In Exercises 13 - 16, find the standard equation of the circle which satisfies the given criteria. And graph it. 13. center (3, 5), passes through (1, 2) 14. center (3, 6), passes through (1, 4) In Exercises 9 - 14, put the equation in standard form. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, the foci and the eccentricity. And graph it. 13. 9x 2 + 4y 2 4y 8 = 0 14. 6x 2 + 5y 2 24x + 20y + 14 = 0 Name_____________________ TRIG QUIZ 5 Date ________________ You must include the following dated statement with your name typed in lieu of a signature. Without this signed statement you will receive a zero. I have completed this quiz myself, working independently and not consulting anyone except the instructor. I have neither given nor received help from any outside source whatsoever on this quiz. Signed _______________________ SHOW ALL WORK !! NO WORK = NO CREDIT !! NO KIDDING !! Submit PDF only - no WORD Check \"Ask the Professor\" to ask any questions and also for any updates or hints on the quiz. 2 problems plus one extra credit problem Leave space put one problem per page #1, 50 pts, pg 958 #11.5.25, solve exactly as requested in the book and then check your solution with desmos, Show all work !! In Exercises 21 - 30, find the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin). 25. r = 2 cos() and r = 2 3 sin() #2, 50 pts, pg 1030 #11.8.62, include a labeled sketch of all forces and the resultant force, Show all work !! 62. Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie 'Ben-Hur' by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N80W, the second points due west and the third points S80W. The force applied to the first rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds otherwise the couch won't move. Does it move? If so, is it heading due west? Ex cr: 10 points, pg 1043 #11.9.16, Show all work !! In Exercises 1 - 20, use the pair of vectors ~v and ~w to find the following quantities. 16. ~v = 5 + 12 and ~w = 3 + 4 General notes: 3 decimal place accuracy everywhere ex: 7856.123 Submit *one* PDF only - no WORD Make sure everything is *clearly visible* and *in order* and nothing is chopped off. Section 11.8: Vectors, from College Trigonometry: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 license. 2013, Carl Stitz. 1012 11.8 Applications of Trigonometry Vectors As we have seen numerous times in this book, Mathematics can be used to model and solve real-world problems. For many applications, real numbers suffice; that is, real numbers with the appropriate units attached can be used to answer questions like \"How close is the nearest Sasquatch nest?\" There are other times though, when these kinds of quantities do not suffice. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. (Foreshadowing the use of bearings in the exercises, perhaps?) To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vectors.1 A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. When referring to vectors in this text, we shall adopt2 the 'arrow' notation, so the symbol ~v is read as 'the vector v'. Below is a typical vector ~v with endpoints P (1, 2) and Q (4, 6). The point P is called the initial point or tail of ~v and the point Q is called the terminal point or head of ~v . Since we can reconstruct ~v completely from P and Q, we write ~v = P Q, where the order of points P (initial point) and Q (terminal point) is important. (Think about this before moving on.) Q (4, 6) P (1, 2) ~v = P Q While it is true that P and Q completely determine ~v , it is important to note that since vectors are defined in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as ~v is considered to be the same vector as ~v , regardless of its initial point. In the case of our vector ~v above, any vector which moves three units to the right and four up3 from its initial point to arrive at its terminal point is considered the same vector as ~v . The notation we use to capture this idea is the component form of the vector, ~v = h3, 4i, where the first number, 3, is called the x-component of ~v and the second number, 4, is called the y-component of ~v . If we wanted to reconstruct ~v = h3, 4i with initial point P 0 (2, 3), then we would find the terminal point of ~v by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point Q0 (1, 7), as seen below. 1 The word 'vector' comes from the Latin vehere meaning 'to convey' or 'to carry.' Other textbook authors use bold vectors such as v. We find that writing in bold font on the chalkboard is inconvenient at best, so we have chosen the 'arrow' notation. 3 If this idea of 'over' and 'up' seems familiar, it should. The slope of the line segment containing ~v is 43 . 2 11.8 Vectors 1013 Q0 (1, 7) up 4 P 0 (2, 3) over 3 ~v = h3, 4i with initial point P 0 (2, 3). The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in our definition below. Definition 11.5. Suppose ~v is represented by a directed line segment with initial point P (x0 , y0 ) and terminal point Q (x1 , y1 ). The component form of ~v is given by ~v = P Q = hx1 x0 , y1 y0 i Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, hv1 , v2 i = hv10 , v20 i if and only if v1 = v10 and v2 = v20 . (Again, think about this before reading on.) We now set about defining operations on vectors. Suppose we are given two vectors ~v and w. ~ The sum, or resultant vector ~v + w ~ is obtained as follows. First, plot ~v . Next, plot w ~ so that its initial point is the terminal point of ~v . To plot the vector ~v + w ~ we begin at the initial point of ~v and end at the terminal point of w. ~ It is helpful to think of the vector ~v + w ~ as the 'net result' of moving along ~v then moving along w. ~ w ~ ~v + w ~ ~v ~v , w, ~ and ~v + w ~ Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines.4 Example 11.8.1. A plane leaves an airport with an airspeed5 of 175 miles per hour at a bearing of N40 E. A 35 mile per hour wind is blowing at a bearing of S60 E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. 4 If necessary, review page 905 and Section 11.3. That is, the speed of the plane relative to the air around it. If there were no wind, plane's airspeed would be the same as its speed as observed from the ground. How does wind affect this? Keep reading! 5 1014 Applications of Trigonometry Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we've seen a few times before in this textbook.6 We let ~v denote the plane's velocity and w ~ denote the wind's velocity in the diagram below. The 'true' speed and bearing is found by analyzing the resultant vector, ~v + w. ~ From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of ~v , which is 175, the magnitude of w, ~ which is 35, and the magnitude of ~v + w, ~ which we'll call c. From the given bearing information, we go through the usual geometry to determine that the angle between the sides of length 35 and 175 measures 100 . N N ~v 35 ~v + w ~ 100 40 175 40 E 60 w ~ c E 60 p From the Law of Cosines, we determine c = 31850 12250 cos(100 ) 184, which means the true speed of the plane is (approximately) 184 miles per hour. To determine the true bearing of the plane, we need to determine the angle . Using the Law of Cosines once more,7 we find 2 cos() = c +29400 so that 11 . Given the geometry of the situation, we add to the given 40 350c and find the true bearing of the plane to be (approximately) N51 E. Our next step is to define addition of vectors component-wise to match the geometric action.8 Definition 11.6. Suppose ~v = hv1 , v2 i and w ~ = hw1 , w2 i. The vector ~v + w ~ is defined by ~v + w ~ = hv1 + w1 , v2 + w2 i Example 11.8.2. Let ~v = h3, 4i and suppose w ~ = P Q where P (3, 7) and Q(2, 5). Find ~v + w ~ and interpret this sum geometrically. Solution. Before can add the vectors using Definition 11.6, we need to write w ~ in component form. Using Definition 11.5, we get w ~ = h2 (3), 5 7i = h1, 2i. Thus 6 See Section 10.1.1, for instance. Or, since our given angle, 100 , is obtuse, we could use the Law of Sines without any ambiguity here. 8 Adding vectors 'component-wise' should seem hauntingly familiar. Compare this with how matrix addition was defined in section 8.3. In fact, in more advanced courses such as Linear Algebra, vectors are defined as 1 n or n 1 matrices, depending on the situation. 7 11.8 Vectors 1015 ~v + w ~ = h3, 4i + h1, 2i = h3 + 1, 4 + (2)i = h4, 2i To visualize this sum, we draw ~v with its initial point at (0, 0) (for convenience) so that its terminal point is (3, 4). Next, we graph w ~ with its initial point at (3, 4). Moving one to the right and two down, we find the terminal point of w ~ to be (4, 2). We see that the vector ~v + w ~ has initial point (0, 0) and terminal point (4, 2) so its component form is h4, 2i, as required. y 4 3 w ~ ~v 2 1 ~v + w ~ 1 2 3 4 x In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our definition of vectors to include a 'zero vector', ~0 = h0, 0i. Geometrically, ~0 represents a point, which we can think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of ~0, since after all, vectors are supposed to have both a magnitude (length) and a direction. While it seems clear that the magnitude of ~0 should be 0, it is not clear what its direction is. As we shall see, the direction of ~0 is in fact undefined, but this minor hiccup in the natural flow of things is worth the benefits we reap by including ~0 in our discussions. We have the following theorem. Theorem 11.18. Properties of Vector Addition Commutative Property: For all vectors ~v and w, ~ ~v + w ~ =w ~ + ~v . Associative Property: For all vectors ~u, ~v and w, ~ (~u + ~v ) + w ~ = ~u + (~v + w). ~ Identity Property: The vector ~0 acts as the additive identity for vector addition. That is, for all vectors ~v , ~v + ~0 = ~0 + ~v = ~v . Inverse Property: Every vector ~v has a unique additive inverse, denoted ~v . That is, for every vector ~v , there is a vector ~v so that ~v + (~v ) = (~v ) + ~v = ~0. 1016 Applications of Trigonometry The properties in Theorem 11.18 are easily verified using the definition of vector addition.9 For the commutative property, we note that if ~v = hv1 , v2 i and w ~ = hw1 , w2 i then ~v + w ~ = = = = hv1 , v2 i + hw1 , w2 i hv1 + w1 , v2 + w2 i hw1 + v1 , w2 + v2 i w ~ + ~v Geometrically, we can 'see' the commutative property by realizing that the sums ~v + w ~ and w ~ + ~v are the same directed diagonal determined by the parallelogram below. ~v w ~+ ~v + ~v w ~ w ~ w ~ ~v Demonstrating the commutative property of vector addition. The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams. The existence and uniqueness of the additive inverse is yet another property inherited from the real numbers. Given a vector ~v = hv1 , v2 i, suppose we wish to find a vector w ~ = hw1 , w2 i so that ~v + w ~ = ~0. By the definition of vector addition, we have hv1 + w1 , v2 + w2 i = h0, 0i, and hence, v1 + w1 = 0 and v2 + w2 = 0. We get w1 = v1 and w2 = v2 so that w ~ = hv1 , v2 i. Hence, ~v has an additive inverse, and moreover, it is unique and can be obtained by the formula ~v = hv1 , v2 i. Geometrically, the vectors ~v = hv1 , v2 i and ~v = hv1 , v2 i have the same length, but opposite directions. As a result, when adding the vectors geometrically, the sum ~v + (~v ) results in starting at the initial point of ~v and ending back at the initial point of ~v , or in other words, the net result of moving ~v then ~v is not moving at all. ~v ~v Using the additive inverse of a vector, we can define the difference of two vectors, ~v w ~ = ~v + (w). ~ If ~v = hv1 , v2 i and w ~ = hw1 , w2 i then 9 The interested reader is encouraged to compare Theorem 11.18 and the ensuing discussion with Theorem 8.3 in Section 8.3 and the discussion there. 11.8 Vectors 1017 ~v w ~ = = = = ~v + (w) ~ hv1 , v2 i + hw1 , w2 i hv1 + (w1 ) , v2 + (w2 )i hv1 w1 , v2 w2 i In other words, like vector addition, vector subtraction works component-wise. To interpret the vector ~v w ~ geometrically, we note w ~ + (~v w) ~ = = = = = w ~ + (~v + (w)) ~ w ~ + ((w) ~ + ~v ) (w ~ + (w)) ~ + ~v ~0 + ~v ~v Definition of Vector Subtraction Commutativity of Vector Addition Associativity of Vector Addition Definition of Additive Inverse Definition of Additive Identity This means that the 'net result' of moving along w ~ then moving along ~v w ~ is just ~v itself. From the diagram below, we see that ~v w ~ may be interpreted as the vector whose initial point is the terminal point of w ~ and whose terminal point is the terminal point of ~v as depicted below. It is also worth mentioning that in the parallelogram determined by the vectors ~v and w, ~ the vector ~v w ~ is one of the diagonals - the other being ~v + w. ~ ~v w ~ ~v w ~ w ~ ~v w ~ w ~ ~v ~v Next, we discuss scalar multiplication - that is, taking a real number times a vector. We define scalar multiplication for vectors in the same way we defined it for matrices in Section 8.3. Definition 11.7. If k is a real number and ~v = hv1 , v2 i, we define k~v by k~v = k hv1 , v2 i = hkv1 , kv2 i Scalar multiplication by k in vectors can be understood geometrically as scaling the vector (if k > 0) or scaling the vector and reversing its direction (if k < 0) as demonstrated below. 1018 Applications of Trigonometry 2~v ~v 1 ~v 2 2~v Note that, by definition 11.7, (1)~v = (1) hv1 , v2 i = h(1)v1 , (1)v2 i = hv1 , v2 i = ~v . This, and other properties of scalar multiplication are summarized below. Theorem 11.19. Properties of Scalar Multiplication Associative Property: For every vector ~v and scalars k and r, (kr)~v = k(r~v ). Identity Property: For all vectors ~v , 1~v = ~v . Additive Inverse Property: For all vectors ~v , ~v = (1)~v . Distributive Property of Scalar Multiplication over Scalar Addition: For every vector ~v and scalars k and r, (k + r)~v = k~v + r~v Distributive Property of Scalar Multiplication over Vector Addition: For all vectors ~v and w ~ and scalars k, k(~v + w) ~ = k~v + k w ~ Zero Product Property: If ~v is vector and k is a scalar, then k~v = ~0 if and only if k = 0 or ~v = ~0 The proof of Theorem 11.19, like the proof of Theorem 11.18, ultimately boils down to the definition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let ~v = hv1 , v2 i. If k and r are scalars then (kr)~v = (kr) hv1 , v2 i = h(kr)v1 , (kr)v2 i Definition of Scalar Multiplication = hk(rv1 ), k(rv2 )i Associative Property of Real Number Multiplication = k hrv1 , rv2 i Definition of Scalar Multiplication = k (r hv1 , v2 i) Definition of Scalar Multiplication = k(r~v ) 11.8 Vectors 1019 The remaining properties are proved similarly and are left as exercises. Our next example demonstrates how Theorem 11.19 allows us to do the same kind of algebraic manipulations with vectors as we do with variables - multiplication and division of vectors notwithstanding. If the pedantry seems familiar, it should. This is the same treatment we gave Example 8.3.1 in Section 8.3. As in that example, we spell out the solution in excruciating detail to encourage the reader to think carefully about why each step is justified. Example 11.8.3. Solve 5~v 2 (~v + h1, 2i) = ~0 for ~v . Solution. 5~v 2 (~v + h1, 2i) 5~v + (1) [2 (~v + h1, 2i)] 5~v + [(1)(2)] (~v + h1, 2i) 5~v + (2) (~v + h1, 2i) 5~v + [(2)~v + (2) h1, 2i] 5~v + [(2)~v + h(2)(1), (2)(2)i] [5~v + (2)~v ] + h2, 4i (5 + (2))~v + h2, 4i 3~v + h2, 4i (3~v + h2, 4i) + ( h2, 4i) 3~v + [h2, 4i + ( h2, 4i)] 3~v + ~0 3~v \u0002 = = = = = = = = = = = = = ~0 ~0 ~0 ~0 ~0 ~0 ~0 ~0 ~0 ~0 + ( h2, 4i) ~0 + (1) h2, 4i ~0 + h(1)(2), (1)(4)i h2, 4i 1 v ) = 31 (h2, 4i) 3 (3~ \u0001 \u0003 1\u0001 \u0001 1 v = (2), 13 (4) 3 (3) ~ 3 1~v = 23 , 43 ~v = 23 , 43 A vector whose initial point is (0, 0) is said to be in standard position. If ~v = hv1 , v2 i is plotted in standard position, then its terminal point is necessarily (v1 , v2 ). (Once more, think about this before reading on.) y (v1 , v2 ) x ~v = hv1 , v2 i in standard position. 1020 Applications of Trigonometry Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector. We can convert the point (v1 , v2 ) in rectangular coordinates to a pair (r, ) in polar coordinates where r p 0. The magnitude of ~v , which we said earlier was length of the directed line segment, is r = v12 + v22 and is denoted by k~v k. From Section 11.4, we know v1 = r cos() = k~v k cos() and v2 = r sin() = k~v k sin(). From the definition of scalar multiplication and vector equality, we get ~v = hv1 , v2 i = hk~v k cos(), k~v k sin()i = k~v k hcos(), sin()i This motivates the following definition. Definition 11.8. Suppose ~v is a vector with component form ~v = hv1 , v2 i. Let (r, ) be a polar representation of the point with rectangular coordinates (v1 , v2 ) with r 0. p The magnitude of ~v , denoted k~v k, is given by k~v k = r = v12 + v22 If ~v 6= ~0, the (vector) direction of ~v , denoted v is given by v = hcos(), sin()i Taken together, we get ~v = hk~v k cos(), k~v k sin()i. A few remarks are in order. First, we note that if ~v 6= 0 then even though there are infinitely many angles which satisfy Definition 11.8, the stipulation r > 0 means that all of the angles are coterminal. Hence, if and 0 both satisfy the conditions of Definition 11.8, then cos() = cos(0 ) and sin() = sin(0 ), and as such, hcos(), sin()i = hcos(0 ), sin(0 )i making v is well-defined.10 If ~v = ~0, then ~v = h0, 0i, and we know from Section 11.4 that (0, ) is a polar representation for the origin for any angle . For this reason, 0 is undefined. The following theorem summarizes the important facts about the magnitude and direction of a vector. Theorem 11.20. Properties of Magnitude and Direction: Suppose ~v is a vector. k~v k 0 and k~v k = 0 if and only if ~v = ~0 For all scalars k, kk ~v k = |k|k~v k. If ~v 6= ~0 then ~v = k~v k v , so that v = \u0010 1 k~v k \u0011 ~v . The proof of the first property p in Theorem 11.20 is a direct consequence of the definition of k~v k. If ~ v = hv , v i, then k~ v k = v12 + v22 which is by definition greater than or equal to 0. Moreover, 1 2 p v12 + v22 = 0 if and only of v12 + v22 = 0 if and only if v1 = v2 = 0. Hence, k~v k = 0 if and only if ~v = h0, 0i = ~0, as required. The second property is a result of the definition of magnitude and scalar multiplication along with a propery of radicals. If ~v = hv1 , v2 i and k is a scalar then 10 If this all looks familiar, it should. The interested reader is invited to compare Definition 11.8 to Definition 11.2 in Section 11.7. 11.8 Vectors 1021 kk ~v k = kk hv1 , v2 i k = k hkv1 , kv2 i k q (kv1 )2 + (kv2 )2 = p = k 2 v 2 + k 2 v22 p 1 k 2 (v 2 + v22 ) = p1 = k 2 v12 + v22 p = |k| v12 + v22 = |k|k~v k Definition of scalar multiplication Definition of magnitude Product Rule for Radicals Since k 2 = |k| The equation ~v = k~v k v in Theorem 11.20 is a consequence of the definitions of k~v k and v and was worked out in the discussion just prior to Definition 11.8 on page 1020. In words, the equation ~v = k~v k v says that any given vector is the product of its magnitude and its direction \u0010 \u0011- an important concept to keep in mind when studying and using vectors. The equation v = k~v1k ~v is a result of solving ~v = k~v k v for v by multiplying11 both sides of the equation by of Theorem 11.19. We are overdue for an example. 1 k~v k and using the properties Example 11.8.4. 1. Find the component form of the vector ~v with k~v k = 5 so that when ~v is plotted in standard position, it lies in Quadrant II and makes a 60 angle12 with the negative x-axis. 2. For ~v = 3, 3 3 , find k~v k and , 0 < 2 so that ~v = k~v k hcos(), sin()i. 3. For the vectors ~v = h3, 4i and w ~ = h1, 2i, find the following. (a) v (b) k~v k 2kwk ~ (c) k~v 2wk ~ (d) kwk Solution. 1. We are told that k~v k = 5 and are given information about its direction, so we can use the formula ~v = k~v k v to get the component form of ~v . To determine v, we appeal to Definition 11.8. We are told that ~v lies in Quadrant II and makes a 60 angle with the negative x-axis, so the polar form of the terminal point of ~v , when plotted in standard position is (5, 120 ). D E v = (See the diagram below.) Thus v = hcos (120 ) , sin (120 )i = 12 , 23 , so ~v = k~v k D D E E 5 12 , 23 = 52 , 5 2 3 . 11 Of course, to go from ~v = k~v k v to v = \u0010 1 k~ vk \u0011 ~v , we are essentially 'dividing both sides' of the equation by the scalar k~v k. The authors encourage the reader, however, to work out the details carefully to gain an appreciation of the properties in play. 12 Due to the utility of vectors in 'real-world' applications, we will usually use degree measure for the angle when giving the vector's direction. However, since Carl doesn't want you to forget about radians, he's made sure there are examples and exercises which use them. 1022 Applications of Trigonometry y 5 4 ~v 3 2 60 3 2 = 120 1 1 1 2 3 x q 2. For ~v = 3, 3 3 , we get k~v k = (3)2 + (3 3)2 = 6. In light of Definition 11.8, we can find the we're after by converting the point with rectangular coordinates (3, 3 3) to polar form (r, ) where r = k~v k > 0. From Section 11.4, we have tan() = 33 3 = 3. Since (3, 3 3) is a point in Quadrant IV, is a Quadrant IV angle. Hence, we pick = 5 3 . We \u0001 \u0001 5 may check our answer by verifying ~v = 3, 3 3 = 6 cos 5 , sin . 3 3 \u0010 \u0011 3. (a) Since we are given the component form of ~v , we'll use the formula v = k~v1k ~v . For ~v = h3, 4i, we have k~v k = 32 + 42 = 25 = 5. Hence, v = 15 h3, 4i = 53 , 45 . (b) We know from our work above that ~ we need only find ~ pk~v k = 5, so to find k~v k2kwk, kwk. 2 2 Since w ~ = h1, 2i, we get kwk ~ = 1 + (2) = 5. Hence, k~v k 2kwk ~ = 5 2 5. (c) In the expression k~v 2wk, ~ notice that the arithmetic on the vectors comes first, then the magnitude. Hence, our first step is to find the component form of the vector ~v 2w. ~ We 2 2 get ~v 2w ~ = h3, 4i 2 h1, 2i = h1, 8i. Hence, k~v 2wk ~ = k h1, 8i k = 1 + 8 = 65. \u0010 \u0011 1 (d) To find kwk, we first need w. Using the formula w = kwk w ~ along with kwk ~ = 5, ~ D E which we found the in the previous problem, we get w = 15 h1, 2i = 15 , 25 = r\u0010 \u0011 D \u0010 \u00112 q E 2 5 5 2 5 2 5 5 20 , . Hence, k wk = + = 1 = 1. + = 5 5 5 5 25 25 The process exemplified by number 1 in Example 11.8.4 above by which we take information about the magnitude and direction of a vector and find the component form of a vector is called resolving a vector into its components. As an application of this process, we revisit Example 11.8.1 below. Example 11.8.5. A plane leaves an airport with an airspeed of 175 miles per hour with bearing N40 E. A 35 mile per hour wind is blowing at a bearing of S60 E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution: We proceed as we did in Example 11.8.1 and let ~v denote the plane's velocity and w ~ denote the wind's velocity, and set about determining ~v + w. ~ If we regard the airport as being 11.8 Vectors 1023 at the origin, the positive y-axis acting as due north and the positive x-axis acting as due east, we see that the vectors ~v and w ~ are in standard position and their directions correspond to the angles 50 and 30 , respectively. Hence, the component form of ~v = 175 hcos(50 ), sin(50 )i = h175 cos(50 ), 175 sin(50 )i and the component form of w ~ = h35 cos(30 ), 35 sin(30 )i. Since we have no convenient way to express the exact values of cosine and sine of 50 , we leave both vectors in terms of cosines and sines.13 Adding corresponding components, we find the resultant vector ~v + w ~ = h175 cos(50 ) + 35 cos(30 ), 175 sin(50 ) + 35 sin(30 )i. To find the 'true' speed of the plane, we compute the magnitude of this resultant vector k~v + wk ~ = p (175 cos(50 ) + 35 cos(30 ))2 + (175 sin(50 ) + 35 sin(30 ))2 184 Hence, the 'true' speed of the plane is approximately 184 miles per hour. To find the true bearing, we need to find the angle which corresponds to the polar form (r, ), r > 0, of the point (x, y) = (175 cos(50 ) + 35 cos(30 ), 175 sin(50 ) + 35 sin(30 )). Since both of these coordinates are positive,14 we know is a Quadrant I angle, as depicted below. Furthermore, tan() = y 175 sin(50 ) + 35 sin(30 ) = , x 175 cos(50 ) + 35 cos(30 ) so using the arctangent function, we get 39 . Since, for the purposes of bearing, we need the angle between ~v + w ~ and the positive y-axis, we take the complement of and find the 'true' bearing of the plane to be approximately N51 E. y (N) y (N) ~v ~v ~v + w ~ 40 50 60 30 w ~ x (E) x (E) w ~ In part 3d of Example 11.8.4, we saw that kwk = 1. Vectors with length 1 have a special name and are important in our further study of vectors. Definition 11.9. Unit Vectors: Let ~v be a vector. If k~v k = 1, we say that ~v is a unit vector. 13 Keeping things 'calculator' friendly, for once! Yes, a calculator approximation is the quickest way to see this, but you can also use good old-fashioned inequalities and the fact that 45 50 60 . 14 1024 Applications of Trigonometry If ~v is a unit vector,\u0010 then v = 1 v = v. Conversely, we leave it as an exercise15 \u0011 necessarily, ~v = k~v k to show that v = k~v1k ~v is a unit vector for any nonzero vector ~v . In practice, if ~v is a unit vector we write it as v as opposed to ~v because we have reserved the '' notation for unit vectors. The process of multiplying a nonzero vector by the factor k~v1k to produce a unit vector is called 'normalizing the vector,' and the resulting vector v is called the 'unit vector in the direction of ~v '. The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take the time to show this.) As a result, we visualize normalizing a nonzero vector ~v as shrinking16 its terminal point, when plotted in standard position, back to the Unit Circle. y ~v 1 v 1 1 x 1 Visualizing vector normalization v = \u0010 1 k~v k \u0011 ~v Of all of the unit vectors, two deserve special mention. Definition 11.10. The Principal Unit Vectors: The vector is defined by = h1, 0i The vector is defined by = h0, 1i We can think of the vector as representing the positive x-direction, while represents the positive y-direction. We have the following 'decomposition' theorem.17 Theorem 11.21. Principal Vector Decomposition Theorem: Let ~v be a vector with component form ~v = hv1 , v2 i. Then ~v = v1 + v2 . The proof of Theorem 11.21 is straightforward. Since = h1, 0i and = h0, 1i, we have from the definition of scalar multiplication and vector addition that v1 + v2 = v1 h1, 0i + v2 h0, 1i = hv1 , 0i + h0, v2 i = hv1 , v2 i = ~v \u0010 \u0011 One proof uses the properties of scalar multiplication and magnitude. If ~v 6= ~0, consider k v k = k~v1k ~v . Use the fact that k~v k 0 is a scalar and consider factoring. 16 . . . if k~v k > 1 . . . 17 We will see a generalization of Theorem 11.21 in Section 11.9. Stay tuned! 15 11.8 Vectors 1025 Geometrically, the situation looks like this: y ~v = hv1 , v2 i v2 x v1 ~v = hv1 , v2 i = v1 + v2 . We conclude this section with a classic example which demonstrates how vectors are used to model forces. A 'force' is defined as a 'push' or a 'pull.' The intensity of the push or pull is the magnitude of the force, and is measured in Netwons (N) in the SI system or pounds (lbs.) in the English system.18 The following example uses all of the concepts in this section, and should be studied in great detail. Example 11.8.6. A 50 pound speaker is suspended from the ceiling by two support braces. If one of them makes a 60 angle with the ceiling and the other makes a 30 angle with the ceiling, what are the tensions on each of the supports? Solution. We represent the problem schematically below and then provide the corresponding vector diagram. 30 60 30 60 T~1 T~2 60 30 50 lbs. w ~ We have three forces acting on the speaker: the weight of the speaker, which we'll call w, ~ pulling ~ the speaker directly downward, and the forces on the support rods, which we'll call T1 and T~2 (for 'tensions') acting upward at angles 60 and 30 , respectively. We are looking for the tensions on the support, which are the magnitudes kT~1 k and kT~2 k. In order for the speaker to remain stationary,19 we require w ~ + T~1 + T~2 = ~0. Viewing the common initial point of these vectors as the 18 19 See also Section 11.1.1. This is the criteria for 'static equilbrium'. 1026 Applications of Trigonometry origin and the dashed line as the x-axis, we use Theorem 11.20 to get component representations for the three vectors involved. We can model the weight of the speaker as a vector pointing directly downwards with a magnitude of 50 pounds. That is, kwk ~ = 50 and w = = h0, 1i. Hence, w ~ = 50 h0, 1i = h0, 50i. For the force in the first support, we get T~1 = kT~1 k hcos (60 ) , sin (60 )i * + kT~1 k kT~1 k 3 = , 2 2 For the second support, we note that the angle 30 is measured from the negative x-axis, so the angle needed to write T~2 in component form is 150 . Hence T~2 = kT~2 k hcos (150 ) , sin (150 )i * + kT~2 k 3 kT~2 k = , 2 2 The requirement w ~ + T~1 + T~2 = ~0 gives us this vector equation. w ~ + T~1 + T~2 = ~0 * + + * kT~1 k kT~1 k 3 kT~2 k 3 kT~2 k h0, 50i + , + , = h0, 0i 2 2 2 2 * + kT~1 k kT~2 k 3 kT~1 k 3 kT~2 k = h0, 0i , + 50 2 2 2 2 Equating the corresponding components of the vectors on each side, we get a system of linear equations in the variables kT~1 k and kT~2 k. ~1 k kT~2 k 3 k T (E1) = 0 2 2 ~ ~ (E2) kT1 k 3 + kT2 k 50 = 0 2 2 ~ ~ From (E1), we get kT~1 k = kT~2 k 3. Substituting that into (E2) gives (kT2 k 2 3) 3 + kT22 k 50 = 0 which yields 2kT~2 k 50 = 0. Hence, kT~2 k = 25 pounds and kT~1 k = kT~2 k 3 = 25 3 pounds. 11.8 Vectors 11.8.1 1027 Exercises In Exercises 1 - 10, use the given pair of vectors ~v and w ~ to find the following quantities. State whether the result is a vector or a scalar. ~v + w ~ w ~ 2~v k~v + wk ~ k~v k + kwk ~ k~v kw ~ kwk~ ~ v kwk ~ v Finally, verify that the vectors satisfy the Parallelogram Law k~v k2 + kwk ~ 2= \u0003 1\u0002 k~v + wk ~ 2 + k~v wk ~ 2 2 1. ~v = h12, 5i, w ~ = h3, 4i 2. ~v = h7, 24i, w ~ = h5, 12i 3. ~v = h2, 1i, w ~ = h2, 4i 4. ~v = h10, 4i, w ~ = h2, 5i 5. ~v = 3, 1 , w ~ = 2 3, 2 D D E E 7. ~v = 22 , 22 , w ~ = 22 , 22 ,w ~ = 45 , 35 D E 8. ~v = 12 , 23 , w ~ = 1, 3 9. ~v = 3 + 4 , w ~ = 2 6. ~v = 10. ~v = 4 5, 5 3 1 2 ( + ), w ~= 1 2 ( ) In Exercises 11 - 25, find the component form of the vector ~v using the information given about its magnitude and direction. Give exact values. 11. k~v k = 6; when drawn in standard position ~v lies in Quadrant I and makes a 60 angle with the positive x-axis 12. k~v k = 3; when drawn in standard position ~v lies in Quadrant I and makes a 45 angle with the positive x-axis 13. k~v k = 32 ; when drawn in standard position ~v lies in Quadrant I and makes a 60 angle with the positive y-axis 14. k~v k = 12; when drawn in standard position ~v lies along the positive y-axis 15. k~v k = 4; when drawn in standard position ~v lies in Quadrant II and makes a 30 angle with the negative x-axis 16. k~v k = 2 3; when drawn in standard position ~v lies in Quadrant II and makes a 30 angle with the positive y-axis 17. k~v k = 27 ; when drawn in standard position ~v lies along the negative x-axis 18. k~v k = 5 6; when drawn in standard position ~v lies in Quadrant III and makes a 45 angle with the negative x-axis 19. k~v k = 6.25; when drawn in standard position ~v lies along the negative y-axis 1028 Applications of Trigonometry 20. k~v k = 4 3; when drawn in standard position ~v lies in Quadrant IV and makes a 30 with the positive x-axis 21. k~v k = 5 2; when drawn in standard position ~v lies in Quadrant IV and makes a 45 with the negative y-axis 22. k~v k = 2 5; when drawn in standard position ~v lies in Quadrant I and makes an measuring arctan(2) with the positive x-axis 23. k~v k = 10; when drawn in standard position ~v lies in Quadrant II and makes an measuring arctan(3) with the negative x-axis angle angle angle angle 24. k~v k = 5; when drawn in standard position ~v lies in Quadrant III and makes an angle measuring \u0001 arctan 43 with the negative x-axis 25. k~v k = 26; when drawn in standard position ~v lies in Quadrant IV and makes an angle \u0001 5 with the positive x-axis measuring arctan 12 In Exercises 26 - 31, approximate the component form of the vector ~v using the information given about its magnitude and direction. Round your approximations to two decimal places. 26. k~v k = 392; when drawn in standard position ~v makes a 117 angle with the positive x-axis 27. k~v k = 63.92; when drawn in standard position ~v makes a 78.3 angle with the positive x-axis 28. k~v k = 5280; when drawn in standard position ~v makes a 12 angle with the positive x-axis 29. k~v k = 450; when drawn in standard position ~v makes a 210.75 angle with the positive x-axis 30. k~v k = 168.7; when drawn in standard position ~v makes a 252 angle with the positive x-axis 31. k~v k = 26; when drawn in standard position ~v makes a 304.5 angle with the positive x-axis In Exercises 32 - 52, for the given vector ~v , find the magnitude k~v k and an angle with 0 < 360 so that ~v = k~v k hcos(), sin()i (See Definition 11.8.) Round approximations to two decimal places. 32. ~v = 1, 3 35. ~v = 2, 2 33. ~v = h5, 5i D E 36. ~v = 22 , 22 34. ~v = 2 3, 2 D E 37. ~v = 12 , 23 38. ~v = h6, 0i 39. ~v = h2.5, 0i 40. ~v = 0, 7 41. ~v = 10 42. ~v = h3, 4i 43. ~v = h12, 5i 11.8 Vectors 1029 44. ~v = h4, 3i 45. ~v = h7, 24i 46. ~v = h2, 1i 47. ~v = h2, 6i 48. ~v = + 49. ~v = 4 50. ~v = h123.4, 77.05i 51. ~v = h965.15, 831.6i 52. ~v = h114.1, 42.3i 53. A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of S68 W. The river is flowing due east at 8 miles per hour. What is the boat's true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 54. The HMS Sasquatch leaves port with bearing S20 E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 5 miles per hour with a bearing of N60 E, find the HMS Sasquatch's true speed and bearing. Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 55. If the captain of the HMS Sasquatch in Exercise 54 wishes to reach Chupacabra Cove, an island 100 miles away at a bearing of S20 E from port, in three hours, what speed and heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If ~v denotes the velocity of the HMS Sasquatch and w ~ denotes the velocity of the current, what does ~v + w ~ need to be to reach Chupacabra Cove in three hours? 56. In calm air, a plane flying from the Pedimaxus International Airport can reach Cliffs of Insanity Point in two hours by following a bearing of N8.2 E at 96 miles an hour. (The distance between the airport and the cliffs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree. 57. The SS Bigfoot leaves Yeti Bay on a course of N37 W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 30 miles from the bay and his bearing back to the bay is S40 E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 58. A 600 pound Sasquatch statue is suspended by two cables from a gymnasium ceiling. If each cable makes a 60 angle with the ceiling, find the tension on each cable. Round your answer to the nearest pound. 59. Two cables are to support an object hanging from a ceiling. If the cables are each to make a 42 angle with the ceiling, and each cable is rated to withstand a maximum tension of 100 pounds, what is the heaviest object that can be supported? Round your answer down to the nearest pound. 1030 Applications of Trigonometry 60. A 300 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 72 angle with the ceiling while the right hand cable makes a 18 angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places. 61. Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N77 E and the other pulls at a heading of S68 E. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound. 62. Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie 'Ben-Hur' by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N80 W, the second points due west and the third points S80 W. The force applied to the first rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds otherwise the couch won't move. Does it move? If so, is it heading due west? 63. Let ~v = hv1 , v2 i be any non-zero vector. Show that 1 ~v has length 1. k~v k 64. We say that two non-zero vectors ~v and w ~ are parallel if they have same or opposite directions. ~ ~ That is, ~v 6= 0 and w ~ 6= 0 are parallel if either v = w or v = w. Show that this means ~v = k w ~ for some non-zero scalar k and that k > 0 if the vectors have the same direction and k < 0 if they point in opposite directions. 65. The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line y = 2x 4. Let ~v0 = h0, 4i and let ~s = h1, 2i. Let t be any real number. Show that the vector defined by ~v = ~v0 + t~s, when drawn in standard position, has its terminal point on the line y = 2x 4. (Hint: Show that ~v0 + t~s = ht, 2t 4i for any real number t.) Now consider the non-vertical line y = mx + b. Repeat the previous analysis with ~v0 = h0, bi and let ~s = h1, mi. Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of h0, bi (the position vector of the y-intercept) and a scalar multiple of the slope vector ~s = h1, mi. 66. Prove the associative and identity properties of vector addition in Theorem 11.18. 67. Prove the properties of scalar multiplication in Theorem 11.19. 11.8 Vectors 11.8.2 1. 1031 Answers ~v + w ~ = h15, 1i, vector k~v + wk ~ = 2. 3. 4. 6. 226, scalar kwk v= ~v + w ~ = h12, 12i, vector w ~ 2~v = h9, 60i, vector k~v + wk ~ = 12 2, scalar k~v k + kwk ~ = 38, scalar k~v kw ~ kwk~ ~ v = h34, 612i, vector 312 kwk v = 91 , 25 25 , vector ~v + w ~ = h0, 3i, vector w ~ 2~v = h6, 6i, vector k~v + wk ~ = 3, scalar k~v k + kwk ~ = 3 5, scalar k~v kw ~ kwk~ ~ v = 6 5, 6 5 , vector kwk v = h4, 2i, vector ~v + w ~ = h8, 9i, vector w ~ 2~v = h22, 3i, vector 145, scalar 60 25 13 , 13 , vector k~v k + kwk ~ = 3 29, scalar k~v kw ~ kwk~ ~ v = 14 29, 6 29 , vector kwk v = h5, 2i, vector ~v + w ~= w ~ 2~v = 4 3, 0 , vector 3, 3 , vector k~v + wk ~ = 2 3, scalar k~v k + kwk ~ = 6, scalar k~v kw ~ kwk~ ~ v = 8 3, 0 , vector kwk v = 2 3, 2 , vector ~v + w ~ = 15 , 57 , vector w ~ 2~v = h2, 1i, vector k~v + wk ~ = 2, scalar k~v kw ~ kwk~ ~ v = 75 , 51 , vector 7. k~v k + kwk ~ = 18, scalar k~v kw ~ kwk~ ~ v = h21, 77i, vector k~v + wk ~ = 5. w ~ 2~v = h21, 14i, vector ~v + w ~ = h0, 0i, vector k~v + wk ~ = 0, scalar k~v kw ~ kwk~ ~ v = 2, 2 , vector k~v k + kwk ~ = 2, scalar kwk v= 3 4 5, 5 , vector D E w ~ 2~v = 3 2 2 , 3 2 2 , vector k~v k + kwk ~ = 2, scalar D E kwk v = 22 , 22 , vector 1032 8. 9. Applications of Trigonometry D E ~v + w ~ = 12 , 23 , vector w ~ 2~v = 2, 2 3 , vector k~v + wk ~ = 1, scalar k~v k + kwk ~ = 3, scalar k~v kw ~ kwk~ ~ v = 2, 2 3 , vector kwk v = 1, 3 , vector ~v + w ~ = h3, 2i, vector w ~ 2~v = h6, 10i, vector k~v + wk ~ = 10. k~v k + kwk ~ = 7, scalar 13, scalar k~v kw ~ kwk~ ~ v = h6, 18i, vector kwk v= ~v + w ~ = h1, 0i, vector w ~ 2~v = 12 , 32 , vector k~v + wk ~ = 1, scalar D E k~v kw ~ kwk~ ~ v = 0, 22 , vector k~v k + kwk ~ = kwk v= 6 8 5, 5 1 1 2, 2 , vector 2, scalar , vector 11. ~v = 3, 3 3 12. ~v = 14. ~v = h0, 12i 15. ~v = 2 3, 2 16. ~v = 3, 3 17. ~v = 72 , 0 18. ~v = 5 3, 5 3 19. ~v = h0, 6.25i 20. ~v = 6, 2 3 21. ~v = h5, 5i 22. ~v = h2, 4i 23. ~v = h1, 3i 24. ~v = h3, 4i 25. ~v = h24, 10i 26. ~v h177.96, 349.27i 27. ~v h12.96, 62.59i 28. ~v h5164.62, 1097.77i 29. ~v h386.73, 230.08i 30. ~v h52.13, 160.44i 31. ~v h14.73, 21.43i 32. k~v k = 2, = 60 33. k~v k = 5 2, = 45 34. k~v k = 4, = 150 35. k~v k = 2, = 135 36. k~v k = 1, = 225 37. k~v k = 1, = 240 38. k~v k = 6, = 0 39. k~v k = 2.5, = 180 40. k~v k = 41. k~v k = 10, = 270 42. k~v k = 5, 53.13 43. k~v k = 13, 22.62 44. k~v k = 5, 143.13 45. k~v k = 25, 106.26 46. k~v k = D E 3 2 3 2 , 2 2 13. ~v = D 3 1 3 ,3 E 7, = 90 5, 206.57 11.8 Vectors 1033 47. k~v k = 2 10, 251.57 48. k~v k = 50. k~v k 145.48, 328.02 51. k~v k 1274.00, 40.75 2, 45 49. k~v k = 17, 284.04 52. k~v k 121.69, 159.66 53. The boat's true speed is about 10 miles per hour at a heading of S50.6 W. 54. The HMS Sasquatch's true speed is about 41 miles per hour at a heading of S26.8 E. 55. She should maintain a speed of about 35 miles per hour at a heading of S11.8 E. 56. She should fly at 83.46 miles per hour with a heading of N22.1 E 57. The current is moving at about 10 miles per hour bearing N54.6 W. 58. The tension on each of the cables is about 346 pounds. 59. The maximum weight that can be held by the cables in that configuration is about 133 pounds. 60. The tension on the left hand cable is 285.317 lbs. and on the right hand cable is 92.705 lbs. 61. The weaker student should pull about 60 pounds. The net force on the keg is about 153 pounds. 62. The resultant force is only about 296 pounds so the couch doesn't budge. Even if it did move, the stronger force on the third rope would have made the couch drift slightly to the south as it traveled down the street. Section 11.9: The Dot Product and Projection, from College Trigonometry: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative Commons Attribution-NonCommercialShareAlike 3.0 license. 2013, Carl Stitz. 1034 11.9 Applications of Trigonometry The Dot Product and Projection In Section 11.8, we learned how add and subtract vectors and how to multiply vectors by scalars. In this section, we define a product of vectors. We begin with the following definition. Definition 11.11. Suppose ~v and w ~ are vectors whose component forms are ~v = hv1 , v2 i and w ~ = hw1 , w2 i. The dot product of ~v and w ~ is given by ~v w ~ = hv1 , v2 i hw1 , w2 i = v1 w1 + v2 w2 For example, let ~v = h3, 4i and w ~ = h1, 2i. Then ~v w ~ = h3, 4i h1, 2i = (3)(1) + (4)(2) = 5. Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity ~v w ~ is often called the scalar product of ~v and w. ~ The dot product enjoys the following properties. Theorem 11.22. Properties of the Dot Product Commutative Property: For all vectors ~v and w, ~ ~v w ~ =w ~ ~v . Distributive Property: For all vectors ~u, ~v and w, ~ ~u (~v + w) ~ = ~u ~v + ~u w. ~ Scalar Property: For all vectors ~v and w ~ and scalars k, (k~v ) w ~ = k(~v w) ~ = ~v (k w). ~ Relation to Magnitude: For all vectors ~v , ~v ~v = k~v k2 . Like most of the theorems involving vectors, the proof of Theorem 11.22 amounts to using the definition of the dot product and properties of real number arithmetic. To show the commutative property for instance, let ~v = hv1 , v2 i and w ~ = hw1 , w2 i. Then ~v w ~ = hv1 , v2 i hw1 , w2 i = v 1 w1 + v 2 w2 Definition of Dot Product = w1 v 1 + w2 v 2 Commutativity of Real Number Multiplication = hw1 , w2 i hv1 , v2 i Definition of Dot Product = w ~ ~v The distributive property is proved similarly and is left as an exercise. For the scalar property, assume that ~v = hv1 , v2 i and w ~ = hw1 , w2 i and k is a scalar. Then (k~v ) w ~ = (k hv1 , v2 i) hw1 , w2 i = hkv1 , kv2 i hw1 , w2 i Definition of Scalar Multiplication = (kv1 )(w1 ) + (kv2 )(w2 ) Definition of Dot Product = k(v1 w1 ) + k(v2 w2 ) Associativity of Real Number Multiplication = k(v1 w1 + v2 w2 ) Distributive Law of Real Numbers = k hv1 , v2 i hw1 , w2 i Definition of Dot Product = k(~v w) ~ We leave the proof of k(~v w) ~ = ~v (k w) ~ as an exercise. 11.9 The Dot Product and Projection 1035 For the last property, we note that if ~v = hv1 , v2 i, then ~v ~v = hv1 , v2 i hv1 , v2 i = v12 + v22 = k~v k2 , where the last equality comes courtesy of Definition 11.8. The following example puts Theorem 11.22 to good use. As in Example 11.8.3, we work out the problem in great detail and encourage the reader to supply the justification for each step. Example 11.9.1. Prove the identity: k~v wk ~ 2 = k~v k2 2(~v w) ~ + kwk ~ 2. Solution. We begin by rewriting k~v wk ~ 2 in terms of the dot product using Theorem 11.22. k~v wk ~ 2 = (~v w) ~ (~v w) ~ = (~v + [w]) ~ (~v + [w]) ~ = (~v + [w]) ~ ~v + (~v + [w]) ~ [w] ~ = ~v (~v + [w]) ~ + [w] ~ (~v + [w]) ~ = ~v ~v + ~v [w] ~ + [w] ~ ~v + [w] ~ [w] ~ = ~v ~v + ~v [(1)w] ~ + [(1)w] ~ ~v + [(1)w] ~ [(1)w] ~ = ~v ~v + (1)(~v w) ~ + (1)(w ~ ~v ) + [(1)(1)](w ~ w) ~ = ~v ~v + (1)(~v w) ~ + (1)(~v w) ~ +w ~ w ~ = ~v ~v 2(~v w) ~ +w ~ w ~ = k~v k2 2(~v w) ~ + kwk ~ 2 Hence, k~v wk ~ 2 = k~v k2 2(~v w) ~ + kwk ~ 2 as required. If we take a step back from the pedantry in Example 11.9.1, we see that the bulk of the work is needed to show that (~v w)(~ ~ v w) ~ = ~v ~v 2(~v w)+ ~ w ~ w. ~ If this looks familiar, it should. Since the dot product enjoys many of the same properties enjoyed by real numbers, the machinations required to expand (~v w) ~ (~v w) ~ for vectors ~v and w ~ match those required to expand (v w)(v w) for real numbers v and w, and hence we get similar looking results. The identity verified in Example 11.9.1 plays a large role in the development of the geometric properties of the dot product, which we now explore. Suppose ~v and w ~ are two nonzero vectors. If we draw ~v and w ~ with the same initial point, we define the angle between ~v and w ~ to be the angle determined by the rays containing the vectors ~v and w, ~ as illustrated below. We require 0 . (Think about why this is needed in the definition.) ~v ~v w ~ w ~ ~v w ~ =0 0<< = The following theorem gives us some insight into the geometric role the dot product plays. Theorem 11.23. Geometric Interpretation of Dot Product: If ~v and w ~ are nonzero vectors then ~v w ~ = k~v kkwk ~ cos(), where is the angle between ~v and w. ~ 1036 Applications of Trigonometry We prove Theorem 11.23 in cases. If = 0, then ~v and w ~ have the same direction. It follows1 that there is a real number k > 0 so that w ~ = k~v . Hence, ~v w ~ = ~v (k~v ) = k(~v ~v ) = kk~v k2 = kk~v kk~v k. Since k > 0, k = |k|, so kk~v k = |k|k~v k = kk~v k by Theorem 11.20. Hence, kk~v kk~v k = k~v k(kk~v k) = k~v kkk~v k = k~v kkwk. ~ Since cos(0) = 1, we get ~v w ~ = kk~v kk~v k = k~v kkwk ~ = k~v kkwk ~ cos(0), proving that the formula holds for = 0. If = , we repeat the argument with the difference being w ~ = k~v where k < 0. In this case, |k| = k, so kk~v k = |k|k~v k = kk~v k = kwk. ~ Since cos() = 1, we get ~v w ~ = k~v kkwk ~ = k~v kkwk ~ cos(), as required. Next, if 0 < < , the vectors ~v , w ~ and ~v w ~ determine a triangle with side lengths k~v k, kwk ~ and k~v wk, ~ respectively, as seen below. k~v wk ~ ~v w ~ w ~ ~v kwk ~ k~v k The Law of Cosines yields k~v wk ~ 2 = k~v k2 + kwk ~ 2 2k~v kkwk ~ cos(). From Example 11.9.1, 2 2 2 we know k~v wk ~ = k~v k 2(~v w) ~ + kwk ~ . Equating these two expressions for k~v wk ~ 2 gives k~v k2 +kwk ~ 2 2k~v kkwk ~ cos() = k~v k2 2(~v w)+k ~ wk ~ 2 which reduces to 2k~v kkwk ~ cos() = 2(~v w), ~ or ~v w ~ = k~v kkwk ~ cos(), as required. An immediate consequence of Theorem 11.23 is the following. Theorem 11.24. Let ~v and w ~ be nonzero vectors and let the angle between ~v and w. ~ Then \u0012 \u0013 ~v w ~ = arccos = arccos( v w) k~v kkwk ~ We obtain the formula in Theorem 11.24 by solving the equation given in Theorem 11.23 for . Since ~v and w ~ are nonzero, so are k~v k and kwk. ~ Hence, we may divide both sides of ~v w ~ = k~v kkwk ~ cos() ~v w ~ by k~v kkwk ~ to get cos() = k~vkkwk the values of exactly match the ~ . Since 0 \u0010by definition, \u0011 ~v w ~ range of the arccosine function. Hence, = arccos k~vkkwk . Using Theorem 11.22, we can rewrite ~ \u0010 \u0011 \u0010 \u0011 ~v w ~ 1 1 v kwk ~ = v w, giving us the alternative formula = arccos( v w). k~v kkwk ~ = k~v k ~ ~ w We are overdue for an example. Example 11.9.2. Find the angle between the following pairs of vectors. ~ = 3, 1 1. ~v = 3, 3 3 , and w 2. ~v = h2, 2i, and w ~ = h5, 5i 3. ~v = h3, 4i, and w ~ = h2, 1i Solution. We use the formula = arccos 1 \u0010 ~v w ~ k~v kkwk ~ \u0011 from Theorem 11.24 in each case below. Since ~v = k~v k v and w ~ = kwk ~ w, if v = w then w ~ = kwk ~ v= kwk ~ (k~v k v) k~ vk = kwk ~ ~v . k~ vk In this case, k = kwk ~ k~ vk > 0. 11.9 The Dot Product and Projection 1037 q 1. We have ~v w ~ = 3, 3 3 3, 1 = 3 33 3 = 6 3. Since k~v k = 32 + (3 3)2 = q \u0010 \u0011 \u0010 \u0011 36 = 6 and kwk ~ = ( 3)2 + 12 = 4 = 2, = arccos 612 3 = arccos 23 = 5 6 . 2. For ~v = h2, 2i and w ~ = h5, 5i, we find ~v w ~\u0010= h2, 2i\u0011 h5, 5i = 10 10 = 0. Hence, it doesn't 2 = arccos(0) = 2 . matter what k~v k and kwk ~ are, = arccos k~v~vkkw~wk ~ p 3. We find ~v w ~ = h3, 4i h2, 1i = \u00106 4\u0011 = 2. Also k~v k\u0011 = 32 + (4)2 = 25 = 5 and \u0010 2 5 2 5 2 w ~ = 22 + 12 = 5, so = arccos 5 = arccos . Since isn't the cosine of one 25 25 5 \u0010 \u0011 of the common angles, we leave our answer as = arccos 2255 . The vectors ~v = h2, 2i, and w ~ = h5, 5i in Example 11.9.2 are called orthogonal and we write ~v w, ~ because the angle between them is 2 radians = 90 . Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. w ~ ~v ~v and w ~ are orthogonal, ~v w ~ We state the relationship between orthogonal vectors and their dot product in the following theorem. Theorem 11.25. The Dot Product Detects Orthogonality: Let ~v and w ~ be nonzero vectors. Then ~v w ~ if and only if ~v w ~ = 0. To prove Theorem 11.25, we first assume ~v and w ~ are nonzero vectors with ~v ~ By definition, \u0001 w. the angle between ~v and w ~ is 2 . By Theorem 11.23, ~v w ~ = k~v kkwk ~ cos 2 = 0. \u0010Conversely, \u0011 if ~v and w ~ are nonzero vectors and ~v w ~ = 0, then Theorem 11.24 gives = arccos k~v~vkkw~wk = ~ \u0010 \u0011 arccos k~vkk0 wk = arccos(0) = 2 , so ~v w. ~ We can use Theorem 11.25 in the following example ~ to provide a different proof about the relationship between the slopes of perpendicular lines.3 Example 11.9.3. Let L1 be the line y = m1 x + b1 and let L2 be the line y = m2 x + b2 . Prove that L1 is perpendicular to L2 if and only if m1 m2 = 1. Solution. Our strategy is to find two vectors: v~1 , which has the same direction as L1 , and v~2 , which has the same direction as L2 and show v~1 v~2 if and only if m1 m2 = 1. To that end, we substitute x = 0 and x = 1 into y = m1 x + b1 to find two points which lie on L1 , namely P (0, b1 ) 2 3 Note that there is no 'zero product property' for the dot product since neither ~v nor w ~ is ~0, yet ~v w ~ = 0. See Exercise 2.1.1 in Section 2.1. 1038 Applications of Trigonometry and Q(1, m1 + b1 ). We let v~1 = P Q = h1 0, (m1 + b1 ) b1 i = h1, m1 i, and note that since v~1 is determined by two points on L1 , it may be viewed as lying on L1 . Hence it has the same direction as L1 . Similarly, we get the vector v~2 = h1, m2 i which has the same direction as the line L2 . Hence, L1 and L2 are perpendicular if and only if v~1 v~2 . According to Theorem 11.25, v~1 v~2 if and only if v~1 v~2 = 0. Notice that v~1 v~2 = h1, m1 i h1, m2 i = 1 + m1 m2 . Hence, v~1 v~2 = 0 if and only if 1 + m1 m2 = 0, which is true if and only if m1 m2 = 1, as required. While Theorem 11.25 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. Consider the two nonzero vectors ~v and w ~ drawn with a common initial point O below. For the moment, assume that the angle between ~v and w, ~ which we'll denote , is acute. We wish to develop a formula for the vector p~, indicated below, which is called the orthogonal projection of ~v onto w. ~ The vector p~ is obtained geometrically as follows: drop a perpendicular from the terminal point T of ~v to the vector w ~ and call the point of intersection R. The vector p~ is then defined as p~ = OR. Like any vector, p~ is determined by its magnitude k~ pk and its direction p according to the formula p~ = k~ pk p. Since we want p to have the same direction as w, ~ we have p = w. To determine k~ pk, we make use of Theorem 10.4 as applied k~ pk to the right triangle 4ORT . We find cos() = k~ pk = k~v k cos(). To get things in terms of v k , or k~ ~ cos() w ~ just ~v and w, ~ we use Theorem 11.23 to get k~ pk = k~v k cos() = k~vkkwk = k~vwk kwk ~ ~ . Using Theorem \u0010 \u0011 1 w ~ v kwk ~ = ~v w. Hence, k~ pk = ~v w, and since p = w, we now have a 11.22, we rewrite k~vwk ~ =~ ~ w formula for p~ completely in terms of ~v and w, ~ namely p~ = k~ pk p = (~v w) w. ~v ~v T T k~v k w ~ w ~ R p~ = OR O O R k~ pk O Now suppose that the angle between ~v and w ~ is obtuse, and consider the diagram below. In this case, we see that p = w and using the triangle 4ORT , we find k~ pk = k~v k cos(0 ). Since +0 = , 0 it follows that cos( ) = cos(), which means k~ pk = k~v k cos(0 ) = k~v k cos(). Rewriting this last equation in terms of ~v and w ~ as before, we get k~ pk = (~v w). Putting this together with p = w, we get p~ = k~ pk p = (~v w)( w) = (~v w) w in this case as well. 11.9 The Dot Product and Projection T 1039 ~v w ~ 0 R O p~ = OR If the angle between ~v and w ~ is 2 then it is easy to show4 that p~ = ~0. Since ~v w ~ in this case, ~ ~v w ~ = 0. It follows that ~v w = 0 and p~ = 0 = 0w = (~v w) w in this case, too. This gives us Definition 11.12. Let ~v and w ~ be nonzero vectors. The orthogonal projection of ~v onto w, ~ denoted projw~ (~v ) is given by projw~ (~v ) = (~v w) w. Definition 11.12 gives us a good idea what the dot product does. The scalar ~v w is a measure of how much of the vector ~v is in the direction of the vector w ~ and is thus called the scalar projection of ~v onto w. ~ While the formula given in Definition 11.12 is theoretically appealing, because of the presence of the normalized unit vector w, computing the projection using the formula w can be messy. We present two other formulas that are often used in practice. projw~ (~v ) = (~v w) Theorem 11.26. Alternate Formulas for Vector Projections: If ~v and w ~ are nonzero vectors then \u0012 \u0013 \u0012 \u0013 ~v w ~ ~v w ~ w ~ = w ~ w = projw~ (~v ) = (~v w) kwk ~ 2 w ~ w ~ The proof of \u0010Theorem 11.26, which we leave to the reader as an exercise, amounts to using the \u0011 1 formula w = kwk w ~ and properties of the dot product. It is time for an example. ~ Example 11.9.4. Let ~v = h1, 8i and w ~ = h1, 2i. Find p~ = projw~ (~v ), and plot ~v , w ~ and p~ in standard position. Solution. We find ~v w ~ = h1, 8i h1, 2i = (1) + 16 = 15 and w ~ w ~ = h1, 2i h1, 2i = 1 + 4 = 5. ~v w ~ 15 w ~ = h1, 2i = h3, 6i. We plot ~ v , w ~ and p ~ below. Hence, p~ = w ~ w ~ 5 4 In this case, the point R coincides with the point O, so p ~ = OR = OO = ~0. 1040 Applications of Trigonometry 8 ~v 7 p ~ 6 5 4 3 2 w ~ 3 2 1 1 Suppose we wanted to verify that our answer p~ in Example 11.9.4 is indeed the orthogonal projection of ~v onto w. ~ We first note that since p~ is a scalar multiple of w, ~ it has the correct direction, so what remains to check is the orthogonality condition. Consider the vector ~q whose initial point is the terminal point of p~ and whose terminal point is the terminal point of ~v . q~ p ~ 8 ~v 7 6 5 4 3 w ~ 3 2 1 2 1 From the definition of vector arithmetic, p~ + ~q = ~v , so that ~q = ~v p~. In the case of Example 11.9.4, ~v = h1, 8i and p~ = h3, 6i, so ~q = h1, 8ih3, 6i = h4, 2i. Then ~q w ~ = h4, 2ih1, 2i = (4)+4 = 0, which shows ~q w, ~ as required. This result is generalized in the following theorem. Theorem 11.27. Generalized Decomposition Theorem: Let ~v and w ~ be nonzero vectors. There are unique vectors p~ and ~q such that ~v = p~ + ~q where p~ = k w ~ for some scalar k, and ~q w ~ = 0. Note that if the vectors p~ and ~q in Theorem 11.27 are nonzero, then we can say p~ is parallel 5 to w ~ and ~q is orthogonal to w. ~ In this case, the vector p~ is sometimes called the 'vector component of ~v parallel to w' ~ and ~q is called the 'vector component of ~v orthogonal to w.' ~ To prove Theorem 11.27, we take p~ = projw~ (~v ) and ~q = ~v p~. Then p~ is, by definition, a scalar multiple of w. ~ Next, we compute ~q w. ~ 5 See Exercise 64 in Section 11.8. 11.9 The Dot Product and Projection 1041 ~q w ~ = (~v p~) w ~ Definition of ~q. = ~v w ~ p~ w ~ Properties of Dot Product \u0012 \u0013 ~v w ~ = ~v w ~ w ~ w ~ Since p~ = projw~ (~v ). ~ w ~\u0013 \u0012w ~v w ~ = ~v w ~ (w ~ w) ~ Properties of Dot Product. w ~ w ~ = ~v w ~ ~v w ~ = 0 Hence, ~q w ~ = 0, as required. At this point, we have shown that the vectors p~ and ~q guaranteed by Theorem 11.27 exist. Now we need to show that they are unique. Suppose ~v = p~ + ~q = p~ 0 + ~q 0 where the vectors p~ 0 and ~q 0 satisfy the same properties described in Theorem 11.27 as p~ and ~q. Then p~ p~ 0 = ~q 0 ~q, so w ~ (~ p p~ 0 ) = w ~ (~q 0 ~q) = w ~ ~q 0 w ~ ~q = 0 0 = 0. Hence, 0 0 w ~ (~ p p~ ) = 0. Now there are scalars k and k so that p~ = k w ~ and p~ 0 = k 0 w. ~ This means 0 0 0 0 w ~ (~ p p~ ) = w ~ (k w ~ k w) ~ =w ~ ([k k ]w) ~ = (k k )(w ~ w) ~ = (k k 0 )kwk ~ 2 . Since w ~ 6= ~0, 2 0 0 2 0 kwk ~ 6= 0, which means the only way w ~ (~ p p~ ) = (k k )kwk ~ = 0 is for k k = 0, or k = k 0 . This means p~ = k w ~ = k 0w ~ = p~ 0 . With ~q 0 ~q = p~ p~ 0 = p~ p~ = ~0, it must be that ~q 0 = ~q as well. Hence, we have shown there is only one way to write ~v as a sum of vectors as described in Theorem 11.27. We close this section with an application of the dot product. In Physics, if a constant force F is exerted over a distance d, the work W done by the force is given by W = F d. Here, we assume the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to find the work done. Consider the scenario below where the constant force F~ is applied to move an object from the point P to the point Q. F~ P F~ Q To find the work W done in this scenario, we need to find how much of the force F~ is in the d direction of the motion P Q. This is precisely what the dot product F~ P Q represents. Since d d the distance the object travels is kP Qk, we get W = (F~ P Q)kP Qk. Since P Q = kP QkP Q, d ~ d ~ ~ ~ W = (F P Q)kP Qk = F (kP QkP Q) = F P Q = kF kkP Qk cos(), where is the angle between the applied force F~ and the trajectory of the motion P Q. We have proved the following. 1042 Applications of Trigonometry Theorem 11.28. Work as a Dot Product: Suppose a constant force F~ is applied along the vector P Q. The work W done by F~ is given by W = F~ P Q = kF~ kkP Qk cos(), where is the angle between F~ and P Q. Example 11.9.5. Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a 30 angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a 30 angle for the duration of the 50 feet. 30 Solution. There are two ways to attack this problem. One way is to find the vectors F~ and P Q mentioned in Theorem 11.28 and compute W = F~ P Q. To do this, we assume the origin is at the point where the handle of the wagon meets the wagon and the positive x-axis lies along the dashed line in the figure above. Since the force applied is a constant 10 pounds, we have kF~ k = 10. Since it is being applied at a constant angle of = 30 with respect to the positive x-axis, Definition ~ 11.8 gives us F = 10 hcos(30 , sin(30 )i = 5 3, 5 . Since the wagon is being pulled along 50 feet in the positive direction, the displacement vector is P Q = 50 = 50 h1, 0i = h50, 0i. We get W = F~ P Q = 5 3, 5 h50, 0i = 250 3. Since force is measured in pounds and distance is measured in feet, we get W = 250 3 foot-pounds. Alternatively, we can use the formulation W = kF~ kkP Qk cos() to get W = (10 pounds)(50 feet) cos (30 ) = 250 3 foot-pounds of work. 11.9 The Dot Product and Projection 11.9.1 1043 Exercises In Exercises 1 - 20, use the pair of vectors ~v and w ~ to find the following quantities. ~v w ~ projw~ (~v ) The angle (in degrees) between ~v and w ~ ~ q = ~v projw~ (~v ) (Show that ~q w ~ = 0.) 1. ~v = h2, 7i and w ~ = h5, 9i 2. ~v = h6, 5i and w ~ = h10, 12i ~ = 1, 3 3. ~v = 1, 3 and w 4. ~v = h3, 4i and w ~ = h6, 8i 5. ~v = h2, 1i and w ~ = h3, 6i 6. ~v = 3 3, 3 and w ~ = 3, 1 7. ~v = h1, 17i and w ~ = h1, 0i 8. ~v = h3, 4i and w ~ = h5, 12i 9. ~v = h4, 2i and w ~ = h1, 5i 10. ~v = h5, 6i and w ~ = h4, 7i 11. ~v = h8, 3i and w ~ = h2, 6i 12. ~v = h34, 91i and w ~ = h0, 1i 13. ~v = 3 and w ~ = 4 14. ~v = 24 + 7 and w ~ = 2 15. ~v = 32 + 32 and w ~ = D E D E ~ = 22 , 22 17. ~v = 12 , 23 and w 16. ~v = 5 + 12 and w ~ = 3 + 4 D E D E 18. ~v = 22 , 22 and w ~ = 12 , 23 19. ~v = D 3 1 2 ,2 E D E and w ~ = 22 , 22 20. ~v = D E 3 1 , 2 2 and w ~= D E 2 2 , 2 2 21. A force of 1500 pounds is required to tow a trailer. Find the work done towing the trailer along a flat stretch of road 300 feet. Assume the force is applied in the direction of the motion. 22. Find the work done lifting a 10 pound book 3 feet straight up into the air. Assume the force of gravity is acting straight downwards. 23. Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a 15 angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to