Partial preview of the text
Download Three-Dimensional Space; Vectors and more Exams Vector Analysis in PDF only on Docsity!
Three-Dimensional Space; Vectors Exercise Set 11.1 1. (a) (0,0, 0), (3,0, 0), (3,5, 0), (0,5, 0), (0,0, 4), (3,0, 4), (3,5,4), (0, 5,4). (b) (0,1,0), (4, 1,0), (4, 6, 0), (0, 6,0), (0, 1, -2), (4, 1, -2), (4, 6, -2), (0, 6, 2). 2. Comers: (2,2, +2), (2, -2, +2), (-2,2, £2), (-2, -2, +2) 3. Comers: (4, 2,—2), (4,2,1), (4,1,1), (4,1, -2), (-6, 1,1), (—6, 2,1), (—6,2, -2), (—6, 1, -2). (-6, 2,1) (-6, 1, -2) (4, 1,-2) 4. (a) (2,41, 21), (2, yo, 21), (@1, 42, 21)(@1, YL, 22); (@2, 41, 22), (@1, Yo, 22)- (b) The midpoint of the diagonal has coordinates which are the coordinates of the midpoints of the edges. The 1 midpoint of the edge (#1, 41,21) and (#2, y1, 21) is (Fe + a2) most): the midpoint of the edge (2,1, 21) and 1 1 (a2, yo, 21) is (x2, 5n+m)a1)s the midpoint of the edge (ar2, y2,21) and (212, yp, 22) is (ae. 5(e1 +2). a+) 5. (a) Assingle point on that line. (b) Aline in that plane. — (c) A plane in 3—space. 1 Thus the coordinates of the midpoint of the diagonal are ( = (a; +22), pica +4. 6. (a) R(1,4,0) and Q lie on the same vertical line, and so does the side of the triangle which connects them. r 519 520 Chapter 11 10. 11. R(1,4,0) and P lie in the plane z = 0. Clearly the two sides are perpendicular, and the sum of the squares of the s is |RQ|? + |RP|? = 4? + (2? + 3%) = 29, so the distance from P to Q is V29. o.4, 4) RU,4, 0) x (b) Clearly, SP is parallel to the y-axis. $(3,4,0) and @ lie in the plane y = 4, and so does SQ. Hence the two sides |SP| and |SQ| are perpendicular, and |PQ| = /|PS[/? + |QSE = /3?+ 22 + £) = V20. Q(1,4,4) PGB.1,0) SG, 4,0) x (c) T(1,1,4) and Q lie on a line through (1,0,4) and is thus parallel to the y-axis, and TQ lies on this line. T and P lie in the same plane y = 1 which is perpendicular to any line which is parallel to the y-axis, thus TP, which lies on such a line, is perpendicular to TQ. Thus |PQ/? = |PT|? + |QT|? = (4+ 16) + 9 = 29. TA, 1,4) Q(1,4,4) x . (a) Let the base of the box have sides a and b and diagonal d;. Then a? +b? = d?, and d, is the base of a rectangular of height ¢ and diagonal d, with @ = d?+ 2? =a? +B? +e. (b) Two unequal points (x1, y1, 21) and («2,9 sides a1 — @,y; — Yo, 21 — 22, and by Part (a) the ) form diagonally opposite corners of a rectangular box with diagonal has length \/(a, — v2)? + (v1 — ye)? + (21 — 22)”. 1 . (a) The vertical plane that passes through (5:0 0) and is perpendicular to the a-axis. 1 1 (b) Equidistant: (« — 3) tyYt2sety+2, or -2vt+1=0ore= > . The diameter is d = \/(1— 3)? + (-2— 4)? + 4+ 12)? = 2986, so the radius is /296/2 = V74. The midpoint (2,1, —4) of the endpoints of the diameter is the center of the sphere. Each side has length V/14 so the triangle is equilateral. (a) The sides have lengths 7, 14, and 75; it is a right triangle because the sides satisfy the Pythagorean theorem, (7V5)? = 7 + 14?. 522 Chapter 11 27. (w — 3/2)? + (y+ 2)? + (2 — 4)? = —11/4; no graph. 28. (% —1)? + (y—3)? + (2 —4)? = 25; sphere, O(1, 3,4), r= 5. 32. (a) Ly (by <4 (©) : 33. (a) —Yy+2=0. (b) -27+2=0. (c) (w-1)? + (y-1)? =1 (a) (w-1)? + (2-1)? =1 34. (a) (x — a)? + (2 —a)? =0?. (b) (w—a)? + (y—a)?=a?. (c) (y—a)? + (2-4)? =a?. Exercise Set 11.1 523 47. Complete the squares to get (a + 1)? +(y — 1)? + (z —2)? = 9; center (—1,1,2 radius 3. The distance between is 3+ V6, the smallest the origin and the center is V6 < 3 so the origin is inside th is 3— 6. sphere. The largest distan 48. (x — 1)? + y? +(z +4)? < 25; all points on and inside the sphere of radius 5 with center at (1, 0,—4). 49. (y +3)? + (2 — 2)? > 16; all points outside the circular cylinder (y + 3)? + (2 — 2)? = 16. Exercise Set 11.2 525 -i+2j+3k (1, -2, 2), } MQ } y 3. (a-b) a 22-1 (ed) / 2i+ 3) -k (-1,3, 2) \* 2) | 4. (ab) A 5. (a) (4-1,1-5) = (3, -4) (b) (0-2, 0-3, 4—0) = (-2, -3, 4) —2i-3j+4k * \ 6. (a) (-3—2, 3-3) =(-5, 0) 526 Chapter 11 10. 11. 12. 13. 14. 15. 16. 17. (b) (0-3, 4-0, 4—4) = (-3, 4, 0) x . (a) (2-3,8—5)=(-1,3) (b) (0-7,0—(-2)) = (-7,2) (c)_ (-3,6,1) » (a) (-4-(-6), . (a) Let (a, y) be the terminal point, then « — 1 = 3, « = 4 and y— (—2) = -2, y= —4. The terminal point is —(-2))=@21) (b) (-L,6,1) (©) (5,0,0) (4,4). (b) Let («,y, 2) be the initial point, then 5 — x = —3, -y = 1, and -1 —y =2 so @ = 8, y =—1, and x = —3. The initial point is (8, —1, —3). (a) Let (w,y) be the terminal point, then « — 2 = 7, «= 9 and y— (1) =6, y= 5. The terminal point is (9,5). (b) Let («,y,2) be the terminal point, then # + 2= 1, y—1=2, and 2-4 =~-3s0 @=~-1, y =3, and z=1. The terminal point is (—1,3, 1). (a) —i+4j-2k (b) 181+12j-6k (ce) -i-5j—2k (d) 40i-4j—-4k (e) —2i-16j-18k (f) —i+13j—2k (a) (1,-2,0) (b) (28,0, -14) + (3,3, 9) = (31, 3, -5) (c) (3,-1, -5) (d) 3((2,-1,3) — (28,0, -14)) = 3(-26, -1,17) = (-78,-3,51) (e) (-12,0,6) — (8,8, 24) = (-20, -8, -18) (f) (8,0, —4) — (3,0,6) = (5, 0,-10) (a) |lv|=VvI+1= v2 (b) |v) =Vv1+49=5v2 (ce) |v) =Vv21 (a) |iv|= Vid (a) lv] =Vv9FIG=5 (b) |IvJ=V247=3 (C) lvl =3 (A) |v =v3 (a) |ju+vil = [21 25+ 2k\| =2V8 (b) lull + |ivl|= V14+ V2 (©) || — 2ul| +2 |v] = 2V14 + 2V2 (d) |[3u —5v + wl) = || —12j + 2k|| =2V37 (e) (1/Vv6)i + (1/v6)j - (2/Vo)k (f) 1 Yes, it is possible. Consider u = i and v =j. False; only if one vector is a positive scalar multiple of the other. If one vector is a positive multiple of the other, say u= av with a > 0, then u,v and u+v are parallel and ||u + v|| = (1 + @)||v|| = |lul] + |Iv|- 528 Chapter 11 30. (a) v = (—10,2) by inspection, sou —v +w =u+vt+w— 2v= (—2,5) + (20, —4) = (18,1). 31 y (81) (b) v y {9,=24) . 6x = 2u—v-—w = (-4,6),x = (—2/3,1). 32. u—2x = x-— w+ 3v, 3x =u w- 3v, x= =(u+ w- 38v) = (2/3, 2/3). 5.2, 1 8 1. 4 33. u= Zit Fit 7k v= zi 7 zk 34, 3ut+2v— ut v)=u=(-5,8),v=utv—u=(7,-11). 35. 36. 37. 38. » G+5) + G = 2))|| = [128 — i] = V5. [1G+ 5) — G— 25)|] = [|] = 3. 3,8) by inspection, so u— v + w =u+v-+ w- 2v = (3, —8) + (6, —16) = (9, -24). a . Let A, B, C be the vertices (0,0), (1,3), (2,4) and D the fourth vertex («, y). For the parallelogram ABCD, AD = Let BC, (v,y) = (1,1) 90 @= 1 y= aes and D is at (1,1). For the parallelogram ACBD, AD=CB, (x,y) = (-1,—1) sow =—1, y = Land D is at (-1,-1). Por the parallelogram ABDC, AC=BD, (w — 1,y — 3) = (2,4), so x =3,y=7and Dis at (3,7). . (a) 5 =||kv|| = [Al||vl] = 43h, so & = £5/3. (b) 6 = |||] = |All[v|| = 2IIvI], so [Iv] = 3. . If ||kv|] = 0 then {&||Iv|] = 0 so either & = 0 or |'v|| = 0; in the latter case, by (9) or (10), v =0 Exercise Set 11.2 529 39. (a) Choose two points on the line, for example P,(0,2) and P2(1,5); then PjP)= (1,3) is parallel to the line, {|(1,3)|] = V0, so (1/V10, 3/10) and (—1/\/10, —3//10) are unit vectors parallel to the line. (b) Choose two points on the line, for example P;(0, 4) and P,(1,3); then P,P) = (1,—1) is parallel to the line, (1, -1)] = V2 so (1/V2, -1/V2) and (-1/V2, 1/V2) are unit vectors parallel to the line. (c) Pick any line that is perpendicular to the line y = —5a + 1, for example y = a/5; then P:(0,0) and P(5, 1) are on the line, so Pi P= (5, 1) is perpendicular to the line, so +4 (h:1) are unit veotors perpendicular to the 26 line. 40. (a) +k (b) 4j (c) +i 41. (a) The circle of radius 1 about the origin. (b) The closed disk of radius 1 about the origin. (c) All points outside the closed disk of radius 1 about the origin. 42. (a) The circle of radius 1 about the tip of ro. (b) The closed disk of radius 1 about the tip of ro (c) All points outside the closed disk of radius 1 about the tip of ro. 43. (a) The (hollow) sphere of radius 1 about the origin. (b) The closed ball of radius 1 about the origin. (c) All points outside the closed ball of radius 1 about the origin. 44. The sum of the distan all points on the elli between (a, y) and the points (#1, y1), (a2, y2) is the constant k, so the set consists of with foci at (@1,y1) and (2, yo), and major axis of length k. 45. Since ¢ = m/2, from (14) we get |Fi + F2|)? = (Fil)? + Fall? = 3600 + 900, so ||F) + F2|| = 30V5 Ib, and F. sina = Fall ing = 26.57°,0 = a & 26.57°. Fi + Fol] 3075" 1 46. ||Fi+Fo|? = |[Fil|? + [P2||?+2|/ Fs ||[P2l| eos @ = 14,400+ 10,000 +2(120)(100); = 36,400, so ||P1 +Fo|| = 20/91 F 5V3 N, sina = a nal sine = 5 gp in 0" = a. © 27.00°, 0 = a 27.00". 47. Fi + Fol? = |Full? + [E22 + 2|[F1||[F2|| cos ¢ = 160,000 + 160,000 — 2(400)(400), so ||F + Fol] © 207.06 N, 2 Fall ox 10 (3) , a = 75.00°, 6 = a — 30° = 45.00°. and sina = 71 _ ging» ——_ Fi +Fo|| 207.06 \2 48. ||F, + Po||? = Fil? + |[Fol|? + 2||F1|||\F2l|cos o = 16 + 4 + 2(4)(2) cos77°, so ||F1 + Fo|| ~ 4.86 Ib, and sina = [Fol gg — 2 sin 77° a #93,.64°. 0 =a 27° & —3.36° TPF Fo 189 = gg it 77s © 28.64, 8 = 0 — 27 3.30". 49. Let Fi,F2,F3 be the forces in the diagram with magnitudes 40,50,75 respectively. Then F) + F) + F3 = (Fi + F2) + F3. Following the examples, F; + Fz has magnitude 45.83 N and makes an angle 79.11° with the positive x-axis. Then ||(F1 + F2) + F3\l? + 45.83? + 75? + 2(45.83)(75) cos79.11°, so F; + F2 + Fs has magnitude & 94.995 N and makes an angle 0 = a = 28.28° with the positive x-axis. Exercise Set 11.3 531 59. 61. 62. 63. 64. (b) Equate corresponding components to get the system of equations c, + 3¢2 + 4c3 = 2, —c1 — cz = 1, and c2 +¢3 = —1. From the equation to get —4 = 2, which is false so the system has no solution. Plac cond and third equations, c, = —1 — cg and cp = —1 — cg; substitute these into the first u and v tip to tail so that u+v is the vector from the initial point of u to the terminal point of v. The shortest distance between two points is along the line joining these points so ||u+ v|| < ||ul + ||v||- a): ut v= (wi turj) + (mit vj) = (mit vej)+ (mituej)=v+u. :ut+0 = (witu2j) + 014+ 0j = wit wjsu © (lu) = k(U(upi + uj) = h(luyi + lug j) = hlegi + blugj = (kl)u. (a): w+ (—u) = (wi + weg) + (—ui — ue j) = (wy — ui )i t+ (ws — ui) § = 0. (g): (e+ Du = (k +1) + u2j) = kui + kuz j + lit lug j = ku+lu. (h): Ia = (ai + up J) Ini + lug j =wyituj=u Draw the triangles with sides formed by the vectors u, v, u + v and ku, kv, ku + kv. By similar triangles, k(u+v) =hutkv. 1 1 Let a, b, ¢ - vectors along the sides of the triangle and A,B the midpoints of a and b, then u = 337 =b= 1 5 —b)= be so u is parallel to ¢ and half as long. 2 Let * b, ¢, d be vectors along the sides of the quadrilateral pee A, B,C, D the corre sponding midpoints, then 1 ua gbsgoandv= Exercise Set 11.3 1. 2. 3. (a) (1)(6) + (2)(-8) = -10; cos @ = (10) /[(v5)(10)] = -1/ V5 (b) (—7)(0) +(—3)(1) = -8; cos 6 = (—8)/[(V58)(1)] = —3/ v58. (c) (1)(8) + (-3)(-2) + (7)(—2) = 0; cos 0 = 0. (d) (-8)(4) + Q)(2) + (2) (5) = —20; cos@ = (—20)/[(V14)(v45)] = -20/(8V70) (a) u-v = (1)(2) cos(x/6) = v3 (b) w+ v = (2)(3) cos 135° = -3/2. (a) u+v=—34 <0, obtuse. (b) u+v =6 >0, acute. (c) u+ V=—1 <0, obtuse. sa—fabntd=atbtesov= gat b+e)—ja=zb+5 1G =u thus ABCD i isa (d) u+v =0, orthogonal. 532 Chapter 11 10. 11. 12. 13. . Proceed as in Exercise . (a) AB= (1,3 . Let the points be P,Q, Rin order, then PQ= (2—(—1), -2—2, 0-3) = (3, —4, 3), QR= (3-2, 1—(-2), 4-0) = (1,3, -4), RP= (-1-3,2-1,3— (-4)} = (-4,1,7): since QP » QR=—3(1) + 4(3) + 3(—4) = -3 < 0,ZPQRis Seto? £43) +3 obtuse: since RP - RQ= —4(—1) +(—3)+7(4) = 29 > 0, ZPRQis acute; since PR » PQ= 4(3)—1(~4)-7(-3) = 37 > 0,2 RPQ is acute. . Since Vp + Vi = cos 4;, the answers are, in order, 2/2, 0, —V2/2, —1, —V2/2, 0, V2/2. + 25/2, 25/2, -25, 25/2, 25/2. a = a = —2), BC = (4,—-2,-1), AB - BC= 0 s0 AB and BC are orthogonal; it is a right triangle with the right angle at vertex B. (b) Let A, B, and C be the vertices (1,0), (2,—1), and (1,4) with corresponding interior angles a, 8, and 7, aman are aaa AB» AC _ (3-1) + (2,4) _ BA + BO (-3,1) +(-1,5) _ then cos @ = = 1/(5V2), so a & 82°, cos 8 = AB AC —_ ¥10V20 \| BA] | BC | vi0v26 moan CA+CB _ (-2,-4)-(1,-5) 4/V65, so 8 © 60°, cosy = =i" 2! — 9/\/130, so 7 © 38°. \|CAl| |CB| v20V26 - (a) vv, =—ab+ba=0; v+ v2 =ab+b(—a) =0. (b) Let vj = 21+ 3j, vo = —2i — 3); take wy = ee Ilva] . (a) The dot product of a vector u and a scalar v + w is not defined (b) A scalar u- w times a vector w is defined. (c) The sum of a scalar u+v and a vector w is not defined. (d) A scalar u- v plus a scalar k is defined. (e) The dot product of a scalar k and a vector u + v is not defined. (f) A scalar u-v times a scalar Vv: w is defined. p:s 181 + p2S2 + p3s3 gives the total revenue of the store in the given month from these 3 types of flatscreen televisions. t- w= 1.2/1 + 56/r2 + 13.1/vs gives the total time (in hours) the athlete needs to complete the race. (a) (1,2) + ((28, -14) + (6,0)) = (1,2) « (34-14) =6. (b). 6w| = 6lpw]] =36. (©) AVE (a) 24V5 AB» AP= [245+ 2k] [(r —Dit (r+ jt (7 —3)k] = Ar —1) + (7 #1) $20 — 3) = 5 —7 = 0,7 = 7/5. 534 Chapter 11 23. Take i, j, 2 direction is —= —=k. The direction cosines are cos a = 2/ V38, © +7 +s soa 71°, yg 61s, and yx 36°. and k along adie ent cess of the box, then 10i+15j +25k is along a diagonal, and a unit vector in this 8 = 3/38, and cosy = 5/V38 24. (a) TH = (3/5,4/5), so projyv = (6/25, 8/25) and v — proj,v = (44/25, -33/25). y Projphy 2 v— proj, (b) oT = (1/V5,-2/V5), so proj,v = (-6/5, 12/5) and v — proj,v = (26/5, 13/5). y V—projpy b Te (b) = (2/7,3/7, -6/7), so proj,v = (-74/49, —111/49, 222/49) and v — proj, v = (270/49, 62/49, 121/49) 26. (a) proj,v = (—1,-1), so v = (—1,—1) + (3, -3). (b) proj,v = (16/5,0,—8/5), so v = (16/5, 0, —8/5) +(—1/5, 1, —2/5). (c) v=-2b+0. 27. (a) proj,v = (1, 1), so v = (1,1) + (4,4). Exercise Set 11.3 535 28. 29. 30. 31. 32. 33. 34, 35. 36. 37. 38. 39. Al. . The diagonals have lengths |ju + v|| and {ju —-v{] but {Ju +-v||? = (u+v) + (u+v) = |lul]? + 2u- v + |v The di, Is h it h d i > 242 P . |ju tv? =(utv)-(u+v) . |jutv|? =(utv)-(u+v) =|lul? +2u-v+| (b) proj,v = (0, -8/5.4/5 so v = (0,—8/5, 4/5) + (—2, 13/5,26/5). (c) v- b =0, hence proj,v = 0,v =0+v. False, for example a = (1,2),b = (-1,0),¢ = (5, -3). True, because (v-+w) -(v +w) = [v|P? + [jw]? 40. True, by Theorem 11.3.3, u--v = |lull||v||cos 4 = 1- lvl]. (41) = 4|v]]. -b True, projp(v) = pp? isa ‘alar multiple of the vector b and is therefore parallel to b. = a AP= ~i+ 3), AB= 3i + 4j, proj, AP || =| AP - AB |/| AB || = 9/5, || AP || = V0, 10-817 = 13/5. = = = o> = = AP= —4i + 2k, AB= —3i + 2j- 4k, |[projz, AP || = | AP + AB \/|| AB || = 4/v29. || AP | = v20, ¥/20 — 16/29 = \/564/29. Let e; = —(cos27°, sin 27°) and ey = (sin 27°, — cos27°) be the forces parallel to and perpendicular to the slide, and let F be the downward force of gravity on the child. Then ||F\| = 34(9.8) = 333.2 N, and F = F, + Fp (F- exer + (F- es)es. The force parallel to the slide is therefore |[Fl] cos63° ~ 151.27 N, and the force against the slide is ||F|] cos 27° © 296.88 N, so it takes a force of 151.27 N to prevent the child from sliding. Let. denote the magnitude of the force in the direction of Q. Then the force F acting on the childis F = ai—333.9). Let e; = —(cos27°, sin27°) and eg = (sin 27°, — cos 27°) be the unit vectors in the directions along and against the slide. Then the component of F in the direction of e; is F- e; = —acos 27° + 333.2sin 27° and the child is prevented from sliding down if this quantity is negative, ic. « > 333.2 tan 27° © 169.77 N. We will obtain the work in two different ways. First, it is simply 4- 151.27 = 605.08 J. (Force times displacement.) Second, it is the same as the change in potential energy, so it is mgh = 34 -9.8 -4sin(27°) = 605.08 J. W = F-15i = 15 - 50.cos 60° = 375 ft Ib. = = Let P and Q be the points (1,3) and (4,7) then PQ= 3i+ 4j so W=F+ PQ=—12 ft-lb. W =F.(15/V3)(i+j+k) = —-15/V3 N-m=—5y3 J. W =F- PQ= |F\| || PO || cos45° = (500)(100) (v2/2) = 25,000V2 N -m = 25,0002 J. u+v and u—v are vectors along the diagonals, (u+v) - (u—v) =u-u-—u-v+v-u—v-v= lull? — |lv|? so (u+v) + (u—v) = 0 if and only if |lul] = |v}. and \Ju—v|/? = (u—v)-(u—v) = |lul? — 2u-v + |v||?. If the parallelogram is a rectangle then u+ v = 0 so |Ju+v\? = ||u—v|?; the diagonals are equal. If the diagonals are equal, then 4du-v = 0, u-v = 0 s0 uis perpendicular to v and hence the parallelogram is a rectangle. ul? +2u-v + ||v||? and ju — vi]? = (u—v) - (u—v) = jul]? — 2u- v + |lv|?, add to get ||u+v|l? + |ju—v||? = 2\lul|? + 2\|v||?. The sum of the squares of the lengths of the diagonals of a parallelogram is equal to twice the sum of the squares of the lengths of the s des. ? and |u — v||? = (u—v) + (u—v) = jul]? —2u-v+]fv|P, It follows by dividing both sides by 4. Iv Ju + v|? —ju—v|/? =4u-v, the re: subtract to get Exercise Set 11.4 537 (d) (v x w) x (ux v) = (0,56, 392). 9. uxv=(itj) x (4j+k) =k—j—k+i=i-j, the direction cosines are 7 a 10. u x v = 121+ 30j — 6k, so + (a vOj_ Tn) v30 V6 30 di. n= AB x AC= (1,1,-3) « (-1,3,—1) = (8,4,4), unit vectors are ta (2 11). v 12. A vector parallel to the yz-plane must be perpendicular to i; ix (3i—j + 2k) = —2) —k, || — 2) —kl| = V5, the unit vectors are +(2j +k)/V5. 13. True. 14. False; (i x j) x j= -k xj=i,ix (jxj)=0. 15. False; let v = (2, 1,-1),u = (1,3, -1), w = (—5, 0,2), then v x u=v x w= (2,1,5), but u4¢ w. 16. True; by Theorem 11.4.6(b); if one row of a determinant is a linear combination of the other two rows, then the determinant is zero. Equivalently, if u — av + bw then u lies in the plane of v and w and is thus per to their cross product. 17. A= |lux vl] = || -71-j + 3k|| = V59. 18. A = |\u x v|| = || —6i+ 4j + 7k|| = V101. 19. A= 5Pa x PR|| = sll 2) x (2,0,3)|] = 5-19.71 = V374/2. 20. A= SPQ x PR|| = SIK-L4. 8) x (5.2, 12)|| = 51K32, 52, -22)|| = 9VIB. 21. (2i- 3) +k) - (Si— 20j + 4k) = 80. 22. (1,—2,2) + (-11, 8,12) = 29. 23. (2,1, 0) - (—3,3,12) = -3. 24. i-(i-j)=1. 25. V = |u-(v x w)| =| — 16] = 16. 26. V = |u- (v x-w)| = [45] — 45. 27. (a) u-(vx w)=0,5 (b) u- (vx w) =0, yes. (c) u- (v x w) = 245, no. 28. (a) u-(wxv)=-u-(vx w)=-3. (b) (Vx w)-u=u-(vxw)=3. (c) w- (uxv)=u-: (vx w)=3. (d) v- (ux w) =u: (wxv)=—3. (e) (uxw)-v=u-(wxv)=—3. (f) v-(wxw)=v-0=0. 29. (a) V=|u-(vxw)| =|-9/=9. (b) A=||u x wl] = |[3i— 8) + 7kl] = VI22. 538 Chapter 11 30. 31. 32. 33. 34, 35. 36. 37. 38. 39. (c) v x w = —3i —j + 2k is perpendicular to the plane determined by v and w; let # be the angle between u u-(vxw) -9 {|u| ||v x wI| Vi4V14 determined by v and w is 6 = 6 — 7/2 = sin71(9/14). and v x w, then cos? = —9/14, so the acute angle ¢ that u makes with the plane v|| sind xv From the diagram, d= |jul|sing = ullilvilsing _ fe x vi Iv] Ivll P | w H d i eo, v B (a) u=AP=~4i+2k, v=AB=~—3i +2) — dk, ux v = —4i — 22 — 8k; distance = |Ju x v||/||vl] = 2141/29. (b) u=AP=2i +2), v=AB= —21 +j, ux v= 6k; distance = |lu x v|j/|Iv|] = 6/V5. Take v and w as sides of the (triangular) base, then area of base = 5illv x w|| and height = ||projy.wul| = : 1 1 [u-(v xw)l y= 2 (area of base) (height) = Liu (v x w)|. \lv x wl] 3 6 3), PR= (2,-2.1), PS = (4,—4,3), vei PQ + (PR x PS)| =F l-4l = 2/3. u-v 23 (a) eo? = Tg ~~ « y_ llux vil _ si 245) 12VTB (b) sin@= "Tvl (c) 23% | 144-13 2401 _ 4g > a9? > GF > i a Since AC + (AB x AD) =AC + (AB x CD)+ AC + (AB x AC) = 0 +0 =O, the volume of the parallelepiped at ‘y determined by AB, AC, and AD is zero, thus A,B,C, and D are coplanar (lie in the same plane). Since aren AB x CD# 0, the lines are not parallel. Hence they must intersect. The points P lie on the plane determined by A,B and C. From Theorems 11.3.3 and 11.4.5a it follows that sin 8 = cos 0, so 0 = 7/4. ju x v|? = ful?’ |lv|)? sin? = lull? |v||?(1 — cos? 8) = |jul!? iv? — (w+ v)?. — — ijk (a) F = 10j and PQ=i+j+k, so the vector moment of F about P is PQxF=)1 1 1 |=—10i+ 10k, 0 10 0 and the scalar moment is 10/2 lb-ft. The direction of rotation of the cube about P is counterclockwise looking along PQ x F = —10i + 10k toward its initial point. (b) F = 10j and PQ=j +k, so the vector moment of F about P is PQ x F = -10i, and the co - Bee con, 
