Vectors and the Geometry of Space

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1 Vectors and the Geometr of Space. Vectors in the Plane. Space Coordinates and Vectors in Space. The Dot Product of Two Vectors. The Cross Product of Two Vectors in Space.5 Lines and Planes in Space.6 Surfaces in Space.7 Clindrical and Spherical Coordinates Geograph (Eercise 5, p. 80) Torque (Eercise 9, p. 78) Work (Eercise 6, p. 77) Auditorium Lights (Eercise 0, p. 765) Naigation (Eercise 8, p. 757) Clockwise from top left, Denis Tabler/Shutterstock.com; Elena Elisseea/Shutterstock.com; Losesk Photo and Video/Shutterstock.com; Mikael Damkier/Shutterstock.com; Zia_K/iStockphoto.com Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it. 77

2 78 Chapter Vectors and the Geometr of Space. Vectors in the Plane Write the component form of a ector. Perform ector operations and interpret the results geometricall. Write a ector as a linear combination of standard unit ectors. P PQ Initial point A directed line segment Figure. Terminal point Equialent directed line segments Figure. Q Component Form of a Vector Man quantities in geometr and phsics, such as area, olume, temperature, mass, and time, can be characteried b a single real number that is scaled to appropriate units of measure. These are called scalar quantities, and the real number associated with each is called a scalar. Other quantities, such as force, elocit, and acceleration, inole both magnitude and direction and cannot be characteried completel b a single real number. A directed line segment is used to represent such a quantit, as shown in Figure.. The directed line segment PQ \ has initial point P and terminal point Q, and its length (or magnitude) is denoted b PQ \. Directed line segments that hae the same length and direction are equialent, as shown in Figure.. The set of all directed line segments that are equialent to a gien directed line segment PQ \ is a ector in the plane and is denoted b PQ \. In tpeset material, ectors are usuall denoted b lowercase, boldface letters such as u,, and w. When written b hand, howeer, ectors are often denoted b letters with arrows aboe them, such as u,, and w. Be sure ou understand that a ector represents a set of directed line segments (each haing the same length and direction). In practice, howeer, it is common not to distinguish between a ector and one of its representaties. Vector Representation: Directed Line Segments Let be represented b the directed line segment from 0, 0 to,, and let u be represented b the directed line segment from, to,. Show that and u are equialent. Solution Let P0, 0 and Q, be the initial and terminal points of, and let R, and S, be the initial and terminal points of u, as shown in Figure.. You can use the Distance Formula to show that PQ \ and RS \ hae the same length. PQ \ 0 0 RS \ Both line segments hae the same direction, because the both are directed toward the (, ) S upper right on lines haing the same slope. and Slope of Slope of PQ \ 0 0 RS \ Because PQ \ and RS \ hae the same length and direction, ou can conclude that the two ectors are equialent. That is, and u are equialent. (, ) (, ) R P (0, 0) The ectors u and are equialent. Figure. u Q Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

3 (0, 0) P (, ) Q =, A ector in standard position Figure.. Vectors in the Plane 79 The directed line segment whose initial point is the origin is often the most conenient representatie of a set of equialent directed line segments such as those shown in Figure.. This representation of is said to be in standard position. A directed line segment whose initial point is the origin can be uniquel represented b the coordinates of its terminal point Q,, as shown in Figure.. Definition of Component Form of a Vector in the Plane If is a ector in the plane whose initial point is the origin and whose terminal point is,, then the component form of is,. The coordinates and are called the components of. If both the initial point and the terminal point lie at the origin, then is called the ero ector and is denoted b 0 0, 0. This definition implies that two ectors u u, u and, are equal if and onl if u and u. The procedures listed below can be used to conert directed line segments to component form or ice ersa.. If Pp, p and Qq, q are the initial and terminal points of a directed line \ segment, then the component form of the ector represented b PQ is, q p, q p. Moreoer, from the Distance Formula, ou can see that the length (or magnitude) of is q p q p. Length of a ector. If,, then can be represented b the directed line segment, in standard position, from P0, 0 to Q,. The length of is also called the norm of. If, then is a unit ector. Moreoer, 0 if and onl if is the ero ector 0. Q (, 5) P (, 7) Component form of : 5, Figure.5 Component Form and Length of a Vector Find the component form and length of the ector that has initial point, 7 and terminal point, 5. Solution Let P, 7 p, p and Q, 5 q, q. Then the components of, are q p 5 and q p 5 7. So, as shown in Figure.5, 5,, and the length of is Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

4 750 Chapter Vectors and the Geometr of Space Vector Operations Definitions of Vector Addition and Scalar Multiplication Let u u, u and, be ectors and let c be a scalar.. The ector sum of u and is the ector u u, u.. The scalar multiple of c and u is the ector cu cu, cu.. The negatie of is the ector,.. The difference of u and is u u u, u. The scalar multiplication of Figure.6 Geometricall, the scalar multiple of a ector and a scalar c is the ector that is c times as long as, as shown in Figure.6. If c is positie, then c has the same direction as. If c is negatie, then c has the opposite direction. The sum of two ectors can be represented geometricall b positioning the ectors (without changing their magnitudes or directions) so that the initial point of one coincides with the terminal point of the other, as shown in Figure.7. The ector u, called the resultant ector, is the diagonal of a parallelogram haing u and as its adjacent sides. u u u + u + u WILLIAM ROWAN HAMILTON ( ) Some of the earliest work with ectors was done b the Irish mathematician William Rowan Hamilton. Hamilton spent man ears deeloping a sstem of ector-like quantities called quaternions. It wasn t until the latter half of the nineteenth centur that the Scottish phsicist James Mawell (8 879) restructured Hamilton s quaternions in a form useful for representing phsical quantities such as force, elocit, and acceleration. See LarsonCalculus.com to read more of this biograph. To find u, () moe the initial point of () moe the initial point of u to the terminal point of u, or to the terminal point of. Figure.7 Figure.8 shows the equialence of the geometric and algebraic definitions of ector addition and scalar multiplication, and presents (at far right) a geometric interpretation of u. (ku, ku ) (u +, u + ) (u, u ) u + u u (, ) u u ku u ku (u, u ) u u u u + ( ) ku Vector addition Scalar multiplication Vector subtraction Figure.8 The Granger Collection, New York Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

5 . Vectors in the Plane 75 Vector Operations For, 5 and w,, find each of the ectors. a. b. w c. w Solution, 5, 5 a. b. w w, w, 5 5, c. Using w 6, 8, ou hae w, 5 6, 8 6, 5 8,. Vector addition and scalar multiplication share man properties of ordinar arithmetic, as shown in the net theorem. EMMY NOETHER (88 95) One person who contributed to our knowledge of aiomatic sstems was the German mathematician Emm Noether. Noether is generall recognied as the leading woman mathematician in recent histor. FOR FURTHER INFORMATION For more information on Emm Noether, see the article Emm Noether, Greatest Woman Mathematician b Clark Kimberling in Mathematics Teacher. To iew this article, go to MathArticles.com. THEOREM. Properties of Vector Operations Let u,, and w be ectors in the plane, and let c and d be scalars.. u u Commutatie Propert. u w u w Associatie Propert. u 0 u Additie Identit Propert. u u 0 Additie Inerse Propert cdu cdu c du cu du Distributie Propert 7. cu cu c Distributie Propert 8. u u, 0u 0 Proof The proof of the Associatie Propert of ector addition uses the Associatie Propert of addition of real numbers. u w u, u, w, w u, u w, w u w, u w u w, u w u, u w, w u w The other properties can be proed in a similar manner. See LarsonCalculus.com for Bruce Edwards s ideo of this proof. An set of ectors (with an accompaning set of scalars) that satisfies the eight properties listed in Theorem. is a ector space.* The eight properties are the ector space aioms. So, this theorem states that the set of ectors in the plane (with the set of real numbers) forms a ector space. * For more information about ector spaces, see Elementar Linear Algebra, Seenth Edition, b Ron Larson (Boston, Massachusetts: Brooks/Cole, Cengage Learning, 0). The Granger Collection, NYC Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

6 75 Chapter Vectors and the Geometr of Space THEOREM. Length of a Scalar Multiple Let be a ector and let be a scalar. Then c c c. c is the absolute alue of c. Proof Because c c, c, it follows that c c, c c c c c c c c. See LarsonCalculus.com for Bruce Edwards s ideo of this proof. In man applications of ectors, it is useful to find a unit ector that has the same direction as a gien ector. The net theorem gies a procedure for doing this. THEOREM. Unit Vector in the Direction of If is a nonero ector in the plane, then the ector u has length and the same direction as. Proof Because is positie and u, ou can conclude that u has the same direction as. To see that u, note that u. So, u has length and the same direction as. See LarsonCalculus.com for Bruce Edwards s ideo of this proof. In Theorem., u is called a unit ector in the direction of. The process of multipling b to get a unit ector is called normaliation of. Finding a Unit Vector Find a unit ector in the direction of, 5 and erif that it has length. Solution From Theorem., the unit ector in the direction of is, 5 5 9, 5 9, 5. 9 This ector has length, because Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

7 u u + Generall, the length of the sum of two ectors is not equal to the sum of their lengths. To see this, consider the ectors u and as shown in Figure.9. With u and as two sides of a triangle, the length of the third side is u, and u u.. Vectors in the Plane 75 Equalit occurs onl when the ectors u and hae the same direction. This result is called the triangle inequalit for ectors. (You are asked to proe this in Eercise 77, Section..) Triangle inequalit Figure.9 Standard Unit Vectors The unit ectors, 0 and 0, are called the standard unit ectors in the plane and are denoted b i, 0 and j 0, Standard unit ectors j = 0, as shown in Figure.0. These ectors can be used to represent an ector uniquel, as follows.,, 0 0,, 0 0, i j The ector i j is called a linear combination of i and j. The scalars and are called the horiontal and ertical components of. i =, 0 Standard unit ectors i and j Figure.0 Writing a Linear Combination of Unit Vectors Let u be the ector with initial point, 5 and terminal point,, and let i j. Write each ector as a linear combination of i and j. a. u b. w u Solution a. u q p, q p, 5, 8 i 8j b. w u i 8j i j 6i 6j 6i j i 9j cos θ u θ (cos θ, sin θ) sin θ The angle from the positie -ais to the ector u Figure. If u is a unit ector and is the angle (measured counterclockwise) from the positie -ais to u, then the terminal point of u lies on the unit circle, and ou hae u cos, sin cos i sin j Unit ector as shown in Figure.. Moreoer, it follows that an other nonero ector making an angle with the positie -ais has the same direction as u, and ou can write cos, sin cos i sin j. Writing a Vector of Gien Magnitude and Direction The ector has a magnitude of and makes an angle of 0 6 with the positie -ais. Write as a linear combination of the unit ectors i and j. Solution Because the angle between and the positie -ais is ou can write cos i sin j cos i sin 6 6 j 6, i j. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

8 75 Chapter Vectors and the Geometr of Space Vectors hae man applications in phsics and engineering. One eample is force. A ector can be used to represent force, because force has both magnitude and direction. If two or more forces are acting on an object, then the resultant force on the object is the ector sum of the ector forces. Finding the Resultant Force 00 cos( 0 ) F F cos(0 ) 00 sin( 0 ) 00 sin(0 ) The resultant force on the ocean liner that is eerted b the two tugboats Figure. Two tugboats are pushing an ocean liner, as shown in Figure.. Each boat is eerting a force of 00 pounds. What is the resultant force on the ocean liner? Solution Using Figure., ou can represent the forces eerted b the first and second tugboats as F 00cos 0, sin 0 00 cos0i 00 sin0j F 00cos0, sin0 00 cos0i 00 sin0j. The resultant force on the ocean liner is F F F 00 cos0i 00 sin0j 00 cos0i 00 sin0j 800 cos0i 75i. So, the resultant force on the ocean liner is approimatel 75 pounds in the direction of the positie -ais. In sureing and naigation, a bearing is a direction that measures the acute angle that a path or line of sight makes with a fied north-south line. In air naigation, bearings are measured in degrees clockwise from north. W 0 N S E Finding a Velocit See LarsonCalculus.com for an interactie ersion of this tpe of eample. An airplane is traeling at a fied altitude with a negligible wind factor. The airplane is traeling at a speed of 500 miles per hour with a bearing of 0, as shown in Figure.(a). As the airplane reaches a certain point, it encounters wind with a elocit of 70 miles per hour in the direction N 5 E ( 5 east of north), as shown in Figure.(b). What are the resultant speed and direction of the airplane? (a) Direction without wind Solution Using Figure.(a), represent the elocit of the airplane (alone) as 500 cos0i 500 sin0j. The elocit of the wind is represented b the ector 70 cos5i 70 sin5j. N The resultant elocit of the airplane (in the wind) is Wind W θ S E 500 cos0i 500 sin0j 70 cos5i 70 sin5j 00.5i 8.5j. To find the resultant speed and direction, write cos i sin j. Because , ou can write i 5.5 j 5.5 cos.6i sin.6j. (b) Direction with wind Figure. The new speed of the airplane, as altered b the wind, is approimatel 5.5 miles per hour in a path that makes an angle of.6 with the positie -ais. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

9 . Vectors in the Plane 755. Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Sketching a Vector In Eercises, (a) find the component form of the ector and (b) sketch the ector with its initial point at the origin... (5, ).. Equialent Vectors In Eercises 5 8, find the ectors u and whose initial and terminal points are gien. Show that u and are equialent. 5. u:,, 5, 6 6. :,,, 8 7. u: 0,, 6, 8. :, 0, 9, 5 Writing a Vector in Different Forms In Eercises 9 6, the initial and terminal points of a ector are gien. (a) Sketch the gien directed line segment, (b) write the ector in component form, (c) write the ector as the linear combination of the standard unit ectors i and j, and (d) sketch the ector with its initial point at the origin. Terminal Terminal Initial Point Point Initial Point Point 9., 0 5, 5 0., 6, 6. 8, 6,. 0, 5,. 6, 6, 6. 7,, , ,.5, (, ) 6 (, ) (, ) 5, (, ) Sketching Scalar Multiples In Eercises 7 and 8, sketch each scalar multiple of. 7., 5 7 (a) (b) (c) (d) 8., (a) (b) (c) 0 (d) 6 u:, 0,, 8 :,, 7, 7 u:,,, : 0,, 5, 0 Using Vector Operations In Eercises 9 and 0, find (a) u, (b), (c) u, and (d) u u, 9,, 5 0. u, 8, 8, 5 (, ) 5 6 (, ) (, ) Sketching a Vector In Eercises 6, use the figure to sketch a graph of the ector. To print an enlarged cop of the graph, go to MathGraphs.com.. u. u.. 5. u 6. u Finding a Terminal Point In Eercises 7 and 8, the ector and its initial point are gien. Find the terminal point. 7., ; Initial point:, 8., 9; Initial point: 5, Finding a Magnitude of a Vector In Eercises 9, find the magnitude of. 9. 7i 0. i.,., 5. 6i 5j. 0i j Finding a Unit Vector In Eercises 5 8, find the unit ector in the direction of and erif that it has length. 5., 6. 5, ,., 5 Finding Magnitudes In Eercises 9, find the following. (a) u (b) (c) u (d) u u (e) (f) u u 9. u,,, 0. u 0,,,. u,,,. u,, 5, 5 Using the Triangle Inequalit In Eercises and, sketch a graph of u,, and u. Then demonstrate the triangle inequalit using the ectors u and.. u,, 5,. Finding a Vector In Eercises 5 8, find the ector with the gien magnitude and the same direction as u. Magnitude Direction 5. 6 u 0, 6. u, 7. 5 u, 8. u, u u,,, Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

10 756 Chapter Vectors and the Geometr of Space Finding a Vector In Eercises 9 5, find the component form of gien its magnitude and the angle it makes with the positie -ais. Finding a Vector In Eercises 5 56, find the component form of u gien the lengths of u and and the angles that u and make with the positie -ais.,, 0 9., 50. 5, 5., 5., u, 0 5. u, u u, u 56. u 5, WRITING ABOUT CONCEPTS 59. Using a Parallelogram Three ertices of a parallelogram are,,,, and 8,. Find the three possible fourth ertices (see figure). 6 5 (, ) (, ), 5, 5 6 u (8, ) u Scalar and Vector In our own words, state the difference between a scalar and a ector. Gie eamples of each. 58. Scalar or Vector Identif the quantit as a scalar or as a ector. Eplain our reasoning. (a) The mule elocit of a gun (b) The price of a compan s stock (c) The air temperature in a room (d) The weight of a car 60. HOW DO YOU SEE IT? Use the figure to determine whether each statement is true or false. Justif our answer. a b u c d t s w Finding Values In Eercises 6 66, find a and b such that au bw, where u, and w,. 6., 6. 0, 6., 0 6., 65., 66., 7 Finding Unit Vectors In Eercises 67 7, find a unit ector (a) parallel to and (b) perpendicular to the graph of f at the gien point. Then sketch the graph of f and sketch the ectors at the gien point. 67. f,, f 5,, 69. f,, 70. f,, 8 7. f 5,, 7. f tan,, Finding a Vector In Eercises 7 and 7, find the component form of gien the magnitudes of u and u and the angles that u and u make with the positie -ais. 7. u, 5 7. u, 0 u, 90 u 6, Resultant Force Forces with magnitudes of 500 pounds and 00 pounds act on a machine part at angles of 0 and 5, respectiel, with the -ais (see figure). Find the direction and magnitude of the resultant force. 500 lb 0 80 N 5 θ 75 N 00 lb Figure for 75 Figure for Numerical and Graphical Analsis Forces with magnitudes of 80 newtons and 75 newtons act on a hook (see figure). The angle between the two forces is degrees. 0, (a) When find the direction and magnitude of the resultant force. (b) Write the magnitude M and direction of the resultant force as functions of, where (c) Use a graphing utilit to complete the table. M (a) (c) (e) a d a u c a d 0 (b) c s (d) w s (f ) u b t (d) Use a graphing utilit to graph the two functions M and. (e) Eplain wh one of the functions decreases for increasing alues of, whereas the other does not. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

11 . Vectors in the Plane Resultant Force Three forces with magnitudes of 75 pounds, 00 pounds, and 5 pounds act on an object at angles of 0, 5, and 0, respectiel, with the positie -ais. Find the direction and magnitude of the resultant force. 78. Resultant Force Three forces with magnitudes of 00 newtons, 80 newtons, and 50 newtons act on an object at angles of 0, 5, and 5, respectiel, with the positie -ais. Find the direction and magnitude of the resultant force. 79. Think About It Consider two forces of equal magnitude acting on a point. (a) When the magnitude of the resultant is the sum of the magnitudes of the two forces, make a conjecture about the angle between the forces. (b) When the resultant of the forces is 0, make a conjecture about the angle between the forces. (c) Can the magnitude of the resultant be greater than the sum of the magnitudes of the two forces? Eplain. 80. Cable Tension Determine the tension in each cable supporting the gien load for each figure. (a) (b) A 50 0 C 000 lb 8. Projectile Motion A gun with a mule elocit of 00 feet per second is fired at an angle of 6 aboe the horiontal. Find the ertical and horiontal components of the elocit. 8. Shared Load To carr a 00-pound clindrical weight, two workers lift on the ends of short ropes tied to an eelet on the top center of the clinder. One rope makes a 0 angle awa from the ertical and the other makes a 0 angle (see figure). (a) Find each rope s tension when the resultant force is ertical. (b) Find the ertical component of each worker s force lb in. 900 km/hr Figure for 8 Figure for 8 Mikael Damkier/Shutterstock.com B A 0 in. 0 in. W C 5000 lb N S E 00 km/hr 5 B 8. Naigation A plane is fling with a bearing of 0. Its speed with respect to the air is 900 kilometers per hour. The wind at the plane s altitude is from the southwest at 00 kilometers per hour (see figure). What is the true direction of the plane, and what is its speed with respect to the ground? 8. Naigation A plane flies at a constant groundspeed of 00 miles per hour due east and encounters a 50-mile-per-hour wind from the northwest. Find the airspeed and compass direction that will allow the plane to maintain its groundspeed and eastward direction. True or False? In Eercises 85 90, determine whether the statement is true or false. If it is false, eplain wh or gie an eample that shows it is false. 85. If u and hae the same magnitude and direction, then u and are equialent. 86. If u is a unit ector in the direction of, then u. 87. If u ai bj is a unit ector, then a b. 88. If ai bj 0, then a b. 89. If a b, then ai bj a. 90. If u and hae the same magnitude but opposite directions, then u Proof Proe that u cos i sin j and sin i cos j are unit ectors for an angle. 9. Geometr Using ectors, proe that the line segment joining the midpoints of two sides of a triangle is parallel to, and one-half the length of, the third side. 9. Geometr Using ectors, proe that the diagonals of a parallelogram bisect each other. 9. Proof Proe that the ector w u u bisects the angle between u and. 95. Using a Vector Consider the ector u,. Describe the set of all points, such that u 5. PUTNAM EXAM CHALLENGE 96. A coast artiller gun can fire at an angle of eleation between 0 and 90 in a fied ertical plane. If air resistance is neglected and the mule elocit is constant 0, determine the set H of points in the plane and aboe the horiontal which can be hit. This problem was composed b the Committee on the Putnam Prie Competition. The Mathematical Association of America. All rights resered. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

12 758 Chapter Vectors and the Geometr of Space. Space Coordinates and Vectors in Space Understand the three-dimensional rectangular coordinate sstem. Anale ectors in space. -plane -plane -plane The three-dimensional coordinate sstem Figure. Coordinates in Space Up to this point in the tet, ou hae been primaril concerned with the two-dimensional coordinate sstem. Much of the remaining part of our stud of calculus will inole the three-dimensional coordinate sstem. Before etending the concept of a ector to three dimensions, ou must be able to identif points in the three-dimensional coordinate sstem. You can construct this sstem b passing a -ais perpendicular to both the - and -aes at the origin, as shown in Figure.. Taken as pairs, the aes determine three coordinate planes: the -plane, the -plane, and the -plane. These three coordinate planes separate three-space into eight octants. The first octant is the one for which all three coordinates are positie. In this three-dimensional sstem, a point P in space is determined b an ordered triple,,, where,, and are as follows. directed distance from -plane to P directed distance from -plane to P directed distance from -plane to P Seeral points are shown in Figure.5. REMARK The threedimensional rotatable graphs that are aailable at LarsonCalculus.com can help ou isualie points or objects in a three-dimensional coordinate sstem. 8 (, 5, ) (,, ) (, 5, ) (, 6, 0) Points in the three-dimensional coordinate sstem are represented b ordered triples. Figure.5 8 A three-dimensional coordinate sstem can hae either a right-handed or a lefthanded orientation. To determine the orientation of a sstem, imagine that ou are standing at the origin, with our arms pointing in the direction of the positie - and -aes, and with the -ais pointing up, as shown in Figure.6. The sstem is right-handed or left-handed depending on which hand points along the -ais. In this tet, ou will work eclusiel with the right-handed sstem. Right-handed sstem Figure.6 Left-handed sstem Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

13 . Space Coordinates and Vectors in Space 759 P d (,, ) Q Man of the formulas established for the two-dimensional coordinate sstem can be etended to three dimensions. For eample, to find the distance between two points in space, ou can use the Pthagorean Theorem twice, as shown in Figure.7. B doing this, ou will obtain the formula for the distance between the points,, and,,. (,, ) (,, ) d Distance Formula ( ) + ( ) The distance between two points in space Figure.7 Finding the Distance Between Two Points in Space Find the distance between the points,, and, 0,. Solution d 0 Distance Formula 5 7 r (,, ) A sphere with center at 0, 0, 0 and radius r is defined to be the set of all points,, such that the distance between,, and 0, 0, 0 is r. You can use the Distance Formula to find the standard equation of a sphere of radius r, centered at 0, 0, 0. If,, is an arbitrar point on the sphere, then the equation of the sphere is ( 0, 0, 0 ) r Equation of sphere Figure.8 as shown in Figure.8. Moreoer, the midpoint of the line segment joining the points,, and,, has coordinates,,. Midpoint Formula Finding the Equation of a Sphere Find the standard equation of the sphere that has the points 5,, and 0,, as endpoints of a diameter. Solution Using the Midpoint Formula, the center of the sphere is 5 0,, Midpoint Formula 5,, 0. B the Distance Formula, the radius is r Therefore, the standard equation of the sphere is Equation of sphere Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

14 760 Chapter Vectors and the Geometr of Space, 0, 0 0, 0, i k,, 0,, 0 The standard unit ectors in space Figure.9 P(p, p, p ) j Q(q, q, q ) = q p, q p, q p Figure.0 Vectors in Space In space, ectors are denoted b ordered triples,,. The ero ector is denoted b 0 0, 0, 0. Using the unit ectors i, 0, 0, and the standard unit ector notation for is i j k j 0,, 0, k 0, 0, as shown in Figure.9. If is represented b the directed line segment from P p, p, p to Qq, q, q, as shown in Figure.0, then the component form of is written b subtracting the coordinates of the initial point from the coordinates of the terminal point, as follows.,, q p, q p, q p Vectors in Space Let u u, u, u and,, be ectors in space and let c be a scalar.. Equalit of Vectors: u if and onl if u, u, and u.. Component Form: If is represented b the directed line segment from P p, p, p to Qq, q, q, then,, q p, q p, q p.. Length:. Unit Vector in the Direction of : 0,,, 5. Vector Addition: u u, u, u 6. Scalar Multiplication: c c, c, c Note that the properties of ector operations listed in Theorem. (see Section.) are also alid for ectors in space. Finding the Component Form of a Vector in Space See LarsonCalculus.com for an interactie ersion of this tpe of eample. Find the component form and magnitude of the ector haing initial point,, and terminal point 0,,. Then find a unit ector in the direction of. Solution The component form of is q p, q p, q p 0,,, 7, which implies that its magnitude is 7 6. The unit ector in the direction of is u, 7, 6 6, 7 6, 6. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

15 . Space Coordinates and Vectors in Space 76 u u = w = Recall from the definition of scalar multiplication that positie scalar multiples of a nonero ector hae the same direction as, whereas negatie multiples hae the direction opposite of. In general, two nonero ectors u and are parallel when there is some scalar c such that u c. For eample, in Figure., the ectors u,, and w are parallel because u and w. w Definition of Parallel Vectors Two nonero ectors u and are parallel when there is some scalar c such that u c. Parallel ectors Figure. Parallel Vectors Vector w has initial point,, and terminal point, 7, 5. Which of the following ectors is parallel to w? a. u,, b., 6, Solution Begin b writing w in component form. w, 7, 5 6, 8, a. Because u,, 6, 8, w, ou can conclude that u is parallel to w. b. In this case, ou want to find a scalar c such that, 6, c6, 8,. To find c, equate the corresponding components and sole as shown. 6c 6 8c c c c c Note that c for the first two components and c for the third component. This means that the equation, 6, c6, 8, has no solution, and the ectors are not parallel. P (,, ) (,, 0) 8 6 Q (, 7, 6) The points P, Q, and R lie on the same line. Figure. R 6 8 Using Vectors to Determine Collinear Points Determine whether the points P,,, Q,, 0, and R, 7, 6 are collinear. Solution The component forms of PQ \ and PR \ are PQ \,, 0,, and PR \, 7, 6, 9, 9. These two ectors hae a common initial point. So, P, Q, and R lie on the same line if and onl if PQ \ and PR \ are parallel which the are because PR \ PQ \, as shown in Figure.. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

16 76 Chapter Vectors and the Geometr of Space Standard Unit Vector Notation a. Write the ector i 5k in component form. b. Find the terminal point of the ector 7i j k, gien that the initial point is P,, 5. c. Find the magnitude of the ector 6i j k. Then find a unit ector in the direction of. Solution a. Because j is missing, its component is 0 and i 5k, 0, 5. b. You need to find Qq, q, q such that PQ \ 7i j k. This implies that q 7, q, and q 5. The solution of these three equations is q 5, q, and q 8. Therefore, Q is 5,, 8. c. Note that 6,, and. So, the magnitude of is The unit ector in the direction of is 76i j k 6 7 i 7 j 7 k. Measuring Force P (0, 0, ) Q,, 0 ) Q (0,, 0) Q,, 0 ) Figure. ) ) A teleision camera weighing 0 pounds is supported b a tripod, as shown in Figure.. Represent the force eerted on each leg of the tripod as a ector. Solution Let the ectors F, F, and F represent the forces eerted on the three legs. From Figure., ou can determine the directions of F, F, and F to be as follows. PQ \ 0 0, 0, 0 0,, PQ \ 0, 0, 0,, PQ \ 0, 0, 0,, Because each leg has the same length, and the total force is distributed equall among the three legs, ou know that F F F. So, there eists a constant c such that F c0,,, F c,,, Let the total force eerted b the object be gien b F 0, 0, 0. Then, using the fact that and F c,,. F F F F ou can conclude that F, F, and F all hae a ertical component of 0. This implies that c 0 and c 0. Therefore, the forces eerted on the legs can be represented b F 0, 0, 0, F 5, 5, 0, and F 5, 5, 0. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

17 . Space Coordinates and Vectors in Space 76. Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Plotting Points In Eercises, plot the points in the same three-dimensional coordinate sstem.. (a),, (b),,. (a),, 5 (b),,. (a) 5,, (b) 5,,. (a) 0,, 5 (b), 0, 5 Finding Coordinates of a Point the coordinates of the point. In Eercises 5 8, find 5. The point is located three units behind the -plane, four units to the right of the -plane, and fie units aboe the -plane. 6. The point is located seen units in front of the -plane, two units to the left of the -plane, and one unit below the -plane. 7. The point is located on the -ais, units in front of the -plane. 8. The point is located in the -plane, three units to the right of the -plane, and two units aboe the -plane. 7. Center: 0,, 5 8. Center:,, Radius: Radius: 5 9. Endpoints of a diameter:, 0, 0, 0, 6, 0 0. Center:,,, tangent to the -plane 9. Think About It What is the -coordinate of an point in the -plane? 0. Think About It What is the -coordinate of an point in the -plane? Using the Three-Dimensional Coordinate Sstem In Eercises, determine the location of a point,, that satisfies the condition(s) < 0 6. > > 9. > 0, 0. < 0,. < 0. > 0 Finding the Distance Between Two Points in Space In Eercises 6, find the distance between the points.. 0, 0, 0,,, 7. 5.,,, 6,, 6. Classifing a Triangle In Eercises 7 0, find the lengths of the sides of the triangle with the indicated ertices, and determine whether the triangle is a right triangle, an isosceles triangle, or neither. 7. 0, 0,,, 6, 7, 6,, 8 8.,,, 0, 6,,, 5, 6 9., 0,,, 5,,,, 0.,,,, 0,,, 5,,,,, 5,,,,, 5, 6. Think About It The triangle in Eercise 7 is translated fie units upward along the -ais. Determine the coordinates of the translated triangle.. Think About It The triangle in Eercise 8 is translated three units to the right along the -ais. Determine the coordinates of the translated triangle. Finding the Midpoint In Eercises 6, find the coordinates of the midpoint of the line segment joining the points..,, 6,, 8, 0. 7,,, 5,, 5. 5, 9, 7,,, 6., 0, 6, 8, 8, 0 Finding the Equation of a Sphere In Eercises 7 0, find the standard equation of the sphere. Finding the Equation of a Sphere In Eercises, complete the square to write the equation of the sphere in standard form. Find the center and radius Finding the Component Form of a Vector in Space In Eercises 5 8, (a) find the component form of the ector, (b) write the ector using standard unit ector notation, and (c) sketch the ector with its initial point at the origin (,, ) (0,, ) (,, 0) (,, ) 6 6 (, 0, ) (,, ) (,, 0) (0, 5, ) 6 6 Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

18 76 Chapter Vectors and the Geometr of Space Finding the Component Form of a Vector in Space In Eercises 9 and 50, find the component form and magnitude of the ector with the gien initial and terminal points. Then find a unit ector in the direction of. 9. Initial point:,, Initial point:,, Terminal point:,, 6 Terminal point:,, Writing a Vector in Different Forms In Eercises 5 and 5, the initial and terminal points of a ector are gien. (a) Sketch the directed line segment, (b) find the component form of the ector, (c) write the ector using standard unit ector notation, and (d) sketch the ector with its initial point at the origin. 5. Initial point:,, 5. Initial point:,, Terminal point:,, Terminal point:,, 7 Finding a Terminal Point In Eercises 5 and 5, the ector and its initial point are gien. Find the terminal point. 5., 5, 6 5.,, Initial point: 0, 6, Initial point: 0,, 5 Finding Scalar Multiples In Eercises 55 and 56, find each scalar multiple of and sketch its graph. 55.,, 56.,, (a) (b) (a) (b) 5 (c) (d) 0 (c) (d) Finding a Vector In Eercises 57 60, find the ector, gien that u,,,,,, and, 0,. w 57. u w 58. 5u w 59. u w 60. u w 0 Parallel Vectors In Eercises 6 6, determine which of the ectors is (are) parallel to. Use a graphing utilit to confirm our results. 6.,, 5 6. i j k (a) 6,, 0 (a) 6i j 9k (b),, 0 (b) i j k (c) 6,, 0 (c) i 9k (d),, (d) i j 9 8 k 6. has initial point,, and terminal point,, 5. (a) 6i 8j k (b) j k 6. has initial point 5,, and terminal point,,. (a) 7, 6, (b), 6, 6 Using Vectors to Determine Collinear Points In Eercises 65 68, use ectors to determine whether the points are collinear ,, 5,,,,,, 66.,, 7,, 0,, 7,, 9 67.,,,, 5, 0, 0,, , 0, 0,,,,, 6, Verifing a Parallelogram In Eercises 69 and 70, use ectors to show that the points form the ertices of a parallelogram. 69., 9,,,,, 0, 0,,,, 5 70.,,, 9,,,,, 9,,, Finding the Magnitude magnitude of. In Eercises 7 76, find the 7. 0, 0, 0 7., 0, 7. j 5k 7. i 5j k 75. i j k 76. i j 7k Finding Unit Vectors In Eercises 77 80, find a unit ector (a) in the direction of and (b) in the direction opposite of. 77.,, 78. 6, 0, i 5j k 80. 5i j k 8. Using Vectors Consider the two nonero ectors u and, and let s and t be real numbers. Describe the geometric figure generated b the terminal points of the three ectors t, u t, and su t. 8. HOW DO YOU SEE IT? Determine,, for each figure. Then find the component form of the ector from the point on the -ais to the point,,. (a) (b) Finding a Vector In Eercises 8 86, find the ector with the gien magnitude and the same direction as u. Magnitude (, 0, 0) (,, ) (0,, ) (0,, 0) (, 0, 8) Direction u 0,, u,, u,, u, 6, Sketching a Vector In Eercises 87 and 88, sketch the ector and write its component form. 87. lies in the -plane, has magnitude, and makes an angle of 0 with the positie -ais. (, 0, 0) (,, ) (0,, 0) Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

19 . Space Coordinates and Vectors in Space lies in the -plane, has magnitude 5, and makes an angle of 5 with the positie -ais. Finding a Point Using Vectors In Eercises 89 and 90, use ectors to find the point that lies two-thirds of the wa from P to Q P,, 0, Q,, P,, 5, Q6, 8, 9. Using Vectors Let u i j, j k, and w au b. (a) Sketch u and. (b) If w 0, show that a and b must both be ero. (c) Find a and b such that w i j k. (d) Show that no choice of a and b ields w i j k. 9. Writing The initial and terminal points of the ector are,, and,,. Describe the set of all points,, such that. WRITING ABOUT CONCEPTS 9. Describing Coordinates A point in the threedimensional coordinate sstem has coordinates 0, 0, 0. Describe what each coordinate measures. 9. Distance Formula Gie the formula for the distance between the points,, and,,. 95. Standard Equation of a Sphere Gie the standard equation of a sphere of radius r, centered at 0, 0, Parallel Vectors State the definition of parallel ectors. 97. Using a Triangle and Vectors Let A, B, and C be ertices of a triangle. Find AB \ BC \ CA \. 98. Using Vectors Let r,, and r 0,,. Describe the set of all points,, such that r r Diagonal of a Cube Find the component form of the unit ector in the direction of the diagonal of the cube shown in the figure Think About It Suppose the length of each cable in Eercise 0 has a fied length L a, and the radius of each disc is r 0 inches. Make a conjecture about the limit lim T r 0 a and gie a reason for our answer. 0. Load Supports Find the tension in each of the supporting cables in the figure when the weight of the crate is 500 newtons. 5 cm 65 cm 0. Auditorium Lights The lights in an auditorium are -pound discs of radius 8 inches. Each disc is supported b three equall spaced cables that are L inches long (see figure). (a) Write the tension T in each cable as a function of L. Determine the domain of the function. (b) Use a graphing utilit and the function in part (a) to complete the table. D A L T L C 8 in. (c) Use a graphing utilit to graph the function in part (a). Determine the asmptotes of the graph. (d) Confirm the asmptotes of the graph in part (c) analticall. (e) Determine the minimum length of each cable when a cable is designed to carr a maimum load of 0 pounds. 70 cm B 60 cm 5 cm D 6 ft 8 ft B C A 8 ft 0 ft = Figure for 99 Figure for Tower Gu Wire The gu wire supporting a 00-foot tower has a tension of 550 pounds. Using the distances shown in the figure, write the component form of the ector F representing the tension in the wire. Losesk Photo and Video/Shutterstock.com 75 Figure for 0 Figure for 0 0. Construction A precast concrete wall is temporaril kept in its ertical position b ropes (see figure). Find the total force eerted on the pin at position A. The tensions in AB and AC are 0 pounds and 650 pounds. 05. Geometr Write an equation whose graph consists of the set of points P,, that are twice as far from A0,, as from B,, 0. Describe the geometric figure represented b the equation. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

20 766 Chapter Vectors and the Geometr of Space. The Dot Product of Two Vectors Use properties of the dot product of two ectors. Find the angle between two ectors using the dot product. Find the direction cosines of a ector in space. Find the projection of a ector onto another ector. Use ectors to find the work done b a constant force. The Dot Product So far, ou hae studied two operations with ectors ector addition and multiplication b a scalar each of which ields another ector. In this section, ou will stud a third ector operation, the dot product. This product ields a scalar, rather than a ector. REMARK Because the dot product of two ectors ields a scalar, it is also called the scalar product (or inner product) of the two ectors. Definition of Dot Product The dot product of u u, u and, is u u u. The dot product of u u, u, u and,, is u u u u. Eploration Interpreting a Dot Product Seeral ectors are shown below on the unit circle. Find the dot products of seeral pairs of ectors. Then find the angle between each pair that ou used. Make a conjecture about the relationship between the dot product of two ectors and the angle between the ectors THEOREM. Properties of the Dot Product Let u,, and w be ectors in the plane or in space and let c be a scalar.. u u Commutatie Propert. u w u u w Distributie Propert. cu cu u c Proof To proe the first propert, let u u, u, u and,,. Then u u u u u u u u. For the fifth propert, let,,. Then. Proofs of the other properties are left to ou. See LarsonCalculus.com for Bruce Edwards s ideo of this proof Finding Dot Products Let u,, 5, 8, and w,. a. u, 5, b. u w 6,, 8 c. u u 6 d. w w w,, 5 Notice that the result of part (b) is a ector quantit, whereas the results of the other three parts are scalar quantities. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

21 Angle Between Two Vectors. The Dot Product of Two Vectors 767 The angle between two nonero ectors is the angle, 0, between their respectie standard position ectors, as shown in Figure.. The net theorem shows how to find this angle using the dot product. (Note that the angle between the ero ector and another ector is not defined here.) u u θ Origin The angle between two ectors Figure. THEOREM.5 Angle Between Two Vectors If is the angle between two nonero ectors u and, where 0, then cos u u. Proof Consider the triangle determined b ectors u,, and u, as shown in Figure.. B the Law of Cosines, ou can write u u u cos. Using the properties of the dot product, the left side can be rewritten as u u u u u u u u u u u u and substitution back into the Law of Cosines ields u u u u cos u u cos cos u u. See LarsonCalculus.com for Bruce Edwards s ideo of this proof. Note in Theorem.5 that because u and are alwas positie, u and cos will alwas hae the same sign. Figure.5 shows the possible orientations of two ectors. Opposite direction θ u cos Figure.5 u < 0 u θ < < < cos < 0 u = 0 u θ cos 0 u > 0 u θ 0 < < 0 < cos < Same direction u 0 cos Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

22 768 Chapter Vectors and the Geometr of Space From Theorem.5, ou can see that two nonero ectors meet at a right angle if and onl if their dot product is ero. Two such ectors are said to be orthogonal. Definition of Orthogonal Vectors The ectors u and are orthogonal when u 0. REMARK The terms perpendicular, orthogonal, and normal all mean essentiall the same thing meeting at right angles. It is common, howeer, to sa that two ectors are orthogonal, two lines or planes are perpendicular, and a ector is normal to a line or plane. From this definition, it follows that the ero ector is orthogonal to eer ector u, because 0 u 0. Moreoer, for 0, ou know that cos 0 if and onl if So, ou can use Theorem.5 to conclude that two nonero ectors are orthogonal if and onl if the angle between them is.. Finding the Angle Between Two Vectors See LarsonCalculus.com for an interactie ersion of this tpe of eample. REMARK The angle between u and in Eample (a) can also be written as approimatel For u,,,, 0,, w,,, and, 0,, find the angle between each pair of ectors. a. u and b. u and w c. and Solution a. cos u 0 8 u Because u < 0,.069 radians. 70 b. cos u w u w Because u w 0, u and w are orthogonal. So, c. cos Consequentl, Note that and are parallel, with.. arccos. When the angle between two ectors is known, rewriting Theorem.5 in the form u u cos Alternatie form of dot product produces an alternatie wa to calculate the dot product. Alternatie Form of the Dot Product Gien that u 0, 7, and the angle between u and is, find u. Solution Use the alternatie form of the dot product as shown. u u cos 07 cos 5 Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

23 . The Dot Product of Two Vectors 769 REMARK Recall that,, and are the Greek letters alpha, beta, and gamma, respectiel. Direction Cosines For a ector in the plane, ou hae seen that it is conenient to measure direction in terms of the angle, measured counterclockwise, from the positie -ais to the ector. In space, it is more conenient to measure direction in terms of the angles between the nonero ector and the three unit ectors i, j, and k, as shown in Figure.6. The angles,, and are the direction angles of, and cos, cos, and cos are the direction cosines of. Because and i i cos cos i,,, 0, 0 it follows that cos. B similar reasoning with the unit ectors j and k, ou hae i k Direction angles Figure.6 α γ β j cos cos is the angle between and i. is the angle between and j. cos. is the angle between and k. Consequentl, an nonero ector in space has the normalied form i j k cos i cos j cos k and because is a unit ector, it follows that cos cos cos. Finding Direction Angles α = angle between and i β = angle between and j γ = angle between and k γ = i + j + k α β The direction angles of Figure.7 Find the direction cosines and angles for the ector i j k, and show that cos cos cos. Solution Because 9, ou can write the following. cos 9 cos 9 cos 9 Angle between and i Angle between and j Angle between and k Furthermore, the sum of the squares of the direction cosines is cos cos cos See Figure Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

24 770 Chapter Vectors and the Geometr of Space w F w The force due to grait pulls the boat against the ramp and down the ramp. Figure.8 Projections and Vector Components You hae alread seen applications in which two ectors are added to produce a resultant ector. Man applications in phsics and engineering pose the reerse problem decomposing a ector into the sum of two ector components. The following phsical eample enables ou to see the usefulness of this procedure. Consider a boat on an inclined ramp, as shown in Figure.8. The force F due to grait pulls the boat down the ramp and against the ramp. These two forces, w and w, are orthogonal the are called the ector components of F. F w w Vector components of F The forces and w help ou anale the effect of grait on the boat. For eample, w indicates the force necessar to keep the boat from rolling down the ramp, whereas indicates the force that the tires must withstand. w w Definitions of Projection and Vector Components Let u and be nonero ectors. Moreoer, let u w w where w is parallel to and w is orthogonal to, as shown in Figure.9.. w is called the projection of u onto or the ector component of u along, and is denoted b w proj u.. w u w is called the ector component of u orthogonal to. θ is acute. θ is obtuse. w u u w θ θ w w w proj u projection of u onto ector component of u along w ector component of u orthogonal to Figure.9 (, ) w 0 8 u (, ) w (5, 0) 6 8 u w w Figure.0 (8, 6) Finding a Vector Component of u Orthogonal to Find the ector component of u 5, 0 that is orthogonal to,, gien that and w proj u 8, 6 u 5, 0 w w. Solution Because u w w, where w is parallel to, it follows that w is the ector component of u orthogonal to. So, ou hae w u w 5, 0 8, 6,. Check to see that w is orthogonal to, as shown in Figure.0. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

25 . The Dot Product of Two Vectors 77 From Eample 5, ou can see that it is eas to find the ector component w once ou hae found the projection, w, of u onto. To find this projection, use the dot product in the net theorem, which ou will proe in Eercise 78. REMARK Note the distinction between the terms component and ector component. For eample, using the standard unit ectors with u u i u j, u is the component of u in the direction of i and u i is the ector component in the direction of i. THEOREM.6 The projection of u onto can be written as a scalar multiple of a unit ector in the direction of. That is, u u k. The scalar k is called the component of u in the direction of. So, k u u cos. Projection Using the Dot Product If u and are nonero ectors, then the projection of u onto is proj u u. u w Decomposing a Vector into Vector Components Find the projection of u onto and the ector component of u orthogonal to for u i 5j k and 7i j k. u = i 5j + k = 7i + j k 8 6 u w w Figure. w Solution The projection of u onto is w proj u u 7i j k 5 9 i 9 j 9 k. The ector component of u orthogonal to is the ector w u w i 5j k 9 i 9 j 9 k See Figure i j 9 9 k. Finding a Force w 0 F w = proj (F) Figure. A 600-pound boat sits on a ramp inclined at 0, as shown in Figure.. What force is required to keep the boat from rolling down the ramp? Solution Because the force due to grait is ertical and downward, ou can represent the graitational force b the ector F 600j. To find the force required to keep the boat from rolling down the ramp, project F onto a unit ector in the direction of the ramp, as follows. cos 0i sin 0j i j Therefore, the projection of F onto is Unit ector along ramp w proj F F F i j. The magnitude of this force is 00, and therefore a force of 00 pounds is required to keep the boat from rolling down the ramp. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

26 77 Chapter Vectors and the Geometr of Space Work The work W done b the constant force F acting along the line of motion of an object is gien b W magnitude of forcedistance F PQ \ as shown in Figure.(a). When the constant force F is not directed along the line of motion, ou can see from Figure.(b) that the work W done b the force is W proj cos F PQ \ F PQ \ PQ \F. PQ\ F F θ proj PQ F P Q P Q Work = F PQ Work = proj PQ F PQ (a) Force acts along the line of motion. (b) Force acts at angle with the line of motion. Figure. This notion of work is summaried in the net definition. Definition of Work The work W done b a constant force F as its point of application moes along the ector PQ \ is one of the following.. W proj PQ \F Projection form PQ\. W F PQ \ Dot product form Finding Work To close a sliding door, a person pulls on a rope with a constant force of 50 pounds at a constant angle of 60, as shown in Figure.. Find the work done in moing the door feet to its closed position. ft P proj PQ F F 60 Q ft Figure. Solution Using a projection, ou can calculate the work as follows. W proj cos60 F 00 foot-pounds 50 PQ \F PQ\ PQ\ Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

27 . The Dot Product of Two Vectors 77. Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Finding Dot Products In Eercises 8, find (a) u, (b) u u, (c) u, (d) u, and (e) u.. u,,, 5.. u 6,,,. 5. u,,, 0, 6, 5 6. u i, i 7. u i j k 8. u i j k i k Finding the Angle Between Two Vectors In Eercises 9 6, find the angle between the ectors (a) in radians and (b) in degrees. 9. u,,, 0.. u i j, i j. 6 i sin 6 j,. u,,. u i j k,, 5. u i j 6. u i j k j k Alternatie Form of Dot Product In Eercises 7 and 8, use the alternatie form of the dot product to find u. 7. u 8, 5, and the angle between u and is. 8. u 0, 5, and the angle between u and is 56. Comparing Vectors In Eercises 9, determine whether u and are orthogonal, parallel, or neither. 9. u,, 0. u i j, i j. u j 6k. u i j k i j k,. u,,. u cos, sin,,, Classifing a Triangle In Eercises 5 8, the ertices of a triangle are gien. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle. Eplain our reasoning. 5.,, 0, 0, 0, 0,,, 0 6., 0, 0, 0, 0, 0,,, 7., 0,, 0,,, 0.5,.5, 0 8., 7,,, 5, 8,, 6, u, 0,, u, 8, 7, 5 i j k u,,, cos i sin j i j i j k i j k sin, cos, 0 Finding Direction Angles In Eercises 9, find the direction cosines and angles of u, and demonstrate that the sum of the squares of the direction cosines is.. u i j k. u i j 5k. u 0, 6,. u, 5, Finding the Projection of u onto In Eercises 5, (a) find the projection of u onto, and (b) find the ector component of u orthogonal to. 5. u 6, 7,, u i j, 5i j u i j, i j u 0,,,,, u 8,, 0,,, u i j k, j k u i k, i k WRITING ABOUT CONCEPTS u 9, 7,,. Dot Product Define the dot product of ectors u and.. Orthogonal Vectors State the definition of orthogonal ectors. When ectors are neither parallel nor orthogonal, how do ou find the angle between them? Eplain. 5. Using Vectors Determine which of the following are defined for nonero ectors u,, and w. Eplain our reasoning. (a) u w (b) u w (c) u w (d) u w 6. Direction Cosines Describe direction cosines and direction angles of a ector. 7. Projection Gie a geometric description of the projection of u onto. 8. Projection What can be said about the ectors u and when (a) the projection of u onto equals u and (b) the projection of u onto equals 0? 9. Projection When the projection of u onto has the same magnitude as the projection of onto u, can ou conclude that u? Eplain. 50. HOW DO YOU SEE IT? What is known about, the angle between two nonero ectors u and, when (a) u 0? (b) u > 0? (c) u < 0? u θ Origin 9. u i j k 0. u 5i j k Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

28 77 Chapter Vectors and the Geometr of Space 5. Reenue The ector u 0, 50, 5 gies the numbers of hamburgers, chicken sandwiches, and cheeseburgers, respectiel, sold at a fast-food restaurant in one week. The ector.5,.95,.65 gies the prices (in dollars) per unit for the three food items. Find the dot product u, and eplain what information it gies. 5. Reenue Repeat Eercise 5 after increasing prices b %. Identif the ector operation used to increase prices b %. Orthogonal Vectors In Eercises 5 56, find two ectors in opposite directions that are orthogonal to the ector u. (The answers are not unique.) 5. u i j 5. u 9i j 55. u,, 56. u,, Finding an Angle Find the angle between a cube s diagonal and one of its edges. 58. Finding an Angle Find the angle between the diagonal of a cube and the diagonal of one of its sides. 59. Braking Load A 8,000-pound truck is parked on a 0 slope (see figure). Assume the onl force to oercome is that due to grait. Find (a) the force required to keep the truck from rolling down the hill and (b) the force perpendicular to the hill. 6. Work A car is towed using a force of 600 newtons. The chain used to pull the car makes a 5 angle with the horiontal. Find the work done in towing the car kilometers. 6. Work A sled is pulled b eerting a force of 00 newtons on a rope that makes a 5 angle with the horiontal. Find the work done in pulling the sled 0 meters. True or False? In Eercises 65 and 66, determine whether the statement is true or false. If it is false, eplain wh or gie an eample that shows it is false. 65. If u u w and u 0, then w. 66. If u and are orthogonal to w, then u is orthogonal to w. Using Points of Intersection In Eercises 67 70, (a) find all points of intersection of the graphs of the two equations, (b) find the unit tangent ectors to each cure at their points of intersection, and (c) find the angles 0 90 between the cures at their points of intersection. 67., , 70.,, 60. Braking Load A 500-pound sport utilit ehicle is parked on an 8 slope. Assume the onl force to oercome is that due to grait. Find (a) the force required to keep the ehicle from rolling down the hill and (b) the force perpendicular to the hill. 6. Work An object is pulled 0 feet across a floor, using a force of 85 pounds. The direction of the force is 60 aboe the horiontal (see figure). Find the work done. 85 lb 60 0 ft Not drawn to scale 0 Weight = 8,000 lb Figure for 6 Figure for 6 6. Work A to wagon is pulled b eerting a force of 5 pounds on a handle that makes a 0 angle with the horiontal (see figure). Find the work done in pulling the wagon 50 feet. Zia_K/iStockphoto.com 0 7. Proof Use ectors to proe that the diagonals of a rhombus are perpendicular. 7. Proof Use ectors to proe that a parallelogram is a rectangle if and onl if its diagonals are equal in length. 7. Bond Angle Consider a regular tetrahedron with ertices 0, 0, 0, k, k, 0, k, 0, k, and 0, k, k, where k is a positie real number. (a) Sketch the graph of the tetrahedron. (b) Find the length of each edge. (c) Find the angle between an two edges. (d) Find the angle between the line segments from the centroid k, k, k to two ertices. This is the bond angle for a molecule such as CH or PbCl, where the structure of the molecule is a tetrahedron. 7. Proof Consider the ectors u cos, sin, 0 and cos, sin, 0, where >. Find the dot product of the ectors and use the result to proe the identit cos cos cos sin sin. 75. Proof Proe that u u u. 76. Proof Proe the Cauch-Schwar Inequalit, u u. 77. Proof Proe the triangle inequalit u u. 78. Proof Proe Theorem.6. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

29 . The Cross Product of Two Vectors in Space 775. The Cross Product of Two Vectors in Space Find the cross product of two ectors in space. Use the triple scalar product of three ectors in space. Eploration Geometric Propert of the Cross Product Three pairs of ectors are shown below. Use the definition to find the cross product of each pair. Sketch all three ectors in a three-dimensional sstem. Describe an relationships among the three ectors. Use our description to write a conjecture about u,, and u. a. u, 0,,, 0, b. c. u,, 0,,, 0 u u 0,,, 0,, u u The Cross Product Man applications in phsics, engineering, and geometr inole finding a ector in space that is orthogonal to two gien ectors. In this section, ou will stud a product that will ield such a ector. It is called the cross product, and it is most conenientl defined and calculated using the standard unit ector form. Because the cross product ields a ector, it is also called the ector product. Definition of Cross Product of Two Vectors in Space Let u u i u j u k It is important to note that this definition applies onl to three-dimensional ectors. The cross product is not defined for two-dimensional ectors. A conenient wa to calculate u is to use the determinant form with cofactor epansion shown below. (This determinant form is used simpl to help remember the formula for the cross product it is technicall not a determinant because not all the entries of the corresponding matri are real numbers.) i j k u u u u Put u in Row. Put in Row. and i j k be ectors in space. The cross product of u and is the ector u u u i u u j u u k. i j k i j k i j u u u i u u u j u u u i u u j u u k u u i u u j u u k Note the minus sign in front of the j-component. Each of the three determinants can be ealuated b using the diagonal pattern a b d ad bc. c Here are a couple of eamples. and u k u k Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

30 776 Chapter Vectors and the Geometr of Space NOTATION FOR DOT AND CROSS PRODUCTS The notation for the dot product and cross product of ectors was first introduced b the American phsicist Josiah Willard Gibbs (89 90). In the earl 880s, Gibbs built a sstem to represent phsical quantities called ector analsis. The sstem was a departure from Hamilton s theor of quaternions. REMARK Note that this result is the negatie of that in part (a). Finding the Cross Product For u i j k and i j k, find each of the following. a. u b. u c. Solution i j k a. u i j k i j 6k i 5j 7k i j k b. u i j k i j 6 k i 5j 7k i j k c. 0 The results obtained in Eample suggest some interesting algebraic properties of the cross product. For instance, u u, and 0. These properties, and seeral others, are summaried in the net theorem. THEOREM.7 Algebraic Properties of the Cross Product Let u,, and w be ectors in space, and let c be a scalar.. u u. u w u u w. cu cu u c. u 0 0 u 0 5. u u 0 6. u w u w Proof To proe Propert, let u u i u j u k and i j k. Then, and u u u i u u j u u k u u u i u u j u u k which implies that u u. Proofs of Properties,, 5, and 6 are left as eercises (see Eercises 5 5). See LarsonCalculus.com for Bruce Edwards s ideo of this proof. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

31 . The Cross Product of Two Vectors in Space 777 Note that Propert of Theorem.7 indicates that the cross product is not commutatie. In particular, this propert indicates that the ectors u and u hae equal lengths but opposite directions. The net theorem lists some other geometric properties of the cross product of two ectors. REMARK It follows from Properties and in Theorem.8 that if n is a unit ector orthogonal to both u and, then u ±u sin n. THEOREM.8 Geometric Properties of the Cross Product Let u and be nonero ectors in space, and let be the angle between u and.. u is orthogonal to both u and.. u u sin. u 0 if and onl if u and are scalar multiples of each other.. u area of parallelogram haing u and as adjacent sides. Proof To proe Propert, note because cos u u, it follows that u sin u cos u u u u u θ sin θ u The ectors u and form adjacent sides of a parallelogram. Figure.5 To proe Propert, refer to Figure.5, which is a parallelogram haing and u as adjacent sides. Because the height of the parallelogram is sin, the area is Area baseheight u sin u. u u u u u u u u ) u u u u u. Proofs of Properties and are left as eercises (see Eercises 55 and 56). See LarsonCalculus.com for Bruce Edwards s ideo of this proof. Both u and u are perpendicular to the plane determined b u and. One wa to remember the orientations of the ectors u,, and u is to compare them with the unit ectors i, j, and k i j, as shown in Figure.6. The three ectors u,, and u form a right-handed sstem, whereas the three ectors u,, and u form a left-handed sstem. k = i j u j i -plane u Plane determined b u and Right-handed sstems Figure.6 Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

32 778 Chapter Vectors and the Geometr of Space Using the Cross Product See LarsonCalculus.com for an interactie ersion of this tpe of eample. (,, ) u (,, ) u (,, 0) The ector u is orthogonal to both u and. Figure.7 Find a unit ector that is orthogonal to both and Solution The cross product u, as shown in Figure.7, is orthogonal to both u and. 0 i j k u Cross product i j k Because u i j k i j. u a unit ector orthogonal to both u and is u u i j k. In Eample, note that ou could hae used the cross product u to form a unit ector that is orthogonal to both u and. With that choice, ou would hae obtained the negatie of the unit ector found in the eample. Geometric Application of the Cross Product D = (5, 0, 6) A = (5,, 0) C = (,, 7) The area of the parallelogram is approimatel.9. Figure.8 6 B = (, 6, ) The ertices of a quadrilateral are listed below. Show that the quadrilateral is a parallelogram, and find its area. A 5,, 0 C,, 7 Solution From Figure.8, ou can see that the sides of the quadrilateral correspond to the following four ectors. AB \ i j k AD \ 0i j 6k So, AB \ is parallel to CD \ and AD \ is parallel to CB \, and ou can conclude that the quadrilateral is a parallelogram with AB \ and AD \ as adjacent sides. Moreoer, because 6 i j k AB \ AD \ Cross product 0 6i 8j 6k the area of the parallelogram is B, 6, D 5, 0, 6 AB \ AD \ CD \ i j k AB \ \ CB 0i j 6k AD \ Is the parallelogram a rectangle? You can determine whether it is b finding the angle between the ectors AB \ and AD \. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

33 M P F PQ Q. The Cross Product of Two Vectors in Space 779 In phsics, the cross product can be used to measure torque the moment M of a force F about a point P, as shown in Figure.9. If the point of application of the force is Q, then the moment of F about P is M PQ \ F. Moment of F about P The magnitude of the moment M measures the tendenc of the ector PQ \ to rotate counterclockwise (using the right-hand rule) about an ais directed along the ector M. An Application of the Cross Product The moment of F about P Figure.9 P 60 A ertical force of 50 pounds is applied at point Q. Figure.0 Q F A ertical force of 50 pounds is applied to the end of a one-foot leer that is attached to an ale at point P, as shown in Figure.0. Find the moment of this force about the point P when Solution Represent the 50-pound force as F 50k and the leer as 60. PQ \ cos60j sin60k j k. The moment of F about P is i j M PQ \ F k 5i. 50 The magnitude of this moment is 5 foot-pounds. Moment of F about P In Eample, note that the moment (the tendenc of the leer to rotate about its ale) is dependent on the angle. When the moment is 0. The moment is greatest when The Triple Scalar Product For ectors u,, and w in space, the dot product of u and w u w 0., is called the triple scalar product, as defined in Theorem.9. The proof of this theorem is left as an eercise (see Eercise 59). THEOREM.9 The Triple Scalar Product For u u i u j u k, i j k, and w w i w j w k, the triple scalar product is u u w w u w u w. Note that the alue of a determinant is multiplied b when two rows are interchanged. After two such interchanges, the alue of the determinant will be unchanged. So, the following triple scalar products are equialent. u w w u w u Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

34 780 Chapter Vectors and the Geometr of Space w w u If the ectors u,, and w do not lie in the same plane, then the triple scalar product u w) can be used to determine the olume of the parallelepiped (a polhedron, all of whose faces are parallelograms) with u,, and w as adjacent edges, as shown in Figure.. This is established in the net theorem. THEOREM.0 Geometric Propert of the Triple Scalar Product The olume V of a parallelepiped with ectors u,, and w as adjacent edges is V u w. proj w u Area of base w Volume of parallelepiped Figure. u w Proof In Figure., note that the area of the base is w and the height of the parallelpiped is proj w u. Therefore, the olume is V heightarea of base proj w u w u w w w u w. See LarsonCalculus.com for Bruce Edwards s ideo of this proof. Volume b the Triple Scalar Product Find the olume of the parallelepiped shown in Figure. haing and u i 5j k j k w i j k as adjacent edges. Solution B Theorem.0, ou hae V u w 5 0 Triple scalar product 0 (, 5, ) (,, ) u 6 w (0,, ) The parallelepiped has a olume of 6. Figure. A natural consequence of Theorem.0 is that the olume of the parallelepiped is 0 if and onl if the three ectors are coplanar. That is, when the ectors u u, u, u, and hae the same initial point, the lie in the same plane if and onl if,,, w w, w, w u u u w 0. w w u w Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

35 . The Cross Product of Two Vectors in Space 78. Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Cross Product of Unit Vectors In Eercises 6, find the cross product of the unit ectors and sketch our result.. j i. i j. j k. k j 5. i k 6. k i Finding Cross Products (b) u, and (c). In Eercises 7 0, find (a) u, 7. u i j 8. u i 5k i j 5k 9. u 7,, 0. u,,,, 5 Finding a Cross Product In Eercises 6, find u and show that it is orthogonal to both u and.. u,, 0. u,,, 5, 0 0,, 0. u,,. u 0, 0, 6,, 5,, 0 5. u i j k 6. u i 6j i j k Finding a Unit Vector In Eercises 7 0, find a unit ector that is orthogonal to both u and. 7. u,, 8. u 8, 6,, 5, 0,, 9. u i j 5k 0. u k i j k Area In Eercises, find the area of the parallelogram that has the gien ectors as adjacent sides. Use a computer algebra sstem or a graphing utilit to erif our result.. u j. u i j k j k j k. u,,. u,, 0,, i j k, 5, i j k i 6k,, 0 Area In Eercises 5 and 6, erif that the points are the ertices of a parallelogram, and find its area. 5. A0,,, B, 5, 5, C6, 9, 5, D5, 7, 6. A,,, B6, 5,, C7,,, D, 6, Area In Eercises 7 and 8, find the area of the triangle with the gien ertices. Hint: u is the area of the triangle haing u and as adjacent sides. 7. A0, 0, 0, B, 0,, C,, 0 8. A,,, B0,,, C,, 0 Elena Elisseea/Shutterstock.com 9. Torque A child applies the brakes on a biccle b appling a downward force of 0 pounds on the pedal when the crank makes a 0 angle with the horiontal (see figure). The crank is 6 inches in length. Find the torque at P. 0. Torque Both the magnitude and the direction of the force on a crankshaft change as the crankshaft rotates. Find the torque on the crankshaft using the position and data shown in the figure. 0.6 ft lb Figure for 0 Figure for P. Optimiation A force of 80 pounds acts on the bracket shown in the figure. (a) Determine the ector AB \ and the ector F representing the force. ( F will be in terms of.) (b) Find the magnitude of the moment about A b ealuating AB \ F. (c) Use the result of part (b) to determine the magnitude of the moment when (d) Use the result of part (b) to determine the angle when the magnitude of the moment is maimum. At that angle, what is the relationship between the ectors F and AB \? Is it what ou epected? Wh or wh not? (e) Use a graphing utilit to graph the function for the magnitude of the moment about A for Find the ero of the function in the gien domain. Interpret the meaning of the ero in the contet of the problem in. 0 in. F = 0 lb B F θ 80 lb 5 in. A Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

36 78 Chapter Vectors and the Geometr of Space. Optimiation A force of 56 pounds acts on the pipe wrench shown in the figure. A (a) Find the magnitude of 8 in. the moment about O b ealuating OA \ F. Use θ F a graphing utilit to graph 0 the resulting function of. O (b) Use the result of part (a) to determine the magnitude of the moment when (c) Use the result of part (a) to determine the angle when the magnitude of the moment is maimum. Is the answer what ou epected? Wh or wh not? Finding a Triple Scalar Product u w.. u i. u,, j,, 0 w k w 0, 0, 5. u, 0, 6. u, 0, 0 0,, 0,, w 0, 0, w 0,, In Eercises 6, find Volume In Eercises 7 and 8, use the triple scalar product to find the olume of the parallelepiped haing adjacent edges u,, and w. 7. u i j 8. j k w i k w u u,, 0, 6, 6 w, 0, 6 8 w Volume In Eercises 9 and 0, find the olume of the parallelepiped with the gien ertices. 9. 0, 0, 0,, 0, 0, 0, 5,,, 0, 5, 5,, 5, 0, 5,, 5, 6, 5, 5, , 0, 0, 0,, 0,, 0, 0,,, 5,, 0,, 5, 5,,, 5,, 5, 5. Comparing Dot Products Identif the dot products that are equal. Eplain our reasoning. (Assume u,, and w are nonero ectors.) (a) u w (b) w u (c) u w (d) u w (e) u w (f) w u (g) u w (h) w u 6 u 5.. Using Dot and Cross Products When u 0 and u 0, what can ou conclude about u and? WRITING ABOUT CONCEPTS. Cross Product Define the cross product of ectors u and.. Cross Product State the geometric properties of the cross product. 5. Magnitude When the magnitudes of two ectors are doubled, how will the magnitude of the cross product of the ectors change? Eplain. 6. HOW DO YOU SEE IT? The ertices of a triangle in space are,,,,,, and,,. Eplain how to find a ector perpendicular to the triangle. 5 (,, ) 5 (,, ) (, ) True or False? In Eercises 7 50, determine whether the statement is true or false. If it is false, eplain wh or gie an eample that shows it is false. 7. It is possible to find the cross product of two ectors in a two-dimensional coordinate sstem. 8. If u and are ectors in space that are nonero and nonparallel, then u u. 9. If u 0 and u u w, then w. 50. If u 0, u u w, and u u w, then w. Proof In Eercises 5 56, proe the propert of the cross product u w u u w cu cu u c 5. u u 0 5. u w u w 55. u is orthogonal to both u and. 56. u 0 if and onl if u and are scalar multiples of each other. 57. Proof Proe that u u if u and are orthogonal. 58. Proof Proe that u w u w u w. 59. Proof Proe Theorem.9. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

37 .5 Lines and Planes in Space 78.5 Lines and Planes in Space Write a set of parametric equations for a line in space. Write a linear equation to represent a plane in space. Sketch the plane gien b a linear equation. Find the distances between points, planes, and lines in space. Q(,, ) L P(,, ) = a, b, c PQ = t Lines in Space In the plane, slope is used to determine the equation of a line. In space, it is more conenient to use ectors to determine the equation of a line. In Figure., consider the line L through the point P,, and parallel to the ector a, b, c. The ector is a direction ector for the line L, and a, b, and c are direction numbers. One wa of describing the line L is to sa that it consists of all points Q,, for which the ector PQ \ is parallel to. This means that PQ \ is a scalar multiple of and ou can write PQ \ t, where t is a scalar (a real number). PQ \,, at, bt, ct t B equating corresponding components, ou can obtain parametric equations of a line in space. Line L and its direction ector Figure. THEOREM. Parametric Equations of a Line in Space A line L parallel to the ector a, b, c and passing through the point P,, is represented b the parametric equations at, bt, and ct. If the direction numbers a, b, and c are all nonero, then ou can eliminate the parameter t to obtain smmetric equations of the line. a b c Smmetric equations (,, ) Finding Parametric and Smmetric Equations Find parametric and smmetric equations of the line L that passes through the point,, and is parallel to,,, as shown in Figure.. Solution To find a set of parametric equations of the line, use the coordinates,, and and direction numbers a, b, and c. L t, Parametric equations Because a, b, and c are all nonero, a set of smmetric equations is t,. t Smmetric equations =,, The ector is parallel to the line L. Figure. Neither parametric equations nor smmetric equations of a gien line are unique. For instance, in Eample, b letting t in the parametric equations, ou would obtain the point,, 0. Using this point with the direction numbers a, b, and c would produce a different set of parametric equations t, t, and t. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

38 78 Chapter Vectors and the Geometr of Space Parametric Equations of a Line Through Two Points See LarsonCalculus.com for an interactie ersion of this tpe of eample. Find a set of parametric equations of the line that passes through the points,, 0 and,, 5. Solution Begin b using the points P,, 0 and Q,, 5 to find a direction ector for the line passing through P and Q. PQ \,, 5 0,, 5 a, b, c Using the direction numbers a, b, and c 5 with the point P,, 0, ou can obtain the parametric equations t, t, and 5t. REMARK As t aries oer all real numbers, the parametric equations in Eample determine the points,, on the line. In particular, note that t 0 and t gie the original points,, 0 and,, 5. Planes in Space You hae seen how an equation of a line in space can be obtained from a point on the line and a ector parallel to it. You will now see that an equation of a plane in space can be obtained from a point in the plane and a ector normal (perpendicular) to the plane. Consider the plane containing the point P,, haing a nonero normal ector n a, b, c as shown in Figure.5. This plane consists of all points Q,, for which ector PQ \ is orthogonal to n. Using the dot product, ou can write the following. n PQ \ 0 a, b, c,, 0 a b c 0 The third equation of the plane is said to be in standard form. P Q n n PQ = 0 The normal ector n is orthogonal to each ector PQ \ in the plane. Figure.5 THEOREM. Standard Equation of a Plane in Space The plane containing the point,, and haing normal ector n a, b, c can be represented b the standard form of the equation of a plane a b c 0. B regrouping terms, ou obtain the general form of the equation of a plane in space. a b c d 0 General form of equation of plane Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

39 .5 Lines and Planes in Space 785 Gien the general form of the equation of a plane, it is eas to find a normal ector to the plane. Simpl use the coefficients of,, and and write n a, b, c. Finding an Equation of a Plane in Three-Space (,, ) 5 (,, ) (0,, ) u 5 A plane determined b u and Figure.6 Find the general equation of the plane containing the points,,, and Solution To appl Theorem., ou need a point in the plane and a ector that is normal to the plane. There are three choices for the point, but no normal ector is gien. To obtain a normal ector, use the cross product of ectors u and etending from the point,, to the points 0,, and,,, as shown in Figure.6. The component forms of u and are u 0,,,, 0,,, 0, and it follows that n u i j k 0 0 9i 6j k a, b, c 0,,,,,. is normal to the gien plane. Using the direction numbers for n and the point,,,,, ou can determine an equation of the plane to be a b c Standard form General form Simplified general form REMARK the equation In Eample, check to see that each of the three original points satisfies 0. n θ n Two distinct planes in three-space either are parallel or intersect in a line. For two planes that intersect, ou can determine the angle 0 between them from the angle between their normal ectors, as shown in Figure.7. Specificall, if ectors n and n are normal to two intersecting planes, then the angle between the normal ectors is equal to the angle between the two planes and is θ cos n n n n. Angle between two planes The angle between two planes Figure.7 Consequentl, two planes with normal ectors n and n are. perpendicular when n n 0.. parallel when n is a scalar multiple of n. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

40 786 Chapter Vectors and the Geometr of Space Finding the Line of Intersection of Two Planes Find the angle between the two planes 0 and 0. Then find parametric equations of their line of intersection (see Figure.8). REMARK The threedimensional rotatable graphs that are aailable at LarsonCalculus.com can help ou isualie surfaces such as those shown in Figure.8. If ou hae access to these graphs, ou should use them to help our spatial intuition when studing this section and other sections in the tet that deal with ectors, cures, or surfaces in space. Plane θ Line of intersection Figure.8 Plane Solution Normal ectors for the planes are n,, and n,,. Consequentl, the angle between the two planes is determined as follows. cos n n n n This implies that the angle between the two planes is You can find the line of intersection of the two planes b simultaneousl soling the two linear equations representing the planes. One wa to do this is to multipl the first equation b and add the result to the second equation. 0 0 Substituting 7 back into one of the original equations, ou can determine that 7. Finall, b letting t 7, ou obtain the parametric equations t, t, and 7t Line of intersection which indicate that,, and 7 are direction numbers for the line of intersection. Note that the direction numbers in Eample can be obtained from the cross product of the two normal ectors as follows. i j k n n i i j 7k j k This means that the line of intersection of the two planes is parallel to the cross product of their normal ectors. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

41 Sketching Planes in Space If a plane in space intersects one of the coordinate planes, then the line of intersection is called the trace of the gien plane in the coordinate plane. To sketch a plane in space, it is helpful to find its points of intersection with the coordinate aes and its traces in the coordinate planes. For eample, consider the plane. Equation of plane -trace You can find the -trace b letting 0 and sketching the line.5 Lines and Planes in Space 787 in the -plane. This line intersects the -ais at, 0, 0 and the -ais at 0, 6, 0. In Figure.9, this process is continued b finding the -trace and the -trace, and then shading the triangular region ling in the first octant. (0, 0, ) (0, 0, ) (0, 6, 0) (0, 6, 0) (0, 6, 0) (, 0, 0) (, 0, 0) (, 0, 0) -trace 0: -trace 0: -trace 0: Traces of the plane Figure.9 If an equation of a plane has a missing ariable, such as then the plane must be parallel to the ais represented b the missing ariable, as shown in Figure.50. If two ariables are missing from an equation of a plane, such as a d 0 then it is parallel to the coordinate plane represented b the missing ariables, as shown in Figure.5. (, 0, 0 ) (0, 0, ) Plane: + = Plane is parallel to the -ais. Figure.50 d a, 0, 0 ) 0, d, 0 b ) Plane a d 0 is parallel Plane b d 0 is parallel Plane c d 0 is parallel to the -plane. to the -plane. to the -plane. Figure.5 ) ) ) 0, 0, d c ) Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

42 788 Chapter Vectors and the Geometr of Space Distances Between Points, Planes, and Lines proj n PQ n P Q D Consider two tpes of problems inoling distance in space: () finding the distance between a point and a plane, and () finding the distance between a point and a line. The solutions of these problems illustrate the ersatilit and usefulness of ectors in coordinate geometr: the first problem uses the dot product of two ectors, and the second problem uses the cross product. The distance D between a point Q and a plane is the length of the shortest line segment connecting Q to the plane, as shown in Figure.5. For an point P in the plane, ou can find this distance b projecting the ector PQ \ onto the normal ector n. The length of this projection is the desired distance. D = proj n PQ The distance between a point and a plane Figure.5 THEOREM. Distance Between a Point and a Plane The distance between a plane and a point Q (not in the plane) is D proj n PQ \ PQ \ n n where P is a point in the plane and n is normal to the plane. To find a point in the plane a b c d 0, where a 0, let 0 and 0. Then, from the equation a d 0, ou can conclude that the point d a, 0, 0 lies in the plane. Finding the Distance Between a Point and a Plane REMARK In the solution to Eample 5, note that the choice of the point P is arbitrar. Tr choosing a different point in the plane to erif that ou obtain the same distance. Find the distance between the point Q, 5, and the plane 6. Solution You know that n,, is normal to the plane. To find a point in the plane, let 0 and 0, and obtain the point P, 0, 0. The ector from P to Q is PQ \, 5 0, 0, 5,. Using the Distance Formula gien in Theorem. produces D PQ \ n n, 5,,, From Theorem., ou can determine that the distance between the point Q 0, 0, 0 and the plane a b c d 0 is or D a 0 b 0 c 0 a b c D a 0 b 0 c 0 d a b c Distance between a point and a plane where P,, is a point in the plane and d a b c. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

43 + 6 = 0 6 (, 0, 0) D = 0 The distance between the parallel planes is approimatel.. Figure Lines and Planes in Space 789 Finding the Distance Between Two Parallel Planes Two parallel planes, 6 0 and 6 0, are shown in Figure.5. To find the distance between the planes, choose a point in the first plane, such as 0, 0, 0, 0, 0. Then, from the second plane, ou can determine that a 6, b, c, and d, and conclude that the distance is D a 0 b 0 c 0 d a b c The formula for the distance between a point and a line in space resembles that for the distance between a point and a plane ecept that ou replace the dot product with the length of the cross product and the normal ector n with a direction ector for the line. THEOREM. Distance Between a Point and a Line in Space The distance between a point Q and a line in space is D PQ\ u u where u is a direction ector for the line and P is a point on the line. P θ u Point D = PQ sin θ Line The distance between a point and a line Figure.5 D Q = (,, ) 6 5 Q The distance between the point Q and the line is 6.5. Figure.55 5 Proof In Figure.5, let D be the distance between the point Q and the line. Then D PQ \ sin, where is the angle between u and PQ \. B Propert of Theorem.8, ou hae u PQ \ sin u PQ \ PQ \ u. Consequentl, D PQ \ sin PQ\ u u See LarsonCalculus.com for Bruce Edwards s ideo of this proof. Finding the Distance Between a Point and a Line Find the distance between the point Q,, and the line t, and Solution Using the direction numbers,, and, a direction ector for the line is u,,. To find a point on the line, let t 0 and obtain P, 0,. So, PQ \, 0, 5,, and ou can form the cross product i j k PQ \ u 5 i j 7k,, 7. Finall, using Theorem., ou can find the distance to be D PQ\ u u t, 7 9. t See Figure.55. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

44 790 Chapter Vectors and the Geometr of Space.5 Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Checking Points on a Line In Eercises and, determine whether each point lies on the line.. t, t, t (a) 0, 6, 6 (b),, 5. (a) (b) Finding Parametric and Smmetric Equations In Eercises 8, find sets of (a) parametric equations and (b) smmetric equations of the line through the point parallel to the gien ector or line (if possible). (For each line, write the direction numbers as integers.) Point. 0, 0, 0. 0, 0, 0 5., 0, 6., 0, 7., 0, 8., 5, 7 8 7,, 0 Parallel to,, 5,,, 5, i j k 6j k t, 5 t, 7 t Finding Parametric and Smmetric Equations In Eercises 9, find sets of (a) parametric equations and (b) smmetric equations of the line through the two points (if possible). (For each line, write the direction numbers as integers.) 9. 5,,,,, 0. 0,,,,, 5. 7,, 6,, 0, 6. 0, 0, 5, 0, 0, 0 Finding Parametric Equations In Eercises 0, find a set of parametric equations of the line.. The line passes through the point,, and is parallel to the -plane and the -plane.. The line passes through the point, 5, and is parallel to the -plane and the -plane. 5. The line passes through the point,, and is perpendicular to the plane gien b The line passes through the point, 5, and is perpendicular to the plane gien b The line passes through the point 5,, and is parallel to,,. 8. The line passes through the point,, and is parallel to 5i j. 9. The line passes through the point,, and is parallel to the line t, t, t. 0. The line passes through the point 6, 0, 8 and is parallel to the line 5 t, t, 0. Using Parametric and Smmetric Equations In Eercises, find the coordinates of a point P on the line and a ector parallel to the line.. t, t,. t, 5 t, t Determining Parallel Lines In Eercises 5 8, determine whether an of the lines are parallel or identical. 5. L : 6 t, t, 5 t L : 6t, t, 8t L : 0 6t, t, 7 8t L : 6t, t, 5 6t 6. L : t, 6t, t L : t, t, t L : t, 0t, t L : 5 t, t, 8 t Finding a Point of Intersection In Eercises 9, determine whether the lines intersect, and if so, find the point of intersection and the cosine of the angle of intersection. 9. t,, t s, s, s 0. t, t, t s, s, s.. L : L : L : L : L : L : L : L :, , 5 Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

45 .5 Lines and Planes in Space 79 Checking Points on a Plane In Eercises and, determine whether the plane passes through each point.. 0 (a) 7,, (b) 5,,. 6 0 (a), 6, (b), 5, Finding an Equation of a Plane In Eercises 5 0, find an equation of the plane passing through the point perpendicular to the gien ector or line. Point 5.,, ,, 7.,, 8. 0, 0, 0 9.,, 0 0.,, Perpendicular to n j n k n i j k n i k t, 5 t, t Finding an Equation of a Plane In Eercises 5, find an equation of the plane.. The plane passes through 0, 0, 0,, 0,, and,, 5.. The plane passes through,,,,, 5, and,,.. The plane passes through,,,,,, and,,.. The plane passes through the point,, and is parallel to the -plane. 5. The plane passes through the point,, and is parallel to the -plane. 6. The plane contains the -ais and makes an angle of 6 with the positie -ais. 7. The plane contains the lines gien b and. 8. The plane passes through the point,, and contains the line gien b. 9. The plane passes through the points,, and,, and is perpendicular to the plane. 50. The plane passes through the points,, and,, 5 and is perpendicular to the plane The plane passes through the points,, and, 5, 6 and is parallel to the -ais. 5. The plane passes through the points,, and, 5, 7 and is parallel to the -ais. Finding an Equation of a Plane In Eercises 5 56, find an equation of the plane that contains all the points that are equidistant from the gien points. 5.,, 0, 0,, 5., 0,,, 0, 55.,,, 6,, 56. 5,,,,, 6 Comparing Planes In Eercises 57 6, determine whether the planes are parallel, orthogonal, or neither. If the are neither parallel nor orthogonal, find the angle of intersection Sketching a Graph of a Plane In Eercises 6 70, sketch a graph of the plane and label an intercepts Parallel Planes In Eercises 7 7, determine whether an of the planes are parallel or identical. 7. P : P : 8 P : P : 5 6 P : 6 9 P : 8 5 P : P : 6 7. P : 5 0 P : P : 5 8 P : P : P : 6 9 P : P : Intersection of Planes In Eercises 75 and 76, (a) find the angle between the two planes, and (b) find a set of parametric equations for the line of intersection of the planes Intersection of a Plane and a Line In Eercises 77 80, find the point(s) of intersection (if an) of the plane and the line. Also, determine whether the line lies in the plane. 77., 78. 5, 6 Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

46 79 Chapter Vectors and the Geometr of Space , 5 7, 5 Finding the Distance Between a Point and a Plane In Eercises 8 8, find the distance between the point and the plane. 8. 0, 0, , 0, , 8, 8.,, Finding the Distance Between Two Parallel Planes In Eercises 85 88, erif that the two planes are parallel, and find the distance between the planes. WRITING ABOUT CONCEPTS L (continued) 99. Normal Vector Let and L be nonparallel lines that do not intersect. Is it possible to find a nonero ector such that is normal to both L and L? Eplain our reasoning. 00. HOW DO YOU SEE IT? Match the general equation with its graph. Then state what ais or plane the equation is parallel to. (a) a b d 0 (b) a d 0 (c) c d 0 (d) a c d 0 (i) (ii) Finding the Distance Between a Point and a Line In Eercises 89 9, find the distance between the point and the line gien b the set of parametric equations. (iii) (i) 89., 5, ; t,, t ,, ;,, ;,, 5; t, t, t t, t, t, t, t Finding the Distance Between Two Parallel Lines In Eercises 9 and 9, erif that the lines are parallel, and find the distance between them. 9. L : t, t, t L : t, 6t, t 9. L : 6t, 9t, t L : t, 6t, 8t WRITING ABOUT CONCEPTS 95. Parametric and Smmetric Equations Gie the parametric equations and the smmetric equations of a line in space. Describe what is required to find these equations. 96. Standard Equation of a Plane in Space Gie the standard equation of a plane in space. Describe what is required to find this equation. 97. Intersection of Two Planes Describe a method of finding the line of intersection of two planes. 98. Parallel and Perpendicular Planes Describe a method for determining when two planes, a b c d 0 and a b c d 0, are (a) parallel and (b) perpendicular. Eplain our reasoning. 0. Modeling Data Personal consumption ependitures (in billions of dollars) for seeral tpes of recreation from 005 through 00 are shown in the table, where is the ependitures on amusement parks and campgrounds, is the ependitures on lie entertainment (ecluding sports), and is the ependitures on spectator sports. (Source: U.S. Bureau of Economic Analsis) Year A model for the data is gien b (a) Complete a fourth row in the table using the model to approimate for the gien alues of and. Compare the approimations with the actual alues of. (b) According to this model, increases in ependitures on recreation tpes and would correspond to what kind of change in ependitures on recreation tpe? Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

47 .5 Lines and Planes in Space Mechanical Design The figure shows a chute at the top of a grain eleator of a combine that funnels the grain into a bin. Find the angle between two adjacent sides. 8 in. 6 in. 8 in. 6 in. 8 in. 0. Distance Two insects are crawling along different lines in three-space. At time t (in minutes), the first insect is at the point,, on the line 6 t, 8 t, t. Also, at time t, the second insect is at the point,, on the line t, t, t. Assume that distances are gien in inches. (a) Find the distance between the two insects at time t 0. (b) Use a graphing utilit to graph the distance between the insects from t 0 to t 0. (c) Using the graph from part (b), what can ou conclude about the distance between the insects? (d) How close to each other do the insects get? 0. Finding an Equation of a Sphere Find the standard equation of the sphere with center,, that is tangent to the plane gien b Finding a Point of Intersection Find the point of intersection of the plane 7 and the line through 5,, that is perpendicular to this plane. 06. Finding the Distance Between a Plane and a Line Show that the plane is parallel to the line t, t,, and find the distance between them. 07. Finding a Point of Intersection Find the point of intersection of the line through,, and,, and the plane gien b. 08. Finding Parametric Equations Find a set of parametric equations for the line passing through the point, 0, that is parallel to the plane gien b 5 and perpendicular to the line t, t, t. True or False? In Eercises 09, determine whether the statement is true or false. If it is false, eplain wh or gie an eample that shows it is false. 09. If a i b j c k is an ector in the plane gien b a b c d 0, then a a b b c c Eer two lines in space are either intersecting or parallel.. Two planes in space are either intersecting or parallel.. If two lines L and L are parallel to a plane P, then L and L are parallel.. Two planes perpendicular to a third plane in space are parallel.. A plane and a line in space are either intersecting or parallel. Distances in Space You hae learned two distance formulas in this section one for the distance between a point and a plane, and one for the distance between a point and a line. In this project, ou will stud a third distance problem the distance between two skew lines. Two lines in space are skew if the are neither parallel nor intersecting (see figure). (a) Consider the following two lines in space. L : 5t, 5 5t, t L : s, 6 8s, 7 s (i) Show that these lines are not parallel. (ii) Show that these lines do not intersect, and therefore are skew lines. (iii) Show that the two lines lie in parallel planes. (i) Find the distance between the parallel planes from part (iii). This is the distance between the original skew lines. (b) Use the procedure in part (a) to find the distance between the lines. (c) Use the procedure in part (a) to find the distance between the lines. L : t, t, t L : s, s, s (d) Deelop a formula for finding the distance between the skew lines. L : a t, b t, c t L : a s, b s, c s L L L : t, t, 6t L : s, s, s Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

48 79 Chapter Vectors and the Geometr of Space.6 Surfaces in Space Recognie and write equations of clindrical surfaces. Recognie and write equations of quadric surfaces. Recognie and write equations of surfaces of reolution. Clindrical Surfaces The first fie sections of this chapter contained the ector portion of the preliminar work necessar to stud ector calculus and the calculus of space. In this and the net section, ou will stud surfaces in space and alternatie coordinate sstems for space. You hae alread studied two special tpes of surfaces.. Spheres: r Section.. Planes: a b c d 0 Section.5 A third tpe of surface in space is a clindrical surface, or simpl a clinder. To define a clinder, consider the familiar right circular clinder shown in Figure.56. The clinder was generated b a ertical line moing around the circle a in the -plane. This circle is a generating cure for the clinder, as indicated in the net definition. Right circular clinder: + = a Definition of a Clinder Let C be a cure in a plane and let L be a line not in a parallel plane. The set of all lines parallel to L and intersecting C is a clinder. The cure C is the generating cure (or directri) of the clinder, and the parallel lines are rulings. Rulings are parallel to -ais Figure.56 Without loss of generalit, ou can assume that C lies in one of the three coordinate planes. Moreoer, this tet restricts the discussion to right clinders clinders whose rulings are perpendicular to the coordinate plane containing C, as shown in Figure.57. Note that the rulings intersect C and are parallel to the line L. For the right circular clinder shown in Figure.56, the equation of the generating cure in the -plane is a. To find an equation of the clinder, note that ou can generate an one of the rulings b fiing the alues of and and then allowing to take on all real alues. In this sense, the alue of is arbitrar and is, therefore, not included in the equation. In other words, the equation of this clinder is simpl the equation of its generating cure. a Rulings intersect C and are parallel to L. Generating cure C L intersects C. Right clinder: A clinder whose rulings are perpendicular to the coordinate plane containing C Figure.57 Equation of clinder in space Equations of Clinders The equation of a clinder whose rulings are parallel to one of the coordinate aes contains onl the ariables corresponding to the other two aes. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

49 .6 Surfaces in Space 795 Sketching a Clinder Sketch the surface represented b each equation. a. b. sin, 0 Solution a. The graph is a clinder whose generating cure,, is a parabola in the -plane. The rulings of the clinder are parallel to the -ais, as shown in Figure.58(a). b. The graph is a clinder generated b the sine cure in the -plane. The rulings are parallel to the -ais, as shown in Figure.58(b). Generating cure C lies in -plane Generating cure C lies in -plane π Clinder: = Clinder: = sin (a) Rulings are parallel to -ais. Figure.58 (b) Rulings are parallel to -ais. Quadric Surfaces The fourth basic tpe of surface in space is a quadric surface. Quadric surfaces are the three-dimensional analogs of conic sections. Quadric Surface The equation of a quadric surface in space is a second-degree equation in three ariables. The general form of the equation is A B C D E F G H I J 0. There are si basic tpes of quadric surfaces: ellipsoid, hperboloid of one sheet, hperboloid of two sheets, elliptic cone, elliptic paraboloid, and hperbolic paraboloid. The intersection of a surface with a plane is called the trace of the surface in the plane. To isualie a surface in space, it is helpful to determine its traces in some well-chosen planes. The traces of quadric surfaces are conics. These traces, together with the standard form of the equation of each quadric surface, are shown in the table on the net two pages. In the table on the net two pages, onl one of seeral orientations of each quadric surface is shown. When the surface is oriented along a different ais, its standard equation will change accordingl, as illustrated in Eamples and. The fact that the two tpes of paraboloids hae one ariable raised to the first power can be helpful in classifing quadric surfaces. The other four tpes of basic quadric surfaces hae equations that are of second degree in all three ariables. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

50 796 Chapter Vectors and the Geometr of Space a Ellipsoid b c -trace -trace Trace Plane Ellipse Parallel to -plane Ellipse Parallel to -plane Ellipse Parallel to -plane The surface is a sphere when a b c 0. -trace Hperboloid of One Sheet a b c Trace Plane Ellipse Parallel to -plane Hperbola Parallel to -plane Hperbola Parallel to -plane The ais of the hperboloid corresponds to the ariable whose coefficient is negatie. -trace -trace -trace Hperboloid of Two Sheets c a b -trace -trace Trace Plane Ellipse Parallel to -plane Hperbola Parallel to -plane Hperbola Parallel to -plane The ais of the hperboloid corresponds to the ariable whose coefficient is positie. There is no trace in the coordinate plane perpendicular to this ais. parallel to -plane no -trace Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

51 .6 Surfaces in Space 797 a Elliptic Cone b c 0 -trace Trace Plane Ellipse Parallel to -plane Hperbola Parallel to -plane Hperbola Parallel to -plane The ais of the cone corresponds to the ariable whose coefficient is negatie. The traces in the coordinate planes parallel to this ais are intersecting lines. -trace (one point) parallel to -plane -trace Elliptic Paraboloid a b -trace -trace Trace Plane Ellipse Parallel to -plane Parabola Parallel to -plane Parabola Parallel to -plane The ais of the paraboloid corresponds to the ariable raised to the first power. parallel to -plane -trace (one point) Hperbolic Paraboloid b a -trace Trace Plane Hperbola Parallel to -plane Parabola Parallel to -plane Parabola Parallel to -plane The ais of the paraboloid corresponds to the ariable raised to the first power. -trace parallel to -plane Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

52 798 Chapter Vectors and the Geometr of Space To classif a quadric surface, begin b writing the equation of the surface in standard form. Then, determine seeral traces taken in the coordinate planes or taken in planes that are parallel to the coordinate planes. Sketching a Quadric Surface = Hperboloid of two sheets: = Figure.59 = Classif and sketch the surface 0. Solution Begin b writing the equation in standard form. 0 0 From the table on pages 796 and 797, ou can conclude that the surface is a hperboloid of two sheets with the -ais as its ais. To sketch the graph of this surface, it helps to find the traces in the coordinate planes. Write original equation. Diide b. Standard form - trace 0: - trace 0: - trace 0: The graph is shown in Figure.59. Hperbola No trace Hperbola Sketching a Quadric Surface Elliptic paraboloid: = + Classif and sketch the surface 0. = Solution Because is raised onl to the first power, the surface is a paraboloid. The ais of the paraboloid is the -ais. In standard form, the equation is. Standard form Some conenient traces are listed below. 0 = + = -trace 0: Parabola -trace 0: Parabola parallel to - plane : Ellipse Figure.60 The surface is an elliptic paraboloid, as shown in Figure.60. Some second-degree equations in,, and do not represent an of the basic tpes of quadric surfaces. For eample, the graph of 0 Single point is a single point, and the graph of Right circular clinder is a right circular clinder. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

53 .6 Surfaces in Space 799 For a quadric surface not centered at the origin, ou can form the standard equation b completing the square, as demonstrated in Eample. ( ) ( + ) ( ) + + = 5 (,, ) An ellipsoid centered at,, Figure.6 A Quadric Surface Not Centered at the Origin See LarsonCalculus.com for an interactie ersion of this tpe of eample. Classif and sketch the surface 0. Solution Begin b grouping terms and factoring where possible. Net, complete the square for each ariable and write the equation in standard form. From this equation, ou can see that the quadric surface is an ellipsoid that is centered at Its graph is shown in Figure.6.,,. TECHNOLOGY A -D graphing utilit can help ou isualie a surface in space.* Such a graphing utilit ma create a three-dimensional graph b sketching seeral traces of the surface and then appling a hidden-line routine that blocks out portions of the surface that lie behind other portions of the surface. Two eamples of figures that were generated b Mathematica are shown below. Elliptic paraboloid Generated b Mathematica Hperbolic paraboloid 6 6 Generated b Mathematica Using a graphing utilit to graph a surface in space requires practice. For one thing, ou must know enough about the surface to be able to specif a iewing window that gies a representatie iew of the surface. Also, ou can often improe the iew of a surface b rotating the aes. For instance, note that the elliptic paraboloid in the figure is seen from a line of sight that is higher than the line of sight used to iew the hperbolic paraboloid. * Some -D graphing utilities require surfaces to be entered with parametric equations. For a discussion of this technique, see Section 5.5. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

54 800 Chapter Vectors and the Geometr of Space Surfaces of Reolution Circular cross section (,, ) Generating cure = r() r () (0, 0, ) (0, r(), ) The fifth special tpe of surface ou will stud is a surface of reolution. In Section 7., ou studied a method for finding the area of such a surface. You will now look at a procedure for finding its equation. Consider the graph of the radius function in the -plane. When this graph is reoled about the -ais, it forms a surface of reolution, as shown in Figure.6. The trace of the surface in the plane 0 is a circle whose radius is r 0 and whose equation is r Generating cure r 0. Circular trace in plane: 0 Figure.6 Replacing 0 with produces an equation that is alid for all alues of. In a similar manner, ou can obtain equations for surfaces of reolution for the other two aes, and the results are summaried as follows. Surface of Reolution If the graph of a radius function r is reoled about one of the coordinate aes, then the equation of the resulting surface of reolution has one of the forms listed below.. Reoled about the -ais: r. Reoled about the -ais: r. Reoled about the -ais: r Finding an Equation for a Surface of Reolution Find an equation for the surface of reolution formed b reoling (a) the graph of about the -ais and (b) the graph of 9 about the -ais. Solution a. An equation for the surface of reolution formed b reoling the graph of Radius function Surface: + = 9 about the -ais is r. Reoled about the -ais Substitute for r. b. To find an equation for the surface formed b reoling the graph of 9 about the -ais, sole for in terms of to obtain r. Radius function So, the equation for this surface is r Reoled about the -ais Substitute for r. Generating cure 9 = 9. Equation of surface Figure.6 The graph is shown in Figure.6. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

55 The generating cure for a surface of reolution is not unique. For instance, the surface e can be formed b reoling either the graph of e about the -ais or the graph of e about the -ais, as shown in Figure.6..6 Surfaces in Space 80 Surface: + = e Generating cure in -plane = e Generating cure in -plane = e Figure.6 Finding a Generating Cure Find a generating cure and the ais of reolution for the surface 9. Solution The equation has one of the forms listed below. r Reoled about -ais r Reoled about -ais r Reoled about -ais Surface: Because the coefficients of and are equal, + + = 9 ou should choose the third form and write 9. The -ais is the ais of reolution. You can choose a generating cure from either of the traces Generating cure in -plane = 9 Trace in -plan 9 or Trace in -plan 9. For instance, using the first trace, the generating cure is the semiellipse 9. Generating cure The graph of this surface is shown in Figure.65. Figure.65 Generating cure in -plane = 9 Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

56 80 Chapter Vectors and the Geometr of Space.6 Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Matching In Eercises 6, match the equation with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] (a) (c) (e) (b) (d) (f) 6 WRITING ABOUT CONCEPTS 5. Clinder State the definition of a clinder. 6. Trace of a Surface What is meant b the trace of a surface? How do ou find a trace? 7. Quadric Surfaces Identif the si quadric surfaces and gie the standard form of each. 8. Classifing an Equation What does the equation represent in the -plane? What does it represent in three-space? 9. Classifing an Equation What does the equation 6 represent in the -plane? What does it represent in three-space? 0. HOW DO YOU SEE IT? The four figures are graphs of the quadric surface. Match each of the four graphs with the point in space from which the paraboloid is iewed. The four points are 0, 0, 0, 0, 0, 0, 0, 0, 0, and 0, 0, 0. (a) (b) Sketching a Surface in Space In Eercises 7, describe and sketch the surface Sketching a Quadric Surface In Eercises, classif and sketch the quadric surface. Use a computer algebra sstem or a graphing utilit to confirm our sketch (c) Finding an Equation of a Surface of Reolution In Eercises 6, find an equation for the surface of reolution formed b reoling the cure in the indicated coordinate plane about the gien ais. (d) Equation Coordinate Ais of of Cure Plane Reolution. -plane -ais. -plane -ais. -plane -ais Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

57 .6 Surfaces in Space 80 Equation Coordinate Ais of of Cure Plane Reolution. -plane -ais 5. -plane -ais 6. ln -plane -ais Finding a Generating Cure In Eercises 7 and 8, find an equation of a generating cure gien the equation of its surface of reolution cos Finding the Volume of a Solid In Eercises 9 and 0, use the shell method to find the olume of the solid below the surface of reolution and aboe the -plane.. The set of all points equidistant from the point 0,, 0 and the plane. The set of all points equidistant from the point 0, 0, and the -plane 9. The cure in the -plane is reoled about the -ais. 0. The cure sin 0 in the -plane is reoled about the -ais. Analing a Trace when the surface is intersected b the indicated planes. In Eercises and, anale the trace. Find the lengths of the major and minor aes and the coordinates of the foci of the ellipse generated when the surface is intersected b the planes gien b (a) and (b) 8.. Find the coordinates of the focus of the parabola formed when the surface is intersected b the planes gien b (a) and (b). Finding an Equation of a Surface In Eercises and, find an equation of the surface satisfing the conditions, and identif the surface. 6. Machine Design The top of a rubber bushing designed to absorb ibrations in an automobile is the surface of reolution generated b reoling the cure for 0 in the -plane about the -ais. (a) Find an equation for the surface of reolution. (b) All measurements are in centimeters and the bushing is set on the -plane. Use the shell method to find its olume. (c) The bushing has a hole of diameter centimeter through its center and parallel to the ais of reolution. Find the olume of the rubber bushing. 7. Using a Hperbolic Paraboloid Determine the intersection of the hperbolic paraboloid b a with the plane b a 0. ( Assume a, b > Intersection of Surfaces Eplain wh the cure of intersection of the surfaces and 6 lies in a plane. True or False? In Eercises 9 5, determine whether the statement is true or false. If it is false, eplain wh or gie an eample that shows it is false. 9. A sphere is an ellipsoid. 50. The generating cure for a surface of reolution is unique. 5. All traces of an ellipsoid are ellipses. 5. All traces of a hperboloid of one sheet are hperboloids. 5. Think About It Three tpes of classic topological surfaces are shown below. The sphere and torus hae both an inside and an outside. Does the Klein bottle hae both an inside and an outside? Eplain. 5. Geograph Because of the forces caused b its rotation, Earth is an oblate ellipsoid rather than a sphere. The equatorial radius is 96 miles and the polar radius is 950 miles. Find an equation of the ellipsoid. (Assume that the center of Earth is at the origin and that the trace formed b the plane 0 corresponds to the equator.) Sphere Klein bottle Denis Tabler/Shutterstock.com Torus Klein bottle Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

58 80 Chapter Vectors and the Geometr of Space.7 Clindrical and Spherical Coordinates Use clindrical coordinates to represent surfaces in space. Use spherical coordinates to represent surfaces in space. Clindrical Coordinates Rectangular coordinates: = r cos θ = r sin θ = Clindrical coordinates: r = + tan θ = = P (,, ) (r, θ, ) You hae alread seen that some two-dimensional graphs are easier to represent in polar coordinates than in rectangular coordinates. A similar situation eists for surfaces in space. In this section, ou will stud two alternatie space-coordinate sstems. The first, the clindrical coordinate sstem, is an etension of polar coordinates in the plane to three-dimensional space. The Clindrical Coordinate Sstem In a clindrical coordinate sstem, a point P in space is represented b an ordered triple r,,. θ r. r, is a polar representation of the projection of P in the -plane.. is the directed distance from r, to P. Figure.66 To conert from rectangular to clindrical coordinates (or ice ersa), use the conersion guidelines for polar coordinates listed below and illustrated in Figure.66. Clindrical to rectangular: r cos, r sin, Rectangular to clindrical: r, tan, (,, ) = (,, ) P The point 0, 0, 0 is called the pole. Moreoer, because the representation of a point in the polar coordinate sstem is not unique, it follows that the representation in the clindrical coordinate sstem is also not unique. Clindrical-to-Rectangular Conersion r θ 5π (r, θ, ) = (,, 6 Figure.67 ( Conert the point r,,, 56, to rectangular coordinates. Solution. Using the clindrical-to-rectangular conersion equations produces cos 5 6 sin 5 6 So, in rectangular coordinates, the point is,,,,, Figure.67. as shown in Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

59 .7 Clindrical and Spherical Coordinates 805 Rectangular- to- Clindrical Conersion (,, ) = (,, ) r = = π θ = π π (r, θ, ) = (,, or,, ( Figure.68 ( ( Conert the point to clindrical coordinates. Solution Use the rectangular-to-clindrical conersion equations. tan n You hae two choices for r and infinitel man choices for. As shown in Figure.68, two conenient representations of the point are and,,,, r ± ±,,,,. n arctan r > 0 and r < 0 and in Quadrant I in Quadrant III Clindrical coordinates are especiall conenient for representing clindrical surfaces and surfaces of reolution with the -ais as the ais of smmetr, as shown in Figure = 9 r = + = r = + = r = + = r = + Clinder Paraboloid Cone Hperboloid Figure.69 Vertical planes containing the -ais and horiontal planes also hae simple clindrical coordinate equations, as shown in Figure.70. Vertical plane: θ = c Horiontal plane: = c θ = c Figure.70 Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

60 806 Chapter Vectors and the Geometr of Space Rectangular-to-Clindrical Conersion Find an equation in clindrical coordinates for the surface represented b each rectangular equation. Rectangular: + = Clindrical: r = a. b. Solution a. From Section.6, ou know that the graph of 6 6 is an elliptic cone with its ais along the -ais, as shown in Figure.7. When ou replace with r, the equation in clindrical coordinates is Figure.7 Rectangular: = Figure.7 Clindrical: r = csc θ cot θ Rectangular equation r. Clindrical equation b. The graph of the surface is a parabolic clinder with rulings parallel to the -ais, as shown in Figure.7. To obtain the equation in clindrical coordinates, replace with r sin and with r cos, as shown. r sin r cos rr sin cos 0 r sin cos 0 r cos sin r csc cot Rectangular equation Substitute r sin for and r cos for. Collect terms and factor. Diide each side b r. Sole for r. Clindrical equation Note that this equation includes a point for which r 0, so nothing was lost b diiding each side b the factor r. Conerting from clindrical coordinates to rectangular coordinates is less straightforward than conerting from rectangular coordinates to clindrical coordinates, as demonstrated in Eample. Clindrical: r cos θ + + = 0 Clindrical-to-Rectangular Conersion Find an equation in rectangular coordinates for the surface represented b the clindrical equation r cos 0. Solution Rectangular: = Figure.7 r cos 0 Clindrical equation r cos sin 0 Trigonometric identit r cos r sin Replace r cos with and r sin with. Rectangular equation This is a hperboloid of two sheets whose ais lies along the -ais, as shown in Figure.7. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

61 .7 Clindrical and Spherical Coordinates 807 Spherical Coordinates 80 W 0 N Prime meridian In the spherical coordinate sstem, each point is represented b an ordered triple: the first coordinate is a distance, and the second and third coordinates are angles. This sstem is similar to the latitude-longitude sstem used to identif points on the surface of Earth. For eample, the point on the surface of Earth whose latitude is 0 North (of the equator) and whose longitude is 80 West (of the prime meridian) is shown in Figure.7. Assuming that Earth is spherical and has a radius of 000 miles, ou would label this point as 000, 80, 50. Equator Radius 80 clockwise from 50 down from prime meridian North Pole Figure.7 The Spherical Coordinate Sstem In a spherical coordinate sstem, a point P in space is represented b an ordered triple,,, where is the lowercase Greek letter rho and is the lowercase Greek letter phi.. is the distance between P and the origin, 0.. is the same angle used in clindrical coordinates for r 0.. is the angle between the positie -ais and the line segment OP \, 0. Note that the first and third coordinates, and, are nonnegatie. O θ φ ρ Spherical coordinates Figure.75 r = ρ sin φ = + r P ( ρ, θ, φ) (,, ) The relationship between rectangular and spherical coordinates is illustrated in Figure.75. To conert from one sstem to the other, use the conersion guidelines listed below. Spherical to rectangular: Rectangular to spherical: sin cos, sin sin, cos, tan, arccos To change coordinates between the clindrical and spherical sstems, use the conersion guidelines listed below. Spherical to clindrical r 0: r sin,, cos Clindrical to spherical r 0: r,, arccos r Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

62 ) 808 Chapter Vectors and the Geometr of Space The spherical coordinate sstem is useful primaril for surfaces in space that hae a point or center of smmetr. For eample, Figure.76 shows three surfaces with simple spherical equations. φ = c c Sphere: ρ = c θ = c Vertical half-plane: θ = c Half-cone: φ = c 0 < c < π ) Figure.76 Rectangular-to-Spherical Conersion See LarsonCalculus.com for an interactie ersion of this tpe of eample. Find an equation in spherical coordinates for the surface represented b each rectangular equation. a. Cone: b. Sphere: 0 Solution a. Use the spherical-to-rectangular equations sin cos, sin sin, and cos and substitute in the rectangular equation as shown. Rectangular: + + = 0 Spherical: ρ = cos φ sin cos sin sin cos sin cos sin cos So, ou can conclude that sin cos sin cos tan tan ± 0 or. The equation represents the upper half-cone, and the equation represents the lower half-cone. b. Because and the rectangular equation has the following spherical form. cos, Figure.77 cos 0 Temporaril discarding the possibilit that cos 0 or ou hae the spherical equation Note that the solution set for this equation includes a point for which so nothing is lost b discarding the factor. The sphere represented b the equation is shown in Figure.77. cos cos 0 cos. 0, 0, Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

63 .7 Clindrical and Spherical Coordinates Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Clindrical-to-Rectangular Conersion In Eercises 6, conert the point from clindrical coordinates to rectangular coordinates.. 7, 0, 5.,,.,,. 5. 7, 6, 6. 6,, 0.5,, 8 Rectangular-to-Clindrical Conersion In Eercises 7, conert the point from rectangular coordinates to clindrical coordinates. 7. 0, 5, 8.,, 9.,, 0.,, 7.,,.,, 6 Rectangular-to-Clindrical Conersion In Eercises 0, find an equation in clindrical coordinates for the equation gien in rectangular coordinates Rectangular-to-Spherical Conersion In Eercises 8, find an equation in spherical coordinates for the equation gien in rectangular coordinates Spherical-to-Rectangular Conersion In Eercises 9 56, find an equation in rectangular coordinates for the equation gien in spherical coordinates, and sketch its graph Matching In Eercises 57 6, match the equation (written in terms of clindrical or spherical coordinates) with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] (a) 5 cos csc π (b) sec csc sec Clindrical-to-Rectangular Conersion In Eercises 8, find an equation in rectangular coordinates for the equation gien in clindrical coordinates, and sketch its graph.. r... r 6 5. r 5 6. r cos (c) 5 (d) 5 7. r sin 8. r cos Rectangular-to-Spherical Conersion In Eercises 9, conert the point from rectangular coordinates to spherical coordinates , 0, 0 0., 0, 0.,,.,,.,,.,, Spherical-to-Rectangular Conersion In Eercises 5 0, conert the point from spherical coordinates to rectangular coordinates. (e) π (f) 5., 6, 6. 7.,, ,, 0.,, 9,, 6,, r r 6. sec Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

64 80 Chapter Vectors and the Geometr of Space Clindrical-to-Spherical Conersion In Eercises 6 70, conert the point from clindrical coordinates to spherical coordinates. 6., 6.,, 0, 0 65., 66.,,, 67., 68. 6, 6,, 69.,, 5 70.,, Spherical-to-Clindrical Conersion In Eercises 7 78, conert the point from spherical coordinates to clindrical coordinates. 7. 0, 7. 6,, 8, 7. 6,, 7. 8,, 75. 6, 76. 5, 5 6, 6, , 78. 7,, 6, 6 WRITING ABOUT CONCEPTS 79. Rectangular and Clindrical Coordinates Gie the equations for the coordinate conersion from rectangular to clindrical coordinates and ice ersa. 80. Spherical Coordinates Eplain wh in spherical coordinates the graph of is a half-plane and not an entire plane. 8. Rectangular and Spherical Coordinates Gie the equations for the coordinate conersion from rectangular to spherical coordinates and ice ersa. c 8. HOW DO YOU SEE IT? Identif the surface graphed and match the graph with its rectangular equation. Then find an equation in clindrical coordinates for the equation gien in rectangular coordinates. (a) (b) (i) 9 (ii) Conerting a Rectangular Equation In Eercises 8 90, conert the rectangular equation to an equation in (a) clindrical coordinates and (b) spherical coordinates Sketching a Solid In Eercises 9 9, sketch the solid that has the gien description in clindrical coordinates. 9. 0, 0 r, 0 9., 0 r, 0 r cos , 0 r a, r a, r, r 6r 8 Sketching a Solid In Eercises 95 98, sketch the solid that has the gien description in spherical coordinates Think About It In Eercises 99 0, find inequalities that describe the solid, and state the coordinate sstem used. Position the solid on the coordinate sstem such that the inequalities are as simple as possible. 99. A cube with each edge 0 centimeters long 00. A clindrical shell 8 meters long with an inside diameter of 0.75 meter and an outside diameter of.5 meters 0. A spherical shell with inside and outside radii of inches and 6 inches, respectiel 0. The solid that remains after a hole inch in diameter is drilled through the center of a sphere 6 inches in diameter 0. The solid inside both 9 and 9 0. The solid between the spheres and 9, and inside the cone True or False? In Eercises 05 08, determine whether the statement is true or false. If it is false, eplain wh or gie an eample that shows it is false. 05. In clindrical coordinates, the equation r is a clinder. 06. The equations and represent the same surface. 07. The clindrical coordinates of a point,, are unique. 08. The spherical coordinates of a point,, are unique. 09. Intersection of Surfaces Identif the cure of intersection of the surfaces (in clindrical coordinates) sin and r. 0. Intersection of Surfaces Identif the cure of intersection of the surfaces (in spherical coordinates) and. sec, 0 6, 0 a sec,, 0, 0, 0, 0, Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

65 Reiew Eercises 8 Reiew Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Writing Vectors in Different Forms In Eercises and, let u PQ \ and PR \, and (a) write u and in component form, (b) write u and as the linear combination of the standard unit ectors i and j, (c) find the magnitudes of u and, and (d) find u.. P,, Q,, R 5,. P,, Q 5,, R, Finding a Vector In Eercises and, find the component form of gien its magnitude and the angle it makes with the positie -ais ,., 5. Finding Coordinates of a Point Find the coordinates of the point located in the -plane, four units to the right of the -plane, and fie units behind the -plane. 6. Finding Coordinates of a Point Find the coordinates of the point located on the -ais and seen units to the left of the -plane. Finding the Distance Between Two Points in Space In Eercises 7 and 8, find the distance between the points. 7., 6,,,, 5 8.,, 5,,, Finding the Equation of a Sphere In Eercises 9 and 0, find the standard equation of the sphere. 9. Center:,, 6; Diameter: 5 0. Endpoints of a diameter: 0, 0,,, 6, 0 Finding the Equation of a Sphere In Eercises and, complete the square to write the equation of the sphere in standard form. Find the center and radius Writing a Vector in Different Forms In Eercises and, the initial and terminal points of a ector are gien. (a) Sketch the directed line segment, (b) find the component form of the ector, (c) write the ector using standard unit ector notation, and (d) sketch the ector with its initial point at the origin.. Initial point:,,. Initial point: 6,, 0 Terminal point:,, 7 Terminal point:,, 8 Using Vectors to Determine Collinear Points In Eercises 5 and 6, use ectors to determine whether the points are collinear. 5.,,,, 6, 9, 5,, ,, 7, 8, 5, 5,, 6, 5 7. Finding a Unit Vector Find a unit ector in the direction of u,, Finding a Vector Find the ector of magnitude 8 in the direction 6,,. Finding Dot Products In Eercises 9 and 0, let u PQ \ and PR \, and find (a) the component forms of u and, (b) u, and (c). 9. P 5, 0, 0, Q,, 0, R, 0, 6 0. P,,, Q 0, 5,, R 5, 5, 0 Finding the Angle Between Two Vectors In Eercises, find the angle between the ectors (a) in radians and (b) in degrees.. u 5cosi sinj cosi sinj.. u 6i j k, u 0, 5, 5, i 5j,,. u, 0,,,, Comparing Vectors In Eercises 5 and 6, determine whether u and are orthogonal, parallel, or neither. 5. u 7,, 6. u,, 6,, 5 6,, Finding the Projection of u onto In Eercises 7 0, find the projection of u onto u 7, 9,, 5 u i j, i j u,,,, 0, u 5i j k, i j k. Orthogonal Vectors Find two ectors in opposite directions that are orthogonal to the ector u 5, 6,.. Work An object is pulled 8 feet across a floor using a force of 75 pounds. The direction of the force is 0 aboe the horiontal. Find the work done. Finding Cross Products In Eercises 6, find (a) u, (b) u, and (c).. u i j 6k. u 6i 5j k 5i j k i j k 5. u,, 6. u 0,,,,,, 7. Finding a Unit Vector Find a unit ector that is orthogonal to both u, 0, 8 and, 6, Area Find the area of the parallelogram that has the ectors u,, 5 and,, as adjacent sides. Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

66 8 Chapter Vectors and the Geometr of Space 9. Torque The specifications for a tractor state that the torque 7 on a bolt with head sie 8 inch cannot eceed 00 foot-pounds. Determine the maimum force F that can be applied to the wrench in the figure. 0. Volume Use the triple scalar product to find the olume of the parallelepiped haing adjacent edges u i j, j k, and w j k. Finding Parametric and Smmetric Equations In Eercises and, find sets of (a) parametric equations and (b) smmetric equations of the line through the two points. (For each line, write the direction numbers as integers.)., 0,, 9,, 6.,,, 8, 0, 5 Finding Parametric Equations In Eercises 6, find a set of parametric equations of the line.. The line passes through the point,, and is perpendicular to the -plane.. The line passes through the point,, and is parallel to the line gien b. 5. The line is the intersection of the planes 7 and. 6. The line passes through the point 0,, and is perpendicular to u, 5, and,,. Finding an Equation of a Plane In Eercises 7 50, find an equation of the plane. 7. The plane passes through,,,,,, and,,. 8. The plane passes through the point,, and is perpendicular to n i j k. 9. The plane contains the lines gien b 7 8 in. and ft. 50. The plane passes through the points 5,, and,, and is perpendicular to the plane. 5. Distance Find the distance between the point, 0, and the plane Distance Find the distance between the point,, and the plane Distance Find the distance between the planes 5 and F 5. Distance Find the distance between the point 5,, and the line gien b t, t, and 5 t. Sketching a Surface in Space In Eercises 55 6, describe and sketch the surface cos Surface of Reolution Find an equation for the surface of reolution formed b reoling the cure in the -plane about the -ais. 66. Surface of Reolution Find an equation for the surface of reolution formed b reoling the cure in the -plane about the -ais. Conerting Rectangular Coordinates In Eercises 67 and 68, conert the point from rectangular coordinates to (a) clindrical coordinates and (b) spherical coordinates. 67.,, 68. Clindrical-to-Spherical Conersion In Eercises 69 and 70, conert the point from clindrical coordinates to spherical coordinates , 70. 6, 50 Spherical-to-Clindrical Conersion In Eercises 7 and 7, conert the point from spherical coordinates to clindrical coordinates. 7. 5, 7.,,, Conerting a Rectangular Equation In Eercises 7 and 7, conert the rectangular equation to an equation in (a) clindrical coordinates and (b) spherical coordinates Clindrical-to-Rectangular Conersion In Eercises 75 and 76, find an equation in rectangular coordinates for the equation gien in clindrical coordinates, and sketch its graph. 75. r 5 cos 76. Spherical-to-Rectangular Conersion In Eercises 77 and 78, find an equation in rectangular coordinates for the equation gien in spherical coordinates, and sketch its graph ,, 8, 5 6, 7 cos Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.

67 P.S. Problem Soling 8 P.S. Problem Soling See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises.. Proof Using ectors, proe the Law of Sines: If a, b, and c are the three sides of the triangle shown in the figure, then sin A a sin B b sin C c. c A. Using an Equation Consider the function f t dt. 0 B (a) Use a graphing utilit to graph the function on the interal. (b) Find a unit ector parallel to the graph of f at the point 0, 0. (c) Find a unit ector perpendicular to the graph of f at the point 0, 0. (d) Find the parametric equations of the tangent line to the graph of f at the point 0, 0.. Proof Using ectors, proe that the line segments joining the midpoints of the sides of a parallelogram form a parallelogram (see figure).. Proof Using ectors, proe that the diagonals of a rhombus are perpendicular (see figure). 5. Distance (a) Find the shortest distance between the point Q, 0, 0 and the line determined b the points P 0, 0, and P 0,,. (b) Find the shortest distance between the point Q, 0, 0 and the line segment joining the points P 0, 0, and P 0,,. b a C 6. Orthogonal Vectors Let P 0 be a point in the plane with normal ector n. Describe the set of points P in the plane for which n PP \ 0 is orthogonal to n PP \ Volume (a) Find the olume of the solid bounded below b the paraboloid and aboe b the plane. (b) Find the olume of the solid bounded below b the elliptic paraboloid a b and aboe b the plane k, where k > 0. (c) Show that the olume of the solid in part (b) is equal to one-half the product of the area of the base times the altitude, as shown in the figure. 8. Volume (a) Use the disk method to find the olume of the sphere r. (b) Find the olume of the ellipsoid. a b c 9. Proof Proe the following propert of the cross product. u w u w u w 0. Using Parametric Equations Consider the line gien b the parametric equations t, and the point,, s for an real number s. (a) Write the distance between the point and the line as a function of s. (b) Use a graphing utilit to graph the function in part (a). Use the graph to find the alue of s such that the distance between the point and the line is minimum. (c) Use the oom feature of a graphing utilit to oom out seeral times on the graph in part (b). Does it appear that the graph has slant asmptotes? Eplain. If it appears to hae slant asmptotes, find them.. Sketching Graphs Sketch the graph of each equation gien in spherical coordinates. (a) (b) sin t, Base Altitude t cos Copright 0 Cengage Learning. All Rights Resered. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial reiew has deemed that an suppressed content does not materiall affect the oerall learning eperience. Cengage Learning reseres the right to remoe additional content at an time if subsequent rights restrictions require it.