A Em 3 Chapter 7

8/2/2019 A Em 3 Chapter 7

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7.1 Vectors in 2-Space

15.

5

5

P1

P1P2

y

x

P1P2 = 2, 5

16.

5

5

P1

P2

P1P2

y

x

P1P2 = 6, 4

17.

P1

P2

P1P2

5

5

y

x

P1P2 = 2, 2

18.

P1

P2

P1P2

5

5

y

x

P1P2 = 2, 3

19. Since P1P2 = OP2 OP1, OP2 = P1P2 + OP1 = (4i + 8j) + (3i + 10j) = i + 18j, and the terminal point is(1, 18).

20. SinceP1P2 =

OP2 OP1, OP1 = OP2 P1P2 = 4, 75, 1 = 9, 8, and the initial point is (9, 8).

21. a(= a), b(= 14a), c(= 52a), e(= 2a), and f(= 12a) are parallel to a.22. We want 3b = a, so c = 3(9) = 27.23. 6, 1524. 5, 225. a = 4 + 4 = 22 ; (a) u = 1

222, 2 = 1

2, 1

2; (b) u = 1

2, 1

2

26. a = 9 + 16 = 5; (a) u =153, 4 =

35 ,

45; (b) u =

35 ,

45

27. a = 5; (a) u = 150, 5 = 0, 1; (b) u = 0, 1

28. a = 1 + 3 = 2; (a) u = 121,

3 = 12 , 32 ; (b) u = 12 ,

32

29. a + b = 5, 12 = 25 + 144 = 13; u = 1135, 12 = 513 , 1213 30. 2a 3b = 5, 4 = 25 + 16 = 41 ; u = 1

415, 4 = 5

41, 4

41

31. a = 9 + 49 = 58 ; b = 2( 158

)(3i + 7j) = 658

i + 1458

j

32. a =

14 +

14 =

12

; b = 3( 11/2

)(12 i 12j) = 32

2i 3

2

2j

33. 34a = 3, 15/2 34. 5(a + b) = 50, 1 = 0, 5

35.

a

b

3b a 36.

ca

b + cb = a + (b + c)

37. x = (a + b) = a b 38. x = 2(a b) = 2a 2b

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39.

a

bc

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

40.

ab

cd

e

From Problem 39, e + c + d = 0. But b = e aand e = a + b, so (a + b) + c + d = 0.

41. From 2i + 3j = k1b + k2c = k1(i + j) + k2(i j) = (k1 + k2)i + (k1 k2)j we obtain the system of equationsk1 + k2 = 2, k1 k2 = 3. Solving, we find k1 = 52 and k2 = 12 . Then a = 52b 12c.

42. From 2i + 3j = k1b + k2c = k1(2i + 4j) + k2(5i + 7j) = (2k1 + 5k2)i + (4k1 + 7k2)j we obtain the system ofequations 2k1 + 5k2 = 2, 4k1 + 7k2 = 3. Solving, we find k1 = 134 and k2 = 717 .

43. From y = 12x 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 + 12j. Another unit vector tangent to the curve is 1

2i 1

2j.

44. From y = 2x + 3 we see that the slope of the tangent line at (0, 0) is 3. A vector with slope 3 is i + 3j. Aunit vector is (i + 3j)/i + 3j = (i + 3j)/10 = 1

10i + 1

10j. Another unit vector is 1

10i 1

10j.

45. (a) Since Ff = Fg, Fg = Ff = Fn and tan = Fg/Fn = Fn/Fn = .(b) = tan1 0.6

31

46. Since w + F1 + F2 = 0,

200j + F1 cos20i + F1 sin20j F2 cos15i + F2 sin15j = 0or (F1 cos20 F2 cos15)i + (F1 sin20 + F2 sin15 200)j = 0.Thus, F1 cos20 F2 cos15 = 0; F1 sin20 + F2 sin15 200 = 0. Solving this system for F1 andF2, we obtain

F1 = 200cos15

sin15 cos20 + cos 15 sin20=

200cos15

sin(15 + 20)=

200cos15

sin35 336.8 lb

and

F2

=

200cos20

sin15 cos20 + cos 15 sin20=

200cos20

sin35 327.7 lb.

47. Since y/2a(L2 + y2)3/2 is an odd function on [a, a], Fy = 0. Now, using the fact that L/(L2 + y2)3/2 is aneven function, we havea

a

L dy

2a(L2 + y2)3/2=

L

a

a0

dy

(L2 + y2)3/2y = L tan , dy = L sec2 d

=L

a

tan1 a/L0

L sec2 d

L3(1 + tan2 )3/2=

1

La

tan1 a/L0

sec2 d

sec3

=1

La

tan1 a/L0

cos d =1

Lasin

tan1 a/L

0

=1

La

a

L2 + a2=

1

LL2 + a2.

Then Fx = qQ/40L

L2 + a2 and F = (qQ/40L

L2 + a2 )i.

P2

P1 P1

P2

NM

48. Place one corner of the parallelogram at the origin and let two adjacent sides

beOP1 and

OP2. Let M be the midpoint of the diagonal connecting P1 and

P2 and N be the midpoint of the other diagonal. ThenOM = 12(

OP1+

OP2).

SinceOP1 +

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.

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A

B

C

DE

49. By Problem 39,AB +

BC+

CA = 0 and

AD +

DE+

EC+

CA = 0. From the first equation,

AB +

BC = CA. Since D and E are midpoints, AD = 12

AB and

EC = 12

BC. Then,

12

AB +

DE+ 12

BC +

CA = 0 and

DE = CA 1

2 (AB + BC) = CA 1

2 (CA) = 1

2CA.Thus, the line segment joining the midpoints D and E is parallel to the side AC and half its length.

50. We haveOA = 150cos20i +150sin20j,

AB = 200cos 113i +200sin113j,

BC = 240cos 190i +240sin190j.

Then

r = (150 cos 20 + 200cos 113 + 240cos 190)i + (150 sin 20 + 200sin 113 + 240sin 190)j

173.55i + 193.73jand r 260.09 miles.

EXERCISES 7.2Vectors in 3-Space

1. 6.z

(0, 0, 4) (1, 1, 5)

(3, 4, 0)

y

x

(6, 0, 0)(6, 2, 0)

(5, 4, 3)

7. A plane 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

x

y

(1, 3, 7)

(1, 6, 7)

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

(3, 6, 4)(3, 3, 4)

(3, 3, 7)(1, 3, 4)

13. (a) xy-plane: (2, 5, 0), xz-plane: (2, 0, 4), yz-plane: (0, 5, 4); (b) (2, 5, 2)

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(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

21. d =

(3 6)2 + (1 4)2 + (2 8)2 = 70

22. d = (1 0)2 + (3 4)2 + (5 3)2 = 3

6

23. (a) 7; (b) d =

(3)2 + (4)2 = 5

24. (a) 2; (b) d =

(6)2 + 22 + (3)2 = 7

25. d(P1, P2) =

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.

26. d(P1, P2) =

12 + 22 + 42 =

21 ; d(P1, P3) =

32 + 22 + (2

2)2 =

21

d(P2, P3) =

(3 1)2 + (2 2)2 + (22 4)2 =

28 162 The triangle is an isosceles triangle.

27. d(P1, P2) =

(4 1)2

+ (1 2)2

+ (3 3)2

=

10d(P1, P3) =

(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.

28. d(P1, P2) =

(1 1)2 + (1 1)2 + (1 (1))2 = 2d(P1, P3) =

(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.

29. d(P1, P2) =

(2 1)2 + (2 2)2 + (3 0)2 = 34d(P1, P3) =

(7 1)2 + (10 2)2 + (6 0)2 = 234

d(P2, P3) =

(7 (2))2 + (10 (2))2 + (6 (3))2 = 334

Since d(P1, P2) + d(P1, P3) = d(P2, P3), the points P1, P2, and P3 are collinear.30. d(P1, P2) =

(1 2)2 + (4 3)2 + (4 2)2 = 6

d(P1, P3) =

(5 2)2 + (0 3)2 + (4 2)2 = 36d(P2, P3) =

(5 1)2 + (0 4)2 + (4 4)2 = 46

Since d(P1, P2) + d(P1, P3) = d(P2, P3), the points P1, P2, and P3 are collinear.

31.

(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

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32.

(0 x)2 + (3 x)2 + (5 1)2 = 5 = 2x2 6x + 25 = 25 = x2 3x = 0 = x = 0, 3

33.

1 + 7

2,

3 + (2)2

,1/2 + 5/2

2

= (4, 1/2, 3/2)

34. 0 + 4

2

,5 + 1

2

,8 + (6)

2 = (2, 3,

7)

35. (x1 + 2)/2 = 1, x1 = 4; (y1 + 3)/2 = 4, y1 = 11; (z1 + 6)/2 = 8, z1 = 10The coordinates ofP1 are (4, 11, 10).

36. (3 + (5))/2 = x3 = 4; (4 + 8)/2 = y3 = 6; (1 + 3)/2 = z3 = 2.The coordinates ofP3 are (4, 6, 2). (a)

3 + (4)2

,4 + 6

2,

1 + 2

2

= (7/2, 5, 3/2)

(b)

4 + (5)2

,6 + 8

2,

2 + 3

2

= (9/2, 7, 5/2)

37.P1P2 = 3, 6, 1 38. P1P2 = 8, 5/2, 8

39.P1P2 = 2, 1, 1 40. P1P23, 3, 7

41. a + (b + c) = 2, 4, 1242. 2a (b c) = 2, 6, 43, 5, 8 = 5, 1, 1243. b + 2(a 3c) = 1, 1, 1 + 25, 21, 25 = 11, 41, 4944. 4(a + 2c) 6b = 45, 9, 206, 6, 6 = 26, 30, 7445. a + c = 3, 3, 11 = 9 + 9 + 121 = 13946. c2b = (4 + 36 + 81 )(2)(1 + 1 + 1 ) = 223

47.

aa + 5

b|b = 1aa + 5 1bb = 1 + 5 = 6

48. ba + ab = 1 + 1 + 1 1, 3, 2 + 1 + 9 + 4 1, 1, 1 =

3 , 3

3 , 2

3 +

14 ,

14 ,

14 =

3

14 , 3

3 +

14 , 2

3 +

14

49. a = 100 + 25 + 100 = 15; u = 115

10, 5, 10 = 2/3, 1/3, 2/3

50. a = 1 + 9 + 4 = 14 ; u = 114

(i 3j + 2k) = 114

i 314

j +214

k

51. b = 4a = 4i 4j + 4k

52. a = 36 + 9 + 4 = 7; b = 1

21

7 6, 3, 2 = 37 , 314 , 17

53.a

by

z

x

(a + b)12

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EXERCISES 7.3Dot Product

1. a b = 10(5) cos(/4) = 252 2. a b = 6(12) cos(/6) = 363

3. a b = 2(1) + (3)2 + 4(5) = 12 4. b c = (1)3 + 2(6) + 5(1) = 4

5. a c = 2(3) + (3)6 + 4(1) = 16 6. a (b + c) = 2(2) + (3)8 + 4(4) = 4

7. a (4b) = 2(4) + (3)8 + 4(20) = 48 8. b (a c) = (1)(1) + 2(9) + 5(5) = 8

9. a a = 22

+ (3)2

+ 42

= 29 10. (2b) (3c) = (2)9 + 4(18) + 10(3) = 2411. a (a + b + c) = 2(4) + (3)5 + 4(8) = 2512. (2a) (a 2b) = 4(4) + (6)(7) + 8(6) = 10

13.

a bb b

b =

2(1) + (3)2 + 4(5)

(1)2 + 22 + 52

1, 2, 5 = 1230

1, 2, 5 = 2/5, 4/5, 2

14. (c b)a = [3(1) + 6(2) + (1)5]2, 3, 4 = 42, 3, 4 = 8, 12, 1615. a and f, b and e, c and d

16. (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

17. Solving the system of equations 3x1 + y1 1 = 0, 3x1 + 2y1 + 2 = 0 gives x1 = 4/9 and y1 = 1/3. Thus,v = 4/9, 1/3, 1.

18. If a and b represent adjacent sides of the rhombus, then a = b, the diagonals of the rhombus are a + band a b, and

(a + b) (a b) = a a a b + b a b b = a a b b = a2 b2 = 0.Thus, the diagonals are perpendicular.

19. Since

c a =

b a ba2 a

a = b a a ba2 (a a) = b a a ba2 a

2 = b a a b = 0,

the vectors c and a are orthogonal.20. a b = 1(1) + c(1) = c + 1; a = 1 + c2 , b = 2

cos45 =1

2=

c + 11 + c2

2

=

1 + c2 = c + 1 = 1 + c2 = c2 + 2c + 1 = c = 0

21. a b = 3(2) + (1)2 = 4; a = 10 , b = 22cos =

4

(

10)(2

2)=

15

= = cos1 15

1.11 rad 63.43

22. a b = 2(3) + 1(4) = 10; a = 5 , b = 5

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7.3 Dot Product

cos =10

(

5 )5= 2

5= = cos1(2/

5 ) 2.68 rad 153.43

23. a b = 2(1) + 4(1) + 0(4) = 6; a = 25 , b = 32cos =

6(2

5)(3

2)

= 110

= = cos1(1/

10 ) 1.89 rad 108.43

24. a b = 12(2) + 12(4) + 32(6) = 8; a =

11/2, b = 214

cos =8

(

11/2)(2

14 )=

8154

= = cos1(8/

154) 0.87 rad 49.86

25. a = 14 ; cos = 1/14 , 74.50; cos = 2/14 , 57.69; cos = 3/14 , 36.70

26. a = 9; cos = 2/3, 48.19; cos = 2/3, 48.19; cos = 1/3, 109.47

27. a = 2; cos = 1/2, = 60; cos = 0, = 90; cos = 3/2, = 150

28. a = 78 ; cos = 5/78 , 55.52; cos = 7/78 , 37.57; cos = 2/78 , 76.91

29. Let be the angle betweenAD and

AB and a be the length of an edge of the cube. Then

AD = ai + aj + ak,

AB = ai andcos =

AD AB

AD AB=

a23a2

a2

=1

3

so 0.955317 radian or 54.7356. Letting be the angle between AD and AC and noting that AC = ai + ajwe have

cos =

AD AC

AD AC=

a2 + a23a2

2a2

=

2

3

so 0.61548 radian or 35.2644.30. If a and b are orthogonal, then a b = 0 and

cos 1 cos 2 + cos 1 cos 2 + cos 1 cos 2 = a1a b1b + a2a b2b + a3a b3b=

1

a b(a1b1 + a2b2 + a3b3) =1

a b (a b) = 0.

31. a = 5, 7, 4; a = 310 ; cos = 5/310 , 58.19; cos = 7/310 , 42.45; cos = 4/310 , 65.06

32. We want cos = cos = cos or a1 = a2 = a3. Letting a1 = a2 = a3 = 1 we obtain the vector i +j + k. A unit

vector in the same direction is 13

i + 13j + 1

3k.

33. compba = a b/b = 1, 1, 3 2, 6, 3/7 = 5/734. compab = b

a/

a

=

2, 6, 3

1,

1, 3

/

11 = 5/

11

35. b a = 1, 7, 0; compa(b a) = (b a) a/a = 1, 7, 0 1, 1, 3/11 = 6/1136. a + b = 3, 5, 6; 2b = 4, 12, 6; comp2b(a + b) 2b/|2b| = 3, 5, 6 4, 12, 6/14 = 54/737.

OP = 3i + 10j; OP = 109 ; comp

OPa = a OP /OP = (4i + 6j) (3i + 10j)/109 = 72/109

38.OP = 1, 1, 1; OP = 3 ; comp

OPa = a OP /OP = 2, 1, 1 1, 1, 1/3 = 0

39. compba = a b/b = (5i + 5j) (3i + 4j)/5 = 7projba = (compba)b/b = 7(3i + 4j)/5 = 215 i + 285 j

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7.3 Dot Product

40. compba = a b/b = (4i + 2j) (3i +j)/

10 = 10proj

ba = (comp

ba)b/b = 10(3i +j)/10 = 3i j

41. compb

a = a b/b = (i 2j + 7k) (6i 3j 2k)/7 = 2projba = (compba)b/b = 2(6i 3j 2k)/7 = 127 i + 67j + 47k

42. compba = a b/b = 1, 1, 1 2, 2, 1/3 = 1/3projba = (compba)b/b = 132, 2, 1/3 = 2/9, 2/9, 1/9

43. a + b = 3i + 4j; a + b = 5; comp(a+b)a = a (a + b)/a + b = (4i + 3j) (3i + 4j)/5 = 24/5proj(a+b)a = (comp(a+b)a)(a + b)/a + b = 245 (3i + 4j)/5 = 7225 i + 9625j

44. a b = 5i + 2j; a b = 29; comp(ab)b = b (a b)/a b = (i +j) (5i + 2j)/

29 = 3/29proj(ab)b = (comp(ab)b)(a b)/a b = 329(5i + 2j)/

29 = 1529 i 629j

45. We identify F = 20, = 60 and d = 100. Then W = F d cos = 20(100)(12) = 1000 ft-lb.46. We identify d = i + 3j + 8k. Then W = F d = 4, 3, 5 1, 3, 8 = 45 N-m.47. (a) Since w and d are orthogonal, W = w d = 0.

(b) We identify = 0. Then W = F d cos = 30(42 + 32 ) = 150 N-m.48. Using d = 6i + 2j and F = 3( 35 i +

45j), W = F d = 95 , 125 6, 2 = 785 ft-lb.

49. Let a and b be vectors from the center of the carbon atom to the centers of two distinct hydrogen atoms. The

distance between two hydrogen atoms is then

b a =

(b a) (b a) = b b 2a b + a a=

b2 + a2 2a b cos =

(1.1)2 + (1.1)2 2(1.1)(1.1)cos109.5=

1.21 + 1.21 2.42(0.333807) 1.80 angstroms.

50. Using the fact that | cos | 1, we have |a b| = a b| cos | = a b| cos | a b.

51. a + b2

= (a + b) (a + b) = a a + 2a b + b b = a2

+ 2a b + b2

a2 + 2|a b| + b2 since x |x| a2 + 2a b + b2 = (a + b)2 by Problem 50

Thus, since a + b and a + b are positive, a + b a + b.52. Let P1(x1, y1) and P2(x2, y2) be distinct points on the line ax + by = c. Then

n P1P2 = a, b x2 x1, y2 y1 = ax2 ax1 + by2 by1= (ax2 + by2) (ax1 + by1) = c (c) = 0,

and the vectors are perpendicular. Thus, n is perpendicular to the line.

53. Let be the angle between n and P2P1. Thend = P1P2 | cos | = |n

P2P1|

n =|a, b x1 x2, y1 y2|

a2 + b2=

|ax1 ax2 + by1 by2|a2 + b2

=|ax1 + by1 (ax2 + by2)|

a2 + b2=

|ax1 + by1 (c)|a2 + b2

=|ax1 + by1 + c|

a2 + b2.

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7.4 Cross Product

EXERCISES 7.4Cross Product

1. a b =

i j k

1 1 00 3 5

=1 03 5

i 1 00 5j + 1 10 3

k = 5i 5j + 3k

2. a b =

i j k

2 1 0

4 0 1

= 1 00 1

i 2 04 1

j + 2 14 0

k = i + 2j 4k

3. a b =

i j k

1 3 12 0 4

=

3 10 4 i

1 12 4j +

1 32 0k = 12, 2, 6

4. a b =

i j k1 1 1

5 2 3

= 1 12 3

i 1 15 3

j + 1 15 2

k = 1, 8, 7

5. a b =

i j k

2 1 21 3 1

=1 23 1

i 2 21 1

j + 2 11 3

k = 5i + 5k

6. a b =

i j k

4 1 52 3 1

=

1 53 1

i

4 52 1

j +

4 1

2 3

k = 14i 6j + 10k

7. a b =

i j k1/2 0 1/2

4 6 0

= 0 1/26 0

i 1/2 1/24 0

j + 1/2 04 6

k = 3, 2, 3

8. a b =

i j k

0 5 0

2 3 4

= 5 03 4

i 0 02 4

j + 0 52 3

k = 20, 0, 10

9. a b =

i j k

2 2 43 3 6

=

2 43 6

i

2 43 6

j +

2 2

3 3

k = 0, 0, 0

10. a b =

i j k

8 1 61 2 10

= 1 62 10

i 8 61 10

j + 8 11 2

k = 2, 86, 17

11.P1P2 = (2, 2, 4); P1P3 = (3, 1, 1)

P1P2 P1P3 =

i j k

2 2 43 1 1

= 2 41 1

i 2 43 1

j +2 23 1

k = 6i + 14j + 4k

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7.4 Cross Product

12.P1P2 = (0, 1, 1);

P1P3 = (1, 2, 2);

P1P2 P1P3 =

i j k

0 1 1

1 2 2

= 1 12 2

i 0 11 2

j + 0 11 2

k = j k

13. a b = i j k

2 7 41 1 1

= 7

4

1 1 i 2

4

1 1 j + 2 7

1 1

k = 3i 2j 5k

is perpendicular to both a and b.

14. a b =

i j k

1 2 44 1 0

=2 41 0

i 1 44 0

j +1 24 1

k = 4, 16, 9is perpendicular to both a and b.

15. a b =

i j k

5 2 12 0 7

=

2 10 7

i

5 1

2 7

j +

5 22 0

k = 14, 37, 4

a (a b) = 5, 2, 1 14, 37, 4 = 70 74 + 4 = 0; b (a b) = 2, 0, 7 14, 37, 4 = 28 + 0 28 = 0

16. a b =

i j k

1/2 1/4 02 2 6

=1/4 02 6

i 1/2 02 6

j + 1/2 1/42 2

k = 32 i 3j 12ka (a b) = ( 12 i 14j) (32 i 3j 12k) = 34 + 34 + 0 = 0b (a b) = (2i 2j + 6k) (32 i 3j 12k) = 3 + 6 3 = 0

17. (a) b c =

i j k

2 1 1

3 1 1

=

1 1

1 1

i

2 1

3 1

j +

2 1

3 1

k = j k

a (b c) =

i j k1 1 20 1 1

=1 21 1

i 1 20 1

j + 1 10 1

k = i +j + k(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 + 3ka (b c) = (a c)b (a b)c = (8i + 4j + 4k) (9i + 3j + 3k) = i +j + k

18. (a) b c =

i j k

1 2 11 5 8

=

2 15 8 i

1 11 8j +

1 21 5k = 21i 7j + 7k

a (b c) =

i j k

3 0 421 7 7

= 0 47 7

i 3 421 7j + 3 021 7

k = 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 + 56ka (b c) = (a c)b (a b)c = (35i 70j + 35k) (7i + 35j + 56k) = 28i 105j 21k

19. (2i) j = 2(i j) = 2k20. i (3k) = 3(i k) = 3(j) = 3j

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7.4 Cross Product

21. k (2i j) = k (2i) + k (j) = 2(k i) (k j) = 2j (i) = i + 2j22. i (j k) = i i = 023. [(2k) (3j)] (4j) = [2 3(k j) (4j)] = 6(i) 4j = (6)(4)(i j) = 24k24. (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 + 2k27. k (j k) = k i = 028. i [j (k)] = i [(j k)] = i (i) = (i i) = 129. 4j 5(i j) = 4j 5k = 4130. (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) = k35. 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) = 636. (i k) (j i) = (j) (k) = (1)(1)(j k) = j k = i37. a (3b) = 3(a b) = 3(4i 3j + 6k) = 12i 9j + 18k38. b a = a b = (a b) = 4i + 3j 6k39. (a) b = (a b) = 4i + 3j 6k40.

|a

b

|= 42 + (3)

2 + 62 =

61

41. (a b) c =

i j k

4 3 62 4 1

=3 64 1

i 4 62 1

j + 4 32 4

k = 21i + 16j + 22k42. (a b) c = 4(2) + (3)4 + 6(1) = 1043. a (b c) = (a b) c = 4(2) + (3)4 + 6(1) = 1044. (4a) (b c) = (4a b) c = 4(a b) c = 16(2) + (12)4 + 24(1) = 4045. (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

AB AC =

i j k

1 3 01 3 4

= 12i 4j 6k

we find that the area is 12i 4j 6k = 144 + 16 + 36 = 14.46. (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 = i4j+k. Since AB = CD and AC = BD, the quadrilateral is a parallelogram.

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7.4 Cross Product

(b) Computing

AB AC =

i j k

4 0 11 4 1

= 4i + 3j + 16k

we find that the area is 4i + 3j + 16k = 16 + 9 + 256 = 281 16.76.47.

P1P2 = j;

P2P3 = j + k

P1P2 P2P3 =

i j k

0 1 0

0 1 1

= 1 01 1

i 0 00 1

j + 0 10 1

k = i; A = 12i = 12 sq. unit48.

P1P2 = j + 2k;

P2P3 = 2i +j 2k

P1P2 P2P3 =

i j k

0 1 2

2 1 2

=

1 2

1 2

i

0 2

2 2

j +

0 1

2 1

k = 4i + 4j 2k

A = 12 4i + 4j 2k = 3 sq. units49.

P1P2 = 3j k; P2P3 = 2i k

P1P2 P2P3 =

i j k

0 3 12 0 1

=3 10 1

i 0 12 1

j + 0 32 0

k = 3i + 2j 6kA = 123i + 2j 6k = 72 sq. units

50.P1P2 = i + 3k; P2P3 = 2i + 4j k

P1P2

P2P3 =

i j k

1 0 3

2 4 1

=

0 3

4 1 i

1 3

2 1 j +

1 0

2 4k =

12i + 5j

4k

A = 12 12i + 5j 4k =1852 sq. units

51. b c =

i j k

1 4 02 2 2

= 4 02 2

i 1 02 2

j +1 42 2

k = 8i + 2j 10kv = |a (b c)| = |(i +j) (8i + 2j 10k)| = |8 + 2 + 0| = 10 cu. units

52. b c =

i j k

1 4 1

1 1 5

=

4 1

1 5

i

1 1

1 5

j +

1 4

1 1

k = 19i 4j 3k

v = |a (b c)| = |(3i +j + k) (19i 4j 3k)| = |57 4 3| = 50 cu. units

53. b c =

i j k

2 6 65/2 3 1/2

= 6 63 2/2

i 2 65/2 1/2

j + 2 65/2 3

k = 21i 14j 21ka (b c) = (4i + 6j) (21i 14j 21k) = 84 84 + 0 = 0. The vectors are coplanar.

54. The four points will be coplanar if the three vectorsP1P2 = 3, 1, 1, P2P3 = 3, 5, 13, and P3P4 =

8, 7, 6 are coplanar.

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7.4 Cross Product

P2P3 P3P4 =

i j k

3 5 138 7 6

=5 137 6

i 3 138 6

j +3 58 7

k = 61, 122, 61P1P2 (P2P3 P3P4) = 3, 1, 1 61, 122, 61 = 183 + 122 + 61 = 0

The four points are coplanar.55. (a) Since = 90, a b = a b | sin90| = 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 bya and b as shown in Figure 7.54 in the text. It makes an angle of 30 with the positive x-axis.

(c) We identify n = (

3 i j)/2. Then a b = 32n = 163 i 16j.56. Using Definition 7.4, a b = 27(8)sin120n = 243 (3/2)n = 36n. By the right-hand rule, n = j or

n = j. Thus, a b = 36j or 36j.57. (a) We note first that a b = k, b c = 12(i k), c a = 12 (j k), a (b c) = 12 , b (c a) = 12 , and

c (a b) = 12 . Then

A =12(i k)

12

= i k, B = 12(j k)12

= j k, and C = k12

= 2k.

(b) We need to compute A (B C). Using formula (10) in the text we have

B C = (c a) (a b)[b (c a)][c (a b)] =

[(c a) b]a [(c a) a]b[b (c a)][c (a b)]

=a

c (a b) since (c a) a = 0.

Then

A (B C) = b ca (b c)

a

c (a b) =1

c (a b)and the volume of the unit cell of the reciprocal latrice is the reciprocal of the volume of the unit cell of

the original lattice.

58. a (b + c) = a2 a3b2 + c2 b3 + c3

i a1 a3b1 + c1 b3 + c3

j + a1 a2b1 + c1 b2 + c2

k= (a2b3 a3b2)i + (a2c3 a3c2)i [(a1b3 a3b1)j + (a1c3 a3c1)j] + (a1b2 a2b1)k + (a1c2 a2c1)k= (a2b3 a3b2)i (a1b3 a3b1)j + (a1b2 a2b1)k + (a2c3 a3c2)i (a1c3 a3c1)j + (a1c2 a2c1)k= a b + a c

59. b c = (b2c3 b3c2)i (b1c3 b3c1)j + (b1c2 b2c1)ka (b c) = [a2(b1c2 b2c1) + a3(b1c3 b3c1)]i [a1(b1c2 b2c1) a3(b2c3 b3c2)]j

+ [a1(b1c3 b3c1) a2(b2c3 b3c2)]k= (a2b1c2 a2b2c1 + a3b1c3 a3b3c1)i (a1b1c2 a1b2c1 a3b2c3 + a3b3c2)j

(a1b1c3 a1b3c1 + a2b2c3 a2b3c2)k(a c)b (a b)c = (a1c1 + a2c2 + a3c3)(b1i + b2j + b3k) (a1b1 + a2b2 + a3b3)(c1i + c2j + c3k)

= (a2b1c2 a2b2c1 + a3b1c3 a3b3c1)i (a1b1c2 a1b2c1 a3b2c3 + a3b3c2)j (a1b1c3 a1b3c1 + a2b2c3 a2b3c2)k

60. The statement is false since i (i j) = i k = j and (i i) j = 0 j = 0.

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7.4 Cross Product

61. Using equation 9 in the text,

a (b c) =

a1 a2 a3

b1 b2 b3

c1 c2 c3

and (a b) c = c (a b) =

c1 c2 c3

a1 a2 a3

b1 b2 b3

.

Expanding these determinants out we obtain a (b c) = a1b2c3 + a2b3c1 + a3b1c2 a3b2c1 a1b3c2 a2b1c3and c (a b) = a2b3c1 + a3b1c2 + a1b2c3 a2b1c3 a3b2c1 a1b3c2. These are equal so a (b c) = (a b) c.

62. a (b c) + b (c a) + c (a b)= (a c)b (a b)c + (b a)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

63. Sincea b2 = (a2b3 a3b2)2 + (a1b3 a3b1)2 + (a1b2 a2b1)2

= a22b23 2a2b3a3b2 + a23b22 + a21b23 2a1b3a3b1 + a23b21 + a21b22 2a1b2a2b1 + a22b21

and

a2

b2

(a

b)2 = (a2

1

+ a2

2

+ a2

3

)(b2

1

+ b2

2

+ b2

3

)

(a1b1 + a2b2 + a3b3)2

= a21a22 + a

21b

22 + a

21b

23 + a

22b

21 + a

22b

22 + a

22b

23 + a

23b

21 + a

23b

22 + a

23b

23

a21b21 a22b22 a23b23 2a1b1a2b2 2a1b1a3b3 2a2b2a3b3= a21b

22 + a

21b

23 + a

22b

21 + a

22b

23 + a

23b

21 + a

23b

22 2a1a2b1b2 2a1a3b1b3 2a2a3b2b3

we see that a b2 = a2b2 (a b)2.64. No. For example i (i +j) = i j by the distributive law (iii) in the text, and the fact that i i = 0. But i +j

does not equal j.

65. By the distributive law (iii) in the text:

(a + b) (a b) = (a + b) a (a + b) b = a a + b a a b b b = 2b asince a a = 0, b b = 0, and a b = b a.

EXERCISES 7.5Lines and Planes in 3-Space

The equation of a line throughP1 and P2 in 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 notunique. (See the alternate solution to Problem 1.)

1. a = 1 3, 2 5, 1 (2) = 2, 3, 3; x, y, z = 1, 2, 1 + t2, 3, 3Alternate Solution: a = 3 1, 5 2, 2 1 = 2, 3, 3; x,y,z = 3, 5, 2 + t2, 3, 3

2. a = 0 (2), 4 6, 5 3 = 2, 2, 2; x, y, z = 0, 4, 5 + t2, 2, 23. a = 1/2 (3/2), 1/2 5/2, 1 (1/2) = 2, 3, 3/2; x, y, z = 1/2, 1/2, 1 + t2, 3, 3/24. a = 10 5, 2 (3), 10 5 = 5, 5, 15; x, y, z = 10, 2, 10 + t5, 5, 15

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7.5 Lines and Planes in 3-Space

5. a = 1 (4), 1 1, 1 (1) = 5, 0, 0; x, y, z = 1, 1, 1 + t5, 0, 06. a = 3 5/2, 2 1, 1 (2) = 1/2, 1, 3; x, y, z = 3, 2, 1 + t1/2, 1, 37. a = 2 6, 3 (1), 5 8 = 4, 4, 3; x = 2 4t, y = 3 + 4t, z = 5 3t8. a =

2

0, 0

4, 0

9

=

2,

4,

9

; x = 2 + 2t, y =

4t, z =

9t

9. a = 1 3, 0 (2), 0 (7) = 2, 2, 7; x = 1 2t, y = 2t, z = 7t10. a = 0 (2), 0 4, 5 0 = 2, 4, 5; x = 2t, y = 4t, z = 5 + 5t11. a = 4 (6), 1/2 (1/4), 1/3 1/6 = 10, 3/4, 1/6; x = 4 + 10t, y = 1

2+

3

4t, z =

1

3+

1

6t

12. a = 3 4, 7 (8), 9 (1) = 7, 15, 10; x = 3 7t, y = 7 + 15t, z = 9 + 10t

13. a1 = 10 1 = 9, a2 = 14 4 = 10, a3 = 2 (9) = 7; x 109

=y 14

10=

z + 2

7

14. a1 = 1 2/3 = 1/3, a2 = 3 0 = 3, a3 = 1/4 (1/4) = 1/2; x 11/3

=y 3

3=

z 1/41/2

15. a1 = 7 4 = 11, a2 = 2 2 = 0, a3 = 5 1 = 4; x + 711 = z 54 , y = 2

16. a1 = 1 (5) = 6, a2 = 1 (2) = 3, a3 = 2 (4) = 6; x 16

=y 1

3=

z 26

17. a1 = 5 5 = 0, a2 = 10 1 = 9, a3 = 2 (14) = 12; x = 5, y 109

=z + 2

12

18. a1 = 5/6 1/3 = 1/2; a2 = 1/4 3/8 = 5/8; a3 = 1/5 1/10 = 1/10x 5/6

1/2=

y + 1/4

5/8 =z 1/5

1/10

19. parametric: x = 4 + 3t, y = 6 + t/2, z = 7 3t/2; symmetric: x 43

=y 61/2

=z + 7

3/2

20. parametric: x = 1 7t, y = 8 8t, z = 2; symmetric: x 17 =y 88 , z = 2

21. parametric: x = 5t, y = 9t, z = 4t; symmetric:x

5=

y

9=

z

4

22. parametric: x = 12t, y = 3 5t, z = 10 6t; symmetric: x12

=y + 3

5 =z 10

623. Writing the given line in the form x/2 = (y 1)/(3) = (z 5)/6, we see that a direction vector is 2, 3, 6.

Parametric equations for the line are x = 6 + 2t, y = 4 3t, z = 2 + 6t.24. A direction vector is 5, 1/3, 2. Symmetric equations for the line are (x4)/5 = (y +11)/(1/3) = (z +7)/(2).25. A direction vector parallel to both the xz- and xy-planes is i = 1, 0, 0. Parametric equations for the line are

x = 2 + t, y = 2, z = 15.26. (a) Since the unit vector j = 0, 1, 0 lies along the y-axis, we have x = 1, y = 2 + t, z = 8.

(b) since the unit vector k = 0, 0, 1 is perpendicular to the xy-plane, we have x = 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. a and f are parallel since 9, 12, 6 = 33, 4, 2. c and d are orthogonal since 2, 3, 4 1, 4, 5/2 = 0.29. 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 = 42(1/2) = 5 and z = 9+3(1/2) = 15/2.

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w = (x + 1)i + (y + 1)j + (z 4)k. Then, a normal vector is

u v =

i j k

1 2 1

3 4 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 = 2.46. 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 t he p oints ( 1, 3, 1) and (x, y, z) we obtain the vectorw = (x 1)i + (y 3)j + (z + 1)k. Then, a normal vector is

u v =

i j k

0 0 1

1 2 2

= 2i +j.A vector equation of the plane is 2(x 1) + (y 3) + 0(z + 1) = 0 or 2x + y = 1.

47. 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 vectorw = (x 3)i + (y 2)j + (z + 1)k. Then, a normal vector is

u v =

i j k

1 1 1

2 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.

48. The three points are not colinear and all satisfy x = 0, which is the equation of the plane.

49. 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 colinear.

50. 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 vectorw = (x 5)i + yj + (z + 5)k. Then, a normal vector is

u v =

i j k

2 0 21 1 5

= 2i + 8j 2k.A vector equation of the plane is 2(x 5) + 8y 2(z + 5) = 0 or x 4y + z = 0.

51. A normal vector to x + y 4z = 1 is 1, 1, 4. The equation of the parallel plane is(x

2) + (y

3)

4(z + 5) = 0 or x + y

4z = 25.

52. A normal vector to 5xy+z = 6 is 5, 1, 1, . The equation of the parallel plane is 5(x0)(y0)+(z0) = 0or 5x y + z = 0.

53. A normal vector to the xy-plane is 0, 0, 1. The equation of the parallel plane is z 12 = 0 or z = 12.54. A normal vector is 0, 1, 0. The equation of the plane is y + 5 = 0 or y = 5.55. Direction vectors of the lines are 3, 1, 1. and 4, 2, 1. A normal vector to the plane is 3, 1, 1 4, 2, 1 =

3, 1, 10. A point on the first line, and thus in the plane, is 1, 1, 2. The equation of the plane is3(x 1) + (y 1) + 10(z 2) = 0 or 3x + y + 10z = 18.

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56. Direction vectors of the lines are 2, 1, 6 and 1, 1, 3. A normal vector to the plane is 2, 1, 61, 1, 3 =3, 12, 3. A point on the first line, and thus in the plane, is (1 , 1, 5). The equation of the plane is3(x 1) + 12(y + 1) + 3(z 5) = 0 or x + 4y + z = 0.

57. A direction vector for the two lines is 1, 2, 1. Points on the lines are (1, 1, 3) and (3, 0, 2). Thus, anothervector parallel to the plane is

1

3, 1

0, 3+2

=

2, 1, 5. A normal vector to the plane is

1, 2, 1

2, 1, 5

=

9, 7, 5. Using the point (3, 0, 2) in the plane, the equation of the plane is 9( x 3) 7(y 0)+5(z + 2) = 0or 9x 7y + 5z = 17.

58. A direction vector for the line is 3, 2, 2. Letting t = 0, we see that the origin is on the line and hence in theplane. Thus, another vector parallel to the plane is 4 0, 0 0, 6 0 = 4, 0, 6. A normal vector to theplane is 3, 2, 24, 0, 6 = 12, 10, 8. The equation of the plane is 12(x0)+10(y 0)8(z 0) = 0or 6x 5y + 4z = 0.

59. A direction vector for the line, and hence a normal vector to the plane, is 3, 1, 1/2. The equation of theplane is 3(x 2) + (y 4) 12(z 8) = 0 or 3x + y 12z = 6.

60. A normal vector to the plane is 2 1, 6 0, 3 + 2 = 1, 6, 1. The equation of the plane is

(x 1) + 6(y 1) (z 1) = 0 or x + 6y z = 6.61. Normal vectors to the planes are (a) 2, 1, 3, (b) 1, 2, 2, (c) 1, 1, 3/2, (d) 5, 2, 4,

(e) 8, 8, 12, (f) 2, 1, 3. Parallel planes are (c) and (e), and (a) and (f). Perpendicular planesare (a) and (d), (b) and (c), (b) and (e), and (d) and (f ).

62. A normal vector to the plane is 7, 2, 3. This is a direction vector for the line and the equations of the lineare x = 4 7t, y = 1 + 2t, z = 7 + 3t.

63. A direction vector of the line is 6, 9, 3, and the normal vectors of the planes are (a) 4, 1, 2, (b) 2, 3, 1,(c) 10, 15, 5, (d) 4, 6, 2. Vectors (c) and (d) are multiples of the direction vector and hence thecorresponding planes are perpendicular to the line.

64. A direction vector of the line is

2, 4, 1

, and normal vectors to the planes are (a)

1,

1, 3

,

(b) 6, 3, 0, (c) 1, 2, 5, (d) 2, 1, 2. Since the dot product of each normal vector with the direc-tion vector is non-zero, none of the planes are parallel to the line.

65. Letting z = t in both equations and solving 5x 4y = 8 + 9t, x + 4y = 4 3t, we obtain x = 2 + t, y = 12 t,z = t.

66. Letting y = t in both equations and solving x z = 2 2t, 3x + 2z = 1 + t, we obtain x = 1 35 t, y = t,z = 1 + 75 t or, letting t = 5s, x = 1 3s, y = 5s, z = 1 + 7s.

67. Letting z = t in both equations and solving 4x 2y = 1 + t, x + y = 1 2t, we obtain x = 12 12 t, y = 12 32 t,z = t.

68. Letting z = t and using y = 0 in the first equation, we obtain x = 12 t, y = 0, z = t.

69. Substituting the parametric equations into the equation of the plane, we obtain 2(1+2 t)3(2t)+2(3t) = 7or t = 3. Letting t = 3 in the equation of the line, we obtain the point of intersection (5, 5, 9).

70. Substituting the parametric equations into the equation of the plane, we obtain (3 2t)+(1+6t)+4(2 12 t) = 12or 2t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (3, 1, 2).

71. 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).

72. Substituting the parametric equations into the equation of the plane, we obtain 4 + t 3(2 + t) + 2 ( 1 + 5t) = 0or t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (4, 2, 1).

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In Problems 73 and 74, 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.

73. Normal vectors are 1, 1, 4 and 2, 1, 1. A direction vector is

1, 1,

4

2,

1, 1

=

3,

9,

3

=

31, 3, 1

.

Equations of the line are x = 5 + t, y = 6 + 3t, z = 12 + t.74. Normal vectors are 2, 0, 1 and 1, 3, 1. A direction vector is

2, 0, 11, 3, 1 = 3, 3, 6 = 31, 1, 2.Equations of the line are x = 3 + t, y = 5 + t, z = 1 2t.

In Problems 75 and 76, 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.

75. A direction vector of the line is

3,

1, 5

and a normal vector to the given plane is

1, 1, 1

. A normal vector

to the desired plane is 3, 1, 5 1, 1, 1 = 6, 2, 4. A point on the line, and hence in the plane, is (4, 0, 1).The equation of the plane is 6(x 4) + 2(y 0) + 4(z 1) = 0 or 3x y 2z = 10.

76. A direction vector of the line is 3, 5, 2 and a normal vector to the given plane is 2, 4, 1. A normal vector tothe desired plane is 3, 5, 2 2, 4, 1 = 3, 1, 2. A point on the line, and hence in the plane, is (2 , 2, 8).The equation of the plane is 3(x 2) + (y + 2) + 2(z 8) = 0 or 3x + y + 2z = 20.

77.

x

y

z

10

2 6

78.

y

x

z

6

6

79. z

y6

2

x

80.

x

z

y4

4

6

81. z

y

x

12

4

82.

x

y

z

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EXERCISES 7.6Vector Spaces

1. Not a vector space. Axiom (vi) is not satisfied. 2. Not a vector space. Axiom (i) is not satisfied.

3. Not a vector space. Axiom (x) is not satisfied. 4. A vector space

5. A vector space 6. A vector space

7. Not a vector space. Axiom (ii) is not satisfied. 8. A vector space

9. A vector space10. Not a vector space. Axiom (i) is not satisfied.

11. A subspace 12. Not a subspace. Axiom (i) is not satisfied.

13. Not a subspace. Axiom (ii) is not satisfied. 14. A subspace

15. A subspace 16. A subspace

17. A subspace 18. A subspace

19. Not a subspace. Neither axioms (i) nor (ii) are satisfied.

20. A subspace

21. Let (x1, y1, z1) and (x2, y2, z2) be in S. Then

(x1, y1, z1) + (x2, y2, z2) = (at1, bt1, ct1) + (at2, bt2, ct2) = (a(t1 + t2), b(t1 + t2), c(t1 + t2))

is in S. Also, for (x, y, z) in S then k(x, y, z) = (kx,ky,kz) = (a(kt), b(kt), c(kt)) is also in S.

22. Let (x1, y1, z1) and (x2, y2, z2) be in S. Then ax1 + by1 + cz1 = 0 and ax2 + by2 + cz2 = 0. Adding gives

a(x1 + x2) + b(y1 + y2) + c(z1 + z2) = 0 and so (x1, y1, z1) + (x2, y2, z2) = (x1 + x2, y1 + y2, z1 + z2) is in S. Also,

for (x, y, z) then ax + by + cz = 0 implies k(ax + by + cz) = k 0 = 0 and a(kx) + b(ky) + c(kz) = 0. this meansk(x, y, z) = (kx,ky,kz) is in S.

23. (a) c1u1 + c2u2 + c3u3 = 0 if and only if c1 + c2 + c3 = 0, c2 + c3 = 0, c3 = 0. The only solution of this system

is c1 = 0, c2 = 0, c3 = 0.

(b) Solving the system c1 + c2 + c3 = 3, c2 + c3 = 4, c3 = 8 gives c1 = 7, c2 = 12, c3 = 8. Thusa = 7u1 12u2 + 8u3.

24. (a) The assumption c1p1 + c2p2 = 0 is equivalent to (c1 + c2)x + (c1 c2) = 0. Thus c1 + c2 = 0, c1 c2 = 0.The only solution of this system is c1 = 0, c2 = 0.

(b) Solving the system c1 + c2 = 5, c1 c2 = 2 gives c1 = 72 , c2 = 32 . Thus p(x) = 72p1(x) + 32p2(x)25. Linearly dependent since 6, 12 = 324, 826. Linearly dependent since 21, 1 + 30, 1 + (1)2, 5 = 0, 0

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7.6 Vector Spaces

27. Linearly independent

28. Linearly dependent since for all x (1) 1 + (2)(x + 1) + (1)(x + 1)2 + (1)x2 = 0.29. f is discontinuous at x = 1 and at x = 3.

30. (x, sin x) = 2

0

x sin x dx = (

x cos x + sin x) 2

0

=

2

31. x2 =20

x2 dx =1

3x32

0

=8

33 and so x = 2

23

3. Now

sin x2 =20

sin2 x dx =1

2

20

(1 cos2x) dx = 12

x 1

2sin2x

20

=

and so sin x = .32. A basis could be 1, x, ex cos3x, ex sin3x.

33. We need to show that Span{x1, x2, . . . , xn} is closed under vector addition and scalar multiplication. Supposeu and v are in Span{x1, x2, . . . , xn}. Then u = a1x1 + a2x2 + + anxn and v = b1x1 + b2x2 + + bnxn, so

thatu + v = (a1 + b1)x1 + (a2 + b2)x2 + + (an + bn)xn,

which is in Span{x1, x2, . . . , xn}. Also, for any real number k,ku = k(a1x1 + a2x2 + + anxn) = ka1x1 + ka2x2 + + kanxn,

which is in Span{x1, x2, . . . , xn}. Thus, Span{x1, x2, . . . , xn} is a subspace of V.34. R2 is not a subspace of either R3 or R4 and R3 is not a subspace of R4. The vectors in R2 are ordered pairs,

while the vectors in R3 are ordered triples.

35. Since a basis for M22 is

B = 1 00 0

, 0 10 0

, 0 01 0

, 0 00 1

,the dimension of M22 is 4.

36. To show that the set of nonzero orthogonal vectors is linearly independent we set c1v1 + c2v2 + + cnvn = 0.For 0 i n,

(c1v1 + c2v2 + + civi + cnvn) vi = c1v1 vi + c2v2 vi + + civi vi + cnvn vi = ci||vi||2,

so ci||vi||2 = 0 because(c1v1 + c2v2 + + civi + cnvn) vi = 0 vi = 0.

Since vi is a nonzero vector, ci = 0. Thus, the assumption that c1v1 + c2v2 + + cnvn = 0 leads toc1 = c2 =

= cn = 0, and the set is linearly independent.

37. We verify the four properties:

(i) (u, v) = u1v1 + 4u2v2 = v1u1 + 4v2u2 = (v, u)

(ii) (ku, v) = (ku1)v1 + 4(ku2)v2 = k(u1v1 + 4u2v2) = k(u, v)

(iii) (u, u) = u21 + 4ku22 > 0 for u = 0. Furthermore, u21 + 4ku22 = 0 if and only if u1 = 0 and u2 = 0, or

equivalently, u = 0.

(iv) (u, v + w) = u1(v1 + w1) + 4u2(v2 + w2) = (u1v1 + 4u2v2) + (u1w1 + 4u2w2) = (u, v) + (u, w)

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7.6 Vector Spaces

38. (a) Let u = 2, 1 and v = 2, 1 be nonzero vectors in R2. With respect to the standard inner or dot producton R2,

u v = 2, 1 2, 1 = 2 2 + 1 (1) = 3.We see that u and v are not orthogonal with respect to that inner product. But using the inner product

in Problem 37, we have(u, v) = 2 2 + 4(1) (1) = 0,

and so u and v are orthogonal with respect to that inner product.

(b) Consider f(x) = sin x and g(x) = cos x in C[0, 2]. Since20

sin x cos x dx =1

2

20

sin2x dx = 14

cos2x20

= 14

(1 1) = 0,

these functions are orthogonal in C[0, 2].

EXERCISES 7.7Gram-Schmidt Orthogonalization Process

1. Letting w1 = 1213 , 513 and w2 = 513 , 1213 , we have

w1 w2 =

12

13

5

13

+

5

13

12

13

= 0,

so the vectors are orthogonal. Also,

||w1|| =

12

13

2+

5

13

2= 1 and ||w2|| =

5

13

2+

12

13

2= 1,

so the basis is orthonormal. To express u = 4, 2 in terms of w1 and w2 we compute

u w1 = 4, 2

12

13,

5

13

= (4)

12

13

+ (2)

5

13

=

58

13

u w2 = 4, 2

5

13, 12

13

= (4)

5

13

+ (2)

12

13

= 4

13,

so

u =58

13

w1

4

13

w2.

2. Letting w1 = 1/

3, 1/

3, 1/3, w2 = 0, 1/

2, 1/2, and w3 = 2/

6, 1/

6, 1/6, we have

w1 w2 =

13

(0) +

1

3

1

2

+

1

3

1

2

= 0

w1 w3 =

13

2

6

+

1

3

1

6

+

1

3

1

6

= 0

w2 w3 = (0)

26

+

1

2

1

6

+

1

2

1

6

= 0,

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7.7 Gram-Schmidt Orthogonalization Process

so the vectors are orthogonal. Also,

||w1|| =

13

2+

1

3

2+

1

3

2= 1, ||w2|| =

02 +

1

2

2+

1

2

2= 1,

and ||w3|| = 262

+

162

+ 16

2

= 1,

so the basis is orthonormal. To express u = 5, 1, 6 in terms of w1, w2, and w3 we compute

u w1 = 5, 1, 6

13

,1

3, 1

3

= (5)

1

3

+ (1)

1

3

+ (6)

1

3

= 2

3

u w2 = 5, 1, 6

0, 12

, 12

= (5)(0) + (1)

1

2

+ (6)

1

2

= 5

2

u w3 = 5, 1, 6

26

,1

6, 1

6

= (5)

2

6

+ (1)

1

6

+ (6)

1

6

= 17

6

so

u = 23

w1 52

w2 176

w3.

Since the basis vectors in Problems 3 and 4 are orthogonal but not orthonormal, the result of Theorem 7.5 must be

slightly modified to read

u =u w1||w1||2 w1 +

u w2||w2||2 w2 + +

u wn||wn||2 wn.

The proof is very similar to that given in the text for Theorem7.5.

3. Letting w1 =

1, 0, 1

, w2 =

0, 1, 0

, and w3 =

1, 0, 1

we have

w1 w2 = (1)(0) + (0)(1) + (1)(0) = 0w1 w3 = (1)(1) + (0)(0) + (1)(1) = 0w2 w3 = (0)(1) + (1)(0) + (0)(1) = 0

so the vectors are orthogonal. We also compute

||w1||2 = 12 + 02 + 12 = 2||w2||2 = 02 + 12 + 02 = 1||w3||2 = (1)2 + 02 + 12 = 2

and, with u = 10, 7, 13,u w1 = (10)(1) + (7)(0) + (13)(1) = 3u w2 = (10)(0) + (7)(1) + (13)(0) = 7u w3 = (10)(1) + (7)(0) + (13)(1) = 23.

Then, using the result given before the solution to this problem, we have

u = 32

w1 + 7w2 232

w3.

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25/34


4. Letting w1 = 2, 1, 2, 0, w2 = 1, 2, 2, 1, w3 = 3, 4, 1, 3, and w4 = 5, 2, 4, 9 we havew1 w2 = (2)(1) + (1)(2) + (2)(2) + (0)(1) = 0w1 w3 = (2)(3) + (1)(4) + (2)(1) + (0)(3) = 0w1 w4 = (2)(5) + (1)(2) + (2)(4) + (0)(9) = 0w2 w3 = (1)(3) + (2)(4) + (2)(1) + (1)(3) = 0w2 w4 = (1)(5) + (2)(2) + (2)(4) + (1)(9) = 0w3 w4 = (3)(5) + (4)(2) + (1)(4) + (3)(9) = 0

so the vectors are orthogonal. We also compute

||w1||2 = 22 + 12 + (2)2 + 02 = 9||w2||2 = 12 + 22 + 22 + 12 = 10||w3||2 = 32 + (4)2 + 12 + 32 = 35||w4||2 = 52 + (2)2 + 42 + (9)2 = 126

and, with u = 1, 2, 4, 3,u w1 = (1)(2) + (2)(1) + (4)(2) + (3)(0) = 4u w2 = (1)(1) + (2)(2) + (4)(2) + (3)(1) = 16u w3 = (1)(3) + (2)(4) + (4)(1) + (3)(3) = 8u w4 = (1)(5) + (2)(2) + (4)(4) + (3)(9) = 10.

Then, using the result given before the solution to this problem, we have

u = 49

w1 +8

5w2 +

8

35w3 5

63w4.

5. (a) We have u1 = 3, 2 and u2 = 1, 1. Taking v1 = u1 = 3, 2, and using u2 v1 = 1 and v1 v1 = 13we obtain

v2 = u2 u2 v1v1 v1 v1 = 1, 1

1

133, 2 =

10

13, 15

13

.

Thus, an orthogonal basis is {3, 2, 1013 , 1513} and an orthonormal basis is {w1, w2}, where

w1 =1

||3, 2|| 3, 2 =113

3, 2 =

313

,213

and

w2 =1

||1013 , 1513||

1013

, 1513

=

1

5/

13

10

13, 15

13

=

2

13, 3

13

.

(b) We have u1 = 3, 2 and u2 = 1, 1. Taking v1 = u2 = 1, 1, and using u1 v1 = 1 and v1 v1 = 2

we obtainv2 = u1 u1 v1

v1 v1 v1 = 3, 2 1

21, 1 =

5

2,

5

2

.

Thus, an orthogonal basis is {1, 1, 52 , 52} and an orthonormal basis is {w3 , w4}, where

w3 =1

||1, 1|| 1, 1 =1

21, 1 =

1

2, 1

2

and

w4 =1

||52 , 52||

5

2,

5

2

=

1

5/

2

5

2,

5

2

=

1

2,

12

.

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(c)

4 2 2 4

4

2

2

4

1 0.5 0.5 1

1

0.5

0.5

1

1 0.5 0.5 1

1

0.5

0.5

1

u

v

w1

w2

w4

w3


v2 = u2 u2 v1v1 v1 v1 = 1, 0

3

253, 4 =

16

25, 12

25

.


w1 =1

||3, 4|| 3, 4 =1

53, 4 =

3

5,

4

5

and

w2 =1

||1625 , 1225||

1625

, 1225

=

14/5

16

25, 12

25

=

45

, 35

.

(b) We have u1 = 3, 4 and u2 = 1, 0. Taking v1 = u2 = 1, 0, and using u1 v1 = 3 and v1 v1 = 1we obtain

v2 = u1 u1 v1v1 v1 v1 = 3, 4

3

11, 0 = 0, 4 .

Thus, an orthogonal basis is {1, 0, 0, 4} and an orthonormal basis is {w3 , w4}, where

w3 =1

||1, 0|| 1, 0 =1

11, 0 = 1, 0

and

w4 =1

||0, 4|| 0, 4 =1

4 0, 4 = 0, 1 .(c)

4 2 2 4

4

2

2

4

1 0.5 0.5 1

1

0.5

0.5

1

1 0.5 0.5 1

1

0.5

0.5

1u

v

w1

w2

w4

w3

7. (a) We have u1 = 1, 1 and u2 = 1, 0. Taking v1 = u1 = 1, 1, and using u2 v1 = 1 and v1 v1 = 2 weobtain

v2 = u2 u2 v1v1 v1 v1 = 1, 0

121, 1 =

12

, 12

.


w1 =1

||1, 1|| 1, 1 =1

21, 1 =

1

2,

12

and

w2 =1

|| 12

, 12||

1

2, 1

2

=

1

1

1

2, 1

2

=

1

2, 1

2

.

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27/34


(b) We have u1 = 1, 1 and u2 = 1, 0. Taking v1 = u2 = 1, 0, and using u1 v1 = 1 and v1 v1 = 1 weobtain

v2 = u1 u1 v1v1 v1 v1 = 1, 1

1

11, 0 = 0, 1 .

Thus, an orthogonal basis is {1, 0, 0, 1}, which is also an orthonormal basis.(c)

2 1 1 2

2

1.5

1

0.5

0.5

1

1.5

2

1 0.5 0.5 1

1

0.5

0.5

1

1 0.5 0.5 1

1

0.5

0.5

1

u

v

w1

w2

w4

w3


v2 = u2 u2 v1v1

v1

v1 = 1, 2 974

5, 7 = 119

74, 85

74 .Thus, an orthogonal basis is {5, 7, 11974 , 8574} and an orthonormal basis is {w1, w2}, where

w1 =1

||5, 7|| 5, 7 =174

5, 7 =

574

,774

and

w2 =1

|| 11974 , 8574||

119

74, 85

74

=

1

17/

74

119

74, 85

74

=

774

, 574

.

(b) We have u1 = 5, 7 and u2 = 1, 2. Taking v1 = u2 = 1, 2, and using u1 v1 = 9 and v1 v1 = 5we obtain

v2 = u1 u1 v1v1

v1

v1 = 5, 7 951, 2 =

34

5,

17

5 .Thus, an orthogonal basis is {1, 2, 345 , 175 } and an orthonormal basis is {w3 , w4}, where

w3 =1

||1, 2|| 1, 2 =1

51, 2 =

1

5, 2

5

and

w4 =1

|| 345 , 175 ||

34

5,

17

5

=

1

17/

5

34

5,

17

5

=

2

5,

15

.

(c)

7.552.5 2.5 5 7.5

8

6

4

2

2

4

6

8

1 0.5 0.5 1

1

0.5

0.5

1

1 0.5 0.5 1

1

0.5

0.5

1u

v

w1

w2

w4

w3

9. We have u1 = 1, 1, 0, u2 = 1, 2, 2, and u3 = 2, 2, 1. Taking v1 = u1 = 1, 1, 0 and using u2 v1 = 3 andv1 v1 = 2 we obtain

v2 = u2 u2 v1v1 v1 v1 = 1, 2, 2

3

21, 1, 0 =

1

2,

1

2, 2

.

386


28/34


Next, using u3 v1 = 4, u3 v2 = 2, and v2 v2 = 92 , we obtain

v3 = u3 u3 v1v1 v1 v1

u3 v2v2 v2 v2 = 2, 2, 1

4

21, 1, 0 2

9/2

1

2,

1

2, 2

=

2

9, 2

9,

1

9

.

Thus, an orthogonal basis is

B =

1, 1, 0 ,

12

, 12

, 2

,

29

, 29

, 19

,

and an orthonormal basis is

B =

1

2,

12

, 0

,

1

3

2,

1

3

2,

4

3

2

,

2

3, 2

3,

1

3

.

10. We have u1 = 3, 1, 1, u2 = 1, 1, 0, and u3 = 1, 4, 1. Taking v1 = u1 = 3, 1, 1 and using u2 v1 = 2and v1 v1 = 11 we obtain

v2 = u2 u2 v1v1 v1 v1 = 1, 1, 0

211

3, 1, 1 =

5

11,

13

11,

2

11

.

Next, using u3 v1 = 8, u3 v2 =49

11 , and v2 v2 =18

11 , we obtain

v3 = u3 u3 v1v1 v1 v1

u3 v2v2 v2 v2 = 1, 4, 1

8

113, 1, 1 49/11

18/11

5

11,

13

11,

2

11

=

1

18,

1

18, 2

9

.


B =

3, 1, 1 ,

5

11,

13

11,

2

11

,

1

18,

1

18, 2

9

,


B =

3

11,

111

,111

,

5

3

22,

13

3

22,

2

3

22

,

1

3

2,

1

3

2,

4

3

2

.

11. We have u1 = 12 ,

12 , 1, u2 = 1, 1,

12, and u3 = 1,

12 , 1. Taking v1 = u1 =

12 ,

12 , 1 and using u2 v1 =

12

and v1 v1 = 32 we obtain

v2 = u2 u2 v1v1 v1 v1 =

1, 1, 1

2

1/2

3/2

1

2,

1

2, 1

=

5

6,

7

6, 1

6

.

Next, using u3 v1 = 34 , u3 v2 = 54 , and v2 v2 = 2512 , we obtain

v3 = u3 u3 v1v1 v1 v1

u3 v2v2 v2 v2 =

1, 1

2, 1

3/4

3/2

1

2,

1

2, 1

5/4

25/12

5

6,

7

6, 1

6

=

3

4, 9

20,

3

5

.


B = 12

,1

2

, 1 ,5

6

,7

6

,

1

6 ,

3

4

,

9

20

,3

5 ,


B =

1

6,

16

,2

6

,

1

3,

7

5

3, 1

5

3

,

1

2, 3

5

2,

4

5

2

.

12. We have u1 = 1, 1, 1, u2 = 9, 1, 1, and u3 = 1, 4, 2. Taking v1 = u1 = 1, 1, 1 and using u2 v1 = 9and v1 v1 = 3 we obtain

v2 = u2 u2 v1v1 v1 v1 = 9, 1, 1

9

31, 1, 1 = 6, 4, 2 .

387


29/34


Next, using u3 v1 = 1, u3 v2 = 18, and v2 v2 = 56, we obtain

v3 = u3 u3 v1v1 v1 v1

u3 v2v2 v2 v2 = 1, 4, 2

1

31, 1, 1 18

566, 4, 2 =

25

42,

50

21, 125

42

.


B =1, 1, 1 , 6, 4, 2 ,25

42, 50

21, 125

42

,


B =

1

3,

13

,1

3

,

314

, 214

, 114

,

142

,442

, 542

.

13. We have u1 = 1, 5, 2, and u2 = 2, 1, 1. Taking v1 = u1 = 1, 5, 2 and using u2 v1 = 5 and v1 v1 = 30we obtain

v2 = u2 u2 v1v1 v1 v1 = 2, 1, 1

5

301, 5, 2 =

13

6,

1

6,

2

3

.

Thus, an orthogonal basis is B =

1, 5, 2 ,

136 , 16 , 23

, and an orthonormal basis is

B =

130

, 530

, 230

, 13

186, 1

186, 4

186

.

14. We have u1 = 1, 2, 3, and u2 = 3, 4, 1. Taking v1 = u1 = 1, 2, 3 and using u2 v1 = 14 and v1 v1 = 14 weobtain

v2 = u2 u2 v1v1 v1 v1 = 3, 4, 1

14

141, 2, 3 = 2, 2, 2 .

Thus, an orthogonal basis is B = {1, 2, 3 , 2, 2, 2} , and an orthonormal basis is

B =

114

,214

,314

,

1

3,

13

, 13

.

15. We have u1 = 1, 1, 1, 1, and u2 = 1, 3, 0, 1. Taking v1 = u1 = 1, 1, 1, 1 and using u2 v1 = 3 and

v1 v1 = 4 we obtainv2 = u2 u2 v1

v1 v1 v1 = 1, 3, 0, 1 34

1, 1, 1, 1 =

7

4,

9

4,

3

4,

1

4

.

Thus, an orthogonal basis is B =1, 1, 1, 1 , 74 , 94 , 34 , 14 , and an orthonormal basis is

B =

1

2, 1

2,

1

2, 1

2

,

7

2

35,

9

2

35,

3

2

35,

1

2

35

.

16. We have u1 = 4, 0, 2, 1, u2 = 2, 1, 1, 1, and u3 = 1, 1, 1, 0. Taking v1 = u1 = 4, 0, 2, 1 and usingu2 v1 = 5 and v1 v1 = 21 we obtain

v2 = u2 u2 v1v1 v1 v1 = 2, 1, 1, 1

5

214, 0, 2, 1 =

22

21, 1, 31

21,

26

21

.

Next, using u3 v1 = 2, u3 v2 = 7421 , and v2 v2 = 12221 , we obtain

v3 = u3 u3 v1v1 v1 v1

u3 v2v2 v2 v2

= 1, 1, 1, 0 221

4, 0, 2, 1 74/21122/21

22

21, 1, 31

21,

26

21

=

1

61,

24

61, 18

61, 40

61

.


B =

4, 0, 2, 1 ,

22

21, 1, 31

21,

26

21

,

1

61,

24

61, 18

61, 40

61

,

388


30/34



B =

421

, 0,221

, 121

,

222562

,212562

, 312562

,262562

,

1

2501,

24

2501,

18

2501,

40

2501.17. We have u1 = 1, u2 = x, and u3 = x

2. Taking v1 = u1 = 1 and using

(u2, v1) =

11

1 x2 dx = 0 and (v1, v1) =11

x x dx = 2

we obtain

v2 = u2 (u2, v1)(v1, v1)

v1 = x 02

x = x.

Next, using

(u3, v1) = 1

1x2

1 dx =

2

3, (u3, v2) =

1

1x2

x dx = 0, and (v2, v2) =

1

1x

x dx =

2

3,

we obtain

v3 = u3 (u3, v1)(v1, v1)

v1 (u3, v2)(v2, v2)

v2 = x2 2/3

21 0

2/3x = x2 1

3.


1, x , x2 13

.

18. We have u1 = x2 x, u2 = x2 + 1, and u3 = 1 x2. Taking v1 = u1 = x2 x and using

(u2, v1) =

11

(x2 + 1)(x2 x)dx = 1615

and (v1, v1) =

11

(x2 x)(x2 x)dx = 1615

we obtain

v2 = u2 (u2, v1)

(v1, v1) v1 = x2

+ 1 16/15

16/15 (x2

x) = x + 1.Next, using

(u3, v1) =

11

(1 x2)(x2 x)dx = 415

, (u3, v2) =

11

(1 x2)(x + 1)dx = 43

,

and

(v2, v2) =

11

(x + 1)(x + 1)dx =8

3,

we obtain

v3 = u3 (u3, v1)(v1, v1)

v1 (u3, v2)(v2, v2)

v2 = 1 x2 4/1516/15

(x2 x) 4/38/3

(x + 1) = 54

x3 14

x +1

2.


x2 x, x + 1, 54 x3 14 x + 12

.

19. Using the solution of Problem 17 and computing

||v1||2 = (v1, v1) =11

1 1 dx = 2, ||v2||2 = (v2, v2) =11

x x dx = 23

,

and

||v3||2 = (v3, v3) =11

x2 1

3

x2 1

3

dx =

8

45,

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31/34


we see that an orthonormal basis is

B =

1

2,

x2/3

,x2 1/3

8/45

=

1

2,

36

x,15

2

10

x2 1

3

.

20. Using the solution of Problem 18 and computing

||v1||2 = (v1, v1) =

1

1(x2 x)(x2 x)dx = 16

15, ||v2||2 = (v2, v2) =

1

1(x + 1)(x + 1)dx =

83

,

and

||v3||2 = (v3, v3) =11

5

4x3 1

4x +

1

2

5

4x3 1

4x +

1

2

dx =

1

3,

we see that an orthonormal basis is

B =

15

4(x2 x), 3

2

6(x + 1),

3

4(5x2 x + 2)

.

21. Using w1 = 1/

2, w2 = 3x/

6, and w3 = (15/2

10)(x2 1/3), and computing

(p,w1) =11

(9x2 6x + 5) 12

dx = 82,

(p,w2) =

11

(9x2 6x + 5) 36

x dx = 2

6

(p,w3) =

11

(9x2 6x + 5)

15

2

10

x2 1

3

dx =

1210

,

we find from Theorem 7.5

p(x) = 9x2 6x + 5 = (p,w1)w1 + (p,w2)w2 + (p,w3)w3 = 8

2 w1 2

6 w2 +12

10w3.

22. Using w1 = (

15/4)(x2 x), w2 = (3/2

6)(x + 1), and w3 = (

3/4)(5x2 + x 2), and computing

(p,w1) =

1

1(9x2 6x + 5)

154

x2 x dx = 41

15,

(p,w2) =

11

(9x2 6x + 5)

3

2

6(x + 1)

dx = 3

6

(p,w3) =

11

(9x2 6x + 5)

3

4(5x2 + x 2)

dx =

13

,

we find from Theorem 7.5

p(x) = 9x2 6x + 5 = (p,w1)w1 + (p,w2)w2 + (p,w3)w3 = 4115

w1 + 3

6 w2 +1

3w3.

23. Since u3 depends on u1 and u2 we would expect the Gram-Schmidt process to yield a pair of orthogonal vectors

v1 and v2, with a third vector v3 that is 0. This is because u3 lies in the subspace W2 of R3 spanned by u1

and u2, and hence the projection of u3 onto W2 is u3 itself. In other words,

u3 = projW3u3 =u3 v1v1 v1 v1 +

u3 v2v2 v2 v2 so v3 = u3

u3 v1v1 v1 v1 +

u3 v2v2 v2 v2 = 0.

To carry out the orthogonalization process we take v1 = u1 = 1, 1, 3. Then, using u2 v1 = 8 and v1 v1 = 11we obtain

v2 = u2 u2 v1v1 v1 v1 = 1, 4, 1

8

111, 1, 3 =

3

11,

36

11, 13

11

.

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32/34

CHAPTER 7 REVIEW EXERCISES

Next, using u3 v1 = 2, u3 v2 = 40211 , and v2 v2 = 13411 , we obtain

v3 = u3 u3 v1v1 v1 v1

u3 v2v2 v2 v2 = 1, 10, 3

2

111, 1, 3 402/11

134/11

3

11,

36

11, 13

11

= 0, 0, 0 .

In this case {v1, v2} = {1, 1, 3}, 311 , 3611 , 1311} is an orthogonal subset of R3 containing the third vectoru3 = 1, 10, 3.


1. True

2. False; the points must be non-collinear.3. False; since a normal to the plane is 2, 3, 4 which is not a multiple of the direction vector 5, 2, 1 of the

line.

4. True 5. True 6. True 7. True 8. 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 = 0.)

11. 9i + 2j + 2k 12. orthogonal

13. 5(k j) = 5(i) = 5i 14. i (i j) = i k = 0

15. (12)2 + 42 + 62 = 14

16. (1 20)i (2 0)j + (8 0)k = 21i + 2j + 8k17. 6i +j 7k18. The coordinates of (1, 2, 10) satisfy the given equation.19. 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 = 4,y = 2 + 3(3) = 7, z = 1 + 2(3) = 5, or (4, 7, 5).

20. |a| =

42 + 32 + (5)2 = 52 ; u = 15

2(4i + 3j 5k) = 4

5

2i 3

5

2j +

12

k

21. x2 2 = 3, x2 = 5; y2 1 = 5, y2 = 6; z2 7 = 4, z2 = 3; P2 = (5, 6, 3)

22. (5, 1/2, 5/2)

23. (7.2)(10)cos 135 = 36224. 2b = 2, 4, 2; 4c = 0, 8, 8; a (2b + 4c) = 3, 1, 0 2, 4, 10 = 1025. 12, 8, 626. cos =

a b|a||b| =

12

2=

1

2; = 60

27. A =1

2|5i 4j 7k| = 3

10

2

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33/34


28. From 3(x 3) + 0(y 6) + (1)(z (2)) = 0 we obtain 3x + z = 7.29. | 5 (3)| = 230. parallel: 2c = 5, c = 5/2; orthogonal: 1(2) + 3(6) + c(5) = 0, c = 4

31. a b = i j k

1 1 01 2 1

= 1 0

2 1 i 1 01 1 j + 1 11 2 k = i j 3k A unit vector perpendicular to both a

and b isa b

a b =1

1 + 1 + 9(i j 3k) = 1

11i 1

11j 3

11k.

32. a =

1/4 + 1/4 + 1/6 =3

4; cos =

1/2

3/4=

2

3, 48.19; cos = 1/2

3/4=

2

3, 48.19;

cos =1/43/4

= 13

, 109.47

33. compba = a b/b = 1, 2, 2 4, 3, 0/5 = 234. compab = b

a/

a

=

4, 3, 0

1, 2,

2

/3 = 10/3

projab = (compab)a/a = (10/3)1, 2, 2/3 = 10/9, 20/9, 20/935. a + b = 1, 2, 2 + 4, 3, 0 = 5, 5, 2

compa

(a + b) = (a + b) a/1 + 4 + 4 = 13(a a + b a) = 13 [(1 + 4 + 4) + (4 + 6 + 0)] = 193proja(a + b) = [compa(a + b)](a/a) = 193 13 , 23 , 23 = 199 , 389 , 389

36. a b = 1, 2, 2 4, 3, 0 = 3, 1, 2comp

b(a b) = (a b) b/16 + 9 = 15(a b b b) = 15 [(4 + 6 + 0) (16 + 9)] = 3

projb

(a b) = [compb

(a b)](b/b) = 345 , 35 , 0 = 125 , 95 , 037. Let a = a,b,c and r = x, y, z. Then

(a) (r a) r = x a, y b, z c x, y, z = x2 ax + y2 by + z2 zc = 0 implies

(x a2

)2 + (y b2

)2 + (z c2

)2 = a2

+ b2

+ c2

4. The surface is a sphere.

(b) (r a) a = x a, y b, z c a,b,c = a(x a) + b(y b) + c(z c) = 0The surface is a plane.

38. 4, 2, 2 2, 4, 3 = 2, 2, 1; 2, 4, 3 6, 7, 5 = 4, 3, 2; 2, 2, 1 4, 3, 2 = 0The points are the vertices of a right triangle.

39. A direction vector of the given line is 4, 2, 6. A parallel line containing (7, 3, 5) is (x7)/4 = (y3)/(2) =(z + 5)/6.

40. A normal to the plane is 8, 3, 4. The line with this direction vector and through (5, 9, 3) is x = 5 + 8t,y = 9 + 3t, z = 3 4t.

41. The direction vectors are 2, 3, 1 and 2, 1, 1. Since 2, 3, 1 2, 1, 1 = 0, the lines are orthogonal. Solving1 2t = x = 1 + 2s, 3t = y = 4 + s, we obtain t = 1 and s = 1. The point (3, 3, 0) obtained by lettingt = 1 and s = 1 is common to the two lines, so they do intersect.

42. Vectors in the plane are 2, 3, 1 and 1, 0, 2. A normal vector is 2, 3, 11, 0, 2 = 6, 3, 3 = 32, 1, 1.An equation of the plane is 2x y z = 0

43. The lines are parallel with direction vector 1, 4, 2. Since (0, 0, 0) is on the first line and (1, 1, 3) is on the secondline, the vector 1, 1, 3 is in the plane. A normal vector to the plane is thus 1, 4, 2 1, 1, 3 = 14, 5, 3.An equation of the plane is 14x 5y 3z = 0.

392


34/34


44. 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 3, 11, 1 and an equation of the plane is

3(x 1) + 11(y 7) + (z + 1) = 0 or 3x + 11y + z = 79.

45. F = 10a

a=

10

2(i +j) = 5

2 i + 5

2j; d =

7, 4, 0

4, 1, 0

= 3i + 3j

W = F d = 152 + 152 = 302 N-m46. F = 5

2 i + 5

2j + 50i = (5

2 + 50)i + 5

2j; d = 3i + 3j

W = 15

2 + 150 + 15

2 = 30

2 + 150 N-m 192.4 N-m47. Since F2 = 200(i +j)/

2 = 100

2 i + 100

2j, F3 = F2 F1 = (100

2 200)i + 1002j and

F3 =

(100

2 200)2 + (1002)2 = 200

2 2 153 lb.48. Let F1 = F1 and F2 = F2. Then F1 = F1[(cos 45)i + (sin 45)j] and F2 = F2[(cos120)i + (sin 120)j], or

F1 = F1(12

i + 12j) and F2 = F2(12 i +

32

j). Since w + F1 + F2 = 0,

F1(

1

2 i +1

2j) + F2(1

2 i +

3

2 j) = 50j, (

1

2 F1 1

2 F2)i + (

1

2 F1 +

3

2 F2)j = 50j

and1

2F1 1

2F2 = 0,

12

F1 +

3

2F2 = 50.

Solving, we obtain F1 = 25(

6 2 ) 25.9 lb and F2 = 50(

3 1) 36.6 lb.49. Not a vector space. Axiom (viii) is not satisfied.

50. The vectors are linearly independent. The only solution of the system

c1 = 0, c1 + 2c2 + c3 = 0, 2c1 + 3c2 c3 = 0

is c1 = 0, c2 = 0, c3 = 0.51. Let p1 and p2 be in Pn such that

d2p1dx2

= 0 andd2p2dx2

= 0. Since

0 =d2p1dx2

+d2p2dx2

=d2

dx2(p1 +p2) and 0 = k

d2p1dx2

=d2

dx2(kp1)

we conclude that the set of polynomials with the given property is a subspace of Pn. A basis for the subspace

is 1, x.

52. The intersection W1 W2 is a subspace of V. If x and y are in W1 W2 then x and y are in each subspaceand so x + y is in each subspace. That is, x + y is in W1 W2. Similarly, if x is in W1 W2 then x is in eachsubspace and so kx is in each subspace. That is, kx is in W1 W2 for any scalar k.

The union W1W2 is generally not a subspace. For example, W1 = {x, y y = x} and W2 = {x, y y = 2x}are subspaces of R2. Now 1, 1 is in W1 and 1, 2 is in W2 but 1, 1 + 1, 2 = 2, 3 is not in W1 W2.

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A Em 3 Chapter 7