Chapter K

Exercises K.1

  1. 1.

    right hand

  2. 2.

    line; plane

  3. 3.

    curve (a parabola); surface (a cylinder)

  4. 4.

    a hyperbolic paraboloid

  5. 5.

    a hyperboloid of two sheets

  6. 6.

    a hyperboloid of one sheet

  7. 7.

    AB¯=6; BC¯=17; AC¯=11. Yes, it is a right triangle as AB¯2+AC¯2=BC¯2.

  8. 8.

    AB¯=3, AC¯=26, BC¯=3. The triangle is isosceles.

  9. 9.
  10. 10.

    AB¯=14, AC¯=14, BC¯=214. The points lie on a line.

  11. 11.

    AB¯=29, AC¯=105, BC¯=261. The points do not lie on a line.

  12. 12.

    Yes, as opposite sides have equal length. AB¯=21=CD¯; BC¯=6=AD¯.

  13. 13.

    Center at (4,1,0); radius = 3

  14. 14.

    Center at (2,1,2); radius = 5

  15. 15.

    closer to the surface

  16. 16.
  17. 17.

    Interior of a sphere with radius 1 centered at the origin.

  18. 18.

    Region bounded between the planes x=0 (the yz coordinate plane) and x=3.

  19. 19.

    The first octant of space along with its adjacent quarter planes; all points (x,y,z) where each of x, y and z are positive or zero. (Analogous to the first quadrant in the plane.)

  20. 20.

    All points in space where the y value is greater than 3; viewing space as often depicted in this text, this is the region “to the right” of the plane y=3 (which is parallel to the xz coordinate plane.)

  21. 21.
  22. 22.
  23. 23.
  24. 24.
  25. 25.

    x2+z2=1(1+y2)2

  26. 26.

    y2+z2=x4

  27. 27.

    z=(x2+y2)2=x2+y2

  28. 28.

    z=1x2+y2

  29. 29.

    (a)   x=y2+z29

  30. 30.

    (b)   x2y2+z2=0

  31. 31.

    (b)   x2+y29+z24=1

  32. 32.

    (a)   y2x2z2=1

  33. 33.
  34. 34.
  35. 35.
  36. 36.
  37. 37.
  38. 38.
  39. 39.
  40. 40.

Exercises K.2

  1. 1.

    Answers will vary.

  2. 2.

    (1,2) is a point; 1,2 is a vector that describes a displacement of 1 unit in the x-direction and 2 units in the y-direction.

  3. 3.

    A vector with magnitude 1.

  4. 4.

    Direction

  5. 5.

    Their respective unit vectors are parallel; unit vectors u1 and u2 are parallel if u1=±u2.

  6. 6.

    It stretches the vector by a factor of 2, and points it in the opposite direction.

  7. 7.

    PQ=1,6=1ı+6ȷ

  8. 8.

    PQ=4,4=4ı4ȷ

  9. 9.

    PQ=6,1,6=6ıȷ+6k

  10. 10.

    PQ=2,2,0=2ı+2ȷ

  11. 11.

    • u+v=2,1; uv=0,3; 2u3v=1,7.

      x=1/2,2.

  12. 12.

    • u+v=3,2,1; uv=1,0,3; πu2v=π22,π2,π22.

      x=1,0,3.

  13. 13.

    uvu+vuvxy
  14. 14.

    uvu+vuvxy

    Sketch of uv shifted for clarity.

  15. 15.

    uvu+vuvxyz
  16. 16.

    uvu+vuvxyz
  17. 17.

    u=5, v=13, u+v=26, uv=10

  18. 18.

    u=17, v=3, u+v=14, uv=26

  19. 19.

    u=5, v=35, u+v=25, uv=45

  20. 20.

    u=7, v=35, u+v=42, uv=28

  21. 21.

    u=3/58,7/58

  22. 22.

    u=0.6,0.8

  23. 23.

    u=1/3,2/3,2/3

  24. 24.

    u=1/3,1/3,1/3

  25. 25.

    When u and v have the same direction. (Note: parallel is not enough.)

  26. 26.

    u=cos50,sin500.643,0.766.

  27. 27.

    u=cos120,sin120=1/2,3/2.

  28. 28.

    u =cos2θsin2ϕ+sin2θsin2ϕ+cos2ϕ
    =sin2ϕ(cos2θ+sin2θ)+cos2ϕ
    =sin2ϕ+cos2ϕ
    =1.
  29. 29.

    The magnitude of the force on each chain is 100/357.735lb.

  30. 30.

    The magnitude of the force on each chain is 100lb.

  31. 31.

    The magnitude of the force on the chain with angle θ is approx. 45.124lb; the magnitude of the force on the chain with angle φ is approx. 59.629lb.

  32. 32.

    The magnitude of the force on each chain is 50lb.

  33. 33.

    θ=45; the weight is lifted 0.29 ft (about 3.5in).

  34. 34.

    θ=5.71; the weight is lifted 0.005 ft (about 1/16th of an inch).

  35. 35.

    θ=45; the weight is lifted 2.93 ft.

  36. 36.

    θ=84.29; the weight is lifted 9 ft.

  37. 37.
  38. 38.
  39. 39.
  40. 40.

Exercises K.3

  1. 1.

    Scalar

  2. 2.

    The magnitude of a vectors is the square root of the dot product of a vector with itself; that is, v=vv.

  3. 3.

    By considering the sign of the dot product of the two vectors. If the dot product is positive, the angle is acute; if the dot product is negative, the angle is obtuse.

  4. 4.

    “Perpendicular” is one answer.

  5. 5.

    22

  6. 6.

    33

  7. 7.

    3

  8. 8.

    0

  9. 9.

    not defined

  10. 10.

    0

  11. 11.

    Answers will vary.

  12. 12.

    Answers will vary.

  13. 13.

    θ=0.321818.43

  14. 14.

    θ=1.647694.4

  15. 15.

    θ=π/4=45

  16. 16.

    θ=π/2=90

  17. 17.

    Answers will vary; two possible answers are 7,4 and 14,8.

  18. 18.

    Answers will vary; two possible answers are 5,3 and 15,9.

  19. 19.

    Answers will vary; two possible answers are 1,0,1 and 4,5,9.

  20. 20.

    Answers will vary; two possible answers are 2,1,0 and 1,1,1/3.

  21. 21.

    projvu=1/2,3/2.

  22. 22.

    projvu=2,6.

  23. 23.

    projvu=1/2,1/2.

  24. 24.

    projvu=0,0.

  25. 25.

    projvu=1,2,3.

  26. 26.

    projvu=4/3,4/3,2/3.

  27. 27.

    u=1/2,3/2+3/2,1/2.

  28. 28.

    u=2,6+3,1.

  29. 29.

    u=1/2,1/2+5/2,5/2.

  30. 30.

    u=0,0+3,2.

  31. 31.

    u=1,2,3+0,3,2.

  32. 32.

    u=4/3,4/3,2/3+5/3,7/3,4/3.

  33. 33.

    1.96lb

  34. 34.

    5lb

  35. 35.

    141.42ft-lb

  36. 36.

    196.96ft-lb

  37. 37.

    500ft-lb

  38. 38.

    424.26ft-lb

  39. 39.

    500ft-lb

  40. 40.
  41. 41.
  42. 42.
  43. 43.
  44. 44.
  45. 45.
  46. 46.

Exercises K.4

  1. 1.

    vector

  2. 2.

    right hand rule

  3. 3.

    “Perpendicular” is one answer.

  4. 4.

    T

  5. 5.

    Torque

  6. 6.

    T

  7. 7.

    • a(b×c)=a(vector)=scalar

      a×(b×c)=a×(vector)=vector

      (ab)×(cd)=(scalar)×(scalar)=not meaningful

      a×(bc)=a(scalar)=not meaningful

      (a×b)(c×d)=(vector)(vector)=not meaningful

      (a×b)(c×d)=(vector)(vector)=scalar

  8. 8.
  9. 9.

    18

  10. 10.

    5

  11. 11.

    0

  12. 12.

    36

  13. 13.

    u×v=12,15,3

  14. 14.

    u×v=11,1,17

  15. 15.

    u×v=5,31,27

  16. 16.

    u×v=47,36,44

  17. 17.

    u×v=0,2,0

  18. 18.

    u×v=0,0,0

  19. 19.

    ı×ȷ=k

  20. 20.

    ı×k=ȷ

  21. 21.

    Answers will vary.

  22. 22.

    Answers will vary.

  23. 23.

    5

  24. 24.

    21

  25. 25.

    0

  26. 26.

    5

  27. 27.

    14

  28. 28.

    230

  29. 29.

    3

  30. 30.

    6

  31. 31.

    52/2

  32. 32.

    330

  33. 33.

    1

  34. 34.

    5/2

  35. 35.

    7

  36. 36.

    87/2

  37. 37.

    2

  38. 38.

    15

  39. 39.

    ±161,1,2

  40. 40.

    ±1212,1,4

  41. 41.

    0,±1,0

  42. 42.

    any unit vector orthogonal to u works (such as 121,0,1).

  43. 43.

    87.5ft-lb

  44. 44.

    43.75375.78ft-lb

  45. 45.

    200/366.67ft-lb

  46. 46.

    11.58ft-lb

  47. 47.

    With u=u1,u2,u3 and v=v1,v2,v3, we have

    u(u×v) =u1,u2,u3
    (u2v3u3v2,(u1v3u3v1),u1v2u2v1)
    =u1(u2v3u3v2)u2(u1v3u3v1)
    +u3(u1v2u2v1)
    =0.
  48. 48.

    With u=u1,u2,u3, we have

    u×u =u2u3u3u2,(u1u3u3u1),u1u2u2u1)
    =0,0,0
    =0.
  49. 49.
  50. 50.
  51. 51.
  52. 52.
  53. 53.
  54. 54.
  55. 55.
  56. 56.
  57. 57.
  58. 58.
  59. 59.
  60. 60.

Exercises K.5

  1. 1.

    A point on the line and the direction of the line.

  2. 2.

    parallel

  3. 3.

    parallel, skew

  4. 4.

    Answers will vary

  5. 5.

    vector: (t)=2,4,1+t9,2,5

    parametric: x=2+9t, y=4+2t, z=1+5t

    symmetric: (x2)/9=(y+4)/2=(z1)/5

  6. 6.

    vector: (t)=6,1,7+t3,2,5

    parametric: x=63t, y=1+2t, z=7+5t

    symmetric: (x6)/3=(y1)/2=(z7)/5

  7. 7.

    Answers can vary: vector: (t)=2,1,5+t5,3,1

    parametric: x=2+5t, y=13t, z=5t

    symmetric: (x2)/5=(y1)/3=(z5)

  8. 8.

    Answers can vary: vector: (t)=1,2,3+t4,7,2

    parametric: x=1+4t, y=2+7t, z=3+2t

    symmetric: (x1)/4=(y+2)/7=(z3)/2

  9. 9.

    Answers can vary; here the direction is given by d1×d2: vector: (t)=0,1,2+t10,43,9

    parametric: x=10t, y=1+43t, z=2+9t

    symmetric: x/10=(y1)/43=(z2)/9

  10. 10.

    Answers can vary; here the direction is given by d1×d2: vector: (t)=5,1,9+t0,1,0

    parametric: x=5, y=1t, z=9

    symmetric: not defined, as some components of the direction are 0.

  11. 11.

    Answers can vary; here the direction is given by d1×d2: vector: (t)=7,2,1+t1,1,2

    parametric: x=7+t, y=2t, z=1+2t

    symmetric: x7=2y=(z+1)/2

  12. 12.

    Answers can vary; here the direction is given by d1×d2: vector: (t)=2,2,3+t5,1,3

    parametric: x=2+5t, y=2t, z=33t

    symmetric: (x2)/5=(y2)=(z3)/3

  13. 13.

    vector: (t)=1,1+t2,3

    parametric: x=1+2t, y=1+3t

    symmetric: (x1)/2=(y1)/3

  14. 14.

    vector: (t)=2,5+t0,1

    parametric: x=2, y=5+t

    symmetric: not defined

  15. 15.

    parallel

  16. 16.

    intersecting; 1(2)=2(2)=12,3,7

  17. 17.

    intersecting; 1(3)=2(4)=9,5,13

  18. 18.

    same

  19. 19.

    skew

  20. 20.

    parallel

  21. 21.

    same

  22. 22.

    skew

  23. 23.

    41/3

  24. 24.

    32

  25. 25.

    52/2

  26. 26.

    5

  27. 27.

    3/2

  28. 28.

    2

  29. 29.

    Since both P and Q are on the line, PQ is parallel to d. Thus PQ×d=0, giving a distance of 0.

  30. 30.

    (Note: this solution is easier once one has studied Section 11.6.) Since the two lines intersect, we can state P2=P1+ad1+bd2 for some scalars a and b. (Here we abuse notation slightly and add points to vectors.) Thus P1P2=ad1+bd2. Vector c is the cross product of d1 and d2, hence is orthogonal to both, and hence is orthogonal to P1P2. Thus P1P2c=0, and the distance between lines is 0.

  31. 31.

    • The distance formula cannot be used because since d1 and d2 are parallel, c is 0 and we cannot divide by 0.

      Since d1 and d2 are parallel, P1P2 lies in the plane formed by the two lines. Thus P1P2×d2 is orthogonal to this plane, and c=(P1P2×d2)×d2 is parallel to the plane, but still orthogonal to both d1 and d2. We desire the length of the projection of P1P2 onto c, which is what the formula provides.

      Since the lines are parallel, one can measure the distance between the lines at any location on either line (just as to find the distance between straight railroad tracks, one can use a measuring tape anywhere along the track, not just at one specific place.) Let P=P1 and Q=P2 as given by the equations of the lines, and apply the formula for distance between a point and a line.

Exercises K.6

  1. 1.

    A point in the plane and a normal vector (i.e., a direction orthogonal to the plane).

  2. 2.

    A normal vector is orthogonal to the plane.

  3. 3.

    Answers will vary.

  4. 4.

    Answers will vary.

  5. 5.

    Answers will vary.

  6. 6.

    Answers will vary.

  7. 7.

    Standard form: 3(x2)(y3)+7(z4)=0

    general form: 3xy+7z=31

  8. 8.

    Standard form: 2(y3)+4(z5)=0

    general form: 2y+4z=26

  9. 9.

    Answers may vary;

    Standard form: 8(x1)+4(y2)4(z3)=0

    general form: 8x+4y4z=4

  10. 10.

    Answers may vary;

    Standard form: 5(x5)+3(y3)+2(z8)=0

    general form: 5x+3y+2z=0

  11. 11.

    Answers may vary;

    Standard form: 7(x2)+2(y1)+(z2)=0

    general form: 7x+2y+z=10

  12. 12.

    Answers may vary;

    Standard form: 3(x5)+3(z3)=0

    general form: 3x+3z=24

  13. 13.

    Answers may vary;

    Standard form: 2(x1)(y1)=0

    general form: 2xy=1

  14. 14.

    Answers may vary;

    Standard form: 2(x1)+(y1)3(z1)=0

    general form: 2x+y3z=0

  15. 15.

    Answers may vary;

    Standard form: 2(x2)(y+6)4(z1)=0

    general form: 2xy4z=6

  16. 16.

    Answers may vary;

    Standard form: 4(x5)2(y7)2(z3)=0

    general form: 4x2y2z=0

  17. 17.

    Answers may vary;

    Standard form: (x5)+(y7)+(z3)=0

    general form: x+y+z=15

  18. 18.

    Answers may vary;

    Standard form: 4(x4)+(y1)+(z1)=0

    general form: 4x+y+z=18

  19. 19.

    Answers may vary;

    Standard form: 3(x+4)+8(y7)10(z2)=0

    general form: 3x+8y10z=24

  20. 20.

    Standard form: x1=0

    general form: x=1

  21. 21.

    Answers may vary:

    ={x=14ty=110tz=28t

  22. 22.

    Answers may vary:

    ={x=1+20ty=3+2tz=3.526t

  23. 23.

    (3,7,5)

  24. 24.

    (3,1,1)

  25. 25.

    No point of intersection; the plane and line are parallel.

  26. 26.

    The plane contains the line, so every point on the line is a “point of intersection.”

  27. 27.

    5/7

  28. 28.

    8/21

  29. 29.

    1/3

  30. 30.

    3

  31. 31.

    If P is any point in the plane, and Q is also in the plane, then PQ lies parallel to the plane and is orthogonal to n, the normal vector. Thus nPQ=0, giving the distance as 0.

  32. 32.

    The intersecting lines define a plane with normal vector n=c=d1×d2. Since points P1 and P2 lie in the plane, c is orthogonal to P1P2, hence P1P2c=0, giving a distance of 0. Knowing the principles of planes, especially their normal vectors, makes this simpler.

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