calculus one and several variables 10e salas solutions manual ch13

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-13 JWDD027-Salas-v1 November 30, 2006 13:43

SECTION 13.1 687

CHAPTER 13

SECTION 13.1

1.

length AB : 2√

5

midpoint: (1, 0,−2)

2.

length AB : 2√

10

midpoint: (0,−1, 3)

3.

length AB : 5√

2

midpoint:(2,− 1

2 ,52

)

4.

length AB : 9

midpoint: (1, 32 , 3)

5. z = −2 6. y = 1 7. y = 1

8. z = −2 9. x = 3 10. x = 3

11. x2 + (y − 2)2 + (z + 1)2 = 9 12. (x− 1)2 + y2 + (z + 2)2 = 16

13. (x− 2)2 + (y − 4)2 + (z + 4)2 = 36 14. x2 + y2 + z2 = 9

15. (x− 3)2 + (y − 2)2 + (z − 2)2 = 13 16. (x− 2)2 + (y − 3)2 + (z + 4)2 = 16



688 SECTION 13.1

17. x2 + y2 + z2 + 4x− 8y − 2z + 5 = 0

x2 + 4x + 4 + y2 − 8y + 16 + z2 − 2z + 1 = −5 + 4 + 16 + 1

(x + 2)2 + (y − 4)2 + (z − 1)2 = 16

center: (−2, 4, 1), radius: 4

18. Rewrite as x2 − 4x + 4 + y2 + z2 − 2z + 1 = −1 + 4 + 1 = 4

=⇒ (x− 2)2 + y2 + (z − 1)2 = 4 center (2, 0, 1); radius 2

19. (2, 3,−5) 20. (2,−3, 5) 21. (−2, 3, 5)

22. (2,−3,−5) 23. (−2, 3,−5) 24. (−2,−3, 5)

25. (−2,−3,−5) 26. (0, 3, 5) 27. (2,−5, 5)

28. (2, 3, 3) 29. (−2, 1,−3) 30. (6,−3,−3)

31. d(PR) =√

14, d(QR) =√

45, d(PQ) =√

59; [d(PR)]2 + [d(QR)]2 = [d(PQ)]2

32. Let the vertices be (xi, yi, zi), i = 1, 2, 3. Then

(x1 + x2

2,y1 + y2

2,z1 + z2

2

)= (5,−1, 3);

(x2 + x3

2,y2 + y3

2,z2 + z3

2

)= (4, 2, 1);

(x1 + x3

2,y1 + y3

2,z1 + z3

2

)= (2, 1, 0)

Solving simultaneously gives vertices (3,−2, 2), (7, 0, 4), (1, 4,−2).

33. The sphere of radius 2 centered at the origin, together with its interior.

34. The exterior of the sphere of radius 3 centered at the origin.

35. A rectangular box in the first octant with sides on the coordinate planes and dimensions 1 × 2 × 3,

together with its interior.

36. A cube of side length 4, together with its interior; the origin in at the center of the cube.

37. A circular cylinder with base the circle x2 + y2 = 4 and height 4, together with its interior.

38. x2 + y2 + z2 = 4 and x2 + y2 + z2 = 9 are concentric spheres; Ω is the region between the two

spheres.



SECTION 13.1 689

39. Let B = (x, y, z). Thenx + 2

2= 1 =⇒ x = 0,

y + 32

= 2 =⇒ y = 1,z + 4

2= 3 =⇒ z = 2.

Therefore B = (0, 1, 2).

41. Let P1 = (x, y, z) be the trisection point closest to A. Then−→AP1= 1

3

−→AB=⇒ (x− a1, y − a2, z − a3) = 1

3 (b1 − a1, b2 − a2, b3 − a3).

Solving for x, y, z gives (x, y, z) =(

2a1 + b13

,2a2 + b2

3,2a3 + b3

3

).

Similarly, if P2 = (x, y, z) is the trisection point closest to B, then

(x, y, z) =(a1 + 2b1

3,a2 + 2b2

3,a3 + 2b3

3

).

42. The points on the line segment AB are given by x = 1 + t, y = −2 + 3t, z =√

2 −√

2 t, 0 ≤ t ≤ 1.

The line segment AP has length 3 if√t2 + (3t)2 + (−

√2 t)2 =

√12t2 = 2t

√3 = 3 =⇒ t = 1

2

√3.

Thus, the point P on the line segment AB that is 3 units from A has coordinates:

1 +12

√3, −2 +

32

√3,

√2 −

√2

2

√3.

43. Substituting the coordinates of the points into the equation Ax + By + Cz + D = 0, we get the

equations

Ax0 + D = 0, By0 + D = 0, Cz0 + D = 0 which implies Ax0 = By0 = Cz0.

Therefore, we have

Ax +Ax0

y0y +

Ax0

z0z + D = 0 or

x

x0+

y

y0+

z

z0+

D

Ax0= 0.

Substituting the point (x0, 0, 0) into this equation givesx

x0+

y

y0+

z

z0= 1.

44. Substituting the coordinates of the points into the equation Ax + By + Cz + D = 0, we get the

equations

Ax0 + By0 + D = 0, Ax0 + Cz0 + D = 0, By0 + Cz0 + D = 0

which implies A = Cz0x0

, B = Cz0y0

. Therefore, we have

Cz0

x0x +

Cz0

y0y + Cz + D = 0 or

x

x0+

y

y0+

z

z0+

D

Cz0= 0.

Substituting the point (x0, y0, 0) into this equation givesx

x0+

y

y0+

z

z0= 2.



690 SECTION 13.2

45. (i) a3 �= 0 The line through the origin and (a1, a2, a3) is given by x = a1t, y = a2t, z = a3t, t

any real number. The line intersects the plane z = z0 at the point Q where t = z0/a3. The

coordinates of Q are: a1a3z0,

a2a3z0, z0.

(ii) a3 = 0 If z0 �= 0, the line does not intersect the plane. If z0 = 0, the line lies in the plane.

46. Using the arguments in Exercise 45 we have (i) a1 �= 0: Q =(x0,

a2a1x0,

a3a1x0

). (ii) a1 = 0:

If x0 �= 0, the line does not intersect the plane. If x0 = 0, the line lies in the plane.

47. The ray that emanates from the origin and passes through the point (a1, a2, a3) is given by x = a1t,

y = a2t, z = a3t, t ≥ 0. The ray intersects the sphere x2 + y2 + z2 = 1 at the point Q where

a21t

2 + a22t

2 + a23t

2 = 1 =⇒ t =1√

a21 + a2

2 + a23

.

The coordinates of Q are:a1√

a21 + a2

2 + a23

,a2√

a21 + a2

2 + a23

,a3√

a21 + a2

2 + a23

.

48. It follows from Exercise 47 that the points where the line x = a1t, y = a2t, z = a3t, intersects the

sphere x2 + y2 + z2 = 1 are:±a1√

a21 + a2

2 + a23

,±a2√

a21 + a2

2 + a23

,±a3√

a21 + a2

2 + a23

.

SECTION 13.2

1.−→PQ= (3, 4,−2); ‖

−→PQ ‖ =

√29 2.

−→PQ= (−2, 6, 0); ‖

−→PQ ‖ = 2

√10

3.−→PQ= (0,−2,−1); ‖

−→PQ ‖ =

√5 4.

−→PQ= (4, 3,−8); ‖

−→PQ ‖ =

√89

5. 2a − b = (2 · 1 − 3, 2 · [−2] − 0, 2 · 3 + 1) = (−1,−4, 7)

6. 2b + 3c = (6, 0,−2) + (−12, 6, 3) = (−6, 6, 1)

7. −2a + b − c = [−(2a − b)] − c = (1 + 4, 4 − 2,−7 − 1) = (5, 2,−8)

8. a + 3b − 2c = (1,−2, 3) + 3(3, 0,−1) − 2(−4, 2, 1) = (18,−6,−2).

9. 3 i − 4 j + 6k 10. 3 i + 5 j + k

11. −3 i − j + 8k 12. 14 i + 4 j − 12k

13. 5 14.√

2 15. 3

16.√

41 17.√

6 18.√

2

19. (a) a, c, and d since a = 13c = − 1

2d

(b) a and c since a = 13c

(c) a and c both have direction opposite to d



SECTION 13.2 691

20. Let R be the point R (x, y, z). Then−→QR= (x− 3, y + 1, z − 1) and

−→OP= (1, 4,−2).

−→QR=

−→OP=⇒ x− 3 = 1, y + 1 = 4, z − 1 = −2 =⇒ x = 4, y = 3, z = −1.

21.−→RQ= (3 − x,−1 − y, 1 − z) and

−→OP= (1, 4,−2).

−→RQ=

−→OP=⇒ 3 − x = 1, −1 − y = 4, 1 − z = −2 =⇒ x = 2, y = −5, z = 3.

22.−→RQ= (3 − x,−1 − y, 1 − z) = 3

−→OP= (3, 12,−6) =⇒ 3 − x = 3, −1 − y = 12, 1 − z = −6

=⇒ x = 0, y = −13, z = 7.

23.−→RQ= (3 − x,−1 − y, 1 − z) = −2

−→OP= (−2,−8, 4) =⇒ 3 − x = −2, −1 − y = −8, 1 − z = 4

=⇒ x = 5, y = 7, z = −3.

24. ‖a‖ − ‖b‖ ≤ ‖a − b‖ since

‖a‖ = ‖(a − b) + b‖ ≤ ‖a − b‖ + ‖b‖.

Similarly ‖b‖ − ‖a‖ ≤ ‖b − a‖ = ‖a − b‖.

25. ‖a‖ = 5;a

‖a‖ =(

35,− 4

5, 0

)26.

( −2√13

,3√13

)

27. ‖a‖ = 3;a

‖a‖ =13

i − 23

j +23

k 28.(

23,13,23

)

29. ‖a‖ =√

14; − a‖a‖ =

1√14

i − 3√14

j − 2√14

k 30. − 2√5

i +1√5

k

31. (i) a − b (ii) −(a + b) (iii) a − b (iv) b − a

32. (a) 6 i + 3 j + 12k

(b) A(1, 1, 1) + B(−1, 3, 2) + C(−3, 0, 1) = (4,−1, 1).

Solve simultaneously to get A = 267 , B = − 11

7 , C = 37

33. (a) a − 3b + 2c + 4d = (2 i − k) − 3(i + 3 j + 5k) + 2(−i + j + k) + 4(i + j + 6k)

= i − 3 j + 10k

(b) The vector equation

(1, 1, 6) = A(2, 0,−1) + B(1, 3, 5) + C(−1, 1, 1)

implies1 = 2A + B − C,

1 = 3B + C,

6 = −A + 5B + C.

Simultaneous solution gives A = −2, B = 32 , C = − 7

2 .



692 SECTION 13.3

34. α = −12 35. ‖3 i + j‖ = ‖α j − k‖ =⇒ 10 = α2 + 1 so

α = ±336.

13(i − 2 j + 2k)

37.

‖αi + (α− 1) j + (α + 1)k‖ = 2 =⇒ α2 + (α− 1)2 + (α + 1)2 = 4

=⇒ 3α2 = 2 so α = ± 13

√6

38.2√6(i + 2 j − k) =

√6

3(i + 2 j − k)

39. ± 213

√13 (3 j + 2k) since ‖α(3 j + 2k)‖ = 2 =⇒ α = ± 2

13

√13

40. (i) c = a +12(b − a) =

12(a + b) (ii) a + c =

12(a + b) =⇒ c =

12(b − a)

41. (a) Since ‖a − b‖ and ‖a + b‖ are the lengths of the diagonals of the parallelogram, the

parallelogram must be a rectangle.

(b) Simplify

√(a1 − b1)2 + (a2 − b2)2 + (a3 − b3)2 =

√(a1 + b1)2 + (a2 + b2)2 + (a3 + b3)2.

the result is a1b1 + a2b2 + a3b3 = 0.

42. (a) If α > 0, then

‖a + αa‖ = ‖(1 + α)a‖ = (1 + α)‖a‖ = ‖a‖ + α‖a‖ = ‖a‖ + ‖αa‖.

(b) The equation does not necessarily hold if α < 0. For example, if α = −1,

0 = ‖a + (−1)a‖ �= ‖a‖ + ‖(−1)a‖ = 2‖a‖.

43. Let P = (x1, y1, z1), Q = (x2, y2, z2), and M = (xm, ym, zm). Then

(xm, ym, zm) = (x1, y1, z1) +12

(x2 − x1, y2 − y1, z2 − z1) =⇒ m = p +12(q − p).

44. r − p = 2(q − r) =⇒ 3r = p + 2q =⇒ r =13p +

23q

SECTION 13.3

1. a · b = (2)(−2) + (−3)(0) + (1)(3) = −1 2. a · b = (4)(−2) + (2)(2) + (−1)(1) = −5

3. a · b = (2)(1) + (−4)(1/2) + (0)(0) = 0 4. a · b = (−2)(3) + (0)(0) + (5)(1) = −1

5. a · b = (2)(1) + (1)(1) − (2)(2) = −1 6. a · b = (2)(1) + (3)(4) + (1)(0) = 14



SECTION 13.3 693

7. a · b

8. a · (a − b) + b · (b + a) = a · a − a · b + b · b + b · a = ‖a‖2 + ‖b‖2

9. (a − b) · c + b · (c + a) = a · c − b · c + b · c + b · a = a · (b + c)

10. a · (a + 2c) + (2b − a) · (a + 2c) − 2b · (a + 2c) = (a + 2b − a − 2b) · (a + 2c) = 0

11. (a) a · b = (2)(3) + (1)(−1) + (0)(2) = 5

a · c = (2)(4) + (1)(0) + (0)(3) = 8

b · c = (3)(4) + (−1)(0) + (2)(3) = 18

(b) ‖a‖ =√

5, ‖b‖ =√

14, ‖c‖ = 5. Then,

cos <) (a,b) =a · b

‖a‖ ‖b‖ =5(√

5) (√

14) =

114

√70,

cos <) (a, c) =8(√

5)(5)

=825

√5,

cos <) (b, c) =18(√

14)(5)

=935

√14.

(c) ub =1√14

(3 i − j + 2k), compb a = a · ub =1√14

(6 − 1) =514

√14,

uc =15(4 i + 3k), compc a = a · uc =

85

(d) projb a = (compb a)ub =514

(3 i − j + 2k), projc a = (compc a)uc =825

(4 i + 3k)

12. (a) a · b = 5, a · c = −3, b · c = 4

(b) cos <) (a,b) = 16

√10, cos <) (a, c) = − 3

10 , cos <) (b, c) = 215

√10

(c) compba = 53 , compca = − 3

10

√10

(d) projba = 59 (2 i − j + 2k), projca = − 3

10 (3 i − k)

13. u = cosπ

3i + cos

π

4j + cos

2π3

k =12

i +12

√2 j − 1

2k

14. v = 2(cosπ

4i + cos

π

4j + cos

π

2k) =

√2 i +

√2 j.

15. cos θ =(3 i − j − 2k) · (i + 2 j − 3k)‖3 i − j − 2k‖ ‖ i + 2 j − 3k‖ =

7√14

√14

=12, θ =

π

3

16. cos θ =(2 i − 3 j + k) · (−3 i + j + 9k)‖2 i − 3 j + k‖ · ‖ − 3 i + j + 9k‖ = 0 =⇒ θ =

π

2



694 SECTION 13.3

17. Since ‖ i − j +√

2k‖ = 2, we have cosα = 12 , cosβ = − 1

2 , cos γ = 12

√2.

The direction angles are 13π,

23π,

14π.

18. α = arccos12

=π

3, β = arccos 0 =

π

2, γ = arccos

−√

32

=56π

19. θ = arccosa · b

‖a‖‖b‖ = arccos( −9√

231

)∼= 2.2 radians or 126.3◦


‖a‖‖b‖ = arccos12√

13√

52∼= 1.09 radians or 62.5◦.


‖a‖‖b‖ = arccos( −13

5√

10

)∼= 2.54 radians or 145.3◦


‖a‖‖b‖ = arccos−4√11√

2∼= 2.59 radians or 148.5◦

23. angles: 38.51◦, 95.52◦, 45.97◦; perimeter : ∼= 15.924

24. ‖a‖ =√

41; cosα =2√41

, cosβ =6√41

, cos γ =−1√41

α ∼= 71.8◦ β ∼= 20.4◦, γ ∼= 99.0◦

25. ‖a‖ =√

12 + 22 + 22 = 3; cosα =13, cosβ =

23, cos γ =

23

α ∼= 70.5◦ β ∼= 48.2◦, γ ∼= 48.2◦

26. ‖a‖ =√

50; cosα =3√50

, cosβ =5√50

, cos γ =−4√50

α ∼= 64.9◦ β ∼= 45◦, γ ∼= 124.4◦

27. ‖a‖ =√

32 + (12)2 + 42 = 13; cosα =313

, cosβ =1213

cos γ =413

α ∼= 76.7◦ β ∼= 22.6◦, γ ∼= 72.1◦

28. cosα ∼= −0.8835, α ∼= 152.067◦; cosβ ∼= −0.3313, β ∼= 109.347◦; cos γ ∼= 0.3313, γ ∼= 70.653◦

29. 2 i + 5 j + 2xk ⊥ 6 i + 4 j − xk =⇒ 12 + 20 − 2x2 = 0 =⇒ x2 = 16 =⇒ x = ±4

30. (x i + 11 j − 3k) · (2x i − x j − 5k) = 0 =⇒ 2x2 − 11x + 15 = 0

=⇒ x = 3, x =52

31. cosπ

3=

c · d‖c‖ ‖d‖ ,

12

=2x + 1x2 + 2

, x2 = 4x; x = 0, x = 4



SECTION 13.3 695

32. (i + x j + k) · (2 i − j + y k) = 0 =⇒ 2 − x + y = 0

1 + x2 + 1 = 4 + 1 + y2 =⇒ x2 − y2 = 3 =⇒ x =74, y = −1

4

33. (a) The direction angles of a vector always satisfy

cos2 α + cos2 β + cos2 γ = 1

and, as you can check,

cos2 14π + cos2 1

6π + cos2 23π �= 1.

(b) The relation

cos2 α + cos2 14π + cos2 1

4π = 1

gives

cos2 α + 12 + 1

2 = 1, cosα = 0, a1 = ‖a‖ cosα = 0.

34. γ = 13π or γ = 2

3π

35. Let θ1, θ2, θ3 be the direction angles of −a. Then

θ1 = arccos[(−a · i)‖ − a‖

]= arccos

[− (a · i)

‖a‖

]= arccos (− cosα) = π − arccos (cosα) = π − α.

Similarly θ2 = π − β and θ3 = π − γ.

36. If v = a i + a j + ak, then α = β = γ = cos−1 a

a√

3= cos−1

(1√3

)∼= 54.7◦.

37. Set u = a i + b j + ck. The relations

(a i + b j + ck) · (i + 2 j + k) = 0 and (a i + b j + ck) · (3 i − 4 j + 2k) = 0

give

a + 2b + c = 0 3a− 4b + 2c = 0

so that b = 18a and c = − 5

4a.

Then, since u is a unit vector,

a2 + b2 + c2 = 1, a2 +(a

8

)2

+(−5a

4

)2

= 1,16564

a2 = 1.

Thus, a = ± 8165

√165 and u = ±

√165

165(8 i + j − 10k).

38. ±k, ±√

1313

(3 i − 2 j)



696 SECTION 13.3

39. We take u = i as an edge and v = i + j + k

as a diagonal of a cube. Then,

cos θ =u · v

‖u‖ ‖v‖ =13

√3,

θ = cos−1(

13

√3

) ∼= 0.96 radians.

40. Take a = i + j, b = i + j + k.

θ = cos−1 a · b‖a‖‖b‖ = cos−1 2√

2 ·√

3= cos−1

(√6

3

)∼= 0.62 radians.

41. (a) projb αa = (αa · ub)ub = α(a · ub)ub = αprojb a

(b) projb (a + c) = [(a + c) · ub]ub

= (a · ub + c · ub)ub

= (a · ub)ub + (c · ub)ub = projb a + projb c

42. (a) If β > 0, then uβb = ub and

projβba = (a · uβb)uβb = (a · ub)ub = projba.

If β < 0, then uβb = −ub and

projβba = (a · uβb)uβb = [a · (−ub)](−ub) = (a · ub)ub = projba.

(b) If β > 0,

compβba = (a · uβb) = (a · ub) = compba.

If β < 0,

compβba = (a · uβb) = [a · (−ub)] = −(a · ub) = −compba.

43. (a) a · b = a · c =⇒ a(b − c) = 0 =⇒ a ⊥ (b − c).

For a �= 0 the following statements are equivalent:

a · b = a · c, b · a = c · a,

b · a‖a‖ = c · a

‖a‖ , b · ua = c · ua

(b · ua)ua = (c · ua)ua,

proja b = proja c.

Thus, a · b = a · c implies only that the projection of b on a equals the projection of c on a.



SECTION 13.3 697

(b) b = (b · i)i + (b · j) j + (b · k)k = (c · i)i + (c · j) j + (c · k)k = c∧

(13.3.14) (13.3.14)∧

44. (a) ‖a + b‖2 = ‖a‖2 + ‖b‖2 + 2a · b = ‖a‖2 + ‖b‖2 =⇒ a ⊥ b.

(b) ‖a − b‖2 = ‖a‖2 + ‖b‖2 − 2a · b = ‖a‖2 + ‖b‖2 =⇒ a ⊥ b.

45. (a) ‖a + b‖2 − ‖a − b‖2 = (a + b) · (a + b) − (a − b) · (a − b)

= [(a · a) + 2(a · b) + (b · b)] − [(a · a) − 2(a · b) + (b · b)] = 4(a · b)

(b) The following statements are equivalent:

a ⊥ b, a · b = 0, ‖a + b‖2 − ‖a − b‖2 = 0, ‖a + b‖ = ‖a − b‖.

(c) By (b), the relation ‖a + b‖ = ‖a − b‖ gives a ⊥ b. The relation a + b ⊥ a − b gives

0 = (a + b) · (a − b) = ‖a‖2 − ‖b‖2 and thus ‖a‖ = ‖b‖.

The parallelogram is a square since it has two adjacent sides of equal length and these meet at

right angles.

46. |a · b| = ‖a‖‖b‖| cos θ| = ‖a‖‖b‖ iff θ = 0 or θ = π

47. ‖a + b‖2 = (a + b) · (a + b) = a · a + 2a · b + b · b = ‖a‖2 + 2a · b + ‖b‖2

‖a − b‖2 = (a − b) · (a − b) = a · a − 2a · b + b · b = ‖a‖2 − 2a · b + ‖b‖2

Add the two equations and the result follows.

48. ||a||2 = ||b||2 =⇒ ||a||2 − ||b||2 = 0 =⇒ (a + b) · (a − b) = 0 =⇒ (a + b) ⊥ (a − b)

49. Let c = ‖b‖a + ‖a‖b. Then

a · c‖a‖ ‖c‖ = ‖a‖ ‖b‖ + a · b =

b · c‖b‖ ‖c‖

50. cos t =a · βb

‖a‖‖βb‖ =(

β

|β|

) (a · b

‖a‖‖b‖

)= − cos θ =⇒ t = π − θ

51. Existence of decomposition:

a = (a · ub)ub + [a − (a · ub)ub].

Uniqueness of decomposition: suppose that

a = a‖ + a⊥ = A‖ + A⊥.

Then the vector a‖ − A‖ = A⊥ − a⊥ is both parallel to b and perpendicular to b. Therefore it is zero.

Consequently A‖ = a‖ and A⊥ = a⊥.

52. (a) ‖ur‖2 = cos2 θ + sin2 θ = 1, ‖uθ‖2 = sin2 θ + cos2 θ = 1.

ur · uθ = − cos θ sin θ + sin θ cos θ = 0, so ur ⊥ uθ.



698 SECTION 13.3

(b) P = (r cos θ, r sin θ) = rur, so−→OP has same direction as ur.

To see that uθ is 90◦ counterclockwise from ur, check the sign of the coefficient in all four

quadrants.

53. Place center of sphere at the origin.−→P1Q ·

−→P2Q = (−a + b) · (a + b)

= −‖a‖2 + ‖b‖2

= 0.

54. Take λ arbitrary, b �= 0.

0 ≤ ‖a − λb‖2 = (a − λb) · (a − λb) = a · a − λ(b · a) − λ(a · b) + λ2(b · b)

= ‖a‖2 − 2λ(a · b) + λ2‖b‖2

Setting λ = (a · b)/‖b‖2 we have

0 ≤ ‖a‖2 − 2|(a · b)|2‖b‖2

+|(a · b)|2‖b‖2

0 ≤ ‖a‖2‖b‖2 − |(a · b)|2.

Thus |(a · b)|2 ≤ ‖a‖2‖b‖2 and |(a · b)| ≤ ‖a‖ · ‖b‖.

PROJECT 13.3

1. (a) W = F · r (b) 0 (c) ‖F‖ i · (b− a) i = ‖F‖(b− a)

2. (a) W = (15 cos 0◦ i + 15 sin 0◦ j) · (50 i) = 15(50) = 750 joules

(b) W = (15 cos 30◦ i + 15 sin 30◦ j) · (50 i) =(

12 15

√3)50 ∼= 649.5 joules

(c) W = (15 cos 45◦ i + 15 sin 45◦ j) · (50 i) =(

12 15

√2)50 ∼= 530.3 joules

3. (a) W1 = F1 · r = ‖F1‖‖r‖ cos θ; W2 = F2 · r = ‖F2‖‖r‖ cos (−θ) = ‖F2‖‖r‖ cos θ

Therefore W2 =‖F2‖‖F1‖

W1.

(b) W1 = ‖F1‖‖r‖ cos π/3 = 12‖F1‖‖r‖; W2 = ‖F2‖‖r‖ cos π/6 = 1

2

√3‖F2‖‖r‖

Therefore W2 =√

3‖F2‖‖F1‖

W1.

4. Since the object returns to its starting point, the total displacement is zero, so the work done is zero.



SECTION 13.4 699

SECTION 13.4

1. (i + j)× (i − j) = [i× (i − j)] + [j× (i − j)] = (0 − k) + (−k − 0) = −2k

2. 0

3. (i − j)× (j − k) = [i× (j − k)] − [j× (j − k)] = (j + k) − (0 − i) = i + j + k

4. j× (2 i − k) = j× 2 i − j×k = −2k − i = −i − 2k

5. (2 j − k)× (i − 3 j) = [2 j× (i − 3 j)] − [k× (i − 3 j)] = (−2k) − (j + 3 i) = −3 i − j − 2k

or

(2 j − k)× (i − 3 j) =

∣∣∣∣∣∣∣∣i j k

0 2 −1

1 −3 0

∣∣∣∣∣∣∣∣= i

∣∣∣∣∣ 2 −1

−3 0

∣∣∣∣∣ − j

∣∣∣∣∣ 0 −1

1 −3

∣∣∣∣∣ + k

∣∣∣∣∣ 0 2

1 −3

∣∣∣∣∣ = −3 i − j − 2k

6. i · (j×k) = i · i = 1 7. j · (i×k) = j · (−j) = −1 8. (j× i) · (i×k) = (−k) · (−j) = 0

9. (i× j)×k = k×k = 0 10. k · (j× i) = k · (−k) = −1 11. j · (k× i) = j · (j) = 1

12. j× (k× i) = j× j = 0

13. (i + 3 j − k)× (i + k) =

∣∣∣∣∣∣∣∣i j k

1 3 −1

1 0 1

∣∣∣∣∣∣∣∣= [(3)(1) − (−1)(0)] i − [(1)(1) − (−1)(1)] j + [(1)0 − (3)(1)]k

= 3 i − 2 j − 3k

14. (3 i − 2 j + k)× (i − j + k) =

∣∣∣∣∣∣∣∣i j k

3 −2 1

1 −1 1

∣∣∣∣∣∣∣∣= −i − 2 j − k

15. (i + j + k)× (2 i + k) =

∣∣∣∣∣∣∣∣i j k

1 1 1

2 0 1

∣∣∣∣∣∣∣∣= [(1)(1) − (1)(0)] i − [(1)(1) − (1)(2)] j + [(1)(0) − (1)(2)]k

= i + j − 2k

16. (2 i − k)× (i − 2 j + 2k) =

∣∣∣∣∣∣∣∣i j k

2 0 −1

1 −2 2

∣∣∣∣∣∣∣∣= −2 i − 5 j − 4k



700 SECTION 13.4

17. [2 i + j] · [(i − 3 j + k)× (4 i + k)] =

∣∣∣∣∣∣∣∣1 −3 1

4 0 1

2 1 0

∣∣∣∣∣∣∣∣=

[(0)(0) − (1)(1)] − (−3)[(4)(0) − (1)(2)] + [(4)(1) − (0)(2)] = −3

18. [(−2 i + j − 3k)× i] × [i + j] = (−3 j − k)× (i + j) =

∣∣∣∣∣∣∣∣i j k

0 −3 −1

1 1 0

∣∣∣∣∣∣∣∣= i − j + 3k

19. [(i − j)× (j − k)] × [i + 5k] = {[i× (j − k)] − [ j× (j − k)]} × [i + 5k]

= [(k + j) − (−i)] × [i + 5k]

= (i + j + k)× (i + 5k)

= [(i + j + k)× i] + [(i + j + k)× 5k]

= (−k + j) + (−5 j + 5i)

= 5 i − 4 j − k

20. (i − j)× [(j − k)× (j + 5k)] = (i − j)×

∣∣∣∣∣∣∣∣i j k

0 1 −1

0 1 5

∣∣∣∣∣∣∣∣= (i − j)× 6 i = 6k

21. a×b =

∣∣∣∣∣∣∣∣i j k

1 3 −1

2 0 1

∣∣∣∣∣∣∣∣= 3 i − 3 j − 6k

a×b‖a×b‖ =

1√6

i − 1√6

j − 2√6

k;b×a‖b×a‖ = − 1√

6i +

1√6

j +2√6

k

22. a×b =

∣∣∣∣∣∣∣∣i j k

1 2 3

2 1 1

∣∣∣∣∣∣∣∣= −i + 5 j − 3k, so take ±

√35

35(−i + 5 j − 3k)

23. Set a =−→PQ = −i + 2k and b =

−→PR = 2 i − k. Then

a × b =

∣∣∣∣∣∣∣∣i j k

−1 0 2

2 0 −1

∣∣∣∣∣∣∣∣= 3 j;

a × b‖a × b‖ = j

and A = 12 ‖a×b ‖ = 1

2 ‖ 3 j ‖ = 32 .



SECTION 13.4 701

24. N =−→PQ ×

−→PR =

∣∣∣∣∣∣∣∣i j k

−2 1 −1

2 −3 −1

∣∣∣∣∣∣∣∣= −4 i − 4 j + 4k;

N‖N‖ = − 1√

3i − 1√

3j +

1√3k

A =12‖N‖ = 2

√3

25. Set a =−→PQ = i + j − 3k and b =

−→PR = −i + 3 j − k. Then

a × b =

∣∣∣∣∣∣∣∣i j k

1 1 −3

−1 3 −1

∣∣∣∣∣∣∣∣= 8 j + 4 j + 4k;

a × b‖a × b‖ =

2√6i +

1√6j +

1√6k

and A = 12 ‖a×b ‖ = 1

2 ‖ 8 i + 4 j + 4k ‖ = 12

√82 + 42 + 42 = 2

√6.

26. N =−→PQ ×

−→PR=

∣∣∣∣∣∣∣∣i j k

2 2 −4

−5 1 2

∣∣∣∣∣∣∣∣= 8 i + 16 j + 12k

N‖N‖ =

2√29

i +4√29

j +3√29

k

Area =12‖N‖ = 2

√29

27. V =∣∣[(i + j)× (2 i − k)] · (3 j + k)

∣∣ =∣∣(−i + j − 2k) · (3 j + k)

∣∣ = 1

28. V = (i − 3 j + k)× (2 j − k)] · (i + j − 2k) =

∣∣∣∣∣∣∣∣1 −3 1

0 2 −1

1 1 −2

∣∣∣∣∣∣∣∣= −2; V = | − 2| = 2

29. V =∣∣∣ −→OP ·

( −→OQ ×

−→OR

)∣∣∣ =

∣∣∣∣∣∣∣∣

∣∣∣∣∣∣∣∣1 2 3

1 1 2

2 1 1

∣∣∣∣∣∣∣∣

∣∣∣∣∣∣∣∣= 2

30.−→PQ ·

( −→PR ×

−→PS

)=

∣∣∣∣∣∣∣∣1 1 −3

−1 3 −1

2 6 3

∣∣∣∣∣∣∣∣= 52; V = 52

31. (a + b)× (a − b) = [a× (a − b)] + [b× (a − b)]

= [a× (−b)] + [b×a]

= −(a×b) − (a×b) = −2(a×b)

32. (a + b)× c = −[c× (a + b)] = −(c×a) − (c×b) = (a× c) + (b× c).

33. a× i = 0, a× j = 0 =⇒ a ‖ i and a ‖ j =⇒ a = 0

34. a×b = (a1b2 − b1a2)k



702 SECTION 13.4

35. By (13.4.4) αa × βb = (αβ)a × b. Therefore, ‖αa × βb‖ = ‖(αβ)a × b‖.

36. (a) The following statements are equivalent:

a×b = a× c

(a×b) − (a× c) = 0

(a×b) + [a× (−c)] = 0

a× (b − c) = 0

(b)

The last equation holds iff a and b − c are parallel. The tip of c must lie on the line which passes

through the tip of b and is parallel to a.

37. (a) a · (b× c): makes sense – this is the dot product of two vectors.

(b) a× (b · c): does not make sense – this is the cross product of a vector with a number.

(c) a · (b · c): does not make sense – this is the dot product of a vector with a number.

(d) a× (b × c): makes sense – this is the cross product of two vectors.

38. Given a vector a. Since a×b ⊥ b, (a×b) · b = 0 for all vectors b. That is, there are no vectors

b such that (a×b) · b �= 0.

39. d · a = d · b =⇒ d ⊥ (a − b); d · a = d · c =⇒ d ⊥ (a − c)

Therefore, d = λ[(a − b) × (a − c)] for some number λ.

40. b× c ⊥ b and b× c ⊥ c, so b× c must be parallel to a. Hence a× (b× c) = 0.

41. a · b = a · c =⇒ a · (b − c) = 0; a is perpendicular to b − c.

a×b = a× c =⇒ a× (b − c) = 0; a is parallel to b − c.

Since a �= 0 it follows that b − c = 0 or b = c.

42. (a) i − component of a× (b× c) = a2(b× c)3 − a3(b× c)2

= a2(b1c2 − b2c1) − a3(b3c1 − b1c3) = (a2c2 + a3c3)b1 − (a2b2 + a3b3)c1

= (a1c1 + a2c2 + a3c3)b1 − (a1b1 + a2b2 + a3b3)c1

= (a · c)b1 − (a · c)c1 = i-component of (a · c)b − (a · c)c

(b) (a×b)× c = −c× (a×b) = − [(c · b)a − (c · a)b] = (c · a)b − (c · b)a

(c) with r = c×d

(a×b) · (c×d) = (a×b) · r = (r×a) · b

= [(c×d)×a] · b

= [(a · c)d − (a · d)c] · b

= (a · c)(d · b) − (a · d)(c · b)

= (a · c)(b · d) − (a · d)(b · c)



SECTION 13.5 703

43. c×a = (a×b)×a = (a · a)b − (a · b)a = (a · a)b = ‖a‖2b

Exercise 42(a) a · b = 0

44. (a · u)u + (u×a)×u = (a · u)u + [(u · u)a − (u · a)u]

Exercise 42(b)

= (a · u)u + a − (a · u)u = a

45. Suppose a �= 0. Then

a · b = 0 =⇒ b ⊥ a; a×b = 0 =⇒ b‖a

Thus b is simultaneously perpendicular to, and parallel to a. It follows that b = 0.

46. D = 12 |

−→PQ ×

−→PR | = 1

2

√a2b2 + a2c2 + b2c2

47. The result is an immediate consequence of Exercise 46.

48. a · (b× c) = (a×b) · c = (c×a) · b = (b× c) · a = (a×− c) · b

a · (c×b) = c · (b×a) = (−a×b) · c

SECTION 13.5

1. P (when t = 0) and Q (when t = −1)

2. l1 and l3 are parallel.

3. Take r0 =−→OP= 3 i + j and d = k. Then, r(t) = (3 i + j) + tk.

4. r(t) = i − j + 2k + t(3 i − j + k)

5. Take r0 = 0 and d =−→OQ . Then, r(t) = t(x1 i + y1 j + z1 k).

6. r(t) = x0 i + y0 j + z0 k + t [(x1 − x0) i + (y1 − y0) j + (z1 − z0)k]

7.−→PQ = i − j + k so direction numbers are 1,−1, 1. Using P as a point on the line, we have

x(t) = 1 + t, y(t) = −t, z(t) = 3 + t.

8. x(t) = x0 + t(x1 − x0), y(t) = y0 + t(y1 − y0), z(t) = z0 + t(z1 − z0).

9. The line is parallel to the y-axis so we can take 0, 1, 0 as direction numbers. Therefore

x(t) = 2, y(t) = −2 + t, z(t) = 3.

10. x(t) = 1 + t, y(t) = 4, z(t) = −3



704 SECTION 13.5

11. Since the line 2(x + 1) = 4(y − 3) = z can be writtenx + 1

2=

y − 31

=z

4,

it has direction numbers 2, 1, 4. The line through P (−1, 2,−3) with direction vector

2 i + j + 4k can be parameterized

r(t) = (−i + 2 j − 3k) + t(2 i + j + 4k).

12.x

x0=

y

y0=

z

z0, so

x− x0

x0=

y − y0

y0=

z − z0

z0provided x0y0z0 �= 0

13. r(t) = (3 i + j + 5k) + t(i − j + 2k) = (3 + t) i + (1 − t) j + (5 + 2t)k

R(u) = (i + 4 j + 2k) + u(j + k) = i + (4 + u) j + (2 + u)k

d = i − j + 2k is a direction vector for l1; D = j + k is a direction vector for l2. Since d is not

a multiple of D, the lines either intersect or are skew. Setting r(t) = R(u) we get the system of

equations:

3 + t = 1, 1 − t = 4 + u, 5 + 2t = 2 + u

This system has the solution t = −2, u = −1. The point of intersection is: (1, 3, 1).

14. d2 = −2 i + 6 j − 4k = −2(i − 3 j + 2k). Therefore, the lines are either parallel or coincident. Since

the point (−1, 2, 1) on l1 does not lie on l2, the lines are parallel.

15. d = 2 i + 4 j − k is a direction vector for l1; D = 2 i + j + 2k is a direction vector for l2. Since d

is not a multiple of D, the lines either intersect or are skew. Equating coordinates, we get the system

of equations:

3 + 2t = 3 + 2u, −1 + 4t = 2 + u, 2 − t = −2 + 2u

From the first two equations, we get t = u = 1. Since these values of t and u do not satisfy the third

equation, the lines are skew.

16. The lines are skew.

17. d = −6 i + 9 j − 3k is a direction vector for l1; D = 2 i − 3 j + k is a direction vector for l2. Since

d = −3D, we conclude that l1 and l2 are either parallel or coincident. The point (1, 2, 0) lies on l1

but does not line on l2. Therefore, the lines are parallel.

18. d = i + 2 j + 3k is a direction vector for l1; D = 3 i + 2 j + 1k is a direction vector for l2. Since d

is not a multiple of D, the lines either intersect or are skew. The system of equations

2 + t = 5 + 3u, −1 + 2t = 1 + 2u, 1 + 3t = 4 + u

does not have a solution. Therefore the lines are skew.



SECTION 13.5 705

19. d = 2 i + 4 j + 3k is a direction vector for l1; D = i + 3 j + 2k is a direction vector for l2. Since d

is not a multiple of D, the lines either intersect or are skew. The system of equations

4 + 2t = 2 + u, −5 + 4t = −1 + 3u, 1 + 3t = 2u

does not have a solution. Therefore the lines are skew.

20. The lines intersect at the point(

32 , 1,

52

).

21. We set r1(t) = r2(u) and solve for t and u:

i + t j = j + u(i + j),

(1 − u) i + (−1 − u + t) j = 0.

Thus,

1 − u = 0 and − 1 − u + t = 0.

These equations give u = 1, t = 2. The point of intersection is P (1, 2, 0).

As direction vectors for the lines we can take u = j and v = i + j. Thus

cos θ =u · v

‖u‖ ‖v‖ =1

(1)(√

2)=

12

√2.

The angle of intersection is 14π radians.

22. Set r1(t) = r2(u) :

i − 4√

3 j + t(i +√

3 j) = 4 i + 3√

3 j + u(i −√

3 j)

(−3 + t− u)i + (−7√

3 + t√

3 + u√

3) j = 0

=⇒ t− u = 3, t + u = 7, =⇒ t = 5, u = 2

r1(5) = r2(2) = 6 i +√

3 j : the point of intersection is: P (6,√

3, 0)

cos θ =|(i +

√3 j) · (i −

√3 j)|

‖i +√

3 j‖‖i −√

3 j‖=

12

=⇒ θ =π

3radians.

23.(x0 −

d1

d3z0, y0 −

d2

d3z0, 0

)

24. The lines meet at (x0, y0, z0), and since d · D = 0, they are perpendicular.

25. The lines are parallel.

26. Note that r(0) = r0 and r(1) = r1, so we need 0 ≤ t ≤ 1

27. r(t) = (2 i + 7 j − k) + t(2 i − 5 j + 4k), 0 ≤ t ≤ 1

28. −1 ≤ t ≤ 2.



706 SECTION 13.5

29. Set u =

−→PQ

‖−→PQ ‖

=−4 i + 2 j + 4k

‖ − 4 i + 2 j + 4k‖ = −23

i +13

j +23

k.

Then r(t) = (6 i − 5 j + k) + tu is−→OP at t = 9 and it is

−→OQ at t = 15. (Check this.)

Answer: u = − 23 i + 1

3 j + 23 k, 9 ≤ t ≤ 15.

30. Since the two lines intersect, there exist numbers t0 and u0 such that

r(t0) = R(u0)

Suppose first that

r(t0) = R(u0) = 0.

Then

r(t) = r(t) − r(t0) = (r0 + td) − (r0 + t0d) = (t− t0)d.

Similarly R(u) = (u− u0)D. Since l1 ⊥ l2, we have d · D = 0 and thus

r(t) · R(u) = (t− t0)(u− u0)(d · D) = 0 for all t, u

Suppose now that

r(t0) = R(u0) �= 0

Then

r(t0) · R(u0) = r(t0) · r(t0) = ‖r(t0)‖2 �= 0

and it is therefore not true that

R(t) · R(u) = 0, for all t, u.

31. The given line, call it l, has direction vector 2 i − 4 j + 6k.

If a i + b j + ck is a direction vector for a line perpendicular to l, then

(2 i − 4 j + 6k) · (a i + b j + ck) = 2a− 4b + 6c = 0.

The lines through P (3,−1, 8) perpendicular to l can be parameterized

X(u) = 3 + au, Y (u) = −1 + bu, Z(u) = 8 + cu

with 2a− 4b + 6c = 0.

32. (a) Since R(0) = R0 is on the line, there exists a number t0 such that r(t0) = r0 + t0d = R0.

(b) Since d and D are both direction vectors for the same line, they are parallel. Since d �= 0,

there exists a scalar α such that D = αd. It follows that, for all real u,

R(u) = R0 + uD = (r0 + t0d) + u(αd) = r0 + (t0 + αu)d.



SECTION 13.5 707

33. d(P, l) =‖(i + 2k)× (2 i − j + 2k)‖

‖2 i − j + 2k‖ = 1

34. d(P, l) =‖(j + k)× (i − 2 j − 2k)‖

‖i − 2 j − 2k‖ =13

√2 ∼= 0.47

35. The line contains the point P0(1, 0, 2). Therefore

d(P, l) =‖(2 j + k)× (i − 2 j + 3k)‖

‖ i − 2 j + 3k‖ =

√6914

∼= 2.22

36. P0 = (1, 0, 0), d = j, d(P, l) =‖ − i× j‖

‖j‖ = 1.

37. The line contains the point P0(2,−1, 0). Therefore

d(P, l) =‖(i − j − k)× (i + j)‖

‖i + j‖ =√

3 ∼= 1.73.

38. The line can be parameterized r(t) = b j + t(i + m j). Since the line contains the point (0, b, 0)

d(P, l) =‖[x0 i + (y0 − b) j + z0 k]× (i + m j)‖

‖i + m j‖

=

√(1 + m2)z0

2 + [y0 − (mx0 + b)]2

1 + m2

If P (x0, y0, z0) lies directly above or below the line, then y0 = mx0 + b and d(P, l) reduces to√z0

2 = |z0|. This is evident geometrically.

39. (a) The line passes through P (1, 1, 1) with direction vector i + j. Therefore

d(0, l) =‖(i + j + k)× (i + j)‖

‖i + j‖ = 1.

(b) The distance from the origin to the line segment is√

3.

Solution. The line segment can be parameterized

r(t) = i + j + k + t(i + j), t ∈ [ 0, 1 ].

This is the set of all points P (1 + t, 1 + t, 1) with t ∈ [ 0, 1 ].

The distance from the origin to such a point is

f(t) =√

2 (1 + t)2 + 1 .

The minimum value of this function is f(0) =√

3.

Explanation. The point on the line through P and Q closest to the origin is not on the line

segment PQ.

40. (a) We want (r0 + t0d) · d = 0, so r0 · d + t0‖d‖2 = 0 =⇒ t0 = −r0 · d‖d‖2

(b) R(t) = r(t0) ± td‖d‖ , where t0 is as in part (a).



708 SECTION 13.5

41. We begin with r(t) = j − 2k + t(i − j + 3k). The scalar t0 for which r(t0) ⊥ l can be found by

solving the equation

[ j − 2k + t0(i − j + 3k)] · [i − j + 3k] = 0.

This equation gives −7 + 11t0 = 0 and thus t0 = 7/11. Therefore

r(t0) = j − 2k + 711 (i − j + 3k) = 7

11 i + 411 j − 1

11 k.

The vectors of norm 1 parallel to i − j + 3k are

± 1√11

(i − j + 3k).

The standard parameterizations are

R(t) =711

i +411

j − 111

k ± t√11

(i − j + 3k)

=111

(7 i + 4 j − k) ± t

[√11

11(i − j + 3k)

].

42. Start with r(t) =√

3 i + t(i + j + k) to get t0 = −√

33

, so

R(t) =√

33

(2 i − j − k) ± t

√3

3(i + j + k).

43. 0 < t < s

By similar triangles, if 0 < s < 1, the tip of−→OA +

s−→AB +s

−→BC falls on AC. If 0 < t < s, then the

tip of−→OA +s

−→AB +t

−→BC falls short of AC and stays

within the triangle. Clearly all points in the interior

of the triangle can be reached in this manner.

A

y

x

OA

O

B

C

sBCsAB

44. d1 ×d2 is a direction vector for the line that is perpendicular to both l1 and l2.∣∣∣ −→PQ · (d1 ×d2)

∣∣∣|d1 ×d2|

is the magnitude of the projection of the vector−→PQ onto the vector d1 ×d2.

45. d = i + 3 j − 2k is a direction vector for l1; D = 4 i − j + 2k is a direction vector for l2. Since d

is not a multiple of D, the lines either intersect or are skew. Equating coordinates, we get the system

of equations:

2 + t = −1 + 4u, −1 + 3t = 2 − u, 1 − 2t = −3 + 2u



SECTION 13.6 709

This system does not have a solution. Therefore the lines are skew. The point P (2,−1, 1) is on l1

and the point Q(−1, 2,−3) is on l2, and−→PQ= −3 i + 3 j − 4k. By Exercise 44, the distance between

l1 and l2 is: ∣∣∣ −→PQ · (d×D)

∣∣∣‖d×D‖ =

|(−3 i + 3 j − 4k) · (4 i − 10 j − 13k)|√285

=10√285

46. D = 2 i + j + 4k is not a multiple of d = i + 3 j − 2k and the system of equations

1 + t = 2u, −2 + 3t = 3 + u, 4 − 2t = −3 + 4u

does not have a solution. Therefore the lines are skew.−→PQ= −i + 5 j − 7k and d×D = 14 i − 8 j − 5k

The distance between the lines is:19√285

SECTION 13.6

1. Q

2. An equation for the plane is: (x− 4) − 3(y − 1) + (z + 1) = 0; R and S lie on the plane.

3. Since i − 4 j + 3k is normal to the plane, we have

(x− 2) − 4(y − 3) + 3(z − 4) = 0 and thus x− 4y + 3z − 2 = 0.

4. N = j + 2k, P (1,−2, 3) =⇒ (y + 2) + 2(z − 3) = 0 or y + 2z − 4 = 0.

5. The vector 3 i − 2 j + 5k is normal to the given plane and thus to every parallel plane:

the equation we want can be written

3(x− 2) − 2(y − 1) + 5(z − 1) = 0, 3x− 2y + 5z − 9 = 0.

6. N = 4 i + 2 j − 7k, P (3,−1, 5) =⇒ 4(x− 3) + 2(y + 1) − 7(z − 5) = 0

7. The point Q (0, 0,−2) lies on the line l; and d = i + j + k is a direction vector for l.

We want an equation for the plane which has the vector

N =−→PQ ×d = (i + 3 j + 3k)× (i + j + k)

as a normal vector:

N =

∣∣∣∣∣∣∣∣i j k

−1 −3 −3

1 1 1

∣∣∣∣∣∣∣∣= −2 j + 2k

An equation for the plane is: −2(y − 3) + 2(z − 1) = 0 or y − z − 2 = 0



710 SECTION 13.6

8. Another point is Q(1, 1, 2), and the plane is parallel to the vectors d = −2 i + 4 j + k and−→PQ= −i + j + k. Thus, N = d×

−→PQ= 3 i + j + 2k is a normal to the plane.

An equation for the plane is: 3(x− 2) + y + 2(z − 1) = 0 or 3x + y + 2z − 8 = 0

9.−→OP0= x0 i + y0 j + z0 k An equation for the plane is:

x0 (x− x0) + y0 (y − y0) + z0 (z − z0) = 0

10. N = 2 i − 3 j + 7k, unit normals: uN = ± 1√62

(2 i − 3 j + 7k)

11. The vector N = 2 i − j + 5k is normal to the plane 2x− y + 5z − 10 = 0. The unit normals are:

N‖N‖ =

1√30

(2 i − j + 5k) and − N‖N‖ = − 1√

30(2 i − j + 5k)

12. (a, 0, 0), (0, b, 0), and (0, 0, c) satisfy the equation.

13. Intercept form:x

15+

y

12− z

10= 1 x-intercept: (15, 0, 0)

y-intercept: (0, 12, 0)

z-intercept: (0, 0,−10)

14.x

−2/3+

y

2+

z

−1/2= 1; (−2

3, 0, 0), (0, 2, 0), (0, 0,−1

2).

15. uN1=

√38

38(5 i − 3 j + 2k), uN2

=√

1414

(i + 3 j + 2k), cos θ =∣∣∣uN1

· uN2

∣∣∣ = 0.

Therefore θ = π/2 radians.

16. cos θ =|(2 i − j + 3k) · (5 i + 5 j − k)|‖2 i − j + 3k‖‖5 i + 5 j − k‖ =

2√14√

51; θ ∼= 1.50 radians.

17. uN1=

√3

3(i − j + k), uN2

=√

1414

(2 i + j + 3k), cos θ =∣∣∣uN1

· uN2

∣∣∣ =221

√42 ∼= 0.617.

Therefore θ ∼= 0.91 radians.

18. cos θ =|(4 i + 4 j − 2k) · (2 i + j + k)|‖4 i + 4 j − 2k‖‖2 i + j + k‖ =

106√

6; θ ∼= 0.82 radian.

19. coplanar since 0(4 j − k) + 0(3 i + j + 2k) + 1(0) = 0

20. s i + t(i − 2 j) + u(3 j + k) = (s + t) i + (−2t + 3u) j + uk = 0 only if s = t = u = 0, so not coplanar.

21. We need to determine whether there exist scalars s, t, u not all zero such that

s(i + j + k) + t(2 i − j) + u(3 i − j − k) = 0

(s + 2t + 3u) i + (s− t− u) j + (s− u)k = 0.



SECTION 13.6 711

The only solution of the system

s + 2t + 3u = 0, s− t− u = 0, s− u = 0

is s = t = u = 0. Thus, the vectors are not coplanar.

22. coplanar since 1(j − k) − 1(3 i − j + 2k) + 1(3 i − 2 j + 3k) = 0

23. By (13.6.5), d(P, p) =|2(2) + 4(−1) − (3) + 1|√

4 + 16 + 1=

2√21

=221

√21.

24. d =|8(3) − 2(−5) + 2 − 5|√

82 + (−2)2 + 12=

31√69

25. By (13.6.5), d(P, p) =|(−3)(1) + 0(−3) + 4(5) + 5|√

9 + 16=

225.

26. d =|1 + 3 − 2 · 4|√12 + 12 + (−2)2

=23

√6

27.−→P1P= (x− 1) i + y j + (z − 1)k,

−→P1P2= i + j − k,

−→P1P3= j.

Therefore

(−→

P1P2 ×−→

P1P3) = (i + j − k)× j = i + k

and

−→P1P · (

−→P1P2 ×

−→P1P3) = [(x− 1) i + y j + (z − 1)k] · [i + k] = x− 1 + z − 1.

An equation for the plane can be written x + z = 2.

28.−→

P1P2= (1,−3,−2),−→

P1P3= (−1, 1, 0), N =−→

P1P2 ×−→

P1P3= (2, 2,−2)

=⇒ 2(x− 1) + 2(y − 1) − 2(z − 1) = 0 or x + y − z − 1 = 0

29.−→P1P= (x− 3) i + (y + 4) j + (z − 1)k,

−→P1P2= 6 j,

−→P1P3= −2 i + 5 j − 3k.

Therefore

(−→

P1P2 ×−→

P1P3) = 6 j× (−4 i + 5 j − 3k) = −18 i + 12k

and

−→P1P · (

−→P1P2 ×

−→P1P3) = [(x− 3) i + (y + 4) j + (z − 1)k] · [−18 i + 12k]

= −18(x− 3) + 12(z − 1)

An equation for the plane can be written −18(x− 3) + 12(z − 1) = 0 or 3x− 2z − 7 = 0.



712 SECTION 13.6

30.−→

P1P2= (0,−4, 5),−→

P1P3= (−2,−3, 4) N =−→

P1P2 ×−→

P1P3= (−1,−10,−8)

=⇒ (x− 3) + 10(y − 2) + 8(z + 1) = 0

31. The line passes through the point P0 (x0, y0, z0) with direction numbers: A, B, C.

Equations for the line written in symmetric form are:

x− x0

A=

y − y0

B=

z − z0

C, provided A �= 0, B �= 0, C �= 0.

32. Take a point P1(x1, y1, z1) on plane 1 (so Ax1 +By1 +Cz1 +D1 = 0) and find the distance to plane 2 :

d =|Ax1 + By1 + Cz1 + D2|√

A2 + B2 + C2=

|D2 −D1|√A2 + B2 + C2

33.x− x0

d1=

y − y0

d2,

y − y0

d2=

z − z0

d3

34. (a) x = t, y = y0, z = z0 (b) x = x0, y = t, z = z0

35. We set x = 0 and find that P0(0, 0, 0) lies on the line of intersection. As normals to the plane we use

N1 = i + 2 j + 3k and N2 = −3 i + 4 j + k.

Note that

N1 ×N2 = (i + 2 j + 3k)× (−3 i + 4 j + k) = −10 i − 10 j + 10k.

We take − 110 (N1 ×N2) = i + j − k as a direction vector for the line through P0(0, 0, 0). Then

x(t) = t, y(t) = t, z(t) = −t.

36. Set x = 0 to find that P (0, 12 ,− 3

2 ) on the line of intersection.

For the direction vector, consider N1 ×N2 = (i + j + k)× (i − j + k) = 2 i − 2k, so we can use

d = i − k. Thus

x(t) = t, y(t) =12, z(t) = −3

2− t.

37. Straightforward computations give us

l : x(t) = 1 − 3t, y(t) = −1 + 4t, z(t) = 2 − t

and

p : x + 4y − z = 6.

Substitution of the scalar parametric equations for l in the equation for p gives

(1 − 3t) + 4(−1 + 4t) − (2 − t) = 6 and thus t = 11/14.

Using t = 11/14, we get x = −19/14, y = 15/7, z = 17/14.



SECTION 13.6 713

38. l : x(t) = 4 − 2t, y(t) = −3 + t, z(t) = 1 + 2t P : x + 4y − z = 6

Note that d · N = (−2 i + j + 2k) · (i + 4 j − k) = 0, so the line is parallel to the plane, and since

P1 does not lie in the plane, l and P do not intersect.

39. Let N = A i + B j + C k be normal to the plane. Then

N · d = (i + B j + C k) · (i + 2 j + 4k) = 1 + 2B + 4C = 0

and

N · D = (i + B j + C k) · (−i − j + 3k) = −1 −B + 3C = 0.

This gives B = −7/10 and C = 1/10. The equation for the plane can be written

1(x− 0) − 710 (y − 0) + 1

10 (z − 0) = 0, which simplifies to 10x− 7y + z = 0.

40. If the two lines intersect, then there exist numbers t0 and u0 such that

r0 + t0d = R0 + u0D.

It follows then that

1(r0 − R0) + t0d − u0D = 0

and therefore the vectors r0 − R0,d, and D are coplanar. Conversely, if r0 − R0, d, D are coplanar,

then there exist numbers α, β, γ not all zero such that

(∗) α(r0 − R0) + βd + γD = 0.

We assert now that α �= 0. [If α were 0, then we would have βd + γD = 0 so that d and D

would have to be parallel, which (by assumption) they are not.] With α �= 0 we can divide equation

(*) by α and obtain

r0 − R0 +β

αd +

γ

αD = 0.

This gives

r0 +β

αd = R0 −

γ

αD,

which means that

r(β

α

)= R

(−γ

α

)and the two lines intersect.

41. N+−→PQ and N−

−→PQ are the diagonals of a rectangle with sides N and

−→PQ. Since the diagonals

are perpendicular, the rectangle is a square; that is ‖N‖ = ‖−→PQ ‖. Thus, the points Q form a circle

centered at P with radius ‖N‖.

42. Given that ‖a‖, ‖b‖, ‖c‖ are nonzero and a · b = a · c = b · c = 0. The vectors a,b, c are

coplanar iff there exist scalars s, t, u not all zero such that sa + tb + uc = 0. Now assume that

sa + tb + uc = 0.



714 SECTION 13.6

The relation

0 = a · (sa + tb + uc) = s‖a‖2 + t(a · b) + u(a · c) = s‖a‖2

gives s = 0. Similarly, we can show that t = 0 and u = 0. The vectors a,b.c can not be linearly

dependent.

43. If α > 0, then P0 lies on the same side of the plane as the tip of N; if α < 0, then P0 and the tip of N

lie on opposite sides of the plane.

To see this, suppose that N emanates from the point P1(x1, y1, z1) on the plane. Then

N ·−→

P1P0= A(x0 − x1) + B(y0 − y1) + C(z0 − z1) = Ax0 + By0 + Cz0 + D = α.

If α > 0, 0 ≤<)(N,

−→P0P1

)< π/2; if α < 0, then π/2 <<)

(N,

−→P0P1

)< π. Since N is perpendicular

to the plane, the result follows.

45. (a) intercepts:

(4, 0, 0), (0, 5, 0), (0, 0, 2)

(b) traces:

in the x, y-plane: 5x + 4y = 20

in the x, z-plane: x + 2z = 4

in the y, z-plane: 2y + 5z = 10

(c) unit normals: ± 1√141

(5 i + 4 j + 10k)

(d)

46. (a) intercepts:

(6, 0, 0), (0, 3, 0), (0, 0, 2)

(b) traces:

in the x, y-plane: x + 2y = 6

in the x, z-plane: x + 3z = 6

in the y, z-plane: 2y + 3z = 6


(i + 2 j + 3k)

(d)



SECTION 13.6 715

47. (a) intercepts:

(4, 0, 0), no y-intercept, (0, 0, 6)

(b) traces:

in the x, y-plane: x = 4

in the x, z-plane: 3x + 2z = 12

in the y, z-plane: z = 6


(3 i + 2k)

(d)

48. (a) intercepts:

(2, 0, 0), (0, 3, 0), no z-intercept

(b) traces:

in the x, y-plane: 3x + 2y = 6

in the x, z-plane: x = 2

in the y, z-plane: y = 3


(3 i + 2 j)

(d)

49. The normal vectors to the planes are: N1 = 2 i + j + 3k, N2 = i + 5 j − 2k.

The cosine of the angle θ between the planes is: cos θ =|N1 · N2|‖N1‖ ‖N2‖

=1

2√

105;

θ ∼= 1.52 radians ∼= 87.2◦.

−20

2

−2

0

2

−4

−2

0

2

4



716 SECTION 13.6

50. The normal vectors to the planes are: N1 = −i − 2 j + 4k, N2 = 2 i − j − 3k.

The cosine of the angle θ between the planes is: cos θ =|N1 · N2|‖N1‖ ‖N2‖

=12

7√

6;

θ ∼= 0.796 radians ∼= 45.6◦.

− 2

02 2

0

2

4

2

0

2

4

− 20

2

−−0

51. An equation of the plane that passes through (2, 7,−3) with normal vector N = 3 i + j + 4k is:

3(x− 2) + (y − 7) + 4(z + 3) = 0 or 3x + y + 4z = 1.

52. An equation of the plane that passes through (5,−3,−4) with normal vector N = −i + 2 j − 3k is:

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



REVIEW EXERCISES 717

53.x

2+

y

5+

z

4= 1

10x + 4y + 5z = 20

54. 4x + 3y + 6z − 12 = 0

55.x

3+

y

5= 1

5x + 3y = 15

56. 3y + 4z − 12 = 0

REVIEW EXERCISES

1. (a) PQ =√

(7 − 3)2 + (−5 − 2)2 + (−4 − [−1])2 =√

16 + 49 + 25 = 3√

10

(b) Midpoint of QR :(

7 + 52

,−5 + 6

2,4 − 3

2

)=

(6,

12,12

)

(c) Let X = (x, y, z). Then

(7,−5, 4) =(

3 + x

2,2 + y

2,−1 + z

2

)=⇒ (x, y, z) = (11,−12, 9).

(d) Midpoint of PR : (4, 4,−2); radius of the sphere:

r =12|PR| =

12

√(5 − 3)2 + (6 − 2)2 + (−3 + 1)2 =

√6.

The equation of the sphere is: (x− 4)2 + (y − 4)2 + (z + 2)2 = 6.

2. (a) PQ =√

62 + 12 + (−7)2 =√

36 + 1 + 49 =√

86

(b) Midpoint of QR : (−2 + 1

2,1 − 1

2,4 − 6

2) = (− 1

2 , 0,−1)

(c) Let X = (x, y, z). Then

(−2, 1, 4) =(

4 + x

2,2 + y

2,−3 + z

2

)=⇒ (x, y, z) = (−8, 0, 11).

(d) Midpoint of PR :(

52 ,

12 ,− 9

2

); radius of sphere:.

r =12|PR| =

12

√(4 − 1)2 + (2 + 1)2 + (−3 + 6)2 = 1

2

√27.

The equation of the sphere is: (x− 52 )2 + (y − 1

2 )2 + (z + 92 )2 = 27

4 .

3. radius:√

22 + (−3)2 + 12 =√

14

equation: (x− 2)2 + (y + 3)2 + (z − 1)2 = 14

4. radius: 12

√42 + 62 + 42 =

√17

midpoint:(−1 + 3

2,4 − 2

2,2 + 6

2

)= (1, 1, 4)

equation: (x− 1)2 + (y − 1)2 + (z − 4)2 = 17



718 REVIEW EXERCISES

5. By completing the square, the equation can be written as

(x + 1)2 + (y + 2)2 + (z − 4)2 = 4 = 22.

center: (−1,−2, 4). radius: 2

6. By completing the square, the equation can be written as

(x− 3)2 + (y + 5)2 + (z − 1)2 = 33.

center: (3,−5, 1) radius:√

33

7. 32 i + j − 1

2 k 8. 29 i + 2 j − 4k

9. b + c = 3 i + 7 j + k, a · (b + c) = (3 i + 2 j − k) · (3 i + 7 j + k) = 22

10. a + b = 8 i + 5 j − k); ‖a + b‖ =√

64 + 25 + 1 = 3√

10

11. ‖c‖2 = (−2)2 + 42 + 12 = 21 12. b × b = 0. Therefore ‖b × b‖ = 0

13. 2a − b = i + j − 2k; (2a − b) · c = (i + j − 2k) · (−2 i + 4 j + k) = 0

14. b + c = 3 i + 7 j + k; a × (b + c) =

∣∣∣∣∣∣∣∣i j k

3 2 −1

3 7 1

∣∣∣∣∣∣∣∣= 9 i − 6 j + 15k

15. ‖a‖ =√

14; ua =1√14

(3 i + 2 j − k) 16. ‖c‖ =√

21; −uc = − 1√21

(−2 i + 4 j + k)

17. cos θ =a · c

‖a‖‖c‖ =1√

14√

21=

17√

6; θ ∼= 1.51 radians

18. cos θ =b · c

‖b‖‖c‖ =2√

34√

21; θ ∼= 85.7◦

19. ‖a‖ =√

14; cosα =3√14

, α ∼= 0.64 radians, cosβ =2√14

, β ∼= 1.01, cos γ =−1√14

, γ ≈ 1.84

20. comp ab = b · ua = (5 i + 3 j) · 1√14

(3 i + 2 j − k) =21√14

21. b × c =

∣∣∣∣∣∣∣∣i j k

5 3 0

−2 4 1

∣∣∣∣∣∣∣∣= 3 i − 5 j + 26k;

comp a(b × c) = (b × c) · ua = (3 i − 5 j + 26k) · 1√14

(3 i + 2 j − k) = − 27√14




22. a × c = 6 i − j + 16k, ‖a × c‖ =√

293; u = ±6 i − j + 16k√293

23. V = |(a × b) · c|; (a × b) · c =

∣∣∣∣∣∣∣∣3 2 −1

5 3 0

−2 4 1

∣∣∣∣∣∣∣∣= −27; V = | − 27| = 27

24. A =12‖a × b‖; a × b =

∣∣∣∣∣∣∣∣i j k

3 2 −1

5 3 0

∣∣∣∣∣∣∣∣= 3 i − 5 j − k; A =

12

√35

25. (a) Direction vector:−→QR= (6,−3, 3); scalar parametric equations for the line:

x = 1 + 6t, y = 1 − 3t, z = 1 + 3t.

(b) Normal vector:−→PR= (3,−3, 2); equation of the plane:

3(x− 1) + (−3)(y − 1) + 2(z − 1) = 0.

(c) A normal vector for the plane is:−→QR ×

−→PR= (3,−3,−9) or N = (1,−1,−3);

an equation for the plane: (x− 1) − (y − 1) − 3(z − 1) = 0

26. (a) Direction vector:−→PQ= (4,−7, 5); scalar parametric equations for the line:

x = 5 + 4t, y = 6 − 7t, z = −3 + 5t.

(b) Scalar parametric equations for the line l through P and Q are:

x = 3 + 4t, y = 2 − 7t, z = −1 + 5t.

For any point S on l, the vector−→RS is (4t− 2,−7t− 4, 5t + 2). We want

−→RS ⊥

−→PQ:

(4t− 2,−7t− 4, 5t + 2) · (4,−7, 5) = 0 =⇒ t = −13.

Therefore(− 10

3 ,− 53 ,

13

)is a direction vector for the line that passes through R perpendicular

to PQ. Scalar parametric equations for this line are:

x = 5 − 103t, y = 6 − 5

3t, z = −3 +

13t.

(c) This plane is determined by the points P, Q, R. A normal vector for the plane is:−→PQ ×

−→PR= −6 i + 18 j + 30k or i − 3 j − 5k.

An equation for the plane is: (x− 3) − 3(y − 2) − 5(z + 1) = 0 or x− 3y − 5z = 2.

27. Solve, if possible, the system of equations: t = 1 − u, −t = 1 + 3u, −6 + 2t = 2u. In this case, the

solution is t = 2, u = −1. The lines intersect at the point (2,−2,−2).

28. Solve, if possible, the system of equations: 1 − 2t = 3 + 2u, 3 + 3t = 1 − u, 5t = 6 + 3u. In this case

there is no solution; the lines are skew.




29. The lines l1 and l2 written in scalar parametric form are:

l1 : x = 1 + 2t, y = −2 − t, z = 3 + 4t; l2 : x = −2 + u, y = 3 + 3u, z = u.

Solve, if possible, the system of equations: 1 + 2t = −2 + u, −2 − t = 3 + 3u, 3 + 4t = u. In this case

there is no solution; the lines are skew.

30. Scalar parametric equations for the line l are: x = −1 + t, y = −2 + t, −1 + t and a direction

vector for l is d = (1, 1, 1). The point Q (−1 + t,−2 + t,−1 + t) is a point on l and−→PQ=

(−4 + t,−3 + t, 1 + t) is the vector from P to Q. We want d ·−→PQ= 0:

(−4 + t,−3 + t, 1 + t) · (1, 1, 1) = 0 =⇒ 3t = 6 and t = 2.

Therefore Q = (1, 0, 1).

31. (a) No.−→PQ= (4,−7, 5),

−→PR= (2,−3, 2); the vectors are not parallel; the points are not collinear.

(b)−→PQ= (4,−7, 5),

−→PR= (2,−3, 2),

−→PS= (−2, 0, 1)

(−→PQ ×

−→PR) ·

−→PS=

∣∣∣∣∣∣∣∣4 −7 5

2 −3 2

−2 0 1

∣∣∣∣∣∣∣∣= 0.

The points are coplanar.

32. The scalar parametric equations for the line are:

x = −2 + 3t, y = 1 + 2t, z = −6 + t.

Substituting x, y, z into the equation for the plane gives

2(−2 + 3t) + (1 + 2t) − 3(−6 + t) + 6 = 0 =⇒ t = −215.

Therefore the line intersects the plane at the point(− 73

5 ,− 375 ,− 51

5

).

33.−→PQ ×

−→PR= −10 i + 5k is a normal vector for the plane; so is N = 2 i − k.

An equation for the plane is: 2(x− 1) − (z − 1) = 0 or 2x− z = 1

34. N = 2 i + 3 j − 4k is a normal vector for the plane. An equation for the plane is:

2(x− 2) + 3(y − 1) − 4(z + 3) = 0 or 2x + 3y − 4z = 19.

35. N = 3 i + 2 j − 1k is a normal vector for the plane. An equation for the plane is:

3(x− 1) + 2(y + 2) − (z + 1) = 0 or 3x + 2y − z = 0.

36. The point Q (2,−1, 0) is in the plane since it is on the line. d = 2 i + 3 j − 2k is a direction vector

for the line. The vector−→PQ ×d is a normal vector for the plane.

−→PQ ×d =

∣∣∣∣∣∣∣∣i j k

−1 0 −2

2 3 −2

∣∣∣∣∣∣∣∣= 6 i − 6 j − 3k.




Take N = 2 i − 2 j − k as a normal vector for the plane. An equation for the plane is:

2(x− 3) − 2(y + 1) − (z − 2) = 0 or 2x− 2y − z = 6.

37. Let P be the plane that satisfies the conditions. A direction vector for the given line is d = (3, 2, 4);

a normal vector for the given plane is N = (2, 1,−3). The cross product d × N is a normal vector

for P .

d × N =

∣∣∣∣∣∣∣∣i j k

3 2 4

2 1 −3

∣∣∣∣∣∣∣∣= −10 i + 17 j − k.

The point Q (−1, 1, 2) is on the plane. An equation for P is:

−10(x + 1) + 17(y − 1) − (z − 2) = 0 or 10x− 17y + z + 25 = 0.

38. Let P be the plane that satisfies the given conditions. The vectors N1 = 3 i + j − k and N2 =

2 i + j + 4k are normal vectors for the given planes. The vector d = N1 × N2 = 5 i − 14 j + k is a

direction vector for the line of intersection. Solving the equations 3x + y − z = 2, 2x + y + 4z = 1

simultaneously, we find that Q (1,−1, 0) lies on the line of intersection (set z = 0 and solve for x

and y). Now, the vector

N =−→PQ ×d =

∣∣∣∣∣∣∣∣i j k

−1 −2 3

5 −14 1

∣∣∣∣∣∣∣∣= 40 i + 16 j + 24k or 5 i + 2 j + 3k

is a normal vector for P . An equation for P is: 5(x− 2) + 2(y − 1) + 3(z + 3) = 0.

39. The line l which passes through Q and R has direction vector d =−→QR= (2, 1,−2). By (13.5.6),

the distance from P to l is given by

d(P, l) =‖

−→QP ×d‖‖d‖ =

‖(2, 4,−5) × (2, 1,−2)‖3

=93

= 3.

40. By (13.6.5), the distance from P to the plane is given by

d =|1(2) − 2(1) + 2(−1) + 5|√

1 + 4 + 4=

33

= 1.

41. The normals are: N1 = (2, 1, 1), N2 = (2, 2,−1). The cosine of the angle between the planes is:

cos θ =|N1 · N2|‖N1‖‖N2‖

=5√54

and θ ∼= 0.822 radians.

42. The normals are: N1 = (2,−3, 1), N2 = (1, 4,−5). The cosine of the angle between the planes is:

cos θ =|N1 · N2|‖N1‖‖N2‖

=15

7√

12and θ ∼= 0.904 radians.




43. The normal vectors to the two planes are: N1 = 3 i + 5 j + 2k, N2 = i + 2 j − k. A direction vector

for the line of intersection is:

N1 × N2 =

∣∣∣∣∣∣∣∣i j k

3 5 2

1 2 −1

∣∣∣∣∣∣∣∣= −9 i + 5 j + k.

A solution of the pair of equations 3x + 5y + 2z − 4 = 0 x + 2y − z − 2 = 0 is x = −2, y = 2, z = 0

(set z = 0 and solve for x and y). Scalar parametric equations for the line of intersection are:

x = −2 − 9t, y = 2 + 5t, z = t.

44. The normal vectors to the two planes are: N1 = i − 2 j + 2k, N2 = 3 i − j − k. A direction vector for

the line of intersection is:

N1 × N2 =

∣∣∣∣∣∣∣∣i j k

1 −2 2

3 −1 −1

∣∣∣∣∣∣∣∣= 4 i + 7 j + 5k.

A solution of the pair of equations x− 2y + 2z = 1 3x− y − z = 2 is x = 0, y = −5/4, z = −3/4

(set x = 0 and solve for y and z). Scalar parametric equations for the line of intersection are:

x = 4t, y = − 54 + 7t, z = − 3

4 + 5t.

45. a × b = −5 i + 11 j + 7k is perpendicular to both a and b; ‖a × b‖ =√

195. The vectors are:

± 4√195

(−5 i + 11 j + 7k).

46. Since a × (b × c) = (a · c)b − (a · b)c (13.4.11),

(a × b) × (c × d) = [(a × b) · d]c − [(a × b) · c]d.

47. (‖b‖a − ‖a‖b) · (‖b‖a + ‖a‖b) = ‖a‖2‖b‖2 + ‖a‖‖b‖a · b − ‖a‖‖b‖a · b − ‖a‖2‖b‖2 = 0.

Therefore, (‖b‖a − ‖a‖b) ⊥ (‖b‖a + ‖a‖b)

48. ‖a + b‖2 − ‖a − b‖2 = (a + b) · (a + b) − (a − b) · (a − b) = ‖a‖2 + ‖b‖2 + 2a · b − ‖a‖2 − ‖b‖2 +

2a · b = 4a · b

49. Let a and b be adjacent sides of a parallelogram. Then the diagonals of the parallelogram are

a + b and a − b. By Exercise 48, the diagonals have equal length iff a ⊥ b, which means that the

parallelogram is a rectangle.

50. Let A, B, C, D be the vertices of the quadrilateral, and let E,F,G,H be the midpoints of

AB, BC, CD, DA, respectively. Then, EF ‖ AC ‖ GH and FG ‖ BD ‖ EH. Therefore EFGH is

a parallelogram.




51. Let A, B, C be the vertices of a triangle. Without loss of generality, assume that A (0, 0), B (x1, y1),

C (x2, 0). Let D and E be the midpoints of AB and BC, respectively. Then D(x1

2,y1

2

)and

E

(x1 + x2

2,y1

2

). Now

−→DE=

(x2

2, 0

), and

−→AC= (x2, 0).

Therefore−→DE‖

−→AC and ‖

−→DE ‖ = 1

2‖−→AC ‖.

52. (a) If A �= 0, then (−CA , 0) and (−B+C

A , 1) are two points on l. Therefore (BA ,−1) or (B,−A) are

directional vectors for l. Thus

r(t) = (−C/A) i + (B i −A j)t = (−C/A + Bt) i + (−At) j

is the parametrization of the line.

(b) (A i + B j) · (B i −A j) = 0

(c) P = (−C/A, 0) is a point on l and−→OP= −C/A i. By (13.5.6)

‖−→OP ×(B i −A j)‖‖(B i −A j)‖ =

|C|√A2 + B2

.

calculus one and several variables 10e salas solutions manual ch13

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