Analytic Geometry in Three Dimensions Cheat Sheet: Examples and Exercises, Cheat Sheet of Analytical Geometry

Download Analytic Geometry in Three Dimensions Cheat Sheet: Examples and Exercises and more Cheat Sheet Analytical Geometry in PDF only on Docsity! 11.1 The Three-Dimensional Coordinate System 11.2 Vectors in Space 11.3 The Cross Product of Two Vectors 11.4 Lines and Planes in Space The three-dimensional coordinate system is used in chemistry to help understand the structure of crystals. For example, isometric crystals are shaped like cubes Amold Fisher/Photo Researchers, Inc. SELECTED APPLICATIONS Three-dimensional analytic geometry concepts have many real-life applications. The applications listed below represent a small sample of the applications in this chapter. * Crystals, * Torque, * Moment of a Force, Exercise 63, page 818 Exercise 43, page 833 Exercise 12, page 849 * Geography, * Data Analysis: Milk Consumption, Exercise 66, page 819 Exercise 47, page 842 * Tension, * Mechanical Design, Exercise 43, page 826 Exercise 48, page 842 as = ¥ ! j 811 * » 812 Chapter11__ Analytic Geometry in Three Dimensions The Three-Dimensiona TE LCT What you should learn eee eee The Three-Dimensional Coordinate System dimensional coordinate Recall that the Cartesian plane is determined by two perpendicular number lines system. called the x-axis and the y-axis. These axes, together with their point of intersec- + Find distances between points tion (the origin), allow you to develop a two-dimensional coordinate system for in space and find midpoints identifying points in a plane. To identify a point in space, you must introduce a third dimension to the model. The geometry of this three-dimensional model is called solid analytic geometry. You can construct a three-dimensional coordinate system by passing a z-axis perpendicular to both the x- and y-axes at the origin. Figure 11.1 shows the positive portion of each coordinate axis. Taken as pairs, the axes determine three coordinate planes: the xy-plane, the xz-plane, and the yz-plane. These three coordinate planes separate the three-dimensional coordinate system into eight octants. The first octant is the one in which all three coordinates are positive. In this three-dimensional system, a point P in space is determined by an ordered triple (x,y, z), where x, y, and z are as follows. of line segments joining points x = directed distance from yz-plane to P y = directed distance from xz-plane to P z = directed distance from xy-plane to P yz-plane e & | or2 e ° . y y ye 4 xy-plane x x FIGURE 11.1 Figure 11.2 A three-dimensional coordinate system can have either a left-handed or a right-handed orientation. In this text, you will work exclusively with right- handed systems, as illustrated in Figure 11.2. In a right-handed system, Octants IL, III, and IV are found by rotating counterclockwise around the positive z-axis. Octant V is vertically below Octant I. Octants VI, VII, and VIII are then found by rotating counterclockwise around the negative z-axis. See Figure 11.3. , ko, ee oe, bh. 4g OWS! Octant I Octant IT Octant IIT Octant IV Octant V Octant VI Octant VII Octant VIII Figure 11.3 dag Find the equation of the sphere that has the points (3, —2, 6) and (—1, 4, 2) as endpoints of a diameter. Explain how this problem gives you a chance to use all three formulas discussed so far in this section: the Distance Formula in Space, the Midpoint Formula in Space, and the standard equation of a sphere. Sphere: (x= 12 + (+2)? + @- 3 = (Vo P Figure 11.9 (x2 - 2 + Section 11.1 The Three-Dimensional Coordinate System 815 Finding the Equation of a Sphere Find the standard equation of the sphere with center (2, 4, 3) and radius 3. Does this sphere intersect the xy-plane? Solution (x —hP + (y —kP + -jP = (x= 2P + (y = 4P + = 3P From the graph shown in Figure 11.8, you can see that the center of the sphere lies three units above the xy-plane. Because the sphere has a radius of 3, you can conclude that it does intersect the xy-plane—at the point (2, 4, 0). ~ 2 Standard equation 3? Substitute. Figure 11.8 eneckpont Now try Exercise 39. Finding the Center and Radius of a Sphere Find the center and radius of the sphere given by xt yt+ 2 -2x + dy — 62+ 8 = 0. Solution To obtain the standard equation of this sphere, complete the square as follows. wet yt — 2+ dy — 62+ 8=0 )t(24+4y+ )+(@-6z+ )=-8 (x2 = 2x + 1) + (y2 + dy 4) + 2-62 +9 =-B+ 14449 (IP + (y +2) + @ = 3)? = (YO)? So, the center of the sphere is (1, —2, 3), and its radius is V6. See Figure 11.9. eueckponr ~— Now try Exercise 49. Note in Example 5 that the points satisfying the equation of the sphere are “surface points,” not “interior points.” In general, the collection of points satisfying an equation involving x, y, and z is called a surface in space. 816 Chapter11__ Analytic Geometry in Three Dimensions Finding the intersection of a surface with one of the three coordinate planes (or with a plane parallel to one of the three coordinate planes) helps one visualize the surface. Such an intersection is called a trace of the surface. For example, the xy-trace of a surface consists of all points that are common to both the surface and the xy-plane. Similarly, the xz-trace of a surface consists of all points that are common to both the surface and the xz-plane. IScTtwae Finding a Trace of a Surface Sketch the xy-trace of the sphere given by (x — 3° + (y — 2)? + ( +4P = 8. Solution To find the xy-trace of this surface, use the fact that every point in the xy-plane has a z-coordinate of zero. By substituting z = 0 into the original equation, the resulting equation will represent the intersection of the surface with the xy-plane. fan oeate? (x — 3P + (y — 2h + (2+ 4P = 57 ‘Write original equation. z (x — 3)? + (y — 2)? + (0 + 4)? = 82 Substitute 0 for z. (x — 3) + (y — 2)? + 16 = 25 Simplify. (x — 3)? + (y — 2)? =9 Subtract 16 from each side. (x — 3P +(y- 2? = 3 Equation of circle You can see that the xy-trace is a circle of radius 3, as shown in Figure 11.10. weweckront ~— Now try Exercise 57. Sphere: (x—3)? + (y— 2)? + @ +4)? = 5? FIGURE 11.10 c Technology Most three-dimensional graphing utilities and computer algebra systems represent surfaces by sketching several traces of the surface. The traces are usually taken in equally spaced parallel planes. To graph an equation involving x, y, and z with a three-dimensional “function grapher,” you must first set the graphing mode to three-dimensional and solve the equation for z. After entering the equation, you need to specify a rectangular viewing cube (the three-dimensional analog of a viewing window).For instance, to graph the top half of the sphere from Example 6, solve the equation for z to obtain the solutions Z= —4+ \/25 — (x — 3)? — (y — 2). The equation z = —4 + 25 — (x — 3)? — (y — 2)? represents the top half of the sphere. Enter this equation, as shown in Figure 11.11. Next, use the viewing cube shown in Figure 11.12. Finally, you can display the graph, as shown in Figure 11.13. [reviskznor|es) « [ars +sehva: [risfebory [rocicevomlreaceliedrann|eiatnorsule ens rT lvzt=-4 + [25 - (x -3)7-(y- 9 aS (x-3°-(y # D : : age ZMax= 20 Inconbour=5. Figure 11.11 FIGURE 11.12 FiGuRE 11.13 Section 11.1 The Three-Dimensional Coordinate System 817 [11.1 Exercises VOCABULARY CHECK: Fill in the blanks. LAL a-axis and the y-axis at the origin. __ coordinate system can be formed by passing a z-axis perpendicular to both the 2. The three coordinate planes of a three-dimensional coordinate system are the ___, the and the 3. The coordinate planes of a three-dimensional coordinate system separate the coordinate system into eight 4, The distance between the points (x), y, z1) and (x3, y2, z) can be found using the __ _______ in Space. 5, The midpoint of the line segment joining the points (x,y), 21) and (x2, yo Midpoint Formula in Space is +») given by the 6. A________ is the set of all points (x, y, z) such that the distance between (x, y, z) anda fixed point (h, k,,) is r. TA in 8. The intersection of a surface with one of the three coordinate planes is called a _. ___is the collection of points satisfying an equation involving x, y, and z. _ of the surface. In Exercises 1 and 2, approximate the coordinates of the points. In Exercises 3-6, plot each point in the same three-dimen- sional coordinate system. 3. (a) (2,1,3) (b) (-1,2,1) 5. (a) (3, -1,0) (b) (4,2, 2) 4. (a) (3,0, 0) (b) (—3,-2,- I) 6. (a) (0.4, ~3) (b) (4,0, 4) In Exercises 7-10, find the coordinates of the point. 7. The point is located three units behind the yz-plane, three units to the right of the xz-plane, and four units above the ay-plane. 8. The point is located six units in front of the yz-plane, one unit to the left of the xz-plane, and one unit below the ay-plane. 9. The point is located on the x-axis, 10 units in front of the yz-plane. 10. The point is located in the yz-plane, two units to the right of the xz-plane, and eight units above the xy-plane. In Exercises 11-16, determine the octant(s) in which (x, y, z) is located so that the condition(s) is (are) satisfied. Il. x >0,y <0,2>0 1.x <0,y>0,2<0 B.2>0 My <0 15. xy <0 16. yz > 0 In Exercises 17-24, find the distance between the points. 17. (0,0, 0), (5,2,6 18. (1,0, 0), (7, 0,4 19. (3,25), (7.4.8 20. (4, 1, 9), (2,1, 6 21. (-1, 4, -2), (6,0, -9) 22. 23. 24. ) ) ) ) . (1, 1, =7), (—2, -3, -7) - (0, —3, 0), (1, 0, — 10) - (2, —4,0), (0, 6, 3) 820 Chapter11__ Analytic Geometry in Three Dimensions Vectors in Space What you should learn Find the component forms of Vectors in Space the unit vectors in the same Physical forces and velocities are not confined to the plane, so it is natural to direction of, the magnitudes extend the concept of vectors from two-dimensional space to three-dimensional of, the dot products of, and the angles between vectors space. In space, vectors are denoted by ordered triples V = (V4, Va, V3). Component form The zero vector is denoted by 0 = (0, 0, 0). Using the unit vectors i = (1, 0, 0), j = (0, 1,0), and k = (0,0,1) in the direction of the positive z-axis, the standard unit vector notation for v is v=vit vj + v3k Unit vector form as shown in Figure 11.14. If v is represented by the directed line segment from P(P,, Pos Ps) 0 O(4,, Ga, q3), aS Shown in Figure 11.15, the component form of v is produced by subtracting the coordinates of the initial point from the corresponding coordinates of the terminal point V = (My, Va, V3) = (41 — Pas G2 — Po» G3 ~ Ps) P(Py, Py» P3), FIGURE 11.14 Figure 11.15 SSS Vectors in Space 1. Two vectors are equal if and only if their corresponding components are equal. 2. The magnitude (or length) of u = (u,, u5, u;) is [ul] = /u? + uz + uZ. 3. A unit vector u in the direction of v is u = iw vt. v 4. The sum of u = (uw), U5, U3) and V = (Vy, V3, V3) is uty = (uy + vy, Uy + Vy, U3 + V5). ‘Vector addition 5. The scalar multiple of the real number c and u = (u,, U5, U3) is cu = (cuy, CU, CU). Scalar multiplication 6. The dot product of u = (u,, Wu, U3) and V = (v1, V2, v3) is Us Vv = uy, + ivy + U3V3. Dot product Technology Some graphing utilities have the capability to perform vector operations, such as the dot product. Consult the user’s guide for your graphing utility for specific instructions. Section 11.2 Vectors in Space 821 Finding the Component Form of a Vector Find the component form and magnitude of the vector v having initial point (3, 4, 2) and terminal point (3, 6, 4). Then find a unit vector in the direction of v. Solution The component form of v is v= (3 — 3,6 —4,4 — 2) = (0,2, 2) which implies that its magnitude is |Wh= J+ 2422 = 8 = 2/2. The unit vector in the direction of v is v 1 1 1 4 Tp a= (5-4) (0%, 2 ) -eueckpont Now try Exercise 3. IS ctin wag 6Finding the Dot Product of Two Vectors Find the dot product of (0, 3, —2) and (4, —2, 3). Solution (0, 3, —2) + (4, —2,3) = 0(4) + 3(—2) + (—2)(3) =0-6-6=-12 Note that the dot product of two vectors is a real number, not a vector. Meseceponr Now try Exercise 19. As was discussed in Section 6.4, the angle between two nonzero vectors is the angle 6,0 < @< 7, between their respective standard position vectors, as shown in Figure 11.16. This angle can be found using the dot product. (Note that the angle between the zero vector and another vector is not defined.) yu <> y Origin FIGURE 11.16 Angle Between Two Vectors uv If 6 is the angle between two nonzero vectors u and v, then cos 0 = Ten jul lv If the dot product of two nonzero vectors is zero, the angle between the vectors is 90° (recall that cos 90° = 0). Such vectors are called orthogonal. For instance, the standard unit vectors i, j, and k are orthogonal to each other. 822 FIGURE 11.17 FIGURE 11.18 Analytic Geometry in Three Dimensions iS cliteme Finding the Angle Between Two Vectors Find the angle between u = (1,0, 2) and v = (3, 1,0). Solution usy — (1,02) (3,10) 3 ljuliiv 1.0, 2)1/|K3, 1 oy] 50 This implies that the angle between the two vectors is cos 6 = 3 0 = arccos Yo = 64.9° as shown in Figure 11.17. weneckronr Now try Exercise 23. Parallel Vectors Recall from the definition of scalar multiplication that positive scalar multiples of a nonzero vector v have the same direction as v, whereas negative multiples have the direction opposite that of v. In general, two nonzero vectors u and v are parallel if there is some scalar c such thatu = cy. For example, in Figure 11.18, the vectors u, v, and w are parallel because u = 2v and w = —y. Parallel Vectors Vector w has initial point (1, —2, 0) and terminal point (3, 2, 1). Which of the following vectors is parallel to w? a. u = (4,8, 2) b. v= (4,8, 4) Solution Begin by writing w in component form. w = 3 — 1,2 — (—2),1 — 0) = (2,4, 1) a. Because u = (4,8, 2) = 2(2, 4, 1) = 2w, you can conclude that u is parallel to w. b. In this case, you need to find a scalar ¢ such that (4, 8,4) = c(2, 4, 1). However, equating corresponding components produces c = 2 for the first two components and c = 4 for the third. So, the equation has no solution, and the vectors v and w are not parallel. weneckronr Now try Exercise 27. Section 11.2 Vectors in Space 825 [11.2 Exercises | VOCABULARY CHECK: Fill in the blanks. 1. The ______ vector is denoted by 0 = (0, 0, 0). 2. The standard unit vector notation for a vector V is given by____. 3. The_______ ofa vector v is produced by subtracting the coordinates of the initial point from the corresponding coordinates of the terminal point. 4, If the dot product of two nonzero vectors is zero, the angle between the vectors is 90° and the vectors are called 5, Two nonzero vectors wand v are _ _ if there is some scalar c such that u = cv. In Exercises 1 and 2, (a) find the component form of the vector vand (b) sketch the vector with its initial point at the origin. In Exercises 3 and 4, (a) write the component form of the vector v, (b) find the magnitude of v, and (c) find a unit vector in the direction of v. 3. Initial point: (—6, 4, —2) Terminal point: (1, — 1, 3) 4, Initial point: (—7, 3, 5) Terminal point: (0, 0, 2) In Exercises 5 and 6, sketch each scalar multiple of v. 5. v = (1, 1,3) (a) 2v. (b) -v. ©) avd) OV 6. v = (~1,2,2) (@) -v (b)2v. @av Wav In Exercises 7-10, find the vector z, given u = (—1, 3,2), v = (1, -2, -2), and w = (5,0,—5). 1az=u-2w 8. z=7u+v— sw 9. 2 - 4u=w 10.ut+v+z=0 In Exercises 11-16, find the magnitude of v. IL. v = (7,87) 12. v = (-2,0, -5) 13. v = 4i — 3j — 7k 14, v= 21- j + 6k 15. Initial point: (1, 3,4) Terminal point: (1,0, 1) 16. Initial point: (0, — 1,0) Terminal point: (1,2, —2) In Exercises 17 and 18, find a unit vector (a) in the direction of uand (b) in the direction opposite of u. 17. u= 8i + 3j —k 18. u = —3i + 5j + 10k In Exercises 19-22, find the dot product of u and v. 19. u = (4,4, -1) v= (2,-5,-8) 20. u = (3, -1, 6) v= (4-10, 1) 21. u = 2i- 5j + 3k vy=9+3j-k 22. u = 3j - 6k v= 61 - 4j — 2k In Exercises 23-26, find the angle 6 between the vectors. 23. u = (0,2, 2) 24. u = (1,3, 0) v = (3,0, -4) v= (1,2,-1) 25. u = 101 + 40j 26. u = 8j — 20k y= —3j + 8k v= 10i- 5k 826 Chapter 11 In Exercises 27-30, determine whether u and v are orthog- onal, parallel, or neither. 27. u = (~ 12,6, 15) 28. u = (—1,3,-1) v = (8-4, — 10) v= (2,-1,5) 29, u =3i — 3) + 2k 30. u=-i+4j-k v=4it+ 1j+k v= 8i- 4j + 8k In Exercises 31-34, use vectors to determine whether the points are collinear. 31. (5, 4, 1), (7,3, — 1), (4, 5,3) 32. (~2,7,4), (—4, 8, 1), (0,6, 7) 33. (1, 3,2), (= 1,2, 5), (3,4, -1) 34. (0, 4,4), (= 1, 5, 6), (= 2, 6,7) In Exercises 35-38, the vector v and its initial point are given. Find the terminal point. 35. v = (2, 4,7) 36. v= (4,-1,-1) Initial point: (1, 5, 0) Initial point: (6, —4, 3) 37. v = (4,3, -4) 38. v 4) Initial point: (3,2, Initial point: (2, 1, — 39. Determine the values of ¢ such that ||cul] = 3, where u=i+ 2+ 3k 40. Determine the values of c such that ||cul| = 12, where us 2 + — 4k. In Exercises 41 and 42, write the component form of v. 41. vlies in the yc-plane, has magnitude 4, and makes an angle of 45° with the positive y-axis. 42. v lies in the xz-plane, has magnitude 10, and makes an angle of 60° with the positive z-axis. 43. Tension The weight of a crate is 500 newtons. Find the tension in each of the supporting cables shown in the figure. Analytic Geometry in Three Dimensions Model It 44. Tension The lights in an auditorium are 24-pound disks of radius 18 inches. Each disk is supported by three equally 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 the function from part (a) to complete the table. 20 | 25 | 30 | 35 | 40 | 45 | 50 @S (c) Use a graphing utility to graph the function in part (a). What are the asymptotes of the graph? Interpret their meaning in the context of the problem. (d) Determine the minimum length of each cable if a cable can carry a maximum load of 10 pounds. Synthesis True or False? \n Exercises 45 and 46, determine whether the statement is true or false. Justify your answer. 45. If the dot product of two nonzero vectors is zero, then the angle between the vectors is a right angle. 46. If AB and AC are parallel vectors, then points A, B, and C are collinear. 47. What is known about the nonzero vectors u and v if u+y <0? Explain. 48. Writing Consider the two nonzero vectors u and v. Describe the geometric figure generated by the terminal points of the vectors rv, u + tv, and su + tv, where s and t represent real numbers. Skills Review In Exercises 49-52, find a set of parametric equations for the rectangular equation using (a) t = x and (b) t = x + 1. we y=? x 49, y= 3x42 Sl. y=2-8 52. y = 4x3 Section 11.3 The Cross Product of Two Vectors 827 The Cross Product o Vectors What you should learn Find cross products of vectors The Cross Product in space. Many applications in physics, engineering, and geometry involve finding a + Use geometric properties of vector in space that is orthogonal to two given vectors. In this section, you will cross products of vectors in study a product that will yield such a vector. It is called the cross product, and ec it is conveniently defined and calculated using the standard unit vector form. Use triple scalar products to Definition of Cross Product of Two Vectors in Space Let u= mi t+ uj + uk and v=vir+vj+v3k be vectors in space. The cross product of u and v is the vector UV = (ty — aya — (vs — Ws) + ur» — wav DK. It is important to note that this definition applies only to three-dimensional vectors. The cross product is not defined for two-dimensional vectors. A convenient way to calculate u x v is to use the following determinant form with cofactor expansion. (This 3 x 3 determinant form is used simply to help remember the formula for the cross product—it is technically not a deter- minant because the entries of the corresponding matrix are not all real numbers.) ij k uxve|u uu, “2 Putuin Row 2. vy, v> v3, <22) Putv in Row 3. Ws V2 V3 wy wl |i yy Val jr vy V3; = (ugyy = uri = (urs = gy )E Fury — wrk Note the minus sign in front of the j-component. Recall from Section 8.4 that each of the three 2 x 2 determinants can be evaluated by using the following pattern. f \ a,b) — ab, |} Find each cross product. What can you conclude? aixj bixk a jxk 830 Chapter 11 FIGURE 11.23 aaa If you connect the terminal points of two vectors u and v that have the same initial points, a triangle is formed. Is it possible to use the cross product u x v to determine the area of the triangle? Explain. Verify your conclusion using two. vectors from Example 3. Analytic Geometry in Three Dimensions In Example 2, note that you could have used the cross product v x u to form a unit vector that is orthogonal to both u and y. With that choice, you would have obtained the negative of the unit vector found in the example. The fourth geometric property of the cross product states that |[u x v|| is the area of the parallelogram that has u and v as adjacent sides. A simple example of this is given by the unit square with adjacent sides of i and j. Because ixj=k and ||k|| = 1, it follows that the square has an area of 1. This geometric property of the cross product is illustrated further in the next example. iS<tiiNteee Geometric Application of the Cross Product Show that the quadrilateral with vertices at the following points is a parallelo- gram. Then find the area of the parallelogram. Is the parallelogram a rectangle? A(5, 2, 0), B(2, 6, 1), C(2, 4,7), D(5, 0, 6) Solution From Figure 11.23 you can see that the sides of the quadrilateral correspond to the following four vectors. AB = -3i+ 4 +k CD = 3i-4j- k = —AB AD = 0i — 2j + 6k CB = 0i + 2j — 6k = -AD Because CD = -AB and CB = —AD, you can conclude that AB is parallel to CD and AD is parallel to CB. It follows that the quadrilateral is a parallelogram with AB and AD as adjacent sides. Moreover, because oo i j k AB x AD = |—3 4 1} = 261 + 18j + 6k Oo -2 6 the area of the parallelogram is AB x AD|| = /262 + 18? + © = /1036 ~ 32.19. You can tell whether the parallelogram is a rectangle by finding the angle between the vectors AB and AD. a AB x AD || 4B | AD — _v 1036 26/40 6 = arcsin 0.998 0 =~ 86.4° = 0.998 Because @ # 90°, the parallelogram is not a rectangle. weneckronr Now try Exercise 27. vx w I projy ul Area of base = ||v x w|| Volume of parallelepiped = |u « (v x w)| Figure 11.24 Figure 11.25 Section 11.3 The Cross Product of Two Vectors 831 The Triple Scalar Product For the vectors u, v, and w in space, the dot product of u and v x w is called the triple scalar product of u, v, and w. The Triple Scalar Product For u = wi + uj + uk, v = vi + vj + v3k, and w = w,i + w,j + wk, the triple scalar product is given by uy u-r(vxwe=|ry oy vf hy, Wy Wy If the vectors u, v, and w do not lie in the same plane, the triple scalar product u (v x w) can be used to determine the volume of the parallelepiped (a polyhedron, all of whose faces are parallelograms) with u, v, and w as adjacent edges, as shown in Figure 11.24. Geometric Property of Triple Scalar Product The volume V of a parallelepiped with vectors u, v, and w as adjacent edges is given by V=|u- (vx wl. Volume by the Triple Scalar Product Find the volume of the parallelepiped having u=3i-S5j+k, v=2j-2k and w=3i+jtk as adjacent edges, as shown in Figure 11.25. Solution The value of the triple scalar product is 3-5 1 u-(vxw)=|0 2 -2| 3 1 1 _ a 7 _ cal 7 4 if (| 1 1 3 1 3 1 = 3(4) + 5(6) + I(-6) = 36. So, the volume of the parallelepiped is Ju - (v x w)| = [36] = 36. Mesecxeowr ~— Now try Exercise 39. 832 Chapter 11 Analytic Geometry in Three Dimensions [11.3] Exercises VOCABULARY CHECK: Fill in the blanks. 1, To find a vector in space that is orthogonal to two given vectors, find the 2uxu= 3, |lux vl] = 4, The dot product of u and v x wis called the ___ of the two vectors. _ of u,v, and w. In Exercises 1-4, find the cross product of the unit vectors and sketch the result. 1 jxi 3 ixk kx j 4k xi In Exercises 5-14, find u x v and show thatit is orthogonal to both uand v. 5. u = (3, -2, 5) 6. u = (6,8, 3) v= (0,-1.1) v= (4,-1,-4) 7. u = (10,0, 6) 8. u = (-5,5, 11) v = (7,0,0) 9% u=6i+ 2)+k v=i+3j-2k Il. u= 6k 13. In Exercises 15-20, find a unit vector orthogonal to u and v. 15. u=3i+j 16. u=i + 2j v v=i-3k 17. u= ~3i + 2j — 5k 18. u=7i- 14j + 5k v=hi-ij+pk v= 14i + 28j — 15k 20. u=i— 2j + 2k vy=2-j-2k In Exercises 21-26, find the area of the parallelogram that has the vectors as adjacent sides. 21u=k 22.u + 2j + 2k v=it+k v=it+k 23. u=3i + 4j + 6k 24. w= —21 + 35 + 2k vy=2-j+5k v=i+t 2j+ 4k 25. u = (2,2, -3) 26. u = (4, —3, 2) v = (0,2,3) v = (5,0, 1) In Exercises 27 and 28, (a) verify that the points are the vertices of a parallelogram, (b) find its area, and (c) deter- mine whether the parallelogram is a rectangle. 27. A(2, -1,4), B(3, 1,2), C(0,5, 6), D(-1, 3, 8) 28. A(I, 1, 1), B(2, 3,4), C(6, 5,2), D(7,7.5) In Exercises 29-32, find the area of the triangle with the given vertices. (The area A of the triangle having uand vas adjacent sides is given by A = }||u x vj.) 29. (0,0,0), (1, 2,3), (—3, 0,0) 30. (1, —4, 3), (2, 0, 2), (=2, 2,0) 31. (2,3, —5), (2, -2, 0), (3, 0, 6) 32. (2,4,0), (—2, 4, 0), (0,0, 4) In Exercises 33-36, find the triple scalar product. 33. u = (2,3,3),v = (4, 4,0), w = (0, 0,4) 34. u = (2,0, 1), v = (0, 3,0), w = (0, 0, 1) 35. u= 2+ 3j+kv=i-jw=4i+3j+k 36. u=i + 4j — 7k, v = 21+ 4k, w = —3j + 6k In Exercises 37-40, use the triple scalar product to find the volume of the parallelepiped having adjacent edges u, v, andw. 3u=itj 38. u=i+ j+3k v=jt+k v= 3) +3k w=itk w= 3i+3k Section 11.4 Lines and Planes in Space 835 Finding Parametric and Symmetric Equations Find parametric and symmetric equations of the line L that passes through the point (1, —2, 4) and is parallel to v = (2, 4, —4). Solution To find a set of parametric equations of the line, use the coordinates x, = 1, y, = —2, and z, = 4 and direction numbers a = 2, b = 4, and c = —4 (see Figure 11.27). x=1+2t, y=-24+41, z=4-4t Parametric equations Because a, b, and ¢ are all nonzero, a set of symmetric equations is yt2 s yy mmetric equations 7 7 y eq FIGURE 11.27 weneceront ~— Now try Exercise 1. Neither the parametric equations nor the symmetric equations of a given line are unique. For instance, in Example 1, by letting t= 1 in the parametric equations you would obtain the point (3, 2, 0). Using this point with the direction numbers a = 2, b = 4, and c = —4 produces the parametric equations x=3+ 21, y=2+4t, and z= —41. sclera Parametric and Symmetric Equations of a Line STUDY TIP oy Through Two Points To check the answer to Example . . . . . 2, verify that the two original Find a set of parametric and symmetric equations of the line that passes through points lie on the line. To see the points (—2, 1, 0) and (1, 3, 5). this, substitute ¢ = 0 and t = Solution inithe|perametric equations as Begin by letting P = (—2, 1,0) and Q = (1, 3,5). Then a direction vector for follows. the line passing through P and Q is ok v=P0 * ea =(1 2),3-1,5-0, rene == (-2),3- 15-0) z= 5(0) =0 = (3,2, 5) = (a,b, c). t=1: =-24+3(I)=1 Using the direction numbers a = 3, b = 2, and c = 5 with the initial point * P(—2, 1,0), you can obtain the parametric equations aa a 243, 1+ 2 id 5 =-2+ =1+ =5t. 7 ations z= 4(1)=5 x 1 y t, and z t. Parametric equations Because a, b, and ¢ are all nonzero, a set of symmetric equations is x+2_ 3 Mevecxeont — Now try Exercise 7. Symmetric equations 836 Analytic Geometry in Three Dimensions Planes in Space You have seen how an equation of a line in space can be obtained from a point on the line and a vector 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 vector normal (perpendicular) to the plane. Figure 11.28 Consider the plane containing the point P(x,,y,,z,) having a nonzero normal vector n = (a, b,c), as shown in Figure 11.28. This plane consists of all points Q(x, y, z) for which the vector PO is orthogonal to n. Using the dot product, you can write the following. n-PO=0 P@ is orthogonal to n. (a,b, e) (x = X.Y — Yye— 4) =O a(x — x) + Dy — y,) + ez - 2) =0 The third equation of the plane is said to be in standard form. Standard Equation of a Plane in Space The plane containing the point (x,,y,, z,) and having normal vector n = (a, b, c) can be represented by the standard form of the equation of a plane a(x — x,) + b(y — y,) + e(z — z,) = 0. Regrouping terms yields the general form of the equation of a plane in space ax + by +ez+d=0. General form of equation of plane Given the general form of the equation of a plane, it is easy to find a normal vector to the plane. Use the coefficients of x, y, and z to write n = (a, b,c). Consider the following four planes. wk+3y- z= 2 4x + 6y-22= 5 -2x-3y+ z=-2 6x — 9y + 32 = 11 What are the normal vectors for each plane? What can you say about the relative positions of these planes in space? (-2, 1,4) Figure 11.29 Figure 11.30 Section 11.4 Lines and Planes in Space 837 Sct wew © Finding an Equation of a Plane in Three-Space Find the general form of the equation of the plane passing through the points (2, 1, 1), (0, 4, 1), and (—2, 1, 4). Solution To find the equation of the plane, you need a point in the plane and a vector that is normal to the plane. There are three choices for the point, but no normal vector is given. To obtain a normal vector, use the cross product of vectors u and v extending from the point (2, 1, 1) to the points (0, 4,1) and (—2, 1,4), as shown in Figure 11.29. The component forms of u and v are u=(0-2,4-1,1- 1) = (-2,3,0) v= (-2-2,1-1,4- 1) = (-4,0,3) and it follows that ij « n=uxv=|-2. 3. 0 -4 0 3 = 91 + 6j + 12k = (a,b,c) is normal to the given plane. Using the direction numbers for n and the initial point (x,, y;, z,) = (2, 1, 1), you can determine an equation of the plane to be alx — x) + (yy — y) tee —z) = 0 9(x — 2) + O(y — 1) + 122 - 1) =0 Standard form 9x + 6y + 122 — 36 =0 3x + 2y + 4z— 12 = 0. General form Check that each of the three points satisfies the equation 3x + 2y + 4z — 12 = 0. eneceronr ~— Now try Exercise 25. Two distinct planes in three-space either are parallel or intersect in a line. If they intersect, you can determine the angle 6(0 < @ < 90°) between them from the angle between their normal vectors, as shown in Figure 11.30. Specifically, if vectors n, and n, are normal to two intersecting planes, the angle 6 between the normal vectors is equal to the angle between the two planes and is given by cos 0 = Angle between two planes. Consequently, two planes with normal vectors n, and n, are 1. perpendicular if n, + n, = 0. 2. parallel if n, is a scalar multiple of n,. 840 Chapter 11 D = ||proj,PQ\) FIGURE 11.35 Analytic Geometry in Three Dimensions Distance Between a Point and a Plane 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 11.35. If P is any point in the plane, you can find this distance by projecting the vector PO onto the normal vector n. The length of this projection is the desired distance. Distance Between a Point and a Plane The distance between a plane and a point Q (not in the plane) is D = ||proj,PO|| _ [PQ - nl where P is a point in the plane and n is normal to the plane. To find a point in the plane given by ax + by + cz + d = 0, where a # 0, let y = Oand z = 0. Then, from the equation ax + d = 0, you can conclude that the point (—d/a, 0, 0) lies in the plane. Finding the Distance Between a Point and a Plane Find the distance between the point O(1, 5, —4) and the plane 3x — y + 2z = 6. Solution You know that n = (3, — 1, 2) is normal to the given plane. To find a point in the plane, let y = 0 and z = 0, and obtain the point P(2, 0, 0). The vector from P to Qis PO = (1 -2,5-0,-4-0) = (-1,5, —4). The formula for the distance between a point and a plane produces PO - p - Poa (In| — M15, =4) 3,1, 2)) Meweceponr Now try Exercise 45. The choice of the point P in Example 5 is arbitrary. Try choosing a different point to verify that you obtain the same distance. Section 11.4 Lines and Planes in Space 841 [11.4 Exercises | VOCABULARY CHECK: Fill in the blanks. 1. The _____ vector for a line L is given byv = ____. 2. The ofa line in space are given by x = x, + at, y = y, + bt, andz =z, + ct. 3. If the direction numbers a, b, and c of the vector v = (a, b,c) are all nonzero, you can eliminate the . A vector that is perpendicular to a plane is called __ parameter to obtain the of a line. . The standard form of the equation of a plane is given by____. Exercises 1-6, find a set of (a) parametric equations and (b) symmetric equations for the line through the point and pa Auk wre rallel to the specified vector or line. (For each line, write the direction numbers as integers.) Point Parallel to . (0,0, 0) v = (1,2,3) . (3,-5.1) v = (3,-7,—10) . (—4, 1,0) v=sit4j—-k . (5,0, 10) v=4i + 3k . (2,-3.5) =54+2%y=7-342=-24+1 . (1,0, 1) 34+ 34y=5-2n2=-7+t (b) if possible, a set of symmetri pa: Exercises 7-14, find (a) a set of parametric equations and equations of the line that sses through the given points. (For each line, write the direction numbers as integers.) 7, (2,0, 2), (1, 4, -3) 9. (—3, 8, 15), (1, -2, 16) 8. (2,3, 0), (10, 8, 12) 10. (2,3, —1), (1, — 5,3) 11. 31,2), (-1.1,5) 12. (2,-1.5), (2,1, -3) 13. (-4, 2,4), (1,-4.0) 14. (-3,3,2), 3, -5, -4) In Exercises 15 and 16, sketch a graph of the line. 15.x=2,y=244 c=l+h Exercises 17-22, find the general form of the equation of the plane passing through the point and perpendicular to the specified vector or line. Point Perpendicular to 17. (2, 1,2) n=i 18. (1,0, —3) n=k 19. (5,6, 3) n=~—2i+ j-2k Point Perpendicular to 20. (0.0, 0) n= —3j + 5k 21. (2,0, 0) x=3-ny=2-2%c=4+4+8 22. (0, 0, 6) l-ty=24+42 -2 In Exercises 23-26, find the general form of the equation of the plane passing through the three points. 23. (0, 0, 0), (1, 2,3), (=2, 3, 3) A. (4, = 1,3), (2,5, 1), (- 1,2, 1) 25. (2, 3, -2), (3,4, 2), (1, - 1, 0) 26. (5,-1.4), (1, —1, 2), (2, 1, -3) In Exercises 27 and 28, find the general form of the equation of the plane with the given characteristics. 27. The plane passes through the point (2, 5, 3) and is parallel to the xz-plane. 28. The plane passes through the points (2,2,1) and (-1,1,-1) and is perpendicular to the plane 2x — 3y +2 =3. In Exercises 29-32, determine whether the planes are parallel, orthogonal, or neither. If they are neither parallel nor orthogonal, find the angle of intersection. 30. 4x ty + 82 = 10 -3 5x — Wy — 52 = In Exercises 33-36, (a) find the angle between the two planes and (b) find parametric equations of their line of intersection. —3yt z + 52435 33. 3x — dy + 5c = z=2 34, 2x -2 xt yn 0 842 Chapter 11 36. Wetdy-2= 1 —3x — 6y + 3 = 10 In Exercises 37-42, plot the intercepts and sketch a graph of the plane. 31. xt 2y + 32-6 39. xt 2y=4 AL. 3x + 2y — 38. 2x —y + 4: 40. y + c= 42. x — 3¢ In Exercises 43-46, find the distance between the point and the plane. 43. (0, 0,0) 44. (3,2, 1) &r — dy $2 =8 x-yt22=4 45. (4, -2, -2) 46. (—1, 2,5) woytc=4 Qx+3y t= 12 Model It 47. Data Analysis: Milk Consumption The table shows the per capita milk consumptions (in gallons) of different types of plain milk in the United States from 1999 to 2003. Consumption of light and skim milks, reduced-fat milk, and whole milk are represented by the variables x, y, and z, respectively. (Source: U.S. Department of Agriculture) A model for the data is given by —0.81x — 0.36y + z = 0.2. (a) Complete a fifth column in the table using the model to approximate < for the given values of x and y. (b) Compare the approximations from part (a) with the actual values of z. (c) According to this model, any increases or decreases in consumption of two types of milk will have what effect on the consumption of the third type of milk? KY Analytic Geometry in Three Dimensions 48. Mechanical Design A chute at the top of a grain elevator of a combine funnels the grain into a bin as shown in the figure, Find the angle between two adjacent sides. T(-1,-1.8) 9; S@,0,0 P6,0,0)g y (6, Oe 7 * 2(6,6, 0) Synthesis True or False? _\n Exercises 49 and 50, determine whether the statement is true or false. Justify your answer. 49, Every two lines in space are either intersecting or parallel. 50. Two nonparallel planes in space will always intersect. 51. The direction numbers of two distinct lines in space are 10, —18, 20, and —15, 27, —30. What is the relationship between the lines? Explain. 52. Exploration (a) Describe and find an equation for the surface generated by all points (x, y, z) that are two units from the point (4,-1.1). (b) Describe and find an equation for the surface generated by all points (x, y, z) that are two units from the plane 4x — 3y +z = 10. Skills Review In Exercises 53-56, convert the polar equation to rectangular form. 53. r= 10 54. 0= —- 55. r= 3.cos 0 56.7 =5— In Exercises 57-60, convert the rectangular equation to polar form. 57. + In Exercises 33 and 34, use vectors to determine whether the points are collinear. 33. (5,2, 0), (2,6, 1), (24,7) 34. (6,3, —1), (5, 8, 3), (7. —5) 35. Tension A load of 300 pounds is supported by three cables, as shown in the figure. Find the tension in each of the supporting cables, (-4, -6, 10) B (0, 10, 10) 36. Tension Determine the tension in each of the supporting cables in Exercise 35 if the load is 200 pounds. In Exercises 37 and 38, find u x v. 37. u = (2, 8,2) v=(1,1,-1) v 38. u = (10, 15,5) (5, -3, 0) In Exercises 39 and 40, find a unit vector orthogonal to u and v. 39. u= —3i + 2j — 5k v= 101 — 15j + 2k 40. u = 4k + 12k v In Exercises 41 and 42, verify that the points are the vertices of a parallelogram and find its area. 41. (2,-1.1), (5, 1, 4), 0, 1, 1), (3.3.4) 42. (0,4, 0), (1,4, 1), (0, 6,0), (1, 6, 1) In Exercises 43 and 44, find the volume of the parallelepiped with the given vertices. 43. A(0, 0, 0), BG, 0,0), C0, 5, 1), D3, 5, Ds E(2,0, 5), F(5, 0, 5), G(2, 5, 6), H(5, 5,6 44. A(0, 0,0), B(2, 0,0), C2, 4, 0), DOO, 4, 0 E(0, 0, 6), F(2, 0, 6), G(2, 4, 6), H(0, 4,6 ) ) ). ) In Exercises 45-48, find a set of (a) parametric equations and (b) symmetric equations for the specified line. 45. The line passes through the points (—1,3,5) and (G.6, -1). Review Exercises 845 46. The line passes through the points (0,—10,3) and (5, 10,0). 47. The line passes through the point (0,0,0) and is parallel to the vector y = (~2,$, 1). 48. The line passes through the point (3,2, 1) and is parallel to the line given by x In Exercises 49-52, find the general form of the equation of the specified plane. 49. The plane passes through the points (0, 0, 0), (5, 0, 2), and (2, 3, 8). 50. The plane passes through the points (~ 1, 3, 4), (4, -2, 2), and (2, 8, 6). 51. The plane passes through the point (5, 3, 2) and is parallel to the xy-plane. 52. The plane passes through the point (0, 0, 6) and is perpen- dicular to the line given by x= 1-1, y=2+1, and 2=4-21 In Exercises 53-56, plot the intercepts and sketch a graph of the plane. 53. 3x - 2y + 3¢= 54. Sx —y — 5c 55. 2x — 32 = 56. dy — 3¢ = 12 In Exercises 57-60, find the distance between the point and the plane. 57, (2, 3, 10) 58. (1, 2,3) 2x — 2y + 6 = dxe-yre= 59. (0,0, 0) 60. (0,0, 0) x l0y+3¢=2 Qe + 3y to = 12 Synthesis True or False? In Exercises 61 and 62, determine whether the statement is true or false. Justify your answer. 61. The cross product is commutative. 62. The triple scalar product of three vectors in space is a scalar. In Exercises 63-66, let u = (3, —2, 1), v= (2, —4,—-3), and w =(-1,2,2). 63. Show that u - u = |julf. 64. Show that u x v= —(v x u). 65. Show thatu:(v + w) =u-v+u-w. 66. Show that u x (v + w) = (ux v) + (u x w). 846 Chapter11__ Analytic Geometry in Three Dimensions FIGURE FOR 13, 12447 (2,2, 12) . R (8, 8, 12) S.(0, 0,0) vw P.(10,0,0) @°(10, 10,0) FIGURE FOR 17 Take this testas you would take a testin class. After you are finished, check your work against the answers given in the back of the book. 1. Plot each point in the same three-dimensional coordinate system. (a) (5,-2.3) (b) (~2, —2,3) In Exercises 2-4, use the points A(8, —2, 5), B(6, 4, — 1), and C(—4, 3, 0), to solve the problem. 2. Consider the triangle with vertices A, B, and C. Is it a right triangle? Explain, 3. Find the coordinates of the midpoint of the line segment joining points A and B. 4, Find the standard form of the equation of the sphere for which A and B are the endpoints of a diameter. Sketch the sphere and its xz-trace. In Exercises 5-8, let u and v be the vectors from A(8, —2, 5) to B(6, 4, —1) and A to C(—4, 3, 0), respectively. 5, Write u and v in component form. 6. Find (a) ||v 7. Find the angle between u and v. |, (b) u + v, and (c) u x v. 8. Find a set of (a) parametric equations and (b) symmetric equations for the line through points A and B. In Exercises 9 and 10, determine whether u and vare orthogonal, parallel, or neither. %u=i-%-k v=j+6k 10.u=2-3j+k v -j-k 11. Verify that the points (2, —3, 1), (6, 5, ~ 1), (3, —6, 4), and (7, 2, 2) are the vertices of a parallelogram, and find its area. 12. Find the general form of the equation of the plane passing through the points (-3, —4, 2), (=3,4, 1), and (1, 1, -2). 13. Find the volume of the parallelepiped at the left with the given vertices. A(0,0, 5), BQO, 10, 5), C(4, 10, 5), D(4, 0, 5), E(0, 1, 0), FO, 11, 0), G(4, 11, 0), H(4, 1, 0) In Exercises 14 and 15, plot the intercepts and sketch a graph of the plane. 14, 2x + 3y + de = 12 15. 5x —y — 2z= 10 16. Find the distance between the point (2, — 1, 6) and the plane 3x — 2y + z = 6. 17. A tractor fuel tank has the shape and dimensions shown in the figure. In fabricating the tank, it is necessary to know the angle between two adjacent sides. Find this angle. Proofs in Mathematics Notation for Dot Algebraic Properties of the Cross Product (p. 828) and Cross Products Let u, v, and w be vectors in space and let c be a scalar. The notation for the dot product leuxv=-(v xu) 2. ux (v + w) = (ux v) + (ux w) tint introdeoed by the Amorioan 3. clu x v) = (cu) x v=ux(cv) 4 ux0=0xu=0 physicist Josiah Willard Gibbs 5S uxu=0 6.u-(vxw) =(uxv)-w (1839-1903). In the early 1880s, Gibbs built a system to represent physical quantities called vector Proof analysis. The system was adepat- Let y= wi + unj + ugk, v = vi + v9j + vyk, Ww = wi + Wj + Wyk, ture from William Hamilton’s 9 — 0j + 0j + Ok, and let c be a scalar. theory of quatemnions. 1. ux v= (uyy3 = u5¥3)i = (uyvy — Uv,)j + (uyv, — uv Jk al Ce yeh Oeste hen Oreo So, this implies u x v = —(v x u). . ux (v + w) = [u,(v; + w,) — u4(v, + w,)]i — [uey(v3 + w,) — uv, + wy)]j + [uy(v2 + w.) — wo(v, + w,)]K = (ugvs — uav2)i — (uyv3 — wavy) + (uyv2 — wav) + (usw3 — ugwa)i — (uyws — ugW7,)j + (up — tyw))k = (u x v) + (u x w) nN w - (cu) x V = (cuav3 — cUgv>)i — (cuyv3 — cu3v,)§ + (cuyv2 — cunv,)k = luv, — uyV,)i — (uv; — uj + (uv. — uv] = clu x v) S . ux 0 = (u,* 0 — uz + O)i— (uO — uz + O)j + (U, > 0 — uy - Ok =0i + 0j + 0k=0 0xu=(0-4-0-+u)i-O-u,-0-u)j+ 0-H -0-u)k =0i + 0j + 0k=0 So, this implies u x 0 = 0 x u=0. 5. ux u = (uu; — ustt)i — (Uys — Uguy)j + (Uw — uam)k = 0 My hy 6.us(vxw)=|y, v, v3] and Nat Wo) Wa a Wo) Wa] we(uxv)=|u uw uy; (ux v)-w Vy Vn Vy u+ (v x w) = u4(v,W3 — w3V;) — u,(v;w3 — W1V5) + uy (Vp, — W,V>) = UyV2Ws — UyW2V5 — UgVpWs + UaWyVs + WV! — WW UW V3 — UW V2 — UyWovs + UGV{We + UyV>Ws — WV,Ws Wy (ugvy — Vatts) — welt — vu) + wylyv, — vite) =(uxv)-w 847