CSM_Chapters11.pdf

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Multivariable Calculus Complete Solutions Manual Brian Fulton

Melanie Fulton

Fourth Edition

Contents 11 Vectors and 3-Space 11.1 Vectors in 2-Space . . 11.2 3-Space and Vectors . 11.3 Dot Product . . . . . . 11.4 Cross Product . . . . . 11.5 Lines in 3-Space . . . 11.6 Planes . . . . . . . . . 11.7 Cylinders and Spheres 11.8 Quadric Surfaces . . . Chapter 11 in Review . . . A. True/False . . . . . B. Fill in the Blanks . C. Exercises . . . . . .

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2 2 6 12 17 24 28 35 38 42 42 43 45

12 Vector-Valued Functions 12.1 Vector Functions . . . . . . . 12.2 Calculus of Vector Functions 12.3 Motion on a Curve . . . . . . 12.4 Curvature and Acceleration . Chapter 12 in Review . . . . . . . A. True/False . . . . . . . . . B. Fill in the Blanks . . . . . C. Exercises . . . . . . . . . .

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48 48 55 62 69 73 73 73 74

13 Partial Derivatives 13.1 Functions of Several Variables . . . . 13.2 Limits and Continuity . . . . . . . . 13.3 Partial Derivatives . . . . . . . . . . 13.4 Linearization and Differentials . . . . 13.5 Chain Rule . . . . . . . . . . . . . . 13.6 Directional Derivative . . . . . . . . 13.7 Tangent Planes and Normal Lines . 13.8 Extrema of Multivariable Functions 13.9 Method of Least Squares . . . . . . . 13.10Lagrange Multipliers . . . . . . . . .

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77 77 82 85 92 99 107 112 118 125 127

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ii

CONTENTS Chapter 13 in Review . . A. True/False . . . . B. Fill in the Blanks C. Exercises . . . . .

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132 132 133 134

14 Multiple Integrals 14.1 The Double Integral . . . . . . . . . . . . . . 14.2 Iterated Integrals . . . . . . . . . . . . . . . . 14.3 Evaluation of Double Integrals . . . . . . . . 14.4 Center of Mass and Moments . . . . . . . . . 14.5 Double Integrals in Polar Coordinates . . . . 14.6 Surface Area . . . . . . . . . . . . . . . . . . 14.7 The Triple Integral . . . . . . . . . . . . . . . 14.8 Triple Integrals in Other Coordinate Systems 14.9 Change of Variables in Multiple Integrals . . Chapter 14 in Review . . . . . . . . . . . . . . . . A. True/False . . . . . . . . . . . . . . . . . . B. Fill in the Blanks . . . . . . . . . . . . . . C. Exercises . . . . . . . . . . . . . . . . . . .

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139 139 141 148 161 169 180 185 193 200 208 208 209 209

15 Vector Integral Calculus 15.1 Line Integrals . . . . . . . . . . 15.2 Line Integrals of Vector Fields . 15.3 Independence of the Path . . . 15.4 Green’s Theorem . . . . . . . . 15.5 Parametric Surfaces and Area . 15.6 Surface Integrals . . . . . . . . 15.7 Curl and Divergence . . . . . . 15.8 Stokes’ Theorem . . . . . . . . 15.9 Divergence Theorem . . . . . . Chapter 15 in Review . . . . . . . . A. True/False . . . . . . . . . . B. Fill in the Blanks . . . . . . C. Exercises . . . . . . . . . . .

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218 218 225 232 239 245 252 263 268 274 280 280 281 282

16 Higher-Order Differential Equations 16.1 Exact First-Order Equations . . . . 16.2 Homogeneous Linear Equations . . . 16.3 Nonhomogeneous Linear Equations . 16.4 Mathematical Models . . . . . . . . 16.5 Power Series Solutions . . . . . . . . Chapter 16 in Review . . . . . . . . . . . A. True/False . . . . . . . . . . . . . B. Fill in the Blanks . . . . . . . . . C. Exercises . . . . . . . . . . . . . .

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288 288 291 295 303 307 316 316 316 317

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Chapter 11

Vectors and 3-Space 11.1

Vectors in 2-Space

1.

(a)

6i + 12j

2.

(a)

h3, 3i

3.

(a)

h12, 0i

4.

(a)

1 2i

5.

(a)

−9i + 6j

6.

(a)

h3, 9i

7.

(a)

−6i + 27i

8.

(a)

h21, 30i

9.

(a)

h4, −12i − h−2, 2i = h6, −14i

10.

(a)

(4i + 4j) − (6i − 4j) = −2i + 8j

11.

(a)

(4i − 4j) − (−6i + 8j) = 10i − 12j

12.

(a)

h8, 0i − h0, −6i = h8, 6i

13.

(a)

h16, 40i − h−4, −12i = h20, 52i

14.

(a)

h8, 12i − h10, 6i = h−2, 6i

− 12 j

(b)

i + 8j h3, 4i

(b)

(c) (c)

h4, −5i

(b)

2 3i

(b) (b)

+ 32 j

(b) (b)

(d)

h4, 5i

(d)

− 13 i − j

(c)

−3i + 9j

0

h6, 18i

(c)

h8, 12i

−4i + 18j (c)

h6, 8i

5

(e)



41

(d) (d) (d)

3 √

41

√ 3 10

(e)

(e) (e)

10/3 √

√ 2 85 10

(−3i − 3j) − (−15i − 10j) = 18i − 17j

(b)

(−3i + 3j) − (−15i + 20j) = 12i − 17j

h−6, 0i − h0, −15i = h−6, 15i h−12, −30i − h−10, −30i = h−2, 0i

h−6, −9i − h25, 15i = h−31, −24i 2

34

√ 6 10

(e)

√ 4 13



(e)

√ 4 10 0

5 √

(e)

√ 2 2/3

(d)

(d)

(b)

(b) (b)

(e)

h−3, 9i − h−5, 5i = h2, 4i

(b)

(b)

65

−3i − 5j

(c)

(c)



(d)

h−1, −2i (c)

h−4, −12i

(b)

3i

11.1. VECTORS IN 2-SPACE

3 16.

15. P2 5

-5

P1P2

5

P1 P1P2

P1 -5

5

−−−→ P1 P2 = h2, 5i

P2

−−−→ P1 P2 = h6, −4i

17.

18. 5

P2

5

P1

P1

P1P2 P2

P1P2

5

5

−−−→ P1 P2 = h2, 2i

−−−→ P1 P2 = h2, −3i

−−−→ −−→ −−→ 19. Since P1 P2 = OP2 − OP1 , terminal point is (1, 18) −−−→ −−→ −−→ 20. Since P1 P2 = OP2 − OP1 , point is (9, 8)

−−→ −−−→ −−→ OP2 = P1 P2 + OP1 = (4i + 8j) + (−3i + 10j) = i + 18j, and the −−→ −−→ −−−→ OP1 = OP2 + P1 P2 = h4, 7i − h−5, −1i = h9, 8i, and the initial

21. a(= −a), b = (− 41 a), c(= 52 a), e(= 2a), and f (= − 21 a) are parallel to a. 22. We want −3b = a, so c = −3(9) = −27 23. h6, 15i 24. h5, 2i 25. |a| = 26. |a| =

√ √

√ 4 + 4 = 2 2; (a) 9 + 16 = 5; (a)

27. |a| = 5; (a) 28. |a| =



u=

    1 1 1 1 1 u = √ h2, 2i = √ , √ ; (b) −u = − √ , − √ 2 2 2 2 2 2     1 −3 4 3 4 u = h−3, 4i = , ; (b) −u = ,− 5 5 5 5 5

1 h0, −5i = h0, −1i; 5

1 + 3 = 2; (a)

√ 1 u = h1, − 3i = 2

(b) *

−u = h0, 1i

√ + 1 3 ,− ; 2 2

* (b)

−u =

√ + 1 3 − , 2 2

4

CHAPTER 11. VECTORS AND 3-SPACE 29. |a + b| = |h5, 12i| = 30. |a + b| = |h−5, 4i| = 31. |a| = 32. |a| =



9 + 49 =

q

1 4

+

1 4

=





25 + 144 = 13;



25 + 16 =

58;

√1 2



3b-a

35.

41;

1 13 h5, 12i

u=

b=3



( 1/1√2



( 12 i

=



√1 h−5, 4i 41

  b = 2 ( √158 3i + 7j) =

58;

3 33. − 4 a = h−3 − 15/2i



u=



√6 i 58 1 2 j)

+

5 12 13 , 13



D E = − √541 , √441 √14 j 58

√ √ 3 2 3 2 = i− j 2 2 34. 5(a + b) = 5h0, 1i = h0, 5i 36.

b = a+(b+c)

3b

b+c

a

a

b

c

37. x = −(a + b) = −a − b

38. x = 2(a − b) = 2a − 2b

39.

40. c

-c

b a

b = (−c) − a; (b + c) + a = 0; a+b+c=0

d

e

b

a

From Problem 39, e + c + d = 0. but b = e − a and e = a + b, so (a + b) + c + d = 0.

41. From 2i + 3j = k1 b + k2 c = k1 (i + j) + k2 (i − j) = (k1 + k2 )i + (k1 − k2 )j we obtain the system of equations k1 + k2 = 2, k1 − k2 = 3. Solving, we find k1 = 25 and k2 = − 12 . Then a = 52 b − 12 c. 42. From 2i + 3j = k1 b + k2 c = k1 (−2i + 4j) + k2 (5i + 7j) = (−2k1 + 5k2 )i + (4k1 + 7k2 )j we 1 obtain the system of equations −2k1 + 5k2 = 2, 4k1 + 7k2 = 3. Solving, we find k1 = 34 and 7 k2 = 17 . 43. From y 0 = 12 x we see that the slope of the tangent line at (2, 2) is 1. A vector with slope 1 is √ i + j. A unit vector is (i + j)/|i + j| = (i + j)/ 2 = √12 i + √12 j. Another unit vector tangent to the curve is − √12 i − √12 j. 44. From y 0 = −2x + 3 we see that the slope of the tangent √ line at (0, 0) is 3. A vector with slope 3 is i + 3j. A unit vector is (i + 3j)/|i + 3j| = (i + 3j)/ 10 = √110 i + √110 j. Another unit vector tangent to the curve is − √110 i − √110 j.

11.1. VECTORS IN 2-SPACE 45.

5 P2

(a) Since the shortest distance between two point is a straight line, |a + b| ≤ |a| + |b|.

a P1

b c

P3

(b) When P2 lies on the line segment between P1 and P3 , |a + b| = |a| + |b|. 46. Since y = 2a(L2 + y 2 )3/2 is an odd function on [−a, a], Fy = 0. Now using the fact that L/(L2 + y 2 )3/2 is an even function, we have Z

a

−a

L L dy = 2 2 3/2 a 2a(L + y )

Z 0

a

(L2

dy + y 2 )3/2

y = L tan θ, dy = L sec2 θ dθ

Z tan−1 a/L L sec2 θ dθ 1 sec2 θ dθ = La 0 sec3 θ L3 (1 + tan2 θ)3/2 0 tan−1 a/L Z tan−1 a/L 1 1 = cos θ dθ = sin θ La 0 La 0 1 a 1 √ = = √ La L2 + a2 L L2 + a2 . L = a

Z

tan−1 a/L

√ √ Then Fx = qQ/4π0 L L2 + a2 and F = (qQ/4π0 L L2 + a2 )i. 47. (a) Since Ff = −Fg , |Fg | = |Ff | = µ|Fn | and tan θ = |Fg |/|Fn | = µ|Fn |/|Fn | = µ (b) θ = arctan 0.6 ≈ 31◦ 48. Since w + F1 + F2 = 0, −200j + |F1 | cos 20◦ i + |F2 | sin 20◦ j − |F2 | cos 15◦ i + |F2 | sin 15◦ j = 0

or

(|F1 | cos 20◦ − |F2 | cos 15◦ )i + (|F1 | sin 20◦ − |F2 | sin 15◦ − 200)j = 0.

Thus, |F1 | cos 20◦ − |F2 | cos 15◦ = 0; |F1 | sin 20◦ + |F2 | sin 15◦ = 0. Solving this system for |F1 | and |F2 |, we obtain |F1 | =

200 cos 15◦ 200 cos 15◦ 200 cos 15◦ = = ≈ 336.8 lb sin 15◦ cos 20◦ + cos 15◦ sin20◦ sin(15◦ + 20◦ ) sin 35◦

and |F2 | =

sin 15◦

200 cos 20◦ 200 cos 20◦ = ≈ 327.7 lb. ◦ ◦ ◦ cos 20 + cos 15 sin20 sin 35◦

√ √ √ √ √ 49. Since Fq 2 = 200(i + j)/ 2 = 100 2i + 100 2j, F3 ) = F2 F1 = (100 2 − 200)i + 100 2j and p √ √ √ |F3 | = (100 2 − 200)2 + (100 2)2 = 200 2 − 2 ≈ 153 lb.

6

CHAPTER 11. VECTORS AND 3-SPACE −→ 50. We have OA = 150 cos 20◦ i + 150 sin 20◦ j, 240 cos 190◦ i + 240 sin 190◦ j. Then

−−→ AB = 200 cos 113◦ i + 200 sin 113◦ j,

−−→ BC =

r = (150 cos 20◦ + 200 cos 133◦ + 240 cos 190◦ )i + (150 sin 20◦ + 200 sin 113◦ + 240 sin 190◦ j and ≈ −173.55i + 193.73j |r| ≈ 260.09 miles. 51. P2

P2 M N P1

P1

−−→ Place one corner of the parallelogram at the origin, and let two adjacent sides be OP1 and −−→ OP2 . Let M be the midpoint of the diagonal connecting P1 and P2 and N be the midpoint −−→ −−→ −−→ −−→ −−→ of the other diagonal. By Problem 37, OM = 21 (OP1 + OP2 ). Since OP1 + OP2 is the main −−→ −−→ −−→ −−→ −−→ diagonal of the parallelogram and N is its midpoint, ON = 12 (OP1 + OP2 . Thus, OM = ON and the diagonals bisect each other. −−→ −−→ −→ −−→ −−→ −−→ −→ B 52. By Problem 39, AB + BC + CA = 0 and AD + DE + ED + CA = −−→ −−→ −→ 0. From the first equation AB + BC = −CA. Since D and E −−→ −−→ −−→ −−→ −−→ −−→ D E are midpoint, AD = 21 AB and EC = 12 BC. Then, 12 AB + DE + − − → − → 1 2 BC + CA = 0 and C A −→ 1  −→ −−→ −→ 1 −−→ −−→ 1 −→ AB + BC = −CA − DE = −CA − −CA = − CA. 2 2 2 Thus, the line segment joining the midpoints D and E is parallel to the side AC and half its length.

11.2

3-Space and Vectors

1.- 6.

(0,0,4)

(1,1,5)

(5,-4,3) (3,4,0)

(6,-2,0)

(6,0,0)

11.2. 3-SPACE AND VECTORS

7

7. A plane is perpendicular to the z-axis, 5 units above the xy-plane. 8. A plane perpendicular to the x-axis, 1 unit in front of the yz-plane. 9. A line perpendicular to the xy-plane at (2, 3, 0). 10. A single point located at (4, −1, 7). 11. (2, 0, 0), (2, 5, 0), (2, 0, 8), (2, 5, 8), (0, 5, 0), (0, 5, 8), (0, 0, 8), (0, 0, 0) 12. z

(-1,3,7)

(-1,6,7)

(3,3,7)

(3,6,7) (-1,3,4)

(3,3,4)

(-1,6,4)

(3,6,4)

y

x

13. (a) xy-plane: (−2, 5, 0), xz-plane: (−2, 0, 4), yz-plane: (0, 5, 4); (b) (−2, 5, −2) (c) Since the shortest distance between a point and a plane is a perpendicular line, the point in the plane x = 3 is (3, 5, 4). 14. We find planes that are parallel to coordinate planes. (a) z = −5; (b) x = 1 and y = −1; (c)

z=2

15. The union of the planes x = 0, y = 0, and z = 0. 16. The origin (0, 0, 0). 17. The point (−1, 2, −3). 18. The union of the planes x = 2 and z = 8. 19. The union of the planes z = 5 and z = −5. 20. The line through the points (1, 1, 1), (−1, −1, −1), and the origin. p √ 21. d = (3 − 6)2 + (−1 − 4)2 + (2 − 8)2 = 70

8

CHAPTER 11. VECTORS AND 3-SPACE p √ (−1 − 0)2 + (−3 − 4)2 + (5 − 3)2 = 3 6 p (a) 7; (b) d = (−3)2 + (−4)2 = 5 p (a) 2; (b) d = (−6)2 + 22 + (−3)2 = 7 p √ d(P1 , P2 ) = p(32 + 62 + (−6)2 = 9; d(P1 , P3 ) = 22 + 12 + 22 = 3 √ d(P2 , P3 ) = ((2 − 3)2 + (1 − 6)2 + (2 − (−6))2 = 90. The triangle is a right triangle. q p √ √ √ d(P1 , P2 ) = (12 + 22 + 42 = 21; d(P1 , P3 ) = 32 + 22 + (2 2)2 = 21 q p √ √ d(P2 , P3 ) = ((3 − 1)2 + (2 − 2)2 + (2 2 − 4)2 = 28 − 16 2. The triangle is an isosceles triangle. p √ d(P1 , P2 ) = p((4 − 1)2 + (1 − 2)2 + (3 − 3)2 = 10 √ d(P1 , P3 ) = p(4 − 1)2 + (6 − 2)2 + (4 − 3)2 = 26 √ d(P2 , P3 ) = ((4 − 4)2 + (6 − 1)2 + (4 − 3)2 = 26; The triangle is an isosceles triangle. p d(P1 , P2 ) = p((1 − 1)2 + (1 − 1)2 + (1 − (−1))2 = 2 d(P1 , P3 ) = p(0 − 1)2 + (−1 − 1)2 + (1 − (−1))2 = 3 √ d(P2 , P3 ) = ((0 − 1)2 + (−1 − 1)2 + (1 − 1)2 = 5; The triangle is a right triangle. p √ d(P1 , P2 ) = p((−2 − 1)2 + (−2 − 2)2 + (−3 − 0)2 = 34 √ d(P1 , P3 ) = p(7 − 1)2 + (10 − 2)2 + (6 − 0)2 = 2 34 √ d(P2 , P3 ) = ((7 − (−2))2 + (10 − (−2))2 + (6 − (−3))2 = 3 34 Since d(P1 , P2 ) + d(P1 , P3 ) = d(P2 , P3 ), the points P1 , P2 , and P3 are collinear. p √ d(P1 , P2 ) = p((0 − 1)2 + (3 − 2)2 + (2 − (−1))2 = 11 √ d(P1 , P3 ) = p(1 − 1)2 + (1 − 2)2 + ((−3) − (−1))2 = 5 √ d(P2 , P3 ) = ((1 − 0)2 + (1 − 3)2 + ((−3) − 2)2 = 30 Since adding any two of the above distances will not result in the third, the points cannot be collinear. p √ d(P1 , P2 ) = p((−4) − 1)2 + ((−3) − 0)2 + (5 − 4)2 = 35 √ d(P1 , P3 ) = p((−7) − 1)2 + ((−4) − 0)2 + (8 − 4)2 = 96 √ d(P2 , P3 ) = ((−7) − (−4))2 + ((−4) − (−3))2 + (8 − 5)2 = 19 Since adding any two of the above distances will not result in the third, the points cannot be collinear. p √ d(P1 , P2 ) = p(1 − 2)2 + (4 − 3)2 + (4 − 2)2 = 6 √ d(P1 , P3 ) = p(5 − 2)2 + (0 − 3)2 + (−4 − 2)2 = 3 6 √ d(P2 , P3 ) = (5 − 1)2 + (0 − 4)2 + (−4 − 4)2 = 4 6 Since d(P1 , P2 ) + d(P1 , P3 ) = d(P2 , P3 ), the points P1 , P2 , and P3 are collinear. p √ (2 − x)2 + (1 − 2)2 + (1 − 3)2 = 21 −→ x2 − 4x + 9 = 21 −→ x2 − 4x + 4 = 16 −→ (x − 2)2 = 16 −→ x = 2 + −4 or x = 6, −2 p (0 − x)2 + (3 − x)2 + (5 − 1)2 = 5 −→ 2x2 − 6x + 25 = 25 −→ x2 − 3x = 0 −→ x = 0, 3

22. d = 23. 24. 25.

26.

27.

28.

29.

30.

31.

32.

33. 34.

11.2. 3-SPACE AND VECTORS

9

 1 + 7 3 + (−2) 1/2 + 5/2 , , = (4, 1/2, 3/2) 35. 2 2 2   0 + 4 5 + 1 −8 + (−6) 36. , , = (2, 3, −7) 2 2 2 

37. (x1 + 2)/2 = −1, x1 = −4, (y1 + 3)/2 = −4, y1 = −11; The coordinates of P1 are (−4, −11, 10).

(z1 + 6)/2 = 8, z1 = 10

38. (−3 + (−5))/2 = x3 = −4, (4 + 8)/2 = y3 = 6; (1 + 3)/2 = z3 = 2. The coordinates of P3 are (−4, 6, 2).   −3 + (−4) 4 + 6 1 + 2 , , = (−7/2, 5, 3/2) (a) 2 2 2   −4 + (−5) 6 + 8 2 + 3 (b) , , = (−9/2, 7, 5/2) 2 2 2 −−−→ 39. P1 P2 = h−3, −6, 1i −−−→ 41. P1 P2 = h2, 1, 1i

−−−→ 40. P1 P2 = h8, −5/2, 8i −−−→ 42. P1 P2 = h−3, −3, 7i

43.

44. z

z

y y

x x

45.

46. z

z

y

y x

x

10

CHAPTER 11. VECTORS AND 3-SPACE

47. Since the k component is zero, while the i and j components are nonzero, the vector lies in the xy-plane. 48. Since the j components is the only nonzero component, the vector lies on the y-axis. 49. Since the vector is a scalar multiple of k, the vector lies on the z-axis. 50. Since the i component is zero while the j and k components are nonzero, the vector lies in the yz-plane. 51. a + (b + c) = h2, 4, 12i 52. 2a − (b − c) = h2, −6, 4i − h−3, −5, −8i = h5, −1, 12i 53. b + 2(a − 3c) = h−1, 1, 1i + 2h−5, −21, −25i = h−11, −41, −49i 54. 4(a + 2c) − 6b = 4h5, 9, 20i − h−6, 6, 6i = h26, 30, 74i √ √ 55. |a + c| = |h3, 3, 11i| = 9 + 9 + 121 = 139 √ √ √ 56. |c||2b| = ( 4 + 36 + 81)(2)( 1 + 1 + 1) = 22 3 a b 1 57. + 5 = |b| = 1 + 5 = 6 |a| |b| |b| √ √ 58. |b|a + |a|b = 1 + 1 + 1h1, −3, 2i + 1 + 9 + 4h−1, 1, 1i √ √ √ √ √ √ = h 3, −3 3, 2 3i + h− 14, 14, 14i √ √ √ √ √ √ = h 3 − 14, −3 3 + 14, 2 3 + 14i 59. |a| = 60. |a| =

√ √

u=−

100 + 25 + 100 = 15; 1+9+4=



14;

1 h10, −5, 10i = h−2/3, 1/3, −2/3i 15

1 1 3 2 u = √ (i − 3j + 2k) = √ i − √ j + √ k 14 14 14 14

61. b = 4a = 4i − 4j + 4k 62. |a| =



36 + 9 + 4 = 7;

b=−

63. z

a

1 (a+b) 2 b y x

1 2

    1 3 3 1 h−6, 3, −2i = ,− , 7 7 14 7

11.2. 3-SPACE AND VECTORS

11

64. Following the hint, we first complete the following: s

2  2  2 y1 + y2 z1 + z2 x1 + x2 − x1 + − y1 + − z1 2 2 2 s 2  2  2 y1 − y2 z1 − z2 x1 − x2 + + = 2 2 2 p 1 = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 2 s 2  2  2 y1 + y2 z1 + z2 x1 + x2 d(M, P2 ) = + y2 − + z2 − x2 − 2 2 2 s 2  2  2 x1 − x2 y1 − y2 z1 − z2 = + + 2 2 2 p 1 = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 2 p d(P1 , P2 ) = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 From the above, we see that d(P1 , P2 ) = d(P1 , M ) + d(M, P2 ). Therefore, M is collinear with P1 and P2 and lies between them. Since d(P1 , M ) = d(M, P2 ), we also have that M is equidistant from P1 and P2 . Therefore, M is the midpoint of the line segment joining P1 and P2 . √ 3 1 √ 65. xP = 1, yP = cos 30◦ + sin 30◦ = + ( 3 + 1), 2 √2 3 1 1 √ ◦ ◦ xP = − sin 30 + cos 30 = − + = ( 3 − 1), 2 2√ 2 √ √ 1 √ 1 √ 2 1 √ 2 1 √ ◦ ◦ xR = cos 45 − ( 3 − 1) sin 45 = − ( 3 − 1) = (3 2 − 6), yr = ( 3 + 1), 2 2 4 2 √2 √2 √ 1 √ 1 √ 2 1 √ 2 ◦ ◦ + ( 3 − 1) = ( 2 + 6), zr = sin 45 + ( 3 − 1) cos 45 = 2 2 2 2 4 √ √ √ 1 1 √ 1 √ 1 √ 1 √ 3 ◦ ◦ xS = (3 2 − 6) cos 60 + ( 3 + 1) sin 60 = (3 2 − 6) + ( 3 + 1) 4 2 4 2 2 2 √ √ 1 √ = (3 2 − 6 + 6 + 2 3), 8 √ √ √ 1 √ 1 √ 1 √ 3 1 √ 1 ◦ ◦ yS = − (3 2 − 6) sin 60 + ( 3 + 1) cos 60 = − (3 2 − 6) + ( 3 + 1) 4 2 4 2 2 2 √ √ √ 1 = (−3 6 + 3 2 + 2 3 + 2), 8 √ 1 √ zs = ( 2 + 6) 4 d(P1 , M ) =

Thus, xS ≈ 1.4072, yS ≈ 0.2948, 1 0 0 66. (a) MP = 0 cos α sin α , 0 − sin α cos α

zS ≈ 0.9659. cos β MR = 0 sin β

0 − sin β 1 0 0 cos β



12

CHAPTER 11. VECTORS AND 3-SPACE 

     x xP xS (b) Mγ MR MP  y  = Mγ MR  yP  =  yS  z zP zS           1 √0 0 1 1 1 1 0 0 1 √  3 1  1  =  21 (√3 + 1)  sin 30◦   1  =  0 (c) MP  1  =  0 cos 30◦ 2 √2  ◦ ◦ 1 3 1 1 0 − sin 30 cos 30 1 0 − 12 2 ( 3 − 1) 2 √ √         2 1 1 1 0 − 22 cos 45◦ 0 − sin 45◦ 2 √ √    1 ( 3 + 1)  =  0   21 (√3 + 1)  0 1 0 MR MP  1  =   √0 1 2 √ √ ◦ ◦ 1 1 2 2 1 sin 45 0 cos 45 0 2 ( 3 − 1) 2 ( 3 − 1) 2 2 √   1 √ 6) 4 (3 √2 − =  12 √ ( 3 +√1)  1 6) 4( 2 + √      1 √ ◦ 6) 1 cos 60 sin 60◦ 0 4 (3 √2 − Mγ MR MP  1  =  sin sin 60◦ cos sin 60◦ 0   12 √ ( 3 +√1)  1 1 0 0 1 6) 4( 2 +   √ √ √ √ √ √    1 1 3 1 6) 6 + 6 +√2 3) 0 4 (3 √2 − 8 (3 √2 − √ 2  2√3  1 1 = − ( 3 +√1)  =  18 (−3 6 +√3 2 √ + 2 3 + 2)  0   2√ 2 2 1 1 6) 6) 0 0 1 4( 2 + 4( 2 +

11.3

Dot Product

1. a · b = 2(−1) + (−3)2 + 4(5) = 12 2. b · c = (−1)3 + 2(6) + 5(−1) = 4 3. a · c = 2(3) + (−3)6 + 4(−1) = −16 4. a · (b + c) = 2(2) + (−3)8 + 4(4) = −4 5. a · (4b) = 2(−4) + (−3)8 + 4(20) = 48 6. b · (a − c) = (−1)(−1) + 2(−9) = 5(5) = 8 7. a · a = 22 + (−3)2 + 42 = 29 8. (2b) · (3c) = (−2)9 + 4(18) + 10(−3) = 24 9. a · (a + b + c) = 2(4) + (−3)5 + 4(8) = 25 10. (2a) · (a − 2b) = 4(4) + (−6)(−7) + 8(−6) = 10     2(−1) + (−3)2 + 4(5) 12 a·b b= h−1, 2, 5i = h−1, 2, 5i = h−2/5, 4/5, 2i 11. b·b (−1)2 + 22 + 52 30 12. (c · b)a = [3(−1) + 6(2) + (−1)5]h2, −3, 4i = 4h2, −3, 4i = h8, −12, 16i √ 13. a · b = 10(5) cos(π/4) = 25 2

11.3. DOT PRODUCT

13

√ 14. a · b = 6(12) cos(π/6) = 36 3 15. a · b = |a||b| cos θ = (2)(3) cos(2π/3) = 6(−1/2) = −3  √  √ 16. a · b = |a||b| cos(θ) = (4)(1) cos(5π.6) = 4 − 23 = −2 3 √ √ 5a · b = 3(2) + (−1)2 = 4; |a| = 10, |b| = 2 2 4 1 1 √ = √ −→ θ = arccos √ ≈ 1.11 rad ≈ 63.45◦ cos θ = √ ( 10)(2 2 5 5 √ √ 18. 5a · b = 2(−3) + 1(−4) = −10; |a| = 5, |b| = 5 √ −10 2 cos θ = √ = − √ −→ θ = arccos(−2/ 5) ≈ 2.68 rad ≈ 153.43◦ ( 5)5 5 √ √ √ 19. 5a · b = 2(−1) + 4(−1) + 0(4) = −6; |a| = 2 5, |b| = 3 2 √ −6 1 √ = − √ −→ θ = arccos(−1/ 10) ≈ 1.89 rad ≈ 108.43◦ cos θ = √ (2 5)(3 2 10 √ √ √ 20. 5a · b = 12 (2) + 12 (−4) + 32 (6) = 8; |a| = 11/2, |b| = 2 14 √ 8 8 √ cos θ = √ =√ −→ θ = arccos(8/ 154) ≈ 0.87 rad ≈ 49.86◦ ( 11/2)(2 14) 154

17.



21. a and f, b and e, c and d 22. (a) a × b = 2 · 3 + (−c)2 + 3(4) = 0 −→ c = 9 (b) a · b = c(−3) + 12 (4) + c2 = c2 − 3c + 2 = (c − 2)(c − 1) = 0 −→ c = 1, 2 23. Solving the system of equations 3x1 + y1 − 1 = 0, −3x1 + 2y1 + 2 = 0 gives x1 = 4/9 and y1 = −1/3. Thus, v = h4/9, −1/3, 1i. 24. If a and b represent adjacent sides of the rhombus, then |a| = |b|, then diagonals of the rhombus are a + b and a − b, and (a + b) · (a − b) = a · a − a · b + b · a − b · b = |a|2 − |b|2 = 0. Thus, the diagonals are perpendicular. 25. Since  c·a=

b−

 a·b a·b a·b 2 a ·a=a·a− (a · a) = b · a − |a| = b · a − a · b = 0, 2 2 |a| |a| |a|2

the vectors c and a are orthogonal. √ √ 26. a · b = 1(1) + c(1) = c + 1; |a| = 1 + c2 , |b| = 2 √ 1 c+1 √ =⇒ 1 + c2 = c + 1 −→ 1 + c2 = c2 + 2c + 1 =⇒ c = 0 cos 45◦ = √ = √ 2 2 1+c 2 √ √ √ √ 27. |a| = 14; cos α − 1/ 14, α ≈ 74.50◦ ; cos β = 2/ 14, β ≈ 57.69◦ ; cos γ = 3/ 14, γ = 36.70◦

14

CHAPTER 11. VECTORS AND 3-SPACE

28. |a| = 9; cos α = 2/3, α ≈ 48.19◦ ; cos β = 2/3, β ≈ 48.19◦ ; cos γ = −1/3, γ = 109.47◦ √ 29. |a| = 2; cos α = 1/2, α ≈ 60◦ ; cos β = 0, β ≈ 90◦ ; cos γ = − 3/2, γ = 150◦ √ √ √ √ 30. |a| = 78; cos α = 5/ 78, α ≈ 55.52◦ ; cos β = 7/ 78, β ≈ 37.57◦ ; cos γ = 2/ 78, γ = 76.91◦ −−→ −−→ 31. Let θ be the angle between AD and AB and a be the length of an edge of an edge of the −−→ −−→ cube. Then AD = ai + aj + ak, AB = ai and −−→ −−→ AD · AB a2 1 cos θ −−→ −−→ = √ √ = √ 2 2 3 3a a |AD||AB| −−→ −→ so θ ≈ 0.955317 radian or 54.7356◦ . Letting φ be the angle between AD and AC and noting −→ that AC = ai + aj we have r −−→ −→ a2 + a2 2 AD · AC = cos θ −−→ −→ = √ √ 2 2a2 3 3a |AD||AC| so φ ≈ 0.61548 radian or 35.2644◦ √ √ √ 32. a = h5, 7, 4i;√ |a| = 3 10; cos α = 5/3 10, α = 58.19◦ ; cos β = 7/3 10, β = 42.45◦ ; cos γ = 4/3 10, γ ≈ 65.06◦ 33. compb a = a · b/|b| = h1, −1, 3i · h2, 6, 3i/7 = 5/7 √ √ 34. compa b = b · a/|a| = h2, 6, 3i · h1, −1, 3i/ 11 = 5/ 11 √ √ 35. b − a = h1, 7, 0i; compa (b − a) = (b − a) · a/|a| = h1, 7, 0i · h1, −1, 3i/ 11 = −6/ 11 36. a + b = h3, 5, 6i; 2b = h4, 12, 6i; comp2b (a + b) · 2a/|2b| = h3, 5, 6i · h4, 12, 6i/14 = 54/7 √ √ −−→ −−→ −−→ −−→ − → a = a · OP /|OP | = (4i + 6j) · (3i + 10j)/ 109 = 37. OP = 3i + 10j; |OP | = 109; comp− OP √ 72/ 109 √ √ −−→ −−→ −−→ −−→ − → a = a · OP /|OP | = h2, 1, −1i · h1, −1, 1i/ 3 = 0 38. OP = h1, −1, 1i; |OP | = 3; comp− OP 39. (a) compb a = a · b/|b| = (−5i + 5j) · (−3i + 4j)/5 = 7 projb a = (compb a)b/|b| = 7(−3i + 4j)/5 = − 21 5 i+

28 5 j = − 45 i

28 (b) projb⊥ a = a − projb a = (−5i + 5j) − (− 21 − 35 j 5 + 5 j) √ √ 40. (a) compb a = a · b/|b| = (4i + 2j)√· (−3i + j)/ √ 10 = − 10 projb a = (compb a)b/|b| = − 10(−3i + j)/ 10 = 3i − j

(b) projb⊥ a = a − projb a = (4i + 2j) − (3i − j) = i + 3j 41. (a) compb a = a · b/|b| = (−i − 2j + 7k) · (6i − 3j − 2k)/7 = −2 6 4 projb a = (compb a)b/|b| = −2(6i − 3j − 2k)/7 = − 12 7 i + 7j + 7k 6 4 5 (b) projb⊥ a = a − projb a = (−i − 2j + 7k) − (− 12 7 i − 7 j + 7 k) = 7 i −

20 7 j

+

45 7 k

11.3. DOT PRODUCT

15

42. (a) compb a = a · b/|b| = h1, 1, 1i · h−2, 2, −1i/3 = −1/3 projb a = (compb a)b/|b| = − 13 h−2, 2, −1i/3 = h2/9, −2/9, 1/9i (b) projb⊥ a = a − projb a = h1, 1, 1i − h2/9, −2/9, 1/9i = h7/9, 11/9, 8/9i 43. a + b = 3i + 4j; |a + b| = 5; compa+b a = a · (a + b)/|a + b| = (4i + 3j) · (3i + 4j)/5 = 72 96 24/5; proj(a+b) a = (comp(a+b) a)(a + b)/|a + b| = 24 5 (3i + 4j)/5 = 25 i + 25 j √ √ √ 44. a − b = 5i + 2j; |a − b| = 29; comp(a−b) b = b · (a − b)/|a − b| = (−i + j)/ 29 = −3/ 29 √ 3 15 6 proj(a−b) b = (comp(a−b) )(a − b)/|a − b| = − √ (5i + 2j)/ 29 = − i − j 29 29 29   15 6 14 35 proj(a−b)⊥ b = b − proj(a−b) b = (−i + j) − − i − j = − i + j 29 29 29 29 45. We identify F = 29, θ = 60◦ and |d| = 100. Then W = |F||d| cos θ = 20(100)( 12 ) = 1000 ft-lb. 46. W = F · d = |F||d| cos θ Since the force is acting at a 45◦ angle to the direction of motion, we have θ = 45◦ . Therefore, √ W = (3000)(400) cos 45◦ = 600, 000 2 ft-lb. 47. We identify d = −i + 3j + 8k. Then W = F · d = h4, 3, 5i · h−1, 3, 8i = 45 N-m. 48. (a) Since w and d are orthogonal, W = w · d = 0.

√ (b) We identify θ = 0◦ . Then W = |F||d| cos θ = 30( 42 + 32 ) = 150 N-m. 

78 49. Using d = 6i + 2j and F = 3 53 i + 45 j , W = F · d = 95 , 12 5 · h6, 2i = 5 ft-lb. 50. Let a and b be vectors from the center of the carbon atom to the center of two distinct hydrogen atoms. The distance between two hydrogen atoms is then p √ |b − a| = (b − a) · (b − a) = b · b − 2a · b + a · a p = |b|2 + |a|2 − 2|a||b| cos θ p = (1.1)2 + (1.1)2 − 2(1.1)(1.1) cos 109.5◦ p = 1.21 + 1.21 − 2.42(−0.333807) ≈ 1.80 angstroms. 51. If a and b are orthogonal, then a · b = 0 and

a 1 b1 a2 b2 a3 b3 + + a b a b a b 1 1 = (a1 b1 + a2 b2 a3 b3 ) = (a · b) = 0. |a||b| |a||b|

cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 =

52. We want cos α = cos β = cos γ or a1 = a2 = a3 . Letting a1 = a2 = a3 = 1 we obtain the 1 1 1 vector i + j + k. A unit vector in the same direction is √ i + √ j + √ k. 3 3 3

16

CHAPTER 11. VECTORS AND 3-SPACE

53. For the following, let a = ha1 , a2 , a3 i, b = hb1 , b2 , b3 i, and let k be any scalar. Proof of (i): If a = 0 = h0, 0, 0i, then a · b = (0)b1 + (0)b2 + (0)b3 = 0. Similarly, if b = 0, then a · b = a1 (0) + a2 (0) + a3 (0) = 0 Proof of (ii): Using the Commutative Property of real numbers, we have a · b = a1 b1 + a2 b2 + a3 b3 = b1 a1 + b2 a2 + b3 a3 =b·a Proof of (iv): a · (kb) = ha1 , a2 , a3 i · hkb2 , kb2 , kb3 i = a1 kb1 + a2 kb2 + a3 kb3 = k(a1 b1 + a2 b2 + a3 b3 ) = k(a · b) = hla1 , ka2 , ka3 i · hb1 , b2 , b3 i = ka1 b1 + ka2 b2 + ka3 b3 = k(a1 b1 + a2 b2 + a3 b3 ) = k(a · b) Therefore, a · (kb) = (ka) · b = k(a · b) Proof of (v): Since x2 ≥ 0 for any real number x, we have a · a = a21 + a22 + a23 ≥ 0. 54. Using the fact that | cos θ| < 1, we have |a · b| = |a||b|| cos θ| ≤ |a||b|. 55. |a + b|2 = (a + b) · (a + b) = a · a + 2a · b + b · b = |a|2 + 2a · b + |b|2 ≤ |a|2 + 2|a · b| + |b|2

since x ≤ |x|

≤ |a|2 + 2|a||b| + |b|2 = (|a| + |b|)2 by Problem 54 Thus, since |a + b| and |a| + |b| are positive, |a + b| ≤ |a| + |b|. 56. Let P1 (x1 , y1 ) and P2 (x2 , y2 ) be distinct points on the line ax + by = −c. Then −−−→ n·P1 P2 = ha, bi·hx2 −x1 , y2 −y1 i = ax2 −ax1 = by2 −by1 (ax2 +by2 )−(ax1 +by1 ) = −c−(−c) = 0, and the vectors are perpendicular. Thus, n is perpendicular to the line. −−−→ 57. Let θ be the angle between n and P2 P1 . Then −−−→ −−−→ |ha, bi · hx1 − x2 , y1 − y2 i| |ax1 − ax2 + by1 = by2 | |n · P2 P1 | √ √ = = d = ||P1 P2 | cos θ| = 2 2 |n| a +b a2 + b2 |ax1 + by1 − (ax2 + by2 )| |ax1 + by1 − (−c)| |ax1 + by1 + c| √ √ √ = = = . 2 2 2 2 a +b a +b a 2 + b2 58. (a) Since N = hx, yi is a unit normal, T = h−y, xi is a unit tangent at P (x, y). Now we

11.4. CROSS PRODUCT

17

compute −−→ N · P O = hx, yi · hc − x, d − yi = cx + dy − (x2 + y 2 ) = cx + dy − 1 −−→ T · OP = h−y, xi · hc − x, d − yi = dx − cy −→ N · P S = hx, yi · ha − x, b − yi = ax + by − (x2 + y 2 ) = ax + by − 1 −→ −T · P S = hy, −xi · ha − x, b − yi = ay − bx. Now, since |N| = |T| = 1, −−→ −→ −−→ −→ N · PO N · PS T · PO −T · P S = cos θ = , and = cos φ = −−→ −→ −−→ −→ , |P O| |P S| |P O| |P S| we have

−−→ −→ N · PO N · PS cx + dy − 1 ax + by − 1 = −−→ −→ or dx − cy = ay − bx . TP O −TP S

(b) With a = 2, b = 0, c = 0, and d = 3 the equation in (a) becomes (3y − 1)/3x = 2 2 (2x − 1)/2y or − 2y. Substituting y 2 = 1 − x2 we obtain 6x2 − 3x = √ 6x − 3x = 6y √ 2 2 2 6(1 − x ) − 2 1 − x or 12x − 3x − 6 = −2 1 − x2 . Squaring both sides, we obtain 144x4 − 72x3 + 36x + 32 = 0. (c) Newton’s method gives us the roots −0.6742, −0.48302, ; 0.76379, and 0.89343. Since S and O are on the√positive x- and y-axes, respectively,√we can ignore the negative roots. Computing y = 1 − 0.763792 ≈ 0.645465 and y = 1 − 0.893432 ≈ 0.449202 we see that only P (x, y) = (0.76379, 0.645465) satisfies (3y − 1)/3x = (2x − 1)/2y.

11.4 1.

2.

3.

4.

5.

Cross Product

a × b = a × b = a × b = a × b = a × b =

−1 0 i − 1 0 j + 1 −1 k = −5i − 5j + 3k = 0 3 0 5 3 5 i j k 2 1 2 0 1 0 k = −i + 2j − 4k 2 1 0 = j+ i− 4 0 4 −1 0 −1 4 0 −1 i j k 1 1 1 −3 −3 1 k = h−12, −2, 6i 1 −3 1 = i− j+ 0 4 2 4 2 0 2 0 4 i j k 1 1 1 1 1 1 k = h1, −8, 7i 1 1 1 = i− j+ 2 3 −5 3 −5 2 −5 2 3 i j k 2 2 −1 −1 2 2 2 −1 2 = i − j + −1 −1 −1 3 k = −5i + 5k 3 −1 −1 3 −1

i 1 0

j −1 3

k 0 5

18

CHAPTER 11. VECTORS AND 3-SPACE i 6. a × b = 4 2

j 1 3

k −5 −1

i 9. a × b = 2 −3 i 10. a × b = 8 1

4 −5 i − 2 −1

j k 0 0 1/2 = 6 6 0

i 7. a × b = 1/2 4 i 8. a × b = 0 2

1 = 3

j 5 −3

k 0 4

j 2 −3 j 1 −2

5 = −3 k −4 6

k −6 10

1/2 1/2 j + 4 0

1/2 1/2 i − 4 0

0 0 i − 2 4

0 0 j + 2 4

2 −4 i − 2 = −3 −3 6

1 = −2

8 −6 i − 1 10

−−−→ P1 P3 = (1, 2, 2) i j k 1 0 1 1 = 2 1 2 2

i j 13. a × b = 2 7 1 1 is perpendicular

k 7 −4 −4 = 1 −1 −1 to both a and b.

i j 14. a × b = −1 −2 4 −1 is perpendicular to

0 1 i − 1 2

i − 2 1

8 −6 j + 1 10

1 k = h−2, −86, −17i −2

2 k = 6i + 14j + 4k 1

1 k=j−k 2

0 1 j + 1 2

4 0

k = h−3, 2, 3i

2 k = h0, 0, 0i −3

−2 −4 j + −3 1

2 −4 j + 1 −1

k −1 −2 4 4 = i − 4 −1 0 0 both a and b.

0 6

5 k = h20, 0, −10i −3

2 −4 j + −3 6

−−−→ −−−→ 11. P1 P2 = (−2, 2, −4); P1 P3 = (−3, 1, 1) i j k 2 −4 −−−→ −−−→ i − −2 P1 P2 × P1 P3 = −2 2 −4 = −3 1 1 −3 1 1 −−−→ 12. P1 P2 = (0, 1, 1); −−−→ −−−→ P1 P2 × P1 P3 =

1 k = 14i − 6j + 10k 3

4 −5 j + 2 −1

7 k = −3i − 2j − 5k 1

j + −1 4

i j k 5 1 5 −2 1 15. a × b = 5 −2 1 = i − 2 −7 j + 2 0 −7 2 0 −7 a · (a × b) = h5, −2, 1i · h14, 37, 4i = 70 − 74 + 4 = 0 b · (a × b) = h2, 0, −7i · h14, 37, 4i = 28 + 0 − 28 = 0

−2 k = h4, 16, 9i −1

−2 k = h14, 37, 4i 0

11.4. CROSS PRODUCT

19

i j k 1/2 −1/4 −1/4 −4 k i + 16. a × b = 1/2 −1/4 −4 = 2 −2 −2 6 2 −2 6 19 1 = − i − 11j − k 2 2  1 1 19 1 19 11 a · (a × b) = i − j − 4k) · (− i − 11j − k = − + +2=0 2 4 2 2 4 4  19 1 b · (a × b) = (2i − 2j + 6k) · − i − 11j − k = −19 + 22 − 3 = 0 2 2 i j k 1 1 i − 2 1 j + 2 1 k = j − k 17. (a) b × c = 2 1 1 = 3 1 3 1 1 1 3 1 1 i j k 1 2 −1 2 j + 1 −1 k = −i + j + k i − a × (b × c) = 1 −1 2 = 0 1 0 −1 1 −1 0 1 −1 (b) a · c = (i − j + 2k) · (3i + j + k) = 4; (a · c)b = 4(2i + j + k) = 8i + 4j + 4k a · b = (i − j + 2k) · (2i + j + k) = 3; (a · b)c = 3(3i + j + k) = 9i + 3j + 3k a × (b × c) = (a · c)b − (a · b)c = (8i + 4j + 4k) − (9i + 3j + 3k) = −i + j + k i j k 1 −1 2 −1 j + 1 2 k = 21i − 7j + 7k i − 18. (a) b × c = 1 2 −1 = −1 5 −1 8 5 8 −1 5 8 i j k 3 3 −4 0 0 −4 3 0 −4 a × (b × c) = = −7 7 i − 21 7 j + 21 −7 k 21 −7 7 = −28i − 105j − 21k (b) a · c = (3i − 4k) · (−i + 5j + 8k) = −35; (a · c)b = −35(i + 2j − k) = −35i − 70j + 35k a · b = (3i − 4k) · (i + 2j − k) = 7; (a · b)c = 7(−i + 5j + 8k) = −7i + 35j + 56k a×(b×c) = (a·c)b−(a·b)c = (−35i−70j+35k)−(−7i+35j+56k) = −28i−105j−21k 19. (2i) × j = 2(i × j) = 2k 20. i × (−3k) = −3(i × k) = −3(−j) = 3j 21. k × (2i − j) = k × (2i) + k × (−j) = 2(k × i) − (k × j) = 2j − (−i) = i + 2j 22. i × (j × k) = i × i = 0 23. [(2k) × (3j)] × (4j) = [2 · 3(k × j) × (4j)] = 6(−i) × 4j = (−6)(4)(i × j) = −24k 24. (2i − j + 5k) × i = (2i × i) + (−j × i) + (5k × i) = 2(i × i) + (i × j) + 5(k × i) = 5j + k 25. (i + j) × (i + 5k) = [(i + j) × i] + [(i + j) × 5k] = (i × i) + (j × i) + (i × 5k) + (j × 5k) = −k + 5(−j) + 5i = 5i − 5j − k 26. i × k − 2(j × i) = −j − 2(−k) = −j + 2k

20

CHAPTER 11. VECTORS AND 3-SPACE

27. k · (j × k) = k · i = 0 28. i · [j × (−k)] = i · [−(j × k)] = i · (−i) = −(i · i) = −1 √ 29. |4j − 5(i × j)| = |4j − 5k| = 41 30. (i × j) · (3j × i) = k · (−3k) = −3(k · k) = −3 31. i × (i × j) = i × k = −j

32. (i × j) × i = k × i = j

33. (i × i) × j = 0 × j = 0

34. (i · i)(i × j) = 1(k) = k

35. 2j · [i × (j − 3k)] = 2j · [(i × j) + (i × (−3k)] = 2j · [k + 3(k × i)] = 2j · (k + 3j) = 2j · k + 2j · 3j = 2(j · k) + 6(j · j) = 2(0) + 6(1) = 6 36. (i × k) × (j × i) = (−j) × (−k) = (−1)(−1)(j × k) = j × k = i 37. a × (3b) = 3(a × b) = 3(4i − 3j + 6k) = 12i − 9j + 18k 38. b × a = −a × b = −(a × b) = −4i + 3j − 6k 39. (−a) × b = −(a × b) = −4i + 3j − 6k p √ 40. |a × b| = 42 + (−3)2 + 62 = 61 i j k 4 −3 6 41. (a × b) × c = 4 −3 6 = i− 2 4 −1 2 4 −1

4 6 j + 2 −1

−3 k = −21i + 16j + 22k 4

42. (a × b) · c = 4(2) + (−3)4 + 6(−1) = −10 43. a · (b × c) = (a × b) · c = 4(2) + (−3)4 + 6(−1) = −10 44. (4a) · (b × c) = (4a × b) · c = 16(2) + (−12)4 + 24(−1) = −40 −−→ 45. (a) Let A = (1, 3, 0), B = (2, 0, 0), C = (0, 0, 4), and D = (1, −3, 4). Then AB = i − 3j, −→ −−→ −−→ −−→ −−→ AC = −i − 3j + 4k, CD = i − 3j, and BD = −i − 3j + 4k. Since AB = CD and −→ −−→ AC = BD, the quadrilateral is a parallelogram. (b) Computing i −−→ −→ AB × AC = 1 −1

k 0 = −12i − 4j − 6k 4 √ we find that the area is | − 12i − 4j − 6k| = 144 + 16 + 36 = 14. j −3 −3

46. (a) Let A = (3, 4, 1), B = (−1, 4, 2), C = (2, 0, 2), and D = (−2, 0, 3). Then −−→ −→ −−→ −−→ AB = −4i + k, AC = −i − 4j + k, CD = −4i + k, and BD = −i − 4j + k. Since −−→ −−→ −→ −−→ AB = CD and AC = BD, the quadrilateral is a parallelogram.

11.4. CROSS PRODUCT

21

(b) Computing j k 0 1 = 4i + 3j + 16k −4 1 √ √ we find that the area is |4i + 3j + 16k| = 16 + 9 + 256 = 281 ≈ 16.76. i −−→ −→ AB × AC = −4 −1

−−−→ −−−→ 47. P1 P2 = j; P2 P3 = −j + k i j k −−−→ −−−→ P 1 P2 × P2 P 3 = 0 1 0 0 −1 1 A = 21 |i| = 12 sq. unit

1 = −1

0 0 i − 0 1

−−−→ −−−→ 48. P1 P2 = j + 2k; P2 P3 = 2i + j − 2k i j k 1 2 −−−→ −−−→ P1 P2 × P2 P3 = 0 1 2 = 1 −2 2 1 −2 A = 21 | − 4i + 4j − 2k| = 3 sq. units −−−→ −−−→ 49. P1 P2 = −3j − k; P2 P3 = −2i − k i j k −−−→ −−−→ −3 P1 P2 × P2 P3 = 0 −3 −1 = 0 −2 0 −1 A = 21 |3i + 2j − 6k| = −−−→ 50. P1 P2 = −i + 3k; −−−→ −−−→ P 1 P2 × P2 P3 =

7 2

i − 0 2

0 0 j + 0 1

0 2 j + 2 −2

−1 0 i− −1 −2

1 k=i −1

1 k = −4i + 4j − 2k 1

−3 k = 3i+2j−6k 0

−1 0 j+ −1 −2

sq. units

−−−→ P2 P3 = 2i + 4j − k i j k 0 −1 0 3 = 4 2 4 −1

A = 12 | − 12i + 5j − 4k| =

√ 185 2

−1 3 i − 2 −1

−1 3 j + 2 −1

0 4

k = −12i + 5j − 4k

sq. units

i j k −1 0 4 0 j + −1 4 k = 8i + 2j − 10k i − 51. b × c = −1 4 0 = 2 2 2 2 2 2 2 2 2 v = |a · (b × c)| = |(i + j) · (8i + 2j − 10k)| = |8 + 2 + 0| = 10 cu. units i j k 4 1 i − 1 1 j + 1 4 k = 19i − 4j − 3k 52. b × c = 1 4 1 = 1 5 1 5 1 1 1 1 5 v = |a · (b × c)| = |(3i + j + k) · (19i − 4j − 3k)| = |57 − 4 − 3| = 50 cu. units i j k −2 −6 −2 6 6 −6 k = 21i − 14j − 21k i− 5 53. b × c = −2 6 −6 = 1 j + 5 3 3 12 2 2 2 5 3 1 2 2 a · (b × c) = (4i + 6j) · (21i − 14j − 21k) = 84 − 84 + 0 = 0. The vectors are coplanar.

22

CHAPTER 11. VECTORS AND 3-SPACE

i j k −2 1 1 1 j + −2 13 k = − 7 i − 4j − 3k i − 54. b × c = −2 1 1 = 3 2 0 −2 0 −2 3 2 2 0 −2 2 a · (b × c) = (i + 2j − 4k) · (− 27 i − 4j − 3k) = − 27 − 8 + 12 = 0. The vectors are not coplanar. −−−→ −−−→ 55. The four points will be coplanar if the three vectors P1 P2 = h3, −1, −1i, P2 P3 = h−3, −5, 13i, −−−→ and P3 P4 = h−8, 7, −6i are coplanar. i j k −−−→ −−−→ −5 13 −3 13 −3 −5 k = h−61, −122, −61i j+ i− P2 P3 ×P3 P4 = −3 −5 13 = 7 −6 −8 −6 −8 7 −8 7 −6 −−−→ −−−→ −−−→ P1 P2 · (P2 P3 × P3 P4 ) = h3, −1, −1i · h−61, −122, −61i = −183 + 122 + 61 = 0 The four points are coplanar. −−−→ −−−→ 56. The four points will be coplanar if the three vectors P1 P2 = h−3, 3, −1i, P2 P3 = h1, 2, −6i, −−−→ and P3 P4 = h4, −6, 5i are coplanar. i j k −−−→ −−−→ 2 −6 1 −6 1 2 k = h−26, −29, −14i j+ i− P2 P3 × P3 P4 = 1 2 −6 = −6 5 4 5 4 −6 4 −6 5 −−−→ −−−→ −−−→ P1 P2 · (P2 P3 × P3 P4 ) = h−3, 3, −1i · h−26, −29, −14i = 78 − 87 + 14 = 5 The four points are not coplanar. 57. (a) Since θ = 90◦ , |a × b| = |a||b|| sin 90◦ | = 6.4(5) = 32. (b) The direction of a × b is into the fourth quadrant of the xy-plane or to the left of the plane determined by a and b as shown in Figure 11.4.9 in the text. It makes an angle of 30◦ with the positive x-axis. √ √ (c) We identify n = ( 3i − j)/2. Then a × b = 32n = 16 3i − 16j. √ √ √ 58. Using Definition 11.4, a × b = 27(8) sin 120◦ n = 24 3( 3/2)n = 36n. By the right-hand rule, n = j or n = −j. Thus, a × b = 36j or −36j. i j k 5 6 i − 4 6 j + 4 5 k = h−30, 30, −3i 59. b × c = 4 5 6 = 7 8 7 3 8 3 7 8 3 i j k 1 3 i − 1 3 j + 1 2 k = h−3, 6, −3i a × b = 1 2 3 = 5 6 4 6 4 5 4 5 6 i j k 2 3 1 3 1 2 2 3 = a×(b×c) = 1 i− j+ k = h−96, −93, 96i 30 −3 −33 −3 −30 30 −33 30 −3 i j k −3 −3 −3 6 6 −3 k = h42, −12, −66i (a × b) × c = −3 6 −3 = i− j+ 8 3 7 3 7 8 7 8 3 Therefore, a × (b × c) 6= (a × b) × c

11.4. CROSS PRODUCT

23

60. For the following, let a = ha1 , a2 , a3 i, b = hb1 , b2 , b3 i, and c = hc1 , c2 , c3 i be arbitrary vectors. Proof of (iv): (a + b) × c = −c × (ab )

by property (ii)

= (−c × a) + (−c × b) = (a × c) + (b × c) Proof of (v): Let k be any scalar. Then i j k a2 a3 a × (kb) = a1 kb1 kb2 kb3

by property (iii)

by property (ii)



= (a2 kb3 − a3 kb2 )i − (a1 kb3 − a3 kb1 )j + (a1 kb2 − a3 kb1 )k = k(a2 b3 − a3 b2 )i − k(a1 b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k i j k (ka) × b = ka1 ka2 ka3 b1 b2 b3 = (ka2 b3 − ka3 b2 )i − (ka1 b3 − ka3 b1 )j + (ka1 b2 − ka2 b1 )k = k(a2 b3 − a3 b2 )i − k(a − 1b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k Using Definition 11.4.1, we have k(a × b) = k[(a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k] = k(a2 b3 − a3 b2 )i − k(a1 b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k Therefore, a × (kb) = (ka) × b = k(a × b). Proof of (vii): From Equation 11.4.7, we have a1 a · (a × b) = a1 b1

a2 a2 b2

a3 a3 b3



However, using Property (ii) of determinants in Appendix I in the text, we see that a·(a×b = 0.) Proof of (viii): From Equation 11.4.7, we have b1 b2 b3 b · (a × b) = a1 a2 a3 b1 b2 b3 However, using Property (ii) of determinants in Appendix I in the text, we see that b × (a × b) = 0. 61. Using equation 9 in the text, a1 a2 a · (b × c) = b1 b2 c1 c2

a3 b3 c3

c1 and (a × b) · c = c · (a × b) = a1 b1

c2 a2 b2

c3 a3 b3

.

24

CHAPTER 11. VECTORS AND 3-SPACE The second determinant can be obtained from the first by an interchange of the second and third rows followed by an interchange of the new first and second rows. Using Property (iii) of determinates in Appendix I in the text, we see that a · (b × c) = (a × b) · c.

62. b × c = (b2 c3 − b3 c2 )i − (b1 c3 − b3 c1 )j + (b1 c2 − b2 c1 )k a × (b × c) = [a2 (b1 c2 − b2 c1 ) + a3 (b1 c3 − b3 c1 )]i − [a1 (b1 c2 − b2 c1 ) − a3 (b2 c3 − b3 c2 )]j + [−a1 (b1 c3 − b3 c1 ) − a2 (b2 c3 − b3 c2 )]k = (a2 b1 c2 − a2 b2 c1 + a3 b3 c1 )i − (a1 b1 c2 − a2 b2 c1 − a3 b2 c3 + a3 b3 c2 )j − (a1 b1 c3 − a1 b3 c1 + a2 b2 c3 − a2 b3 c2 )k (a × c)b − (a · b)c = (a1 c1 + a2 c2 + a3 c3 )(b1 i + b2 j + b3 k) − (a1 b1 + a2 b2 + a3 b3 )(c1 i + c2 j + c3 k) = (a2 b1 c2 − a2 b2 c1 + a3 b1 c3 − a3 b3 c1 )i − (a1 b1 c2 − a2 b2 c1 − a3 b2 c3 + a3 b3 c2 )j − (a1 b1 c3 − a1 b3 c1 + a2 b2 c3 − a2 b3 c2 )k 63. a × (b × c) + b × (c × a) + c × (a × b) = (a · c)b − (a · b)c − (b · c)a + (c · b)a − (c · a)b = [(a · c)b − (c · a)b] + [(b · a)c − (a · b)c] + [(c · b)a − (b · c)a] = 0 64. If either a, b or c is the zero vector, the result is trivial. Therefore, assume all three are nonzero. If b is a scalar multiple of c, then b × c = 0 and a · (b × c) = a · 0 = 0 If b is not a scalar multiple of c, then b × c is orthogonal to the plane containing b and c. This implies b × c is orthogonal to a since a lies in the same plane as b and c. Hence, by Theorem 11.3.3, we have a · (b × c) = 0. 65. (a) We first note that a × b = k, b × c = 21 (i − k), c × a = 1 1 1 2 , b · (c × a) = 2 , and c · (a × b) = 2 . Then A=

1 2 (i

− k) 1 2

= i − k, B =

1 2 (j

− k) 1 2

= j − k,

1 2 (j

− k), a · (b × c) =

and C =

k 1 2

= 2k.

(b) We need to compute A · (B × C). Using the formula from Problem 62 we have (c × a) × (a × b) ([c × a) · b]a − [(c × a) · a]b = [b · (c × a)][c × (a × b)] [b · (c × a)][c × (a × b)] a = since (c × a) · a = 0. c · (a × b

B×C=

Then A · (B × C) =

b×c a 1 · = a · (b × c) c · (a × b) c · (a × b)

and the volume of the unit cell of the reciprocal lattice is the reciprocal of the volume of the unit cell of the original lattice.

11.5

Lines in 3-Space

1. hx, y, zi = h4, 6, −7i + th3, 12 , − 32 i

11.5. LINES IN 3-SPACE

25

2. hx, y, zi = h1, 8, −2i + th−7, −7, 0i 3. hx, y, zi = h0, 0, 0i + th5, 9, 4i 4. hx, y, zi = h0, −3, 10i + th12, −5, −6i −−→ −−→ The equation of a line through P1 and P2 is 3-space with r1 = OP1 and r2 = OP2 can be expressed as r = r1 + t(ka) or r = r2 + t(ka) where a = r2 − r1 and k is any non-zero scalar. Thus, the form of the equation of a line is not unique. (See the alternative solution to Problem 5.) 5. a = h1 − 3, 2 − 5, 1 − (−2)i = h−2, −3, 3i;

hx, y, zi = h1, 2, 1i + th−2, −3, 3i

Alternate Solution: a = h−31, 5 − 2, −2 − 1i = h2, 3, −3i; 6. a = h0 − (−2), 4 − 6, 5 − 3i = h2, −2, 2i;

hx, y, zi = h3, 5, −2i + th2, 3, −3i

hx, y, zi = h0, 4, 5i + th−2, −2, 2i

7. a = h1/2 − (−3/2), −1/2 − 5/2, 1 − (−1/2)i = h2, −3, 3/2i; hx, y, zi = h1/2, −1/2, 1i + th2, −3, 3/2i 8. a = h10 − 5, 2 − (−3), 10 − 5i = h5, 5, −15i;

hx, y, zi = h10, 2, −10i + th5, 5, −15i

9. a = h1 − (−4), 1 − 1, −1 − (−1)i = h5, 0, 0i;

hx, y, zi = h1, 1, −1i + th5, 0, 0i

10. a = h3 − 5/2, 2 − 1, 1 − (−2)i = h1/2, 1, 3i;

hx, y, zi = h3, 2, 1i + th1/2, 1, 3i

11. a = h2 − 6, 3 − (−1), 5 − 8i = h−4, 4, −3i; x = 2 − 4t, y = 3 + 4t, z = 5 − 3t 12. a = h2 − 0, 0 − 4, 0 − 9i = h2, −4, −9i; x = 2 + 2t, y = −4t, z = −9t 13. a = h1 − 3, 0 − (−2), 0 − (−7)i = h−2, 2, 7i; x = 1 − 2t, y = 2t, z = 7t 14. a = h0 − (−2), 0 − 4, 5 − 0i = h2, −4, 5i; x = 2t, y = −4t, z = 5 + 5t 15. a = h4−(−)6, 1/2−(−1/4), 1/3−1/6i = h10, 3/4, 1/6i; x = 4+10t, y =

1 3 1 1 + t, z = + t 2 4 3 6

16. a = h−3 − 4, 7 − (−8), 9 − 1(−1)i = h−7, 15, 10i; x = −3 − 7t, y = 7 + 15t, z = 9 + 10t 17. a1 = 10 − 1 = 9, a2 = 14 − 4 = 10, a3 = −2 − (−9) = 7;

x − 10 y − 14 z+2 = = 9 10 7

18. a1 = 1 − 2/3 = 1/3, a2 = 3 − 0 = 3, a3 = 1/4 − (1/4) = 1/2; 19. a1 = −7 − 4 = −11, a2 = 2 − 2 = 0, a3 = 5 − 1 = 4; 20. a1 = 1 − (−5) = 6, a2 = 1 − (−2), a3 = 2 − (−4);

x−1 y−3 z − 1/4 = = 1/3 3 1/2

x−7 z−5 = , y=2 −11 4

x−1 y−1 z−2 = = 6 3 6

21. a1 = 5 − 5 = 0, a2 = 10 − 1 = 9, a3 = −2 − (−14) = 12; x = 5,

y − 10 z+2 = 9 12

26

CHAPTER 11. VECTORS AND 3-SPACE

22. a1 = 5/6 − 1/3 = 1/2, a2 = −1/4 − 3/8 = 5/8, a3 = 1/5 − 1/10 = 1/10; x − 5/6 y + 1/4 z − 1/5 = = 1/2 −5/8 1/10 23. Writing the given line in the form x/2 = (y − 1)/(−3) = (z − 5)/6, we see that a direction vector is h2, −3, 6i. Parametric equations for the lines are x = 6+2t, y = 4−3t, z = −2+6t. 24. A direction vector is h5, 1/3, −2i. Symmetric equations for the line are (x − 4)/5 = (y + 11)(1/3) = (z + 7)/(−2). 25. A direction vector parallel to both the xy- and xy-planes is i = h1, 0, 0i. Parametric equations for the line are x = 2 + t, y = −2, z = 15. 26. (a) Since the unit vector j = h0, 1, 0i lies along the y-axis, we have x = 1, y = 2 + t, z = 8. (b) Since the unit vector k = h0, 0, 1i is perpendicular to the xy-plane, we have z = 1, y = 2, z = 8 + t. 27. Both lines go through the points (0, 0, 0) and (6, 6, 6). Since two points determine a line, the lines are the same. 28. The direction vector of line L1 is v1 = h3, 6, −9i. The direction vector of line L2 is v2 = h−1, −2, 3i. Since v1 = −3v2 , lines L1 and L2 are parallel. Hence, if we can find a point that lies on both lines, then they must be parallel. Letting t = 0 for L1 and t = 3 for L2 , we see that the point (2, −5, 4) lies on both lines. Therefore L1 and L2 are the same. −7 − 3 = −5. 29. (a) Equating the x components, we have x = 3 + 2t, = −7, which gives t = 2 We can check our work by plugging this value of t into the y and z components to get y = 4 − (−5) = 9 and z = −1 + 6(−5) = −31 (b) Equating the x components, we have x = 5 − x = −7 which gives s = 5 + 7 = 12. We can check our work by plugging this value of s into the y and z components to get 1 y = 3 + (12) = 9 and z = 5 − 3(12) = −31 2 30. a and f are parallel since h9, −12, 6i = −3h−3, 4, 2i. c and d are orthogonal since h2, −3, 4i · h1, 4, 5/2i = 0. 31. In the xy-plane, z = 9+3t = 0 and t = −3. Then x = 4−2(−3) = 10 and y = 1+2(−3) = −5. The point is (10, −5, 0). In the xz-plane, y = 1+2t = 0 and t = −1/2. Then x = 4−2(−1/2) = 5 and z = 9 + 3(−1/2) = 15/2. The point is (5, 0, 15/2). In the yz-plane, x = 4 − 2t = 0 and t = 2. Then y = 1 + 2(2) = 5 and z = 9 + 3(2) = 15. The point is (0, 5, 15). 32. The parametric equations for the line are x = 1+2t, y = −2+3t, z = 4+2t. In the xy-plane, z = 4 + 2t = 0 and t = −2. Then x = 1 + 2(−2) = −3 and y = −2 + 3(−2) = −8. The point is (−3, −8, 0). In the xz-plane, y = −2 + 3t = 0 and t = 2/3. Then x = 1 + 2(2/3) = 7/3 and z = 4 + 2(2/3) = 16/3. The point is (7/3, 0, 16/3). In the yz-plane, x = 1 + 2t = 0 and t = −1/2. Then y = −2 + 3(−1/2) = −7/2 and z = 4 + 2(−1/2) = 3. The point is (0, −7/2, 3).

11.5. LINES IN 3-SPACE

27

33. Solving the system 4 + t = 6 + 2s, 5 + t = 11 + 4s, −1 + 2t = −3 + s, or t − 2s = 2, t − 4s = 6, 2t − s = −2 yields s = −2 and t = −2 in all three equations. Thus, the lines intersect at the point x = 4 + (−2) = 2, y = 5 + (−2) = 3, z = −1 + 2(−2) = −5, or (2, 3, −5). 34. Solving the system 1 + t = 2 − s, 2 − t = 1 + s, 3t = 6s, or t + s = 1, t + s = 1, t − 2s = 0 yields s = 1/3 and t = 2/3 in all three equations. Thus, the lines intersect at the point x = 1 + 2/3 = 5/3, y = 2 − 2/3 = 4/3, z = 3(2/3) = 2, or (5/3, 4/3, 2). 35. The system of equations 2 − t = 4 + s, 3 + t = 1 + s, 1 + t = 1 − s, or t + s = −2, t − s = −2, t + s = 0 has no solution since −2 6= 0. Thus, the lines do not intersect. 36. Solving the system 3 + t = 2 + 2s, 2 + t = −2 + 3s, 8 + 2t = −2 + 8s, or t + 2s = 1, t − 3s = −4, 2t − 8s = −10 yields s = 1 and t = −1 in all three equations. Thus, the lines intersect at the point x = 3 − (−1) = 4, y = 2 + (−1) = 1, z = 8 + 2(−1) = 6, or (4, 1, 6). 37. Using the first two points, we determine the line x = 4 + 6t, y = 3 + 12t, z = −5 − 6t. Letting t = −5/6 we see that (−1, −7, 0) is on the line. Thus, the points lie on the same line. 38. Using the first two points, we determine the line x = −1 − 12t, y = 6 + 4t, z = 6 − 8t.  6 5 when Setting x = −2 in the first equation, we obtain t = 1/4. Since z = 6 − 8 41 = 4 = t = 1/4, the points do not lie on the same line. 39. A direction vector for the line is h6−2, −1−5, 3−9i = h4, −6, −6i. Thus, parametric equations for the line segment are x = 2 + 4t, y = 5 − 6t, z = 9 − 6t, where 0 ≤ t ≤ 1. 40. The midpoint of the first line segment, obtained by letting t = 3/2, is (4, 1/2, −1/2). The midpoint of the second line segment, obtained by letting t = 0, is (−2, 6, 5). A direction vector for the line segment connecting the midpoints is h−2−4, 6−1/2, 5−(−1/2)i = h−6, 11/2, 11/2i. Thus, parametric equations for the line segment are x = 4 − 6t, y = 1/2 + (11/2)t, z = −1/2 + (11/2)t, where 0 ≤ t ≤ 1. 41. a = h−1, 2, −2i, θ = arccos

b = h2, 3, −6i,

a · b = 16,

|a| = 3,

|b| = 7;

cos θ =

a·b 16 = ; |a||b| 3·7

16 ≈ 40.37◦ 21

√ √ 42. a = h2, 7, −1i, b = h−2, 1, 4i, a · b = −1, |a| = 3  6, |b| = 21;  a·b 1 1 1 cos θ = =− √ √ = − √ ; θ = arccos − √ √ ≈ 91.70◦ |a||b| (3 6)( 21) 9 14 (3 6)( 21) 43. A direction vector perpendicular to the given lines will be h1, 1, 1i × h−1, 1, −5i = h−6, 3, 3i. Equations of the lines are x = 4 − 6t, y = 1 + 3t, z = 6 + 3t. 44. The direction vectors of the given lines are h3, 2, 4i and h6, 4, 8i = 2h3, 2, 4i. These are parallel, so we need a third vector parallel to the plane containing the lines which is not parallel to them. The point (1, −1, 0) is on the first line and (−4, 6, 10) is on the second line. A third vector is then h1, −1, 0i − h−4, 6, 10i = h5, −7, −10i. Now a direction vector perpendicular to the plane is h3, 2, 4i × h5, −, 7−, 10i = h8, 50, −31i. Equations of the line through (1, −1, 0) and perpendicular to the plane are x = 1 + 8t, y = −1 + 50t, z = −31t.

28

CHAPTER 11. VECTORS AND 3-SPACE

45. In the system −3 + t = 4 + s, 7 + 3t = 8 − 2s, 5 + 2t = 10 − 4s, or t − s = 7, 3t + 2s = 1, 2t + 4s = 5, the first and second equations have solution t = 3 and s = −4. Substituting into the third equation, we find 2(3) = 4(−4) = 6 − 16 = −10 6= 5. The direction vectors of the lines are h1, 3, 2i and h1, −2, −4i, so the lines are not parallel. Thus, the lines are skew. 46. In the system 6 + 2t = 7 + 8s, 6t = 4 − 4s, −8 + 10t = 3 − 24s, or 2t − 8s = 1, 6t + 4s = 4, 10t + 24s = 11, the second and third equations have solution t = 1/2 and s = 1/4. Substituting into the first equation, we find 2(1/2) − 8(1/4) = −1 6= 1. The direction vectors of the lines are h2, 6, 10i and h8, −4, , −24i, so the lines are not parallel. Thus, the lines are skew. −−−→ −−−→ −−−→ −−−→ 47. The vector (P1 P2 × P3 P4 )/|P1 P2 × P3 P4 | is a unit vector perpendicular to the two planes. To find the shortest distance between the planes we compute the absolute value of the component −−−→ of P1 P3 on this unit vector. Then −−−→ d = P1 P 3 ·

−−−→ −−−→ −−−→ −−−→ −−−→ P1 P2 × P3 P4 |P1 P3 · P1 P2 × P3 P4 | −−−→ −−−→ = −−−→ −−−→ | P1 P2 × P3 P4 | | P1 P2 × P3 P4 |

−−−→ 48. We take P1 = (−3, 7, 5), P2 = (−2, 10, 7), P3 + (4, 8, 10), and P4 = (5, 6, 6). Then P1 P3 = −−−→ −−−→ −−−→ −−−→ h7, 1, 5i, P1 P2 = h1, 3, 2i, P3 P4 = h1, −2, −4i, and P1 P2 × P3 P4 = h−8, 6, −5i. The distance between the lines is then d=

11.6

√ |h7, 1, 5i · h−8, 6, −5i| 75 = √ = 3 5. |h−8, 6, −5i| 5 5

Planes

1. 2(x − 5) − 3(y − 1) + 4(z − 3) = 0;

2x − 3y + 4z = 19

2. 4(x − 1) − 2(y − 2) + 0(z − 5) = 0;

4x − 2y = 0

3. −5(x − 6) + 0(y − 10) + 3(z + 7) = 0;

−5x + 3z = −51

4. 6x − y + 3z = 0 5. 6(x − 1/2) + 8(y − 3/4) − 4(z − 1/2) = 0; 6. −(x + 1) + (y − 1) − (z − 0) = 0;

6x + 8y − 4z = 11

−x + y − z = 2

7. From the points (3, 5, 2) and (2, 3, 1) we obtain the vector u = i + 2j + k. From the points (2, 3, 1) and (−1, −1, 4) we obtain the vector v = 3i + 4j − 3k. From the points (−1, −1, 4) and (x, y, z) we obtain the vector w = (x + 1)i + (y + 1)j + (z − 4)k. Then, a normal vector is i u × v = 1 3

j 2 4

k 1 −3

= −10i + 6j − 2k

A vector equation of the plane is −10(x + 1) + 6(y + 1) − 2(z − 4) = 0 or 5x − 3y + z = 2.

11.6. PLANES

29

8. From the points (0, 1, 0) and (0, 1, 1) we obtain the vector u = k. From the points (0, 1, 1) and (1, 3, −1) we obtain the vector v = i + 2j − 2k. From the points (1, 3, −1) and (x, y, z) we obtain the vector w = (x − 1)i + (y − 3)j + (z + 1)k. Then, a normal vector is i u × v = 0 1

j 0 2

k 1 −2

= −2i + j

A vector equation of the plane is −2(x − 1) + (y − 3) + 0(z + 1) = 0 or −2x + y = 1. 9. From the points (0, 0, 0) and (1, 1, 1) we obtain the vector u = i + j + k. From the points (1, 1, 1) and (3, 2, −1) we obtain the vector v = 2i + j − 2k. From the points (3, 2, −1) and (x, y, z) we obtain the vector w = (x − 3)i + (y − 2)j + (z + 1)k. Then, a normal vector is i u × v = 1 2

j 1 1

k 1 −2

= −3i + 4j − k

A vector equation of the plane is −3(x − 3) + 4(y − 2) − (z + 1) = 0 or −3x + 4y − z = 0. 10. The three points are not collinear and all satisfy x = 0, which is the equation of the plane. 11. From the points (1, 2, −1) and (4, 3, 1) we obtain the vector u = 3i + j + 2k. From the points (4, 3, 1) and (7, 4, 3) we obtain the vector v = 3i + j + 2k. From the points (7, 4, 3) and (x, y, z) we obtain the vector w = (x − 7)i + (y − 4)j + (z − 3)k. Since u × v = 0, the points are collinear. 12. From the points (2, 1, 2) and (4, 1, 0) we obtain the vector u = 2i−2k. From the points (4, 1, 0) and (5, 0, −5) we obtain the vector v = i − j − 5k. From the points (5, 0, −5) and (x, y, z) we obtain the vector w = (x − 5)i + yj + (z + 5)k. Then, a normal vector is i u × v = 2 1

j 0 −1

k −2 −5

= −2i + 8j − 2k

A vector equation of the plane is −2(x − 5) + 8y − 2(z + 5) = 0 or x − 4y = z = 0. 13. A normal vector to x + y − 4z = 1 is h1, 1, −4i. The equation of the parallel plane is (x − 2) + (y − 3) − 4(z + 5) = 0 or x + y − 4z = 25. 14. A normal vector to 5x − y + z = 6 is h5, −1, 1i. The equation of the parallel plane is 5(x − 0) − (y − 0) + (z − 0) = 0 or 5x − y + z = 0. 15. A normal vector to the xy-plane is h0, 0, 1i. The equation of the parallel plane is z − 12 = 0 or z = 12. 16. A normal vector is h0, 1, 0i. The equation of the plane is y + 5 = 0 or y = −5.

30

CHAPTER 11. VECTORS AND 3-SPACE

17. Direction vectors of the lines are h3, −1, 1i and h4, 2, 1i. A normal vector to the plane is h3, −1, 1i × h4, 2, 1i = h−3, 1, 10i. A point on the first line, and thus in the plane, is (1, 1, 2). The equation of the plane is −3(x − 1) + (y − 1) + 10(z − 2) = 0 or −3x + y + 10z = 18. 18. Direction vectors of the lines are h2, −1, 6i and h1, 1, −3i. A normal vector to the plane is h2, −1, 6i×h1, 1, −3i = h−3, 12, 3i. A point on the first line, and thus in the plane, is (1, −1, 5). The equation of the plane is −3(x − 1) + 12(y + 1) + 3(z − 5) = 0 or −x + 4y + z = 0. 19. A direction vector for the two lines is h1, 2, 1i. Points on the lines are (1, 1, 3) and (3, 0, −2). Thus, another vector parallel to the plane is h1 − 3, 1 − 0, 3 + 2i = h−2, 1, 5i. A normal vector to the plane is h1, 2, 1i × h−2, 1, 5i = h9, −7, 5i. Using the point (3, 0, −2) in the plane, the equation of the plane is 9(x − 3) − 7(y − 0) + 5(z + 2) = 0 or 9x − 7y + 5z = 17. 20. A direction vector for the line is h3, 2, −1i. Letting t = 0, we see that the origin is on the line and hence in the plane. Thus, another vector parallel to the plane is h4 − 0, 0 − 0, −6 − 0i = h4, 0, −6i. A normal vector to the plane is h3, 2, −2i × h4, 0, −6i = h−12, 10, −8i. The equation of the plane is −12(x − 0) + 10(y − 0) − 8(z − 0) = 0 or 6x − 5y + 4z = 0. 21. A direction vector for the line, and hence a normal vector for the plane, is h−3, 1, −1/2i. The equation of the plane is −3(x − 2) + (y − 4) − 21 (z − 8) = 0 or −3x + y − 21 z = −6. 22. A normal vector to the plane is h2 − 1, 6 − 0, −3 + 2i = h1, 6, −1i. The equation of the plane is (x − 1) + 6(y − 1) − (z − 1) = 0 or x + 6y − z = 6. 23. Normal vectors to the plane are (a) h2, −1, 3i, (b) h1, 2, 2i, (c) h1, 1, −3/2i, (d) h−5, 2, 4i, (e) h−8, −8, 12i, (f ) h−2, 1, −2i. Parallel planes are (c) and (e), and (a) and (f ). Perpendicular planes are (a) and (d), (b) and (c), (b) and (e), and (d) and (f ). 24. A normal vector to the plane is h−7, 2, 3i. This is the direction vector for the line and the equations of the line are x − 4 − 7t, y = 1 + 2t, z = 7 + 3t. 25. A direction vector of the line is h−6, 9, 3i, and the normal vectors of the plane are (a) h4, 1, 2i, (b) h2, −3, 1i, (c) h10, −15, −5i, (d) h−4, 6, 2i. Vectors (c) and (d) are multiples of the direction vector and hence the corresponding planes are perpendicular to the line. 26. A direction vector of the line is h−2, 4, 1i, and the normal vectors of the plane are (a) h1, −1, 3i, (b) h6, −3, 0i, (c) h1, −2, 5i, (d) h−2, 1, −2i. Since the dot product of each normal vector with the direction vector is non-zero, none of the planes are parallel to the line. 27. Letting z = t in both equations and solving 5x − 4y = 8 + 9t, x = 2 + t, y = 21 − t, z = t.

x + 4y = 4 − 3t, we obtain

28. Letting y = t in both equations and solving x − z = 2 − 2t, 3x + 2z = 1 + t, we obtain x = 1 − 53 t, y = t, z = −1 + 57 t or, letting t = 5s, x = 1 − 3s, y = 5s, z = −1 + 7s. 29. Letting z = t in both equations and solving 4x − 2y = 1 + t, x = 12 − 12 t, y = 12 − 32 t, z = t.

x + y = 1 − 2t, we obtain

30. Letting z = t and using y = 0 in the first equation, we obtain x = − 21 t, y = 0, z = t.

11.6. PLANES

31

31. Substituting the parametric equations into the equation of the plane, we obtain 2(1+2t)23(2− t) + 2(−3t) = −7 or t = −3. Letting t = −3 in the equation of the line, we obtain the point of intersection (−5, 5, 9). 32. Substituting the parametric equations into the equation of the plane, we obtain (3 − 2t) + (1 + 6t) − 4(2 − 21 ) = 12 or 2t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (3, 1, 2). 33. Substituting the parametric equations into the equation of the plane, we obtain 1+2−(1+t) = 8 or t = −6. Letting t = −6 in the equation of the line, we obtain the point of intersection (1, 2, −5). 34. Substituting the parametric equations into the equation of the plane, we obtain 4 + t − 3(2 + t) + 2(1 + 5t) = 0 or t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (4, 2, 1). In Problems 35 and 26, the cross product of the normal vectors to the two planes will be a vector parallel to both planes, and hence a direction vector for a line parallel to the two planes. 35. Normal vectors are h1, 1, −4i and h2, −1, 1i. A direction vector is h1, 1, −4i × h2, −1, 1i = h−3, −9, −3i = −3h1, 3, 1i. Equations of the line are x = 5 + t,

y = 6 + 3t,

z = −12 + t.

36. Normal vectors are h2, 0, 1i and h−1, 3, 1i. A direction vector is

h2, 0, 1i × h−1, 3, 1i = h−3, −3, 6i = −3h1, 1, −2i. Equations of the line are x = −3 + t, y = 5 + t, z = −1 − 2t. In Problems 37 and 38, the cross product of the direction vector of the line with the normal vector of the given plane will be a normal vector to the desired plane. 37. A direction vector of the line is h3, −1, 5i and a normal vector to the given plane is h1, 1, 1i. A normal vector to the desired plane is h3, −1, 5i × h1, 1, 1i = h−6, 2, 4i. A point on the line, and hence in the plane is h4, 0, 1i. The equation of the plane is −6(x − 4) + 2(y − 0) + 4(z − 1) = 0 or 3x − y − 2x = 10. 38. A direction vector of the line is h3, 5, 2i and a normal vector to the given plane is h2, −4, −1i. A normal vector to the desired plane is h−3, 5, 2i × h2, −4, −1i = h3, 1, 2i. A point on the line, and hence in the plane is h2, −2, 8i. The equation of the plane is 3(x−2)+(y +2)+2(z −8) = 0 or 3x + y + 2x = 20.

32

CHAPTER 11. VECTORS AND 3-SPACE

39.

40. z

z 10

6

y x

6

2

5

y

x

41.

42. z

z

2

4

y 6

4

y

x x

-6

43.

44. z

z

6

4 2 y

1 x

2

y x

45. (a) A direction vector for the line is a = −2i + j − k and a normal vector for the plane is n = i + j − k. Since a · n = −2 + 1 + 1 = 0, the line is perpendicular to n and thus parallel

11.6. PLANES

33

to the plane. Since (0, 0, 0) is on the line and (0, 0, −1) is in the plane, the line is above the plane. (b) A normal vector for the plane is n − 3i − 4j + 2k. Since a · n = 6 − 4 − 2 = 0, the line is parallel to the plane. Since (0, 0, 0) is one the line and (0, 0, 4) is in the plane, the line is below the plane. −−−→ 46. The distance D will be the absolute value of compn P0 P1 . Thus, using ax1 + by1 + cz1 = −d, −−−→ n hx2 − x1 , y2 − y1 , z2 − z1 i · ha, b, ci = √ D = P0 P1 · |n| a2 + b2 + c2 |ax2 + by2 + cz2 − (zx1 + by1 + cz1 )| |ax2 + by2 + cz2 + d| √ √ = = . a2 + b2 + c2 a2 + b2 + c2 47. Using, Problem 46, D =

3 |1(2) − 3(1) + 1(4) − 6| √ =√ . 1+9+1 11

48. (a) The normal vectors are n1 = i − 2j + 3k and n2 = −4i + 8j − 12k = −4n. Since n1 and n2 are parallel, the planes are parallel. (b) To find the distance between the planes we choose (0, 0, 1) on the first plane. Then, using Problem 46, the distance between the planes is D=

19 | − 4(0) + +8(0) − 12(1) − 7| p =√ ≈ 1.27. 2 2 2 224 (−4) + 8 + (−12)

49. Normal vectors are h1, −3, 2i and h−1, 1, 1i. Then cos θ =

−2 2 h1, −3, 2i · h−1, 1, 1i = √ √ = −√ |h1, −3, 2i||h−1, 1, 1i| 14 3 42

√ and θ = arccos(−2/ 42) ≈ 107.98◦ 50. Normal vectors are h, 2, 6, 3i and h4, −2, 4i. Then cos θ =

h2, 6, 3i · h4, −2, 4i 8 4 = = |h2, 6, 3i||h4, −2, 4i| 7(6) 21

and θ = arccos(−4/21) ≈ 79.02◦ 51. Let the bottoms of the table legs be represented by points in 3-space. The rocking of a fourlegged table occurs when these four points are not coplanar. Hence, not all four legs can rest on the plane of the floor simultaneously. However, a three-legged table cannot have this problem. Given any three points in space, a plane can be found passing through them. Therefore, the bottoms of the legs in a three-legged table are coplanar. This implies that they will all rest on the plane of the floor, even if the legs are of uneven lengths.

34

CHAPTER 11. VECTORS AND 3-SPACE

52. Let n1 = h1, −1, 2i which is normal to the plane x − y + 2z = 1. Let n2 = h1, 1, 1i which is normal to the plane x + y + z = 3. Since L is perpendicular to both n1 and n2 , L must be parallel to n1 × n2 . i n1 × n2 = 1 1

j −1 1

k 2 1

− 3i + j + 2k

Therefore, L is parallel to v = h−3, 1, 2i. To completely determine L, we need a point which L passes through. Hence we need a point (x, y, z) which satisfies the equations of both planes. Since it satisfies both equations, it must satisfy their sum: x − y + 2z = 1 x+y+z =3 2x + 3z = 4 So the coordinates of the point satisfy 2x + 3z = 4. Since v as a nonzero k-component, the line L passes through every possible z-value. This implies the existence of a point on the intersection of the two planes with a z-value of zero. Letting z = 0, we must have x = 2 since 2x + 3z = 4 for every point on L. Plugging x = 2 and z = 0 into the equation of the first place, we get y = 1. Therefore (2, 1, 0) lies on the line L. Using this point and the parallel vector v, the parametric equations of L are x = 2 − 3t; y = 1 + t; z = 2t To show that this answer is equivalent to that found in Example 8, first note that both lines 1 pass through (2, 1, 0). Also, the parallel vector used in Example 8 is h−3/2, 1/2, 1i = v. 2 Therefore, the two solutions are the same.

53. (a) The plane should pass through the midpoint of the −2, 3) and   line segment joining (1,  3 3 1 + 2 −2 + 5 3 − 1 , , = , ,1 . (2, 5, −2). This is given in Problem 11.2.64 as M = 2 2 2 2 2 The vector joining (1, −2, 3) and (2, 5, −1) should be perpendicular to the plane. This vector is n = h1, 7, −4i. Using the point (3/2, 3/2, 1) and the normal vector n, the equation of the plane is given by z + 7y − 4z = 8. (b) The distance from the plane to either of the two points is equal to half the length of the 1p 2 1√ line segment joining the two points. This is given by 1 + 72 + (−4)2 = 66 2 2

11.7. CYLINDERS AND SPHERES

11.7

35

Cylinders and Spheres

1.

2.

z

3.

z

z

y

y y

2

x

x

x

4.

5.

6. z

z

z

5

3

y

y y

x

x

x

7.

8.

9. z

z

z

y

y 1 y x

x x

36

CHAPTER 11. VECTORS AND 3-SPACE

10.

11.

12.

z

z

z

6 y

3 y

x

y

1 x x

13.

14.

15.

z

z

z

y

y y

x

x x

16.

17.

z

18. z

z

3 y

y x

y x

x

center: (0, 0, 3) radius: 4

11.7. CYLINDERS AND SPHERES

37

19.

20. z

z

y

x

y

x

center: (−3, −4, 5) radius: 2

center: (1, 1, 1) radius: 1

21. (x2 + 8x + 16) + (y 2 − 6x + 9) + (z 2 − 4z + 4) = 7 + 16 + 9 + 4 (x + 4)2 + (y − 3)2 + (z − 2)2 = 36; center: (−4, 3, 2); radius: 6 22. 4(x2 + x + 1/4) + 4y 2 + 4(z 2 − 3z + 9/4) = −9 + 1 + 9 (x + 1/2)2 + y 2 + (z − 3/2)2 = 1/4; center: (−1/2, 0, 3/2); 23. x2 + y 2 + (z 2 − 16z = 64) = 64; 2

center: (0, 0, 8);

2

2

24. (x − x + 1/4) + (y + y + 1/4)√+ z = 1/4 + 1/4; center: (1/2, −1/2, 0); radius: 2/2

radius: 1/2

radius: 8 (x − 1/2)2 + (y + 1/2)2 + z 2 = 1/2

25. (x + 1)2 + (y − 4)2 + (z − 6)2 = 3 26. x2 + (y − 3)2 + z 2 = 25/16 27. (x − 1)2 + (y − 1)2 + (z − 4)2 = 16 28. (x − 5)2 + (y − 2)2 + (z − 2)2 = 52 29. There are two solutions: one sphere is inside the given sphere and the other is outside. x2 + (y − 8)2 + z 2 = 4 or x2 + (y − 4)2 + z 2 = 4. p 30. (2t)2 + (3t)2 + (6t)2 = 21; t = 3; a = 2t = 6; b = 3t = 9; c = 6t = 18 (x − 6)2 + (y − 9)2 + (z − 18)2 = 25 p √ 31. The center is at (1, 4, 2) and the radius is (1 − 0)2 + (4 + 4)2 (2 − 7)2 = 3 10. The equation is (x − 1)2 + (y − 4)2 + (z − 2)2 = 90. p √ 32. The radius is (−3 − 0)2 + (1 − 0)2 + (2 − 0)2 = 14. The equation is (x + 3)2 + (y − 1)2 + (z − 2)2 = 14. 33. The upper half of the sphere x2 + y 2 + (z − 1)2 = 4; a hemisphere 34. A circle√on the sphere x2 + y 2 + (z − 1)2 = 4; the circle is parallel to the xy-plane and has radius 3. 35. All points on and outside the unit sphere centered at the origin

38

CHAPTER 11. VECTORS AND 3-SPACE

36. All points inside the sphere of radius 1 centered at (1, 2, 3), except the center 37. x2 + y 2 + z 2 = 1 represents a sphere of radius 1 and x2 + y 2 + z 2 = 9 represents a sphere of radius 3. Therefore 1 ≤ x2 + y 2 + z 2 ≤ 9 represents the set of points lying between these two spheres. Thus, the geometric object is a hollowed out ball with outer radius 3 and inner radius 1. 38. This set of points is identical to the found in Problem 11.7.37, with the added restriction z ≥ 0. This restriction will remove points with negative z-coordinates, leaving only the upper half of the hollowed out ball.

11.8

Quadric Surfaces 2. elliptical cone

1. paraboloid

z

z

y x

y x

3. x2 /4 + y 2 + z 2 /9 = 1; ellipsoid

4. −x2 /4 − y 2 /4 + z 2 /4 = 1 hyperboloid of two sheets

z z

y x

x

y

11.8. QUADRIC SURFACES

39 6. x2 /25 + y 2 /25 + z 2 /100 = 1 ellipsoid

5. x2 /4 − y 2 /144 + z 2 /16 = 1 hyperboloid of one sheet z

z 10

4

y

5

x

y

5

x

8. y 2 /9 − x2 /16 = z hyperbolic paraboloid

7. elliptical cone

z

z

y

y

x

x

10. x2 + y 2 = −9z paraboloid

9. hyperbolic paraboloid

z

z

y y

x x

40

CHAPTER 11. VECTORS AND 3-SPACE 12. −z 2 /9 + y 2 + z 2 /9 = 1 hyperboloid of one sheet

11. x2 /4 − y 2 /4 − z 2 /4 = 1 hyperboloid of two sheets

z

z

3

y

y

1

x

x

z2 =x 1/4 paraboloid

13. y 2 +

14. hyperboloid of one sheet z

z

y

y x

x

15.

16. z

z

4 x

3 y x

y

11.8. QUADRIC SURFACES

17.

41

18.

z

z

10 y y

x

x

p 19. The equation can be written as x2 + (± y 2 + z 2 )2 = 1. The surface is generated by revolving the circles x2 + y 2 = 1 or x2 + z 2 = 1 about the x-axis. [Alternatively, the surface is generated by revolving the circles x2 + y 2 = 1 or y 2 + z 2 = 1 about the y-axis, or the circles x2 + z 2 = 1 or y 2 + z 2 = 1 about the z-axis.] p 20. The equation can be written as −9x2 + (±2 y 2 + z 2 02 = 36. The surface is generated by revolving the hyperbolas = 9x2 + 4y 2 = 36 or −9x2 + 4z 2 = 36 about the x-axis. √

2

2 2

21. The equation can be written as y = e± x +z ) . The surface is generated by revolving the 2 2 curves y = ex or y = ez about the y-axis. p 22. The equation can be written as (± x2 + y 2 )2 = sin2 z. The surface is generated by revolving the curves x2 = sin2 z or y 2 = sin2 z about the z-axis. p √ 23. Replacing x by ± x2 + y 2 we have y = ±2 x2 + z 2 or y 2 = 4x2 + 4z 2 . √ √ 24. Replacing z by x2 + z 2 we have y = ( x2 + z 2 )1/2 or y 4 = x2 + z 2 ; y ≥ 0. p p 25. Replacing z by ± y 2 + z 2 we have ± y 2 + z 2 = 9 − x2 or y 2 + z 2 = (9 − x2 )2 , x ≥ 0. 26. Replacing y by

p p x2 + y 2 we have z = 1 + ( x2 + y 2 )2 or z = 1 + x2 + y 2 .

27. Replacing z by

p p y 2 + z 2 we have x2 − (± y 2 + z 2 )2 = 4 or x2 − y 2 − z 2 = 4.

p p 28. Replacing x by ± x2 + y 2 we have 3(± x2 + y 2 )2 + 4z 2 = 12 or 3x2 + 3y 2 + 4z 2 = 12. 29. Replacing y by

p

x2 + y 2 we have z = ln

p

x2 + y 2 .

p p 30. Replacing y by ± y 2 + z 2 we have x(± y 2 + z 2 ) = 1 or x2 (y 2 + z 2 ) = 1. 31. The surface is Problem 11 is a surface of revolution about the x-axis. The surface in Problem 2 is a surface of revolution about the y-axis. The surface is Problems 1, 4, 6, 10, and 14 are surfaces of revolution about the z

42

CHAPTER 11. VECTORS AND 3-SPACE

32. z

π

1

y

x

33. The first equation is the lower nappe of the cone (z + 2)2 = x2 + y 2 whose axis of revolution is the z-axis and whose vertex is at (0, 0, −2). 34. The first equation is the right-hand of the cone (y − 1)2 = x2 + z 2 whose axis of revolution is the y-axis and whose vertex is at (0, 1, 0). 2 2 2 2 35. (a) Writing the equation of the ellipse in√the form √ x /(c − z)a + y /(c − z)b = 1 we see that the area of a cross-section is πa c − zb c − z = πab(c − z). c  Rc 1 (b) V = 0 πab(c − x)dz = πab − 21 (c − z)2 0 = πabc2 2

36. (a) Using the formula for the area of an ellipse given in Problem 35(a) we see that a horizontal cross-sectional area of the ellipsoid is πab(1 − z 2 /c2 ). Then Z

c

V =2 0

z πab 1 − 2 dz = 2πab c 

  c 4 z 3 z − 2 = πabc. 3c 3 0

(b) When a = b = c the volume is 34 πa3 , which is the formula for the volume of a sphere. 37. Expressing the line in the form (x − 2)/4 = (y + 2)/(−6) = (z − 6)/3 we see that parametric equations for the line are x = 2 + 4t, y = −2 = 6t, z = 6 + 3t. Writing the equation of the ellipse as 36x2 + 9y 2 + 4z 2 = 324 and substituting, we obtain 36(2 + 4t)2 + 9(−2 − 6t)2 + 4(6 + 3t)2 = 936t2 + 936t + 324 or 936t(t + 1) = 0. When t = 0 we obtain the point (2, −2, 6), and when t = −1 we obtain the point (−2, 4, 3).

Chapter 11 in Review A. True/False 1. True 2. False; the points must be non-collinear. 3. False; since a normal to the plane is h2, 3, −4i which is not a multiple of the direction vector h5, −2, 2i of the line.

CHAPTER 11 IN REVIEW 4. True 7. True

43 5. True 8. True

6. True 9. True

10. True; since a × b and c × d are both normal to the plane and hence parallel (unless a × b = 0 or c × d.) 11. True. The normal vector of the first plane is h1, 2, −1i while the normal vector of the second plane is h−2, −4, 2i. Since the second vector is a scalar multiple of the first, the planes are parallel. 12. False. Look at Figure 11.5.3 in the text. 13. True. This is a parabolic cylinder similar to that shown in Figure 11.7.6. 14. True. In the yz-plane, we have x = 0. Therefore, the equation of the surface becomes 1 or y 2 + z 2 = 2.

y2 2

2

+ z2 =

15. False. Find the equation of the plane containing the first three points, P1 (0, 1, 2), P2 (1, −1, 1), −−−→ −−−→ and P3 (3, 2, 6). This plane must contain the vectors P1 P2 = h1, −2, 1i and P1 P3 = h3, 1, 4i. i j k −−−→ −−−→ Define n = P1 P2 × P1 P3 = 1 −2 −1 = h−7, −7, −7i. Then n must be normal to the 3 1 4 plane. Using n and the point P1 , the equation of the plane becomes −7x − 7y + 7z = 7 or z +y −z = −1. The fourth point P4 (2, 1, 2) does not lie on the plane since (2)+(1)−(2) 6= −1. 16. True 17. False. The trace in the yz-plane is described by the equation 9y 2 + z 2 = 1 which represents an ellipse. 18. True. This ellipsoid results from revolving the graph of the ellipse x2 + 9y 2 = 1 about the y-axis. 19. True. |a × b| = |a||b|| sin θ| = |a||b| since θ = 90◦ 20. False. Let a = i,

b = j, and c = k. Then a · b = a · c = 0 but b 6= c.

B. Fill in the Blanks 1. 9i + 2j + 2k 2. orthogonal 3. −5(k × j) = −5(−i) = 5i 4. i · (i × j) = i × k = 0 p (−12)2 + 42 + 62 = 14 5. 6. k × (i + 2j − 5k) = k × i + 2(k × j) − 5(k × k) = j − 2i − 5(0) = h−2, 1, 0i

44

CHAPTER 11. VECTORS AND 3-SPACE 2 7. 4

−5 = 2(3) − (−5)(4) = 6 + 20 = 26 3

8. (−1 − 20)i − (−2 − 0)j + (8 − 0)k = −21i + 2j + 8k 9. −6i + j − 7k 10. The smallest component is the j-component with magnitude 3. Therefore, the sphere cannot have a radius larger than 3 or its interior will intersect the xz-plane. Thus, we need a sphere with radius 3 and center (4, 3, 7). The equation is (x − 4)2 + (y − 3)2 + (z − 7)2 = 9 11. Writing the line in parametric form, we have x = 1 + t, y = −2 + 3t z = −1 + 2t. Substituting into the equation of the plane yields (1 + t) + 2(−2 + 3t) − (−1 + 2t) = 13 or t = 3. Thus, the point of intersection is x = 1 + 3, y = −2 + 3(3) = 7, z = −1 + 2(3) = 5, or (4, 7, 5). 12. |a| =

p √ 42 + 32 + (−5)2 = 5 2;

13. x2 − 2 = 3,

x2 = 5;

1 4 3 1 u = − √ (4i + 3j − 5k) = − √ i − √ j + √ k 5 2 5 2 5 2 2

y2 − 1 = 5,

y2 = 6;

z2 − 7 = −4,

z2 = 3;

P2 = (5, 6, 3)

14. (5, 1/2, 5/2) √ 15. (7.2)(10) cos 135◦ = −36 2 16. 2b = h−2, 4, 2i;

4c = h0, −8, 8i;

a · (2b + 4c) = h3, 1, 0i · h−2, −4, 10i = −10

17. 12, −8, 6 18. cos θ = 19. A =

a·b 1 = √ √ = 1/2; |a||b| 2 2

θ = 60◦

√ 1 |5i − 4j − 7k| = 3 10/2 2

20. (x + 5)2 + (y − 7)2 + (z + 9)2 = 6 21. | − 5 − (−3)| = 2 22. parallel: −2c = 5,

c = −5/2; orthogonal: 1(−2) + 3(−6) + c(5) = 0,

c=4

23. The equation can be transformed into something more recognizable by completing the square: x2 + 2y 2 + 2z 2 − 4y − 12z = 0 =⇒ x2 + 2(y 2 − 2y) + 2(z 2 − 6z) = 0 =⇒ x2 + 2(y 2 − 2y + 1) + 2(z 2 − 6z = 9) = 20 =⇒ x2 + 2(y − 1)2 + 2(z − 3)2 = 20 This is the equation of an ellipsoid centered at (0, 1, 3). 24. Letting z = 1, the trace is described by the equation y = x2 − 1, which is a parabola.

CHAPTER 11 IN REVIEW

45

C. Exercises i j k 1 0 1 1 0 + i− 1. a × b = 1 1 0 = 1 1 1 −2 1 1 −2 1 A unit vector perpendicular to both a and b is

1 k = i − j + 3k −2

a×b 1 1 3 1 (i − j − 3k = √ i − √ j − √ k. =√ |a × b| 1+1+9 11 11 11 s  r √  2  2 2 1 1 1 3 3 2. The magnitude of a is given by |a| = + + − = = . Letting 2 2 2 4 2 α, β, and γ represent the angles between a and i, j, and k respectively, we have cos α = 1 1 −1 1 1 1  √2  = √ , cos β =  √2  = √ , and cos γ =  √ 2  = − √ . From this we are able to 3 3 3 3 3 3 2 2  2 1 compute: α = cos−1 √ ≈ 0.95532 3   1 β = cos−1 √ ≈ 0.95532  3  1 γ = cos−1 − √ ≈ 2.18628 3 3. compb a = a · b/|b| = h1, 2, −2i · h4, 3, 0i/5 = 2 4. compa b = b · a/|a| = h4, 3, 0i · h1, 2, −2i/3 = 10/3 proja b = (compa ba/|a| = (10/3)h1, 2, −2i/3 = h10/9, 20/9, −20/9i √ 5. First we compute 2a = h2, 4, −4i, |b| = 16 + 9 = 5, and 2a · b = 20. So projb 2a = 2a · b 20 16 12 b= h4, 3, 0i = h , , 0i. 2 |b| 25 5 5 6. compb (a − b) = (a − b) · b/|b| = h−3, −1, −2i · h4, 3, 0i/5 = −3 projb (a − b) = (compb (a − b))b/|b| = −3h4, 3, 0i/5 = h−12/5, −9/5, 0i projb⊥ (a−b) = (a−b)−projb (a−b) = h−3, −1, −2i−h−12/5, −9/5, 0i = h−3/5, 4/5, −10/5i 7.

x2 y2 + = 1; 16 4

8.

x2 1 + z 2 = − y; 2 4

9. − 10.

elliptical cylinder paraboloid

x2 y2 z2 − + = 1; 9 9/4 9

x2 y2 (z − 5)2 + + = 1; 25 25 25

11. x2 − y 2 = 9z; 12. plane

hyperboloid of two sheets sphere

hyperbolic paraboloid

46

CHAPTER 11. VECTORS AND 3-SPACE

√ √ 13. Replacing x by ± x2 + z 2 we have (± x2 +p z 2 )2 − y 2 = 1 or x2 + z 2 −p y 2 = 1, which is a 2 2 2 hyperboloid of one sheet. Replacing y by (± y + z ) we have x − (± y 2 + z 2 )2 = 1 or x2 − y 2 − z 2 = 1, which is a hyperboloid of two sheets. 14. The surface is generated by revolving y = 1+x, x ≥ 0, about the y-axis or √ y = 1+z, z ≥ 0 about the z-axis. The restrictions on x and z are required since y = 1 + x2 + z 2 ≥ 1. 15. Let a = ha, b, ci and r = hx, y, zi. Then (a) (r − a) · r = hx − a, y − b, z − ci · hx, y, zi = x2 − ax + y 2 − by + z 2 − ac = 0 implies  2   a 2 b c 2 a2 + b2 + c2 x− + y− + z− = . The surface is a sphere. 2 x 2 4 (b) (r − a) · a = hx − a, y − b, z − ci · ha, b, ci = a(x − a) + b(y − b) + c(z − c) = 0 The surface is a plane.

16. h4, 2, −2i − h2, 4, −3i = h2, −2, 1i; h2, 4, −3i − h6, 7, −5i = h−4, −3, 2i; h2, −2, 1i · h−4, −3, 2i = 0. The points are the vertices of a right triangle. 17. A direction vector of the given line is h4, −2, 6i. A parallel line containing (7, 3, −5) is (x − 7)/4 = (y − 3)/(−2) = (z + 5)/6. 18. A normal to the plane is h8, 3, −4i. The line with this direction vector and through (5, −9, 3) is x = 5 + 8t, y = −9 + 3t, z = 3 − 4t. 19. The direction vectors are h−2, 3, 1i and h2, 1, 1i. Since h−2, 3, 1i · h2, 1, 1i = 0, the lines are orthogonal. Solving 1 − 2t = x = 1 + 2s, 3t = y = −4 + s, we obtain t = −1 and s = 1. The point (3, −3, 0) obtained by letting t = −1 and s = 1 is common to the two lines, so they do intersect. 20. Vectors in the plane are h2, 3, 1i and h1, 0, 2i. A normal vector is h2, 3, 1i×h1, 0, 2i = h6, −3, −3i = 3h2, −1, −1i. An equation of the plane is 2x − y − z = 0. 21. The lines are parallel with direction vector h1, 4, −2i. Since (0, 0, 0) is on the first line and (1, 1, 3) is on the second line, the vector h1, 1, 3i is in the plane. A normal vector to the plane is thus h1, 4, −2i × h1, 1, 3i = h14, −5, −3i. An equation of the plane is 14x − 5y − 3z = 0. 22. Letting z = t in the equations of the plane and solving −x + y = 4 + 8t, 3x − y = −2t, we obtain x = 2 + 3t, y = 6 + 11t, z = t. Thus, a normal to the plane is h3, 11, 1i and an equation of the plane is 3(x − 1) + 11(y − 7) + (z + 1) = 0 or 3x + 11y + z = 79. 23. A normal vector is (i − 2j) × (2i + 3k) = −6i − 3j + 4k. Thus, an equation of the plane is −6(z − 1) − 3(y + 1) + 4(z − 2) = 0 or 6x + 3y − 4z = −5. 24. The points at the ends of the diameter, obtained from t = −1 and t = 0 are (2, 4, 2) and (4, 7, 8). The center of the √ sphere is the midpoint of the line segment or (3, 11/2, 5). The diameter of the sphere is 22 + 32 + 62 = 7. The equation is (x−3)2 +(y −11/2)2 +(z −5)2 = 49/4.

CHAPTER 11 IN REVIEW

47

25. We compute (a × b) · c. First a × b = −3i + 3j − 3k. Then (a × b) · c = −3(4) + 3(5) − 3(1) = 0, and the vectors are coplanar. 1 |c|. Since 2 2 2 2 a · b = 0 we have |b − a| = (b − a) · (b − a) = a · a + 2a · b + b · b = |a| + |b| , we have 1p 2 1 1 |d| = 21 |a + b| = |a| + |b|2 = |b − a| = |c|. 2 2 2

26. Let d be the vector from the right angle to M . We want to show that |d| =

27. (a) We have v = vj and B = Bi. Then F = q(v × B = q(vj × Bi) = q(−vBk) = −qvBk. (b) We first note that L = mr × v and r × v = 0. Then r × L = r × (mr × v) = m[r × (r × v)] = m[(r · v)r − (r · r)v] = −m|r|2 v, and so v = −

1 1 (r × L) = (L × r). 2 m|r| m|r|2

√ √ 10 a = √ (i + j) = 5 2i + 5 2j; d = h7, 4, 0i − h4, 1, 0i = 3i + 3j |a| 2√ √ √ W = F · d = 15 2 + 15 2 = 30 2 N-m √ √ √ √ 29. F = 5 2i + 5 2j + 50i = (5 2 + 50)i + 5 2j; d = 3i + 3j 28. F = 10

√ √ √ W = 15 2 + 150 + 15 2 = 30 2 + 150N-m ≈ 192.4 N-m 30. Let |F1 | = F1 and |F2 | = F2 . Then F1 = F1 [(cos 45◦ )i + (sin 45◦ )j] and F2 = F2 [(cos 120◦ )i + √ !   1 1 3 1 (sin 120◦ )j] or F1 = F1 √ i + √ j . Since w + F1 + F2 = 0, and F2 = F2 − i + 2 2 2 2  F1 and

 1 1 √ i + √ + F2 2 2

√ ! 1 3 − i+ j = 50j, 2 2



 1 1 √ F1 − F2 i + 2 2

! √ 3 1 √ F1 + F2 j = 50j 2 2

√ 1 1 1 3 √ F1 − F2 = 0, √ F1 + F2 = 50. 2 2 2 2 √ √ √ Solving, we obtain F1 = 25( 6 − 2) ≈ 25.9lb and F2 = 50( 3 − 1) ≈ 36.6lb.

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