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Solutions To Selected Problems

Chapter 11

Exercises 11.1

  1. 1.

    right hand

  2. 3.

    curve (a parabola); surface (a cylinder)

  3. 5.

    a hyperboloid of two sheets

  4. 7.

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

  5. 9.
  6. 11.

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

  7. 13.

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

  8. 15.

    closer to the surface

  9. 17.

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

  10. 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.)

  11. 21.
  12. 23.
  13. 25.

    x2+z2=1(1+y2)2

  14. 27.

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

  15. 29.

    (a) x=y2+z29

  16. 31.

    (b) x2+y29+z24=1

  17. 33.
  18. 35.
  19. 37.
  20. 39.

Exercises 11.2

  1. 1.

    Answers will vary.

  2. 3.

    A vector with magnitude 1.

  3. 5.

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

  4. 7.

    PQ=1,6=1ı+6ȷ

  5. 9.

    PQ=6,-1,6=6ı-ȷ+6k

  6. 11.
    (a) u+v=2,-1; u-v=0,-3; 2u-3v=-1,-7. (c) x=1/2,2.
  7. 13.

    u

    v

    u+v

    u-v

    x

    y
  8. 15.

    u

    v

    u+v

    u-v

    x

    y

    z
  9. 17.

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

  10. 19.

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

  11. 21.

    u=3/58,7/58

  12. 23.

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

  13. 25.

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

  14. 27.

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

  15. 29.

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

  16. 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.

  17. 33.

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

  18. 35.

    θ=45; the weight is lifted 2.93 ft.

  19. 37.
  20. 39.

Exercises 11.3

  1. 1.

    Scalar

  2. 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.

  3. 5.

    -22

  4. 7.

    3

  5. 9.

    not defined

  6. 11.

    Answers will vary.

  7. 13.

    θ=0.321818.43

  8. 15.

    θ=π/4=45

  9. 17.

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

  10. 19.

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

  11. 21.

    projvu=-1/2,3/2.

  12. 23.

    projvu=-1/2,-1/2.

  13. 25.

    projvu=1,2,3.

  14. 27.

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

  15. 29.

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

  16. 31.

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

  17. 33.

    1.96lb

  18. 35.

    141.42ft-lb

  19. 37.

    500ft-lb

  20. 39.

    500ft-lb

  21. 41.
  22. 43.
  23. 45.

Exercises 11.4

  1. 1.

    vector

  2. 3.

    “Perpendicular” is one answer.

  3. 5.

    Torque

  4. 7.
    (a) a(b×c)=a(vector)=scalar (b) a×(b×c)=a×(vector)=vector (c) (ab)×(cd)=(scalar)×(scalar)=not meaningful (d) a×(bc)=a(scalar)=not meaningful (e) (a×b)(c×d)=(vector)(vector)=not meaningful (f) (a×b)(c×d)=(vector)(vector)=scalar
  5. 9.

    18

  6. 11.

    0

  7. 13.

    u×v=12,-15,3

  8. 15.

    u×v=-5,-31,27

  9. 17.

    u×v=0,-2,0

  10. 19.

    ı×ȷ=k

  11. 21.

    Answers will vary.

  12. 23.

    5

  13. 25.

    0

  14. 27.

    14

  15. 29.

    3

  16. 31.

    52/2

  17. 33.

    1

  18. 35.

    7

  19. 37.

    2

  20. 39.

    ±161,1,-2

  21. 41.

    0,±1,0

  22. 43.

    87.5ft-lb

  23. 45.

    200/366.67ft-lb

  24. 47.
    With u=u1,u2,u3 and v=v1,v2,v3, we have u(u×v) =u1,u2,u3   (u2v3-u3v2,-(u1v3-u3v1),u1v2-u2v1) =u1(u2v3-u3v2)-u2(u1v3-u3v1)     +u3(u1v2-u2v1) =0.
  25. 49.
  26. 51.
  27. 53.
  28. 55.
  29. 57.
  30. 59.

Exercises 11.5

  1. 1.

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

  2. 3.

    parallel, skew

  3. 5.
    vector: (t)=2,-4,1+t9,2,5 parametric: x=2+9t, y=-4+2t, z=1+5t symmetric: (x-2)/9=(y+4)/2=(z-1)/5
  4. 7.
    Answers can vary: vector: (t)=2,1,5+t5,-3,-1 parametric: x=2+5t, y=1-3t, z=5-t symmetric: (x-2)/5=-(y-1)/3=-(z-5)
  5. 9.
    Answers can vary; here the direction is given by d1×d2: vector: (t)=0,1,2+t-10,43,9 parametric: x=-10t, y=1+43t, z=2+9t symmetric: -x/10=(y-1)/43=(z-2)/9
  6. 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=2-t, z=-1+2t symmetric: x-7=2-y=(z+1)/2
  7. 13.
    vector: (t)=1,1+t2,3 parametric: x=1+2t, y=1+3t symmetric: (x-1)/2=(y-1)/3
  8. 15.

    parallel

  9. 17.

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

  10. 19.

    skew

  11. 21.

    same

  12. 23.

    41/3

  13. 25.

    52/2

  14. 27.

    3/2

  15. 29.

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

  16. 31.
    (a) The distance formula cannot be used because since d1 and d2 are parallel, c is 0 and we cannot divide by 0. (b) 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. (c) 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 11.6

  1. 1.

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

  2. 3.

    Answers will vary.

  3. 5.

    Answers will vary.

  4. 7.
    Standard form: 3(x-2)-(y-3)+7(z-4)=0 general form: 3x-y+7z=31
  5. 9.
    Answers may vary; Standard form: 8(x-1)+4(y-2)-4(z-3)=0 general form: 8x+4y-4z=4
  6. 11.
    Answers may vary; Standard form: -7(x-2)+2(y-1)+(z-2)=0 general form: -7x+2y+z=-10
  7. 13.
    Answers may vary; Standard form: 2(x-1)-(y-1)=0 general form: 2x-y=1
  8. 15.
    Answers may vary; Standard form: 2(x-2)-(y+6)-4(z-1)=0 general form: 2x-y-4z=6
  9. 17.
    Answers may vary; Standard form: (x-5)+(y-7)+(z-3)=0 general form: x+y+z=15
  10. 19.
    Answers may vary; Standard form: 3(x+4)+8(y-7)-10(z-2)=0 general form: 3x+8y-10z=24
  11. 21.
    Answers may vary: ={x=14ty=-1-10tz=2-8t
  12. 23.

    (-3,-7,-5)

  13. 25.

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

  14. 27.

    5/7

  15. 29.

    1/3

  16. 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.

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