TRIGONOMETRY MATH 117
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This 66 page Class Notes was uploaded by Sadie Schroeder on Wednesday September 30, 2015. The Class Notes belongs to MATH 117 at Western Kentucky University taught by Jemal Gishe in Fall. Since its upload, it has received 15 views. For similar materials see /class/216737/math-117-western-kentucky-university in Applied Mathematics at Western Kentucky University.
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Date Created: 09/30/15
VECTORS AND DOT PRODUCTS Copyright Cengage Learning All rights reserved hat You Should Learn Find the dot product of two vectors and use the properties of the dot product Find the angle between two vectors and determine whether two vectors are orthogonal Write a vector as the sum of two vector components Use vectors to find the work done by a force he Dot Product of Two Vectors In this section you will study a third vector operation the dot product This product yields a scalar rather than a vector Definition of the Dot Product The do product of u 1117 141 and v v1 v is u v 141134 142v he Dot Product of Two Vectors Properties of the Dot Product Let LL v and w be vectors in Ihe plane or in space and let 6 be a scalar l 11 V v u 2 0 v 0 311 vwuvu w 4 v 4 v Hsz 5 L39l vcu 39 v u L V xample 1 Finding Dot Products Find each dot product a 4 5 2 3 b 2 1 1 2 c 0 3 4 2 Solution a 4 5 2 3 42 53 8 15 23 b 2 1 1 2 21 12 xample 1 Solution cont d I c 0 3 4 2 04 3 2 0 6 6 he Dot Product of Two Vectors In Example 1 be sure you see that the dot product of two vectors is a scalar a real number not a vector Moreover notice that the dot product can be positive zero or negative xample 2 Using Properties of Dot Products Let u 13v 2 4 and w 1 2 Find each dot product au vw b u 2v Solution Begin by finding the dot product of u and v u V 1 32 4 12 34 14 xample 2 Solution cont d I a u vw 141 2 14 28 bu2v2uv 2 14 28 Notice that the product in part a is a vector whereas the product in part b is a scalar he Angle Between Two Vectors The angle between two nonzero quotu vectors is the angle 6 0 S 6 S 7139 u between their respective standard quot position vectors as shown in 0 Figure 333 Figure 333 This angle can be found using the dot product Angle Between Two Vectors If H is the angle between two nonzero vectors 1 and v then n V cm 6 7 M M xample 4 Finding the Angle Between Two Vectors Find the angle Hbetween u 4 3 and v 3 5 Solution The two vectors and Bare shown in Figure 334 Figure 334 xample 4 Solution cont d I iluil HVH 43 35 iilt4 3gtii iilt355gtli 27 534 This implies that the angle between the two vectors is C086 6 27 arccos 5 34 N 222quot he Angle Between Two Vectors Rewriting the expression for the angle between two vectors in the form u V u cos 6 Alternative form of clot product produces an alternative way to calculate the dot product From this form you can see that because and are always positive u v and cos 6will always have the same sign he Angle Between Two Vectors Figure 335 shows the five possible orientations oftwo vectors II H u 9 39 0 41 u vk 6 7r lt e lt Tr u 1 l cos 9 ti Opposite Diremnn obtusemme 900 Anglp u H v v u o lt e lt g 0 0 lt 1 ms 6 7 i Acute Angie Same iiireniun Figure 335 he Angle Between Two Vectors De nition of Orthogonal Vectors The vectors 1 and v are orthogonal if u V 0 The terms orthogonal and perpendicular mean essentially the same thing meeting at right angles Note that the zero vector is orthogonal to every vector u because 0 u 0 inding Vector Components You have already seen applications in which two vectors are added to produce a resultant vector Many applications in physics and engineering pose the reverse problem decomposing a given vector into the sum of two vector components Consider a boat on an inclined ramp as shown in Figure 337 The force F due to gravity pulls the boat down the ramp and against the ramp Figure337 inding Vector Components These two orthogonal forces w1 and wz are vector components of F That is F W1 W2 Vector components of F The negative of component w1 represents the force needed to keep the boat from rolling down the ramp whereas w2 represents the force that the tires must withstand against the ramp inding Vector Components A procedure for finding w1 and w2 is shown as below DEf un ul Veuur Cnmpununts Ler u and v he nnnmn melons such rhm urw w when m and w are cnhogmml and w parallel Io or a scam mumplc on 2 us shmvn m Frgum 338111 vmms w and w are called vecmr mmponems m u The vcclm m in m pmjrction ufu unto v and As dcnmcd by w pm The mm w Is gwrn by w n W 7 5 cu V new 553 c B Is amuse inding Vector Components From the definition of vector components you can see that it is easy to find the component w2 once you have found the projection of u onto v To find the projection you can use the dot product as follows u W1 W2 CV W2 w1 is a scalar multiple of v u v cv wz v Take dot product of each side with v cvvw2v inding Vector Components cl I V I I2 0 w2 and v are orthogonal C W uV w1 p1 olvu v Wv Projection of u onto v Let u and v be nonzero Vectors The projection of u onto v is ll V pl UJv H quotH xample 6 Decomposing a Vector into Components Find the projection of u 3 5 onto v 6 2 Then write u as the sum of two orthogonal vectors one of which is projvu Solution The projection of u onto v is w prnjvu 2mm xample 6 Solution lt gt as shown in Figure 339 cont d I 4 w 5 Figure 339 xample 6 Solution The other component wz is w2uw1 u w1 W2 80 3 lt25 gt lt 33 3 5V cont d I The work Wdone by a constant force F acting along the line of motion of an object is given by W magnitude of forcedistance HFH HE H as shown in Figure 341 p Force acts along the line of motion Figure 341 Ifthe constant force F is not directed along the line of motion as shown in Figure 342 P Q Force acts at angle 9 with the line of motion Figure 3 42 the work Wdone by the force is given by W IPTOJ H F Projection form for work cos 6ilFli WM upmwr mum ork F Alternative form of clot product This notion of work is summarized in the following definition Definition of Work The work W done by 21 mustant force F as its point if application moves ulmig thc vcclar m is given by either of the following iii ll 1 W iiproi p F 2 W F E Projcutiun 1mm Uni iniimici mm xample 8 Finding Work To close a sliding barn door a person pulls on a rope with a constant force of 50 pounds at a constant angle of 60 as shown in Figure 343 Find the work done in moving the barn door 12 feet to its closed position Figure 343 xample 8 Solution Using a projection you can calculate the work as follows W Hproj PQ FH Projection form forwork cos 60 llFllllmll 1 50 12 2 i gt Z 300 foot pounds So the work done is 300 footpounds You can verify this result by finding the vectors F and w and calculating their dot product VECTORS IN THE PLANE Copyright Cengage Learning All rights reserved hat You Should Learn Represent vectors as directed line segments Write the component forms of vectors Perform basic vector operations and represent them graphically Write vectors as linear combinations of unit vectors Find the direction angles of vectors Use vectors to model and solve reallife problems Quantities such as force and velocity involve both magnitude and direction and cannot be completely characterized by a single real number To represent such a quantity you can use a directed line segment as shown in Figure 315 Terminal point A PQ Initial pOiDL Figure 315 The directed line segment W has initial point P and terminal point Q Its magnitude or length is denoted by ll ll and can be found using the Distance Formula Two directed line segments that have the same magnitude and direction are equivalent For example the directed line segments in Figure 316 are all equivalent Figure 316 The set of all directed line segments that are equivalent to the directed line segment E is a vector v in the plane written V PQ Vectors are denoted by lowercase boldface letters such as u v and w uxample 1 Vector Representation by Directed Line Segments Let u be represented by the directed line segment from P0 0 to Q3 2 and let v be represented by the directed line segment from R1 2 to 84 4 as shown in Figure 317 Show that u and v are equivalent Figure 317 xample 1 Solution From the Distance Formula it follows that FAQ and have the same magnitude HPQQH m ll 4 1 4 7 22 m MoreOVer both line segments have the same direction because they are both directed toward the upper right on lines having a slope of 472270 Because and If have the same magnitude and direction u and v are equivalent Component Form of a Vector omponent Form of a Vector The directed line segment whose initial point is the origin is often the most convenient representative of a set of equivalent directed line segments This representative of the vector v is in standard position A vector whose initial point is the origin 0 0 can be uniquely represented by the coordinates of its terminal point v1 v2 This is the component form of a vector v written as v v1 v2 omponent Form of a Vector The coordinates v1 and v2 are the components of v If both the initial point and the terminal point lie at the origin v is the zero vector and is denoted by 0 0 0 Component Form of a Vector The component form of the vector with initial point Plp 12 and terminal point qui I is given by PQ qt 1qu 1I2gt tnw v The magnitude or length of v is given by W T xqi pl 412 i 172 m t 13922 Tt39 4 l V is 1 lmitveetor Moreover HvH i it39 and only it39v 15 the 7am Vector L omponent Form of a Vector Two vectors u U1 u2 and v v1 v2 are equal if and only if u1 v1 and U2 v2 For instance in Example 1 the vector u from P0 0 to Q3 2 is u I 3 02 0 3 2 and the vector v from R1 2 to 84 4 isv 4 1 4 23 2 xample 2 Finding the Component Form of a Vector Find the component form and magnitude ofthe vector v that has initial point 4 7 and terminal point 1 5 Solution Let P4 7 131132 and Q1 5 C11 12 Then the components of v v1 v2 are V1q1 P1 1 4 5 v2 q2 p25 7 12 xample 2 Solution 80 v 5 12 and the magnitude of v is Hvll 169 2 l3 cont d I ector Operations The two basic vector operations are scalar multiplication and vector addition In operations with vectors numbers are usually referred to as scalars In this section scalars will always be real numbers Geometrically the product of a vector v and a scalar k is the vector that is k times as long as v ector Operations Ifk is positive kv has the same direction as v and if k is negative kv has the direction opposite that ofv as shown in Figure 319 To add two vectors u and v geometrically first position them without changing their lengths or directions so that the initial point of the second vector v coincides with the terminal point of the rst vector u Figure 319 ector Operations The sum u v is the vector formed by joining the initial point ofthe first vector u with the terminal point of the second vector v as shown in Figure 320 Figure 320 ector Operations This technique is called the parallelogram law for vector addition because the vector u v often called the resultant of vector addition is the diagonal of a parallelogram having adjacent sides u and v Definitions of Vector Additinn and Scalar Multiplication Let u uh Lil and v tilt v1 he vetlurs and let k be 2 53le a real uumhen Then the mm of u and v is the vector u v i u v ut w hum and the vralur mullisz of k limes u is the Vector ku 1cm 1 km kill s ilr Inulitpln39 ector Operations The negative of v v1 v2 is v 1v V1 V2 Negative and the difference of u and v is u v u v Add v See Figure 321 u1 v1 u2 v2 Difference u vu v Figure 321 18 ector Operations To represent u v geometrically you can use directed line segments with the same initial point The difference u v is the vector from the terminal point of v to the terminal point of u which is equal to u v as shown in Figure 321 The component definitions of vector addition and scalar multiplication are illustrated in Example 3 In this example notice that each of the vector operations can be interpreted geometrically xample 3 Vector Operations Let v 2 5 and w 3 4 and find each of the following vectors a 2v b w v Solution a Because v 2 5 you have 2v 2 2 5 22 25 4 10 A sketch of 2v is shown in Figure 322 cv2w l 4 Ill 444 7 2 Figure 322 xample 3 Solution cont d I b The difference of w and v is w v3 4 2 5 3 2 4 5 5 1 A sketch ofw v is shown in Figure 323 Note that the figure shows the vector difference w v as the sum w v 39 s in Figure323 xample 3 Solution cont d I c The sum of v and 2w is v 2w 2 5 23 4 2 5 23 24 2 5 5 8 2 6 5 8 4 13 A sketch of v 2w is shown in Figure 324 391 4 n Figure 32 ector Operations Vector addition and scalar multiplication share many ofthe properties of ordinary arithmetic Properties of Vector Addition and Scalar Multiplication Let l7 V and W be vectors and 161 C and d be scalars Then the following properties an truc luvvu 2uvwnvw 3u0u 4uu0 5 Cdu Uu 6 C Nu Cu Llu 7 Cu v cu cv 8 1ul u OLul 0 1 llcvll lcl llvll Property 9 can be stated as follows the magnitude of the vector cv is the absolute value of c times the magnitude of v 23 In many applications of vectors it is useful to find a unit vector that has the same direction as a given nonzero vector v To do this you can divide v by its magnitude to obtain u unit vector L Unit vector in direction ofv llvll llvll Note that u is a scalar multiple ofv The vector u has a magnitude of 1 and the same direction as v The vector u is called a unit vector in the direction of v xample 4 Finding a Unit Vector Find a unit vector in the direction ofv 2 5 and verify that the result has a magnitude of 1 Solution The unit vector in the direction of v is L i M 22 52 L2 3 J29 xample 4 Solution This vector has a magnitude of 1 because 72 2 5 2 4 25 W lt gt VEE 29 29 cont d I The unit vectors 1 0 and 0 1 are called the standard unit vectors and are denoted by i1 0andj0 1 as shown in Figure 325 Note that the lowercase letter i is 1 H D written in boldface to distinguish iltiD it from the imaginary number quot 7 1 Figure325 These vectors can be used to represent any vector v v1 v2 as follows VV11V2 V1110 V2411 V1i V2 The scalars v1 and v2 are called the horizontal and vertical components of v respectively nit Vectors The vector sum V1i V2 is called a linear combination ofthe vectors i and j Any vector in the plane can be written as a linear combination of the standard unit vectors i and j irection Angles If u is a unit vector such that His the angle measured counterclockwise from the positive X axis to u the terminal point of u lies on the unit circle and you have u X y cos 6 sin 639 cos 6i sin 6 as shown in Figure 327 The angle His the direction angle ofthe vector u u 1 Figure 327 irection Angles Suppose that u is a unit vector with direction angle 6 va ai bj is any vectorthat makes an angle 6with the positive X axis it has the same direction as u and you can write v vcos 6 sin 6 Vcos 639i Vsin 6 irection Angles Because v ai bj vcos 6i vsin 6 it follows that the direction angle Hfor v is determined from 5m 9 Quotient identity cos 9 tan 6 Multiply numerator and denominator by v H v cos 739 Simplify uxample 7 Finding Direction Angles of Vectors Find the direction angle of each vector a u 3i 3 b v 3i 4 Solution a The direction angle is b 7 2 an 49 7 E 7 3 1 So 6 45 as shown in Figure 328 1 Z 3 Figure 323 ixample 7 Solution mm b The direction angle is I 7 4 t 0 7 39 an a 3 Moreover because v 3i 4 lies in Quadrant IV 6ies in Quadrant IV and its reference angle is lt4 11 All 3 a 753V13 9 53130 xample 7 Solution cont d I So it follows that 6 360 5313quot 30687 as shown in Figure 329 4 r3 74 0 Figure 329 xample 8 Finding the Component Form of a Vector Find the component form of the vector that represents the velocity of an airplane descending at a speed of 150 miles per hour at an angle 20 below the horizontal as shown in Figure 330 Figure 330 xample 8 Solution The velocity vector v has a magnitude of 150 and a direction angle of 6 200 v vcos 6M vsin 6 150cos 200 i 150sin 200 j e 150 09397i 150 03420j a 14096i 5130 14096 5130 xample 8 Solution cont d I You can check that v has a magnitude of 150 as follows HVH m m z 150
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