 1 3 1 4 i (0 (1)) j (8 3) k ((2) 0) i 5 j 2k 1,5,2 Hence, we can say that u v (v u ) 2. u 0 0 u 0 3. u u 0 Example: Let Solution: . Prove that u u 0 3, 1,4 i j k u u 3 1 4 3 1 4 i (4 (4)) j (12 12) k (3 (3)) 0i 0 j 0k 0,0,0 0 4. u (u v) 0 & v (u v ) 0 3, 1,4 Example: Let and 2,0,1 . Prove that u (u v) 0 & v (u v) 0 Solution: We have already calculated above that u v 1,5,2 u (u v) (3)(1) ( 1)(5) ( 4)(2) 0 Now, v (u v) ( 2)(1) (0)(5) (1)(2) 0 Hence, u (u v) 0 & v (u v) 0 5. u (v w) (u w) v (u v) w 6. (u v) w (u w) v (v w) u 7. u (v w) (u v) (u w) 8. u (v w) (u v) (u w) 9. (u v) w (u w) (v w) 10. c (u v) (cu ) v u (cv) 11. Lagrange’s Identity: 2 2 2 u v u v (u v) 2 Directio on Cosiness: Let us start with aa vector , iin three dimensionall spaces. This vector will form angles w with ( ), th he ( ) and tthe ( ). Theese angles are called d direction an ngles and tthe cosiness of these angles aree called direection cosines. Here is a sketch of aa vector an nd the direection anglles. mulas for tthe directio on cosiness of a vecto or The form , . ‖ ‖ ‖ ‖ . ‖ ‖ . , ‖ ‖ ‖ ‖ ‖ ‖ Here . , , . 1 1,0,0 . , , . 0,1,0 . , , . 0 0,0,1 are, Note: , 1. The vector 2. , is a unit vector. 1 Example: Determine the direction cosine and direction angles of the vector 2,1, 4 . Solution: First we will find the magnitude of the vector ‖ ‖ 2 1 4 √21 The direction cosines and the direction angles are given by, 2 ∴ 64.12° ∴ 77.39° ∴ 150.79° √21 1 √21 4 √21 POSITION VELOCITY ACCELERATION 2‐Space r (t ) x(t )i y (t ) j 3‐Space r (t ) x(t )i y (t ) j z (t ) k v(t ) dx dy i j dt dt v(t ) dx dy dz i j k dt dt dt a(t ) d 2x d 2 y i 2 j dt dt 2 a(t ) d 2x d 2 y d 2z k i j dt 2 dt 2 dt 2 SPEED 2 dx dy v(t ) dt dt 2 2 2 dx dy dz v (t ) dt dt dt 2 Example 1: A particle moves along a circular path in such a way that its ‐ and co‐ordinates at time are x 2 cos t & y 2 sin t Find the instantaneous velocity and speed of the particle at time t. Solution: At time , the position vector is r (t ) ( 2 cos t )i ( 2 sin t ) j So the instantaneous velocity and speed are v(t ) dr (2 sin t )i (2 cos t ) j dt v(t ) (2 sin t ) 2 (2 cos t ) 2 2 Example 2: A particle moves through 3‐space in such a way that its velocity is v(t ) i (t ) j (t 2 )k Find the coordinates of the particle at time =1 given that the particle is at the point 1,2,4 at time =0. Solution: First we shall integrate velocity function in order to obtain the position vector. t2 t3 r (t ) v (t ) dt (i tj t k ) dt (t )i 2 j 3 k C ‐‐‐‐‐‐‐‐‐‐‐‐‐ Equation 1 2 Since the coordinates of the particle at time vector at time 0 is 0 are 1,2,4 , the position r (0) i 2 j 4k Substituting 0 in equation 1 gives, r (0) (0)i (0) j i 2 j 4k ( 0) k C C ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Equation 2 Substituting the value of from equation 2 in equation 1, we get t2 t3 r (t ) (t )i j k i 2 j 4k 2 3 t3 t2 r (t ) (t 1)i 2 j 4 k 3 2 Thus, at time t=1 the position vector of the particle is r (1) 0i 5 13 j k 2 3 So its coordinates at that instant are 0, 52 , 133 . Example 3: Suppose that a particle moves along a circular helix in 3‐Space so that the position vector at time is r (t ) ( 4 cos t )i ( 4 sin t ) j tk Find the distance travelled and the displacement of the particle during the time interval 1 t 5 . Solution: We have, v(t ) dr (4 sin t )i (4 cos t ) j k dt v(t ) (4 sin t ) 2 (4 cos t ) 2 1 16 2 1 The distance travelled by the particle from time 1 to 5 is 5 s (t ) 16 2 1dt 4 16 2 1 1 The displacement over the time interval is r r (5) r (1) (4 cos 5i 4 sin 5j 5k ) (4 cos i 4 sin j k ) (4i 5k ) (4i k ) 4k This tells us that the change in the position of the particle over the time interval was 4 units straight up. Microsoft Equation 3.0 ...
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  • Spring '16
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