OpenStax-CNX module: m49412 1 Vectors OpenStax College This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 4.0 In this section you will: Abstract View vectors geometrically. Find magnitude and direction. Perform vector addition and scalar multiplication. Find the component form of a vector. Find the unit vector in the direction of v. Perform operations with vectors in terms of i and j. Find the dot product of two vectors. An airplane is ying at an airspeed of 200 miles per hour headed on a SE bearing of 140. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in Figure 1. What are the ground speed and actual bearing of the plane? Version 1.6: Mar 11, 2015 1:56 pm +0000 http://creativecommons.org/licenses/by/4.0/
OpenStax-CNX module: m49412 2 Figure 1 Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the eect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will nd the airplane's groundspeed and bearing, while investigating another approach to problems of this type. First, however, let's examine the basics of vectors. 1 A Geometric View of Vectors A vector is a specic quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is dened by its magnitude, or the
OpenStax-CNX module: m49412 3 length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities: Lower case, boldfaced type, with or without an arrow on top such as v, u, w, v, u, w. Given initial point P and terminal point Q, a vector can be represented as P Q. The arrowhead on top is what indicates that it is not just a line, but a directed line segment. Given an initial point of (0, 0) and terminal point (a, b), a vector may be represented as< a, b >. This last symbol < a, b > has special signicance. It is called the standard position. The position vector has an initial point (0, 0) and a terminal point< a, b >.To change any vector into the position vector, we think about the change in the x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector CD is C (x 1, y 1 ) and the terminal point is D (x 2, y 2 ), then the position vector is found by calculating In Figure 2, we see the original vector AB = < x 2 x 1, y 2 y 1 > = < a, b > CD and the position vector AB. (1) Figure 2 a general note label: A vector is a directed line segment with an initial point and a terminal point. Vectors are identied by magnitude, or the length of the line, and direction, represented
OpenStax-CNX module: m49412 4 by the arrowhead pointing toward the terminal point. at (0, 0) and is identied by its terminal point< a, b >. The position vector has an initial point Example 1 Find the Position Vector Consider the vector whose initial point is P (2, 3) and terminal point is Q (6, 4). Find the position vector. Solution The position vector is found by subtracting one x-coordinate from the other x-coordinate, and one y-coordinate from the other y-coordinate. Thus v =< 6 2, 4 3 > =< 4, 1 > The position vector begins at (0, 0) and terminates at (4, 1). The graphs of both vectors are shown in Figure 3. (2) Figure 3 We see that the position vector is< 4, 1 >.
OpenStax-CNX module: m49412 5 Example 2 Drawing a Vector with the Given Criteria and Its Equivalent Position Vector Find the position vector given that vector v has an initial point at ( 3, 2) and a terminal point at (4, 5), then graph both vectors in the same plane. Solution The position vector is found using the following calculation: v =< 4 ( 3), 5 2 > =< 7, 3 > (3) Thus, the position vector begins at (0, 0) and terminates at (7, 3). See Figure 4. Figure 4 try it feature:
OpenStax-CNX module: m49412 6 Exercise 3 (Solution on p. 51.) Draw a vector v that connects from the origin to the point (3, 5). 2 Finding Magnitude and Direction To work with a vector, we need to be able to nd its magnitude and its direction. We nd its magnitude using the Pythagorean Theorem or the distance formula, and we nd its direction using the inverse tangent function. a general note label: Given a position vector v=< a, b >,the magnitude is found by v = a2 + b 2.The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tan θ = ( ) ( b a θ = tan 1 b a), as illustrated in Figure 5. Figure 5 Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal. Example 3 Finding the Magnitude and Direction of a Vector Find the magnitude and direction of the vector with initial point P ( 8, 1) and terminal point Q ( 2, 5).Draw the vector. Solution First, nd the position vector. u =< 2, ( 8), 5 1 > =< 6, 6 > We use the Pythagorean Theorem to nd the magnitude. (4)
OpenStax-CNX module: m49412 7 The direction is given as u = (6) 2 + ( 6) 2 = 72 = 6 2 (5) tan θ = 6 6 = 1 θ = tan 1 ( 1) = 45 (6) However, the angle terminates in the fourth quadrant, so we add 360 to obtain a positive angle. Thus, 45 + 360 = 315. See Figure 6. Figure 6 Example 4 Showing That Two Vectors Are Equal Show that vector v with initial point at (5, 3) and terminal point at ( 1, 2) is equal to vector u with initial point at ( 1, 3) and terminal point at ( 7, 2). Draw the position vector on the same grid as v and u. Next, nd the magnitude and direction of each vector.
OpenStax-CNX module: m49412 8 Solution As shown in Figure 7, draw the vector v starting at initial (5, 3) and terminal point ( 1, 2). Draw the vector u with initial point ( 1, 3) and terminal point ( 7, 2). Find the standard position for each. Next, nd and sketch the position vector for v and u. We have v =< 1 5, 2 ( 3) > =< 6, 5 > u =< 7 ( 1), 2 ( 3) > =< 6, 5 > Since the position vectors are the same, v and u are the same. An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem. v = ( 1 5) 2 + (2 ( 3)) 2 = ( 6) 2 + (5) 2 = 36 + 25 = 61 u = ( 7 ( 1)) 2 + (2 ( 3)) 2 = ( 6) 2 + (5) 2 = 36 + 25 = 61 As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives (7) (8) tan θ = 5 6 θ = ( ) tan 1 5 6 (9) = 39.8 However, we can see that the position vector terminates in the second quadrant, so we add 180. Thus, the direction is 39.8 + 180 = 140.2.
OpenStax-CNX module: m49412 9 Figure 7 3 Performing Vector Addition and Scalar Multiplication Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector u=< x, y >as an arrow or directed line segment from the origin to the point (x, y), vectors can be situated anywhere in the plane. The sum of two vectors u and v, or vector addition, produces a third vector u+ v, the resultant vector. To nd u + v, we rst draw the vector u, and from the terminal end of u, we drawn the vector v. In other words, we have the initial point of v meet the terminal end of u. This position corresponds to the
OpenStax-CNX module: m49412 10 notion that we move along the rst vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in Figure 8. Figure 8 Vector subtraction is similar to vector addition. To nd u v, view it as u + ( v). Adding v is reversing direction of v and adding it to the end of u. The new vector begins at the start of u and stops at the end point of v. See Figure 9 for a visual that compares vector addition and vector subtraction using parallelograms. Figure 9 Example 5 Adding and Subtracting Vectors Given u=< 3, 2 >andv=< 1, 4 >,nd two new vectors u + v, and u v. Solution To nd the sum of two vectors, we add the components. Thus,
OpenStax-CNX module: m49412 11 u + v =< 3, 2 > + < 1, 4 > =< 3 + ( 1), 2 + 4 > =< 2, 2 > See Figure 10(a). To nd the dierence of two vectors, add the negative components of v to u. Thus, (10) See Figure 10(b). u + ( v) =< 3, 2 > + < 1, 4 > =< 3 + 1, 2 + ( 4) > =< 4, 6 > (11) Figure 10: (a) Sum of two vectors (b) Dierence of two vectors 4 Multiplying By a Scalar While adding and subtracting vectors gives us a new vector with a dierent magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no eect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector. Scalar multiplication involves the product of a vector and a scalar. A general note label: Each component of the vector is multiplied by the scalar. Thus, to multiply v=< a, b > by k, we have
OpenStax-CNX module: m49412 12 kv =< ka, kb > (12) Only the magnitude changes, unless k is negative, and then the vector reverses direction. Example 6 Performing Scalar Multiplication Given vector v=< 3, 1 >, nd 3v, 1 2v, and v. Solution See Figure 11 for a geometric interpretation. If v=< 3, 1 >,then 3v =< 3 3, 3 1 > =< 9, 3 > 1 2 v =< 1 2 3, 1 2 1 > =< 3 2, 1 2 > v =< 3, 1 > (13) Figure 11
OpenStax-CNX module: m49412 13 Analysis Notice that the vector 3v is three times the length of v, 1 2v is half the length of v, and v is the same length of v, but in the opposite direction. Try it Feature: Exercise 8 (Solution on p. 51.) Find the scalar multiple 3u given u=< 5, 4 >. Example 7 Using Vector Addition and Scalar Multiplication to Find a New Vector Given u=< 3, 2 >andv=< 1, 4 >,nd a new vector w = 3u + 2v. Solution First, we must multiply each vector by the scalar. Then, add the two together. 3u = 3 < 3, 2 > =< 9, 6 > 2v = 2 < 1, 4 > =< 2, 8 > (14) So, w=< 7, 2 >. w = 3u + 2v =< 9, 6 > + < 2, 8 > =< 9 2, 6 + 8 > =< 7, 2 > (15) 5 Finding Component Form In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the x direction, and the vertical component is the y direction. For example, we can see in the graph in Figure 12 that the position vector< 2, 3 >comes from adding the vectors v 1 and v 2. We have v 1 with initial point (0, 0) and terminal point (2, 0). v 1 =< 2 0, 0 0 > =< 2, 0 > (16) We also have v 2 with initial point (0, 0) and terminal point (0, 3).
OpenStax-CNX module: m49412 14 Therefore, the position vector is v 2 =< 0 0, 3 0 > =< 0, 3 > (17) v =< 2 + 0, 3 + 0 > =< 2, 3 > Using the Pythagorean Theorem, the magnitude of v 1 is 2, and the magnitude of v 2 is 3. To nd the magnitude of v, use the formula with the position vector. (18) v = v 1 2 + v 2 2 = 2 2 + 3 2 = 13 (19) The magnitude of v is 13. To nd the direction, we use the tangent function tan θ = y x. tan θ = v2 v 1 tan θ = 3 2 θ = tan ( ) 1 3 2 = 56.3 (20)
OpenStax-CNX module: m49412 15 Figure 12 Thus, the magnitude of v is 13 and the direction is 56.3 o the horizontal. Example 8 Finding the Components of the Vector Find the components of the vector v with initial point (3, 2) and terminal point (7, 4). Solution First nd the standard position. v =< 7 3, 4 2 > =< 4, 2 > (21) See the illustration in Figure 13.
OpenStax-CNX module: m49412 16 Figure 13 The horizontal component is v 1 =< 4, 0 > and the vertical component is v 2 =< 0, 2 >. 6 Finding the Unit Vector in the Direction of v In addition to nding a vector's components, it is also useful in solving problems to nd a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. We can then preserve the direction of the original vector while simplifying calculations. Unit vectors are dened in terms of components. The horizontal unit vector is written as i=< 1, 0 >and is directed along the positive horizontal axis. The vertical unit vector is written asj=< 0, 1 >and is directed along the positive vertical axis. See Figure 14.
OpenStax-CNX module: m49412 17 Figure 14 a general note label: If v is a nonzero vector, then v v is a unit vector in the direction of v. Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar. Example 9 Finding the Unit Vector in the Direction of v Find a unit vector in the same direction as v=< 5, 12 >. Solution First, we will nd the magnitude. v = ( 5) 2 + (12) 2 = 25 + 144 = 169 = 13 Then we divide each component by v, which gives a unit vector in the same direction as v: (22) or, in component form v v = 5 13 i + 12 13 j (23) v v =< 5 13, 12 13 > (24)
OpenStax-CNX module: m49412 18 See Figure 15.
OpenStax-CNX module: m49412 19 Figure 15
OpenStax-CNX module: m49412 20 Verify that the magnitude of the unit vector equals 1. The magnitude of 5 13 i + 12 13j is given as The vector u= 5 13 ( ) 5 2 ( 13 + 12 ) 2 13 = 25 169 + 144 169 169 = 169 = 1 (25) 12 i+ 13 j is the unit vector in the same direction as v=< 5, 12 >. 7 Performing Operations with Vectors in Terms of i and j So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j. a general note label: Q = (x 2, y 2 ),v is written as Given a vector v with initial point P = (x 1, y 1 ) and terminal point v = (x 2 x 1 ) i + (y 1 y 2 ) j (26) The position vector from (0, 0) to (a, b), where (x 2 x 1 ) = a and (y 2 y 1 ) = b, is written as v = ai + bj. This vector sum is called a linear combination of the vectors i and j. The magnitude of v = ai + bj is given as v = a 2 + b 2. See Figure 16. Figure 16
OpenStax-CNX module: m49412 21 Example 10 Writing a Vector in Terms of i and j Given a vector v with initial point P = (2, 6) and terminal point Q = ( 6, 6), write the vector in terms of i and j. Solution Begin by writing the general form of the vector. Then replace the coordinates with the given values. v = (x 2 x 1 ) i + (y 2 y 1 ) j = ( 6 2) i + (6 ( 6)) j = 8i + 12j (27) Example 11 Writing a Vector in Terms of i and j Using Initial and Terminal Points Given initial point P 1 = ( 1, 3) and terminal point P 2 = (2, 7), write the vector v in terms of i and j. Solution Begin by writing the general form of the vector. Then replace the coordinates with the given values. v = (x 2 x 1 ) i + (y 2 y 1 ) j v = (2 ( 1)) i + (7 3) j = 3i + 4j (28) try it feature: Exercise 14 (Solution on p. 51.) Write the vector u with initial point P = ( 1, 6) and terminal point Q = (7, 5) in terms of i and j. 8 Performing Operations on Vectors in Terms of i and j When vectors are written in terms of i and j, we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components. a general note label: Given v = ai + bj and u = ci + dj, then v + u = (a + c) i + (b + d) j v u = (a c) i + (b d) j (29)
OpenStax-CNX module: m49412 22 Example 12 Finding the Sum of the Vectors Find the sum of v 1 = 2i 3j and v 2 = 4i + 5j. Solution According to the formula, we have v 1 + v 2 = (2 + 4) i + ( 3 + 5) j = 6i + 2j (30) 9 Calculating the Component Form of a Vector: Direction We have seen how to draw vectors according to their initial and terminal points and how to nd the position vector. We have also examined notation for vectors drawn specically in the Cartesian coordinate plane using i and j. For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction. Calculating direction follows the same straightforward process we used for polar coordinates. We nd the direction of the vector by nding the angle to the horizontal. We do this by using the basic trigonometric identities, but with v replacing r. a general note label: Given a position vector v =< x, y > and a direction angle θ, cos θ = x v and sin θ = y v x = v cos θ y = v sin θ (31) Thus, v = xi + yj = v cos θi + v sin θj, and magnitude is expressed as v = x 2 + y 2. Example 13 Writing a Vector in Terms of Magnitude and Direction Write a vector with length 7 at an angle of 135 to the positive x-axis in terms of magnitude and direction. Solution Using the conversion formulas x = v cos θi and y = v sin θj, we nd that x = 7cos (135 ) i = 7 2 2 y = 7sin (135 ) j (32) = 7 2 2 This vector can be written as v = 7cos (135 ) i + 7sin (135 ) j or simplied as v = 7 2 2 i + 7 2 2 j (33)
OpenStax-CNX module: m49412 23 trey it feature: Exercise 17 (Solution on p. 51.) A vector travels from the origin to the point (3, 5). Write the vector in terms of magnitude and direction. 10 Finding the Dot Product of Two Vectors As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses. The dot product of two vectors involves multiplying two vectors together, and the result is a scalar. a general note label: The dot product of two vectors v =< a, b > and u =< c, d > is the sum of the product of the horizontal components and the product of the vertical components. To nd the angle between the two vectors, use the formula below. v u = ac + bd (34) cos θ = v v u u Example 14 Finding the Dot Product of Two Vectors Find the dot product of v =< 5, 12 > and u =< 3, 4 >. Solution Using the formula, we have (35) v u =< 5, 12 > < 3, 4 > = 5 ( 3) + 12 4 = 15 + 48 = 33 (36) Example 15 Finding the Dot Product of Two Vectors and the Angle between Them Find the dot product of v 1 = 5i + 2j and v 2 = 3i + 7j. Then, nd the angle between the two vectors. Solution Finding the dot product, we multiply corresponding components.
OpenStax-CNX module: m49412 24 v 1 v 2 =< 5, 2 > < 3, 7 > = 5 3 + 2 7 = 15 + 14 = 29 To nd the angle between them, we use the formula cos θ = v v u u. (37) See Figure 17. v v u u =< 5 29 + 2 29 > < 3 58 + 7 58 > = 5 3 29 58 + 2 7 29 58 = 15 1682 + 14 1682 = 29 1682 = 0.707107 cos 1 (0.707107) = 45 (38)
OpenStax-CNX module: m49412 25 Figure 17 Example 16 Finding the Angle between Two Vectors Find the angle between u =< 3, 4 > and v =< 5, 12 >. Solution Using the formula, we have
OpenStax-CNX module: m49412 26 See Figure 18. ( u u v v ( θ = cos 1 ) u u v v ) = 3i+4j 5 5i+12j 13 = ( 3 5 5 13 = 15 65 + 48 65 = 33 65 θ = cos ( 1 33 65 = 59.5 ) + ( 4 5 12 13 ) ) (39)
OpenStax-CNX module: m49412 27 Figure 18
OpenStax-CNX module: m49412 28 Example 17 Finding Ground Speed and Bearing Using Vectors We now have the tools to solve the problem we introduced in the opening of the section. An airplane is ying at an airspeed of 200 miles per hour headed on a SE bearing of 140. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See Figure 19. Figure 19
OpenStax-CNX module: m49412 29 Solution The ground speed is represented by x in the diagram, and we need to nd the angle α in order to calculate the adjusted bearing, which will be 140 + α. Notice in Figure 19, that angle BCO must be equal to angle AOC by the rule of alternating interior angles, so angle BCO is 140. We can nd x by the Law of Cosines: x 2 = (16.2) 2 + (200) 2 2 (16.2) (200) cos (140 ) x 2 = 45, 226.41 x = 45, 226.41 x = 212.7 The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines. sin α 16.2 = sin(140 ) 212.7 sin α = 16.2sin(140 ) 212.7 = 0.04896 sin 1 (0.04896) = 2.8 Therefore, the plane has a SE bearing of 140 +2.8 =142.8. The ground speed is 212.7 miles per hour. (40) (41) media feature label: vectors. Access these online resources for additional instruction and practice with Introduction to Vectors 1 Vector Operations 2 The Unit Vector 3 11 Key Concepts The position vector has its initial point at the origin. See Example 1. If the position vector is the same for two vectors, they are equal. See Example 2. Vectors are dened by their magnitude and direction. See Example 3. If two vectors have the same magnitude and direction, they are equal. See Example 4. Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See Example 5. Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See Example 6 and Example 7. Vectors are comprised of two components: the horizontal component along the positive x-axis, and the vertical component along the positive y-axis. See Example 8. 1 http://openstaxcollege.org/l/introvectors 2 http://openstaxcollege.org/l/vectoroperation 3 http://openstaxcollege.org/l/unitvector
OpenStax-CNX module: m49412 30 The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude. The magnitude of a vector in the rectangular coordinate system is v = a 2 + b 2. See Example 9. In the rectangular coordinate system, unit vectors may be represented in terms of i and j where i represents the horizontal component and j represents the vertical component. Then, v = ai + bj is a scalar multiple of v by real numbers a and b. See Example 10 and Example 11. Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coecients of i and corresponding coecients of j. See Example 12. A vector v = ai + bj is written in terms of magnitude and direction as v = v cos θi + v sin θj. See Example 13. The dot product of two vectors is the product of the i terms plus the product of the j terms. See Example 14. We can use the dot product to nd the angle between two vectors. Example 15 and Example 16. Dot products are useful for many types of physics applications. See Example 17. 12 Section Exercises 12.1 Verbal Exercise 22 (Solution on p. 51.) What are the characteristics of the letters that are commonly used to represent vectors? Exercise 23 How is a vector more specic than a line segment? Exercise 24 (Solution on p. 51.) What are i and j,and what do they represent? Exercise 25 What is component form? Exercise 26 (Solution on p. 51.) When a unit vector is expressed as< a, b >,which letter is the coecient of the i and which the j? 12.2 Algebraic Exercise 27 Given a vector with initial point (5, 2) and terminal point ( 1, 3), nd an equivalent vector whose initial point is (0, 0). Write the vector in component form< a, b >. Exercise 28 (Solution on p. 51.) Given a vector with initial point ( 4, 2) and terminal point (3, 3), nd an equivalent vector whose initial point is (0, 0). Write the vector in component form< a, b >. Exercise 29 Given a vector with initial point (7, 1) and terminal point ( 1, 7), nd an equivalent vector whose initial point is (0, 0). Write the vector in component form< a, b >. For the following exercises, determine whether the two vectors u and v are equal, where u has an initial point P 1 and a terminal point P 2 and v has an initial point P 3 and a terminal point P 4. Exercise 30 (Solution on p. 51.) P 1 = (5, 1), P 2 = (3, 2), P 3 = ( 1, 3), and P 4 = (9, 4) Exercise 31 P 1 = (2, 3), P 2 = (5, 1), P 3 = (6, 1), and P 4 = (9, 3)
OpenStax-CNX module: m49412 31 Exercise 32 (Solution on p. 51.) P 1 = ( 1, 1), P 2 = ( 4, 5), P 3 = ( 10, 6), and P 4 = ( 13, 12) Exercise 33 P 1 = (3, 7), P 2 = (2, 1), P 3 = (1, 2), and P 4 = ( 1, 4) Exercise 34 (Solution on p. 51.) P 1 = (8, 3), P 2 = (6, 5), P 3 = (11, 8), andp 4 = (9, 10) Exercise 35 Given initial point P 1 = ( 3, 1) and terminal point P 2 = (5, 2), write the vector v in terms of i and j. Exercise 36 (Solution on p. 51.) Given initial point P 1 = (6, 0) and terminal point P 2 = ( 1, 3), write the vector v in terms of i and j. For the following exercises, use the vectors u = i + 5j, v = 2i 3j, and w = 4i j. Exercise 37 Find u + (v w) Exercise 38 (Solution on p. 52.) Find 4v + 2u For the following exercises, use the given vectors to compute u + v, u v, and 2u 3v. Exercise 39 u =< 2, 3 >, v =< 1, 5 > Exercise 40 (Solution on p. 52.) u =< 3, 4 >, v =< 2, 1 > Exercise 41 Let v = 4i + 3j. Find a vector that is half the length and points in the same direction as v. Exercise 42 (Solution on p. 52.) Let v = 5i + 2j. Find a vector that is twice the length and points in the opposite direction as v. For the following exercises, nd a unit vector in the same direction as the given vector. Exercise 43 a = 3i + 4j Exercise 44 (Solution on p. 52.) b = 2i + 5j Exercise 45 c = 10i j Exercise 46 (Solution on p. 52.) d = 1 3 i + 5 2 j Exercise 47 u = 100i + 200j Exercise 48 (Solution on p. 52.) u = 14i + 2j For the following exercises, nd the magnitude and direction of the vector, 0 θ < 2π. Exercise 49 < 0, 4 > Exercise 50 (Solution on p. 52.) < 6, 5 > Exercise 51 < 2, 5 >
OpenStax-CNX module: m49412 32 Exercise 52 (Solution on p. 52.) < 4, 6 > Exercise 53 Given u = 3i 4j and v = 2i + 3j, calculate u v. Exercise 54 (Solution on p. 52.) Given u = i j and v = i + 5j, calculate u v. Exercise 55 Given u =< 2, 4 > and v =< 3, 1 >, calculate u v. Exercise 56 (Solution on p. 52.) Given u=< 1, 6 >and v=< 6, 1 >,calculate u v. 12.3 Graphical For the following exercises, given v, drawv,3v and 1 2 v. Exercise 57 < 2, 1 > Exercise 58 (Solution on p. 52.) < 1, 4 > Exercise 59 < 3, 2 > For the following exercises, use the vectors shown to sketch u + v, u v, and 2u. Exercise 60 (Solution on p. 53.)
OpenStax-CNX module: m49412 33 Exercise 61 Exercise 62 (Solution on p. 53.) For the following exercises, use the vectors shown to sketch 2u + v.
OpenStax-CNX module: m49412 34 Exercise 63
OpenStax-CNX module: m49412 35 Exercise 64 (Solution on p. 54.) For the following exercises, use the vectors shown to sketch u 3v.
OpenStax-CNX module: m49412 36 Exercise 65
OpenStax-CNX module: m49412 37 Exercise 66 (Solution on p. 54.) For the following exercises, write the vector shown in component form. Exercise 67 Exercise 68 (Solution on p. 55.) Exercise 69 Given initial point P 1 = (2, 1) and terminal point P 2 = ( 1, 2), write the vector v in terms of i and j, then draw the vector on the graph.
OpenStax-CNX module: m49412 38 Exercise 70 (Solution on p. 55.) Given initial point P 1 = (4, 1) and terminal point P 2 = ( 3, 2), write the vector v in terms of i and j. Draw the points and the vector on the graph. Exercise 71 Given initial point P 1 = (3, 3) and terminal point P 2 = ( 3, 3), write the vector v in terms of i and j. Draw the points and the vector on the graph. 12.4 Extensions For the following exercises, use the given magnitude and direction in standard position, write the vector in component form. Exercise 72 (Solution on p. 56.) v = 6, θ = 45 Exercise 73 v = 8, θ = 220 Exercise 74 (Solution on p. 56.) v = 2, θ = 300 Exercise 75 v = 5, θ = 135 Exercise 76 (Solution on p. 56.) A 60-pound box is resting on a ramp that is inclined 12. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) component of the force. b. Find the magnitude of the component of the force that is parallel to the ramp. Exercise 77 A 25-pound box is resting on a ramp that is inclined 8. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) component of the force. b. Find the magnitude of the component of the force that is parallel to the ramp. Exercise 78 (Solution on p. 56.) Find the magnitude of the horizontal and vertical components of a vector with magnitude 8 pounds pointed in a direction of 27 above the horizontal. Round to the nearest hundredth. Exercise 79 Find the magnitude of the horizontal and vertical components of the vector with magnitude 4 pounds pointed in a direction of 127 above the horizontal. Round to the nearest hundredth. Exercise 80 (Solution on p. 56.) Find the magnitude of the horizontal and vertical components of a vector with magnitude 5 pounds pointed in a direction of 55 above the horizontal. Round to the nearest hundredth. Exercise 81 Find the magnitude of the horizontal and vertical components of the vector with magnitude 1 pound pointed in a direction of 8 above the horizontal. Round to the nearest hundredth.
OpenStax-CNX module: m49412 39 12.5 Real-World Applications Exercise 82 (Solution on p. 56.) A woman leaves home and walks 3 miles west, then 2 miles southwest. How far from home is she, and in what direction must she walk to head directly home? Exercise 83 A boat leaves the marina and sails 6 miles north, then 2 miles northeast. How far from the marina is the boat, and in what direction must it sail to head directly back to the marina? Exercise 84 (Solution on p. 57.) A man starts walking from home and walks 4 miles east, 2 miles southeast, 5 miles south, 4 miles southwest, and 2 miles east. How far has he walked? If he walked straight home, how far would he have to walk? Exercise 85 A woman starts walking from home and walks 4 miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk? Exercise 86 (Solution on p. 57.) A man starts walking from home and walks 3 miles at 20 north of west, then 5 miles at 10 west of south, then 4 miles at 15 north of east. If he walked straight home, how far would he have to the walk, and in what direction? Exercise 87 A woman starts walking from home and walks 6 miles at 40 north of east, then 2 miles at 15 east of south, then 5 miles at 30 south of west. If she walked straight home, how far would she have to walk, and in what direction? Exercise 88 (Solution on p. 57.) An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees o course will the plane end up ying, and what is the plane's speed relative to the ground? Exercise 89 An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees o course will the plane end up ying, and what is the plane's speed relative to the ground? Exercise 90 (Solution on p. 57.) An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane ies with an airspeed of 550 km/hr. To end up ying due north, how many degrees west of north will the pilot need to y the plane? Exercise 91 An airplane needs to head due north, but there is a wind blowing from the northwest at 80 km/hr. The plane ies with an airspeed of 500 km/hr. To end up ying due north, how many degrees west of north will the pilot need to y the plane? Exercise 92 (Solution on p. 57.) As part of a video game, the point (5, 7) is rotated counterclockwise about the origin through an angle of 35. Find the new coordinates of this point. Exercise 93 As part of a video game, the point (7, 3) is rotated counterclockwise about the origin through an angle of 40. Find the new coordinates of this point.
OpenStax-CNX module: m49412 40 Exercise 94 (Solution on p. 57.) Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 10 mph relative to the car, and the car is traveling 25 mph down the road. If one child doesn't catch the ball, and it ies out the window, in what direction does the ball y (ignoring wind resistance)? Exercise 95 Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 8 mph relative to the car, and the car is traveling 45 mph down the road. If one child doesn't catch the ball, and it ies out the window, in what direction does the ball y (ignoring wind resistance)? Exercise 96 (Solution on p. 57.) A 50-pound object rests on a ramp that is inclined 19. Find the magnitude of the components of the force parallel to and perpendicular to (normal) the ramp to the nearest tenth of a pound. Exercise 97 Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it upward, and 5 pounds acting on it 45 from the horizontal. What single force is the resultant force acting on the body? Exercise 98 (Solution on p. 57.) Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it [U+2500]135 from the horizontal, and 5 pounds acting on it directed 150 from the horizontal. What single force is the resultant force acting on the body? Exercise 99 The condition of equilibrium is when the sum of the forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45 from the horizontal. What single force is needed to produce a state of equilibrium on the body? Exercise 100 (Solution on p. 57.) Suppose a body has a force of 3 pounds acting on it to the left, 4 pounds acting on it upward, and 2 pounds acting on it 30 from the horizontal. What single force is needed to produce a state of equilibrium on the body? Draw the vector. 13 Chapter Review Exercises 13.1 Non-right Triangles: Law of Sines For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. Exercise 101 (Solution on p. 57.) β = 50, a = 105, b = 45 Exercise 102 α = 43.1, a = 184.2, b = 242.8 Exercise 103 (Solution on p. 57.) Solve the triangle.
OpenStax-CNX module: m49412 41 Exercise 104 Find the area of the triangle. Exercise 105 (Solution on p. 57.) A pilot is ying over a straight highway. He determines the angles of depression to two mileposts, 2.1 km apart, to be 25 and 49, as shown in Figure 20. Find the distance of the plane from point A and the elevation of the plane.
OpenStax-CNX module: m49412 42 Figure 20 13.2 Non-right Triangles: Law of Cosines Exercise 106 Solve the triangle, rounding to the nearest tenth, assuming α is opposite side a, β is opposite side b, and γ s opposite sidec : a = 4, b = 6, c = 8. Exercise 107 (Solution on p. 57.) Solve the triangle in Figure 21, rounding to the nearest tenth.
OpenStax-CNX module: m49412 43 Figure 21 Exercise 108 Find the area of a triangle with sides of length 8.3, 6.6, and 9.1. Exercise 109 (Solution on p. 57.) To nd the distance between two cities, a satellite calculates the distances and angle shown in Figure 22 (not to scale). Find the distance between the cities. Round answers to the nearest tenth.
OpenStax-CNX module: m49412 44 Figure 22 13.3 Polar Coordinates Exercise 110 Plot the point with polar coordinates ( ) 3, π 6. Exercise 111 (Solution on p. 57.) Plot the point with polar coordinates ( ) 5, 2π 3 Exercise 112 Convert ( ) 6, 3π 4 to rectangular coordinates. Exercise 113 (Solution on p. 58.) Convert ( ) 2, 3π 2 to rectangular coordinates. Exercise 114 Convert(7, 2)to polar coordinates. Exercise 115 (Solution on p. 58.) Convert( 9, 4) to polar coordinates. For the following exercises, convert the given Cartesian equation to a polar equation. Exercise 116 x = 2 Exercise 117 (Solution on p. 58.) x 2 + y 2 = 64 Exercise 118 x 2 + y 2 = 2y
OpenStax-CNX module: m49412 45 For the following exercises, convert the given polar equation to a Cartesian equation. Exercise 119 (Solution on p. 58.) r = 7cos θ Exercise 120 r = 2 4cos θ+sin θ For the following exercises, convert to rectangular form and graph. Exercise 121 (Solution on p. 58.) θ = 3π 4 Exercise 122 r = 5sec θ 13.4 Polar Coordinates: Graphs For the following exercises, test each equation for symmetry. Exercise 123 (Solution on p. 59.) r = 4 + 4sin θ Exercise 124 r = 7 Exercise 125 (Solution on p. 59.) Sketch a graph of the polar equation r = 1 5sin θ. Label the axis intercepts. Exercise 126 Sketch a graph of the polar equation r = 5sin (7θ). Exercise 127 (Solution on p. 60.) Sketch a graph of the polar equation r = 3 3cos θ 13.5 Polar Form of Complex Numbers For the following exercises, nd the absolute value of each complex number. Exercise 128 2 + 6i Exercise 129 (Solution on p. 61.) 4 3i Write the complex number in polar form. Exercise 130 5 + 9i Exercise 131 (Solution on p. 61.) 1 2 3 2 i For the following exercises, convert the complex number from polar to rectangular form. Exercise 132 z = 5cis ( ) 5π 6 Exercise 133 (Solution on p. 61.) z = 3cis (40 ) For the following exercises, nd the product z 1 z 2 in polar form.
OpenStax-CNX module: m49412 46 Exercise 134 z 1 = 2cis (89 ) z 2 = 5cis (23 ) Exercise 135 (Solution on p. 61.) z 1 = 10cis ( π 6 z 2 = 6cis ( π 3 ) ) For the following exercises, nd the quotient z1 z 2 in polar form. Exercise 136 z 1 = 12cis (55 ) z 2 = 3cis (18 ) Exercise 137 (Solution on p. 61.) z 1 = 27cis ( 5π 3 z 2 = 9cis ( π 3 ) ) For the following exercises, nd the powers of each complex number in polar form. Exercise 138 Find z 4 when z = 2cis (70 ) Exercise 139 (Solution on p. 61.) Find z 2 when z = 5cis ( 3π 4 ) For the following exercises, evaluate each root. Exercise 140 Evaluate the cube root of z when z = 64cis (210 ). Exercise 141 (Solution on p. 61.) Evaluate the square root of z when z = 25cis ( 3π 2 For the following exercises, plot the complex number in the complex plane. Exercise 142 6 2i Exercise 143 (Solution on p. 61.) 1 + 3i ). 13.6 Parametric Equations For the following exercises, eliminate the parameter t to rewrite the parametric equation as a Cartesian equation. Exercise 144 x (t) = 3t 1 { y (t) = t Exercise 145 (Solution on p. 62.) x (t) = cos t { y (t) = 2sin 2 t Exercise 146 Parameterize (write a parametric equation for) each Cartesian equation by using x (t) = acos t and y (t) = bsin t for x2 25 + y2 16 = 1. Exercise 147 (Solution on p. 62.) Parameterize the line from ( 2, 3) to (4, 7) so that the line is at ( 2, 3) at t = 0 and (4, 7) at t = 1.
OpenStax-CNX module: m49412 47 13.7 Parametric Equations: Graphs For the following exercises, make a table of values for each set of parametric equations, graph the equations, and include an orientation; then write the Cartesian equation. Exercise 148 x (t) = 3t2 { y (t) = 2t 1 Exercise 149 (Solution on p. 62.) x (t) = et { y (t) = 2e 5 t Exercise 150 x (t) = 3cos t { y (t) = 2sin t Exercise 151 (Solution on p. 63.) A ball is launched with an initial velocity of 80 feet per second at an angle of 40 to the horizontal. The ball is released at a height of 4 feet above the ground. a. Find the parametric equations to model the path of the ball. b. Where is the ball after 3 seconds? c. How long is the ball in the air? 13.8 Vectors (Section ) For the following exercises, determine whether the two vectors, u and v, are equal, where u has an initial point P 1 and a terminal point P 2, and v has an initial point P 3 and a terminal point P 4. Exercise 152 P 1 = ( 1, 4), P 2 = (3, 1), P 3 = (5, 5)and P 4 = (9, 2) Exercise 153 (Solution on p. 63.) P 1 = (6, 11), P 2 = ( 2, 8), P 3 = (0, 1) and P 4 = ( 8, 2) For the following exercises, use the vectors u = 2i j,v = 4i 3j, and w = 2i+5j to evaluate the expression. Exercise 154 u v Exercise 155 (Solution on p. 63.) 2v u + w For the following exercises, nd a unit vector in the same direction as the given vector. Exercise 156 a = 8i 6j Exercise 157 (Solution on p. 63.) b = 3i j For the following exercises, nd the magnitude and direction of the vector. Exercise 158 < 6, 2 >
OpenStax-CNX module: m49412 48 Exercise 159 (Solution on p. 63.) < 3, 3 > For the following exercises, calculate u v. Exercise 160 u = 2i + j and v = 3i + 7j Exercise 161 (Solution on p. 63.) u = i + 4j and v = 4i + 3j Exercise 162 Given v= 1 3, 42draw v, 2v, and 1 2 v. Exercise 163 (Solution on p. 63.) Given the vectors shown in Figure 23, sketch u + v, u v and 3v. Figure 23 Exercise 164 Given initial point P 1 = (3, 2) and terminal point P 2 = ( 5, 1), write the vector v in terms of i and j. Draw the points and the vector on the graph.
OpenStax-CNX module: m49412 49 14 Practice Test Exercise 165 (Solution on p. 64.) Assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve the triangle, if possible, and round each answer to the nearest tenth, given β = 68, b = 21, c = 16. Exercise 166 Find the area of the triangle in Figure 24. Round each answer to the nearest tenth. Figure 24 Exercise 167 (Solution on p. 64.) A pilot ies in a straight path for 2 hours. He then makes a course correction, heading 15 to the right of his original course, and ies 1 hour in the new direction. If he maintains a constant speed of 575 miles per hour, how far is he from his starting position? Exercise 168 Convert (2, 2) to polar coordinates, and then plot the point. Exercise 169 (Solution on p. 64.) Convert ( ) 2, π 3 to rectangular coordinates. Exercise 170 Convert the polar equation to a Cartesian equation: x 2 + y 2 = 5y. Exercise 171 (Solution on p. 64.) Convert to rectangular form and graph:r = 3csc θ. Exercise 172 Test the equation for symmetry: r = 4sin (2θ). Exercise 173 (Solution on p. 65.) Graph r = 3 + 3cos θ. Exercise 174 Graph r = 3 5sin θ. Exercise 175 (Solution on p. 66.) Find the absolute value of the complex number 5 9i. Exercise 176 Write the complex number in polar form: 4 + i.
OpenStax-CNX module: m49412 50 Exercise 177 (Solution on p. 66.) Convert the complex number from polar to rectangular form: z = 5cis ( ) 2π 3. Given z 1 = 8cis (36 ) and z 2 = 2cis (15 ),evaluate each expression. Exercise 178 z 1 z 2 Exercise 179 (Solution on p. 66.) z 1 z 2 Exercise 180 (z 2 ) 3 Exercise 181 z1 (Solution on p. 66.) Exercise 182 Plot the complex number 5 i in the complex plane. Exercise 183 (Solution on p. 66.) Eliminate the parameter t to rewrite the following parametric equations as a Cartesian equation: { x (t) = t + 1. y (t) = 2t 2 Exercise 184 Parameterize (write a parametric equation for) the following Cartesian equation by using x (t) = acos t and y (t) = bsin t : x2 36 + y2 100 = 1. Exercise 185 (Solution on p. 66.) x (t) = 2sin t Graph the set of parametric equations and nd the Cartesian equation: {. y (t) = 5cos t Exercise 186 A ball is launched with an initial velocity of 95 feet per second at an angle of 52 to the horizontal. The ball is released at a height of 3.5 feet above the ground. a. Find the parametric equations to model the path of the ball. b. Where is the ball after 2 seconds? c. How long is the ball in the air? For the following exercises, use the vectors u = i 3j and v = 2i + 3j. Exercise 187 (Solution on p. 67.) Find 2u 3v. Exercise 188 Calculate u v. Exercise 189 (Solution on p. 67.) Find a unit vector in the same direction as v. Exercise 190 Given vector v has an initial point P 1 = (2, 2) and terminal point P 2 = ( 1, 0), write the vector v in terms of i and j. On the graph, draw v, and v.
OpenStax-CNX module: m49412 51 Solutions to Exercises in this Module Solution to Exercise (p. 5) Solution to Exercise (p. 13) 3u =< 15, 12 > Solution to Exercise (p. 21) u = 8i 11j Solution to Exercise (p. 23) v = 34cos (59 ) i + 34sin (59 ) j Magnitude = 34 θ = tan ( ) 1 5 3 = 59.04 Solution to Exercise (p. 30) lowercase, bold letter, usually u, v, w Solution to Exercise (p. 30) They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1. Solution to Exercise (p. 30) The rst number always represents the coecient of the i, and the second represents the j. Solution to Exercise (p. 30) 17, 52 Solution to Exercise (p. 30) not equal Solution to Exercise (p. 30) equal Solution to Exercise (p. 31) equal
OpenStax-CNX module: m49412 52 Solution to Exercise (p. 31) 7i 3j Solution to Exercise (p. 31) 6i 2j Solution to Exercise (p. 31) u + v = 1 5, 52, u v = 1 1, 32, 2u 3v = 10, 52 Solution to Exercise (p. 31) 10i 4j Solution to Exercise (p. 31) 2 29 29 i + 5 29 29 j Solution to Exercise (p. 31) 2 229 229 i + 15 229 229 j Solution to Exercise (p. 31) 7 2 10 i + 2 10 j Solution to Exercise (p. 31) v = 7.810, θ = 39.806 Solution to Exercise (p. 31) v = 7.211, θ = 236.310 Solution to Exercise (p. 32) 6 Solution to Exercise (p. 32) 12
OpenStax-CNX module: m49412 53 Solution to Exercise (p. 32) Solution to Exercise (p. 32)
OpenStax-CNX module: m49412 54 tion to Exercise (p. 33) Solution to Exercise (p. 35) Solu
OpenStax-CNX module: m49412 55 tion to Exercise (p. 37) Solution to Exercise (p. 37) 14, 12 Solution to Exercise (p. 38) v = 7i + 3j
OpenStax-CNX module: m49412 56 to Exercise (p. 38) 3 2i + 3 2j Solution to Exercise (p. 38) i 3j Solution to Exercise (p. 38) a. 58.7; b. 12.5 Solution to Exercise (p. 38) x = 7.13 pounds, y = 3.63 pounds Solution to Exercise (p. 38) x = 2.87 pounds, y = 4.10 pounds Solution
OpenStax-CNX module: m49412 57 Solution to Exercise (p. 39) 4.635 miles, 17.764 N of E Solution to Exercise (p. 39) 17 miles. 10.318 miles Solution to Exercise (p. 39) Distance: 2.868. Direction: 86.474 North of West, or 3.526 West of North Solution to Exercise (p. 39) 4.924. 659 km/hr Solution to Exercise (p. 39) 4.424 Solution to Exercise (p. 39) (0.081, 8.602) Solution to Exercise (p. 40) 21.801, relative to the car's forward direction Solution to Exercise (p. 40) parallel: 16.28, perpendicular: 47.28 pounds Solution to Exercise (p. 40) 19.35 pounds, 231.54 from the horizontal Solution to Exercise (p. 40) 5.1583 pounds, 75.8 from the horizontal Solution to Exercise (p. 40) Not possible Solution to Exercise (p. 40) C = 120, a = 23.1, c = 34.1 Solution to Exercise (p. 41) distance of the plane from point A : 2.2 km, elevation of the plane: 1.6 km Solution to Exercise (p. 42) B = 71.0, C = 55.0, a = 12.8 Solution to Exercise (p. 43) 40.6 km Solution to Exercise (p. 44)
OpenStax-CNX module: m49412 58 Solution to Exercise (p. 44) (0, 2) Solution to Exercise (p. 44) (9.8489, 203.96 ) Solution to Exercise (p. 44) r = 8 Solution to Exercise (p. 45) x 2 + y 2 = 7x Solution to Exercise (p. 45) y = x
OpenStax-CNX module: m49412 59 to Exercise (p. 45) symmetric with respect to the lineθ = π 2 Solution to Exercise (p. 45) Solution
OpenStax-CNX module: m49412 60 Solution to Exercise (p. 45)
OpenStax-CNX module: m49412 61 Solution to Exercise (p. 45) 5 Solution to Exercise (p. 45) cis ( ) π 3 Solution to Exercise (p. 45) 2.3 + 1.9i Solution to Exercise (p. 46) 60cis ( ) π 2 Solution to Exercise (p. 46) 3cis ( ) 4π 3 Solution to Exercise (p. 46) 25cis ( ) 3π 2 Solution to Exercise (p. 46) 5cis ( ) ( 3π 4, 5cis 7π ) 4
OpenStax-CNX module: m49412 62 Solution to Exercise (p. 46) Solution to Exercise (p. 46) x 2 + 1 2 y = 1 Solution to Exercise (p. 46) x (t) = 2 + 6t { y (t) = 3 + 4t Solution to Exercise (p. 47) y = 2x 5
OpenStax-CNX module: m49412 63 Solution to Exercise (p. 47) a. { x (t) = (80cos (40 )) t y (t) = 16t 2 + (80sin (40 )) t + 4 b. The ball is 14 feet high and 184 feet from where it was launched. c. 3.3 seconds Solution to Exercise (p. 47) not equal Solution to Exercise (p. 47) 4i Solution to Exercise (p. 47) 3 10 10 i 10 10 Solution to Exercise (p. 47) Magnitude: 3 2, Direction:225 Solution to Exercise (p. 48) 16 Solution to Exercise (p. 48) j
OpenStax-CNX module: m49412 64 Solution to Exercise (p. 49) α = 67.1, γ = 44.9, a = 20.9 Solution to Exercise (p. 49) 1712 miles Solution to Exercise (p. 49) ( 1, 3 ) Solution to Exercise (p. 49) y = 3
OpenStax-CNX module: m49412 65 Solution to Exercise (p. 49)
OpenStax-CNX module: m49412 66 Solution to Exercise (p. 49) 106 Solution to Exercise (p. 49) 5 2 + i 5 3 2 Solution to Exercise (p. 50) 4cis (21 ) Solution to Exercise (p. 50) 2 2cis (18 ), 2 2cis (198 ) Solution to Exercise (p. 50) y = 2(x 1) 2 Solution to Exercise (p. 50)
OpenStax-CNX module: m49412 67 Solution to Exercise (p. 50) 4i 15j Solution to Exercise (p. 50) 2 13 13 i + 3 13 13 j Glossary Denition 1: dot product given two vectors, the sum of the product of the horizontal components and the product of the vertical components Denition 2: initial point the origin of a vector Denition 3: magnitude the length of a vector; may represent a quantity such as speed, and is calculated using the Pythagorean Theorem Denition 4: resultant a vector that results from addition or subtraction of two vectors, or from scalar multiplication
OpenStax-CNX module: m49412 68 Denition 5: scalar a quantity associated with magnitude but not direction; a constant Denition 6: scalar multiplication the product of a constant and each component of a vector Denition 7: standard position the placement of a vector with the initial point at (0, 0) and the terminal point (a, b), represented by the change in the x-coordinates and the change in the y-coordinates of the original vector Denition 8: terminal point the end point of a vector, usually represented by an arrow indicating its direction Denition 9: unit vector a vector that begins at the origin and has magnitude of 1; the horizontal unit vector runs along the x-axis and is dened as v 1 =< 1, 0 > the vertical unit vector runs along the y-axis and is dened as v 2 =< 0, 1 >. Denition 10: vector a quantity associated with both magnitude and direction, represented as a directed line segment with a starting point (initial point) and an end point (terminal point) Denition 11: vector addition the sum of two vectors, found by adding corresponding components