11

CHAPTER

Geometry in Space
and Vectors
7. P(2, 1, 6), Q(4, 7, 9), R(8, 5, –6)

11.1 Concepts Review

PQ = (2 – 4) 2 + (1 – 7) 2 + (6 – 9)2 = 7

1. coordinates
( x + 1) + ( y – 3) + ( z – 5)
2

2.

2

PR = (2 – 8)2 + (1 − 5) 2 + (6 + 6)2 = 14

2

QR = (4 – 8) 2 + (7 – 5)2 + (9 + 6) 2 = 245

3. (–1, 3, 5); 4

2

2

2

PQ + PR = 49 + 196 = 245 = QR , so the

4. plane; 4; –6; 3

triangle formed by joining P, Q, and R is a right
triangle, since it satisfies the Pythagorean
Theorem.

Problem Set 11.1
1. A(1, 2, 3), B(2, 0, 1), C(–2, 4, 5), D(0, 3, 0),
E(–1, –2, –3)

8. a.

The distance to the xy-plane is 1 since the
point is 1 unit below the plane.

b. The distance is
(2 – 0)2 + (3 – 3)2 + (–1 – 0) 2 = 5 since
the distance from a point to a line is the
length of the shortest segment joining the
point and the line. Using the point (0, 3, 0)
on the y-axis clearly minimizes the length.

2. A

(


)

1 ⎞
⎛
3, – 3, 3 , B (0, π, – 3), C ⎜ –2, , 2 ⎟ , D(0, 0, e)
3 ⎠
⎝

c.

(2 – 0)2 + (3 − 0) 2 + (–1 – 0) 2 = 14

9. Since the faces are parallel to the coordinate
planes, the sides of the box are in the planes
x = 2, y = 3, z = 4, x = 6, y = –1, and z = 0 and the
vertices are at the points where 3 of these planes
intersect. Thus, the vertices are (2, 3, 4), (2, 3, 0),
(2, –1, 4), (2, –1, 0), (6, 3, 4), (6, 3, 0), (6, –1, 4),
and (6, –1, 0)

3. x = 0 in the yz-plane. x = 0 and y = 0 on the z-axis.
4. y = 0 in the xz-plane. x = 0 and z = 0 on the y-axis.
5. a.

(6 – 1) 2 + (–1 – 2)2 + (0 – 3)2 = 43

b.

(–2 – 2) 2 + (–2 + 2)2 + (0 + 3) 2 = 5


c.

(e + π)2 + (π + 4)2 + 0 – 3

(

)

2

≈ 9.399

6. P(4, 5, 3), Q(1, 7, 4), R(2, 4, 6)
PQ = (4 − 1) + (5 – 7) + (3 – 4)
2

2

2

= 14

10. It is parallel to the y-axis; x = 2 and z = 3. (If it
were parallel to the x-axis, the y-coordinate could
not change, similarly for the z-axis.)
11. a.

( x –1)2 + ( y – 2)2 + ( z – 3) 2 = 25


PR = (4 – 2)2 + (5 – 4) 2 + (3 – 6)2 = 14

b.

( x + 2)2 + ( y + 3)2 + ( z + 6)2 = 5

QR = (1 – 2)2 + (7 – 4)2 + (4 – 6)2 = 14

c.

( x – π)2 + ( y – e)2 + z – 2

(

)

2

=π

Since the distances are equal, the triangle formed
by joining P, Q, and R is equilateral.

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12. Since the sphere is tangent to the xy-plane, the point (2, 4, 0) is on the surface of the sphere. Hence, the radius of
the sphere is 5 so the equation is ( x – 2)2 + ( y – 4) 2 + ( z – 5) 2 = 25.
13. ( x 2 –12 x + 36) + ( y 2 + 14 y + 49) + ( z 2 – 8 z + 16) = –1 + 36 + 49 + 16
( x – 6) 2 + ( y + 7)2 + ( z – 4) 2 = 100
Center: (6, –7, 4); radius 10

14. ( x 2 + 2 x + 1) + ( y 2 – 6 y + 9) + ( z 2 –10 z + 25) = –34 + 1 + 9 + 25
( x + 1)2 + ( y – 3)2 + ( z – 5)2 = 1
Center: (–1, 3, 5); radius 1

15. x 2 + y 2 + z 2 – x + 2 y + 4 z =

13
4

1⎞
13 1
⎛ 2
2
2
⎜ x – x + ⎟ + ( y + 2 y + 1) + ( z + 4 z + 4) = + + 1 + 4
4⎠
4 4
⎝
2

1⎞

⎛
2
2 17
⎜ x – ⎟ + ( y + 1) + ( z + 2) =
2⎠
2
⎝
⎛1
⎞
Center: ⎜ , – 1, – 2 ⎟ ; radius
⎝2
⎠

17
≈ 2.92
2

16. ( x 2 + 8 x + 16) + ( y 2 – 4 y + 4) + ( z 2 – 22 z + 121) = –77 + 16 + 4 + 121
( x + 4)2 + ( y – 2) 2 + ( z – 11)2 = 64
Center: (–4, 2, 11); radius 8

17. x-intercept: y = z = 0 ⇒ 2x = 12, x = 6
y-intercept: x = z = 0 ⇒ 6y = 12, y = 2
z-intercept: x = y = 0 ⇒ 3z = 12, z = 4

18. x-intercept: y = z = 0 ⇒ 3x = 24, x = 8
y-intercept: y = z = 0 ⇒ –4y = 24, y = –6
z-intercept: x = y = 0 ⇒ 2z = 24, z = 12

Instructor’s Resource Manual


19. x-intercept: y = z = 0 ⇒ x = 6
y-intercept: x = z = 0 ⇒ 3y = 6, y = 2
z-intercept: x = y = 0 ⇒ –z = 6, z = –6

20. x-intercept: y = z = 0 ⇒ –3x = 6, x = –2
y-intercept: x = z = 0 ⇒ 2y = 6, y = 3
z-intercept: x = y = 0 ⇒ z = 6

Section 11.1

669

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21. x and y cannot both be zero, so the plane is
parallel to the z-axis.
x-intercept: y = z = 0 ⇒ x = 8
8
y-intercept: x = z = 0 ⇒ 3y = 8, y =
3

b ⎛ dx ⎞2 ⎛ dy ⎞2 ⎛ dz ⎞2
+
+

For problems 25-36, L = ∫a
25.


⎜ ⎟
⎝ dt ⎠

⎜ ⎟
⎝ dt ⎠

⎜ ⎟
⎝ dt ⎠

dt

dx
dy
dz
= 1,
= 1, = 2
dt
dt
dt
2

2

L = ∫0 12 + 12 + 22 dt = ∫0 6 dt =
2

⎡ 6t ⎤ = 2 6 ≈ 4.899
⎣
⎦0


26.

dx 1 dy 1 dz 1
= ,
= , =
dt 4 dt 3 dt 2
3

L = ∫1

22. x and z cannot both be zero, so the plane is
parallel to the y-axis.
x-intercept: y = z = 0 ⇒ 3x = 12, x = 4
z-intercept: x = y = 0 ⇒ 4z = 12, z = 3

⎡
⎣

27.

61
144

61
144

dt =

≈ 1.302


L = ∫1

⎛9 ⎞
⎜ t ⎟ +9+16
⎝4 ⎠

1 136
18 109

u du =

4

1 76
36 40

∫

41

dt = ∫1

9t + 100 dt =

2

u = 9t +100
du = 9 dt


136

1 ⎡ 32 ⎤
u ⎥ ≈ 16.59
27 ⎢⎣
⎦109

dx 3
dy
t,
=
=
dt 2
dt
L = ∫2

29.

61
144

3

dt = ∫1

dx 3
dy
dz
=
= 3, = 4

t,
dt 2
dt
dt

∫

23. This is a sphere with center (0, 0, 0) and radius 3.

3

t⎤ = 2
⎦1

4

28.

2
2
2
⎛1⎞ ⎛1⎞ ⎛1⎞
⎜ ⎟ +⎜ ⎟ +⎜ ⎟
4
3
2
⎝ ⎠ ⎝ ⎠ ⎝ ⎠

3
2


t,

⎛9 ⎞ ⎛9 ⎞
⎜ t ⎟ + ⎜ t ⎟ +1 dt
⎝4 ⎠ ⎝4 ⎠

dz
=1
dt
41

= ∫2

2

18t + 4 dt =

u =18t + 4
du =18 dt

76

u du =

1 ⎡ 32 ⎤
u ⎥ ≈ 7.585
54 ⎢⎣
⎦ 40


dx
dy
dz
= 2t ,
= 2 t, =1
dt
dt
dt
8
8
2
L = ∫0 4t + 4t + 1 dt = ∫0 (2t + 1) 2 dt =
8

24. This is a sphere with center (2, 0, 0) and radius 2.

8
2
∫0 (2t + 1) dt = ⎡⎣t + t ⎤⎦ 0 = 72

30.

dx
dy
dz
= 2t ,
= 2 3t , = 3
dt
dt
dt

4
4
L = ∫1 4t 2 + 12t + 9 dt = ∫1 (2t + 3) 2 dt =
4

4
2
∫1 (2t + 3) dt = ⎡⎣t + 3t ⎤⎦1 = 28 − 4 = 24

31.

dx
dy
dz
= −2sin t ,
= 2 cos t , = 3
dt
dt
dt
π

π

L = ∫−π 4sin 2 t + 4 cos 2 t + 9 dt = ∫−π 13 dt =
π

⎡ 13 t ⎤ = 2π 13 ≈ 22.654
⎣
⎦ −π


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32.

dx
dy
dz 1
= −2sin t ,
= 2 cos t , =
dt
dt
dt 20
8π

L = ∫0

π
5

40

6π


6π

By the Parabolic Rule (n = 10):
i
0
1
2
3
4
5
6
7
8
9
10

dx
1 dy
dz
33.
=
= 1, = 1
,
dt 2 t dt
dt
6

⎛1⎞
⎜ ⎟ +1+1 dt

⎝ 4t ⎠

6

= ∫1

2 + ⎛⎜ 1 ⎞⎟ dt
⎝ 4t ⎠

By the Parabolic Rule (n = 10):
i
0
1
2
3
4
5
6
7
8
9
10

xi
1
1.5
2
2.5
3
3.5

4
4.5
5
5.5
6

ci c i ⋅ f ( x i )
1
1.5000
4
5.8878
2
2.9155
4
5.7966
2
2.8868
4
5.7570
2
2.8723
4
5.7349
2
2.8636
4
5.7208
1
1.4289
approximation 7.2273


f ( xi )
1.5000
1.4720
1.4577
1.4491
1.4434
1.4392
1.4361
1.4337
1.4318
1.4302
1.4289

dx
dy
dz
34.
= 1,
= 2t , = 3t 2
dt
dt
dt
2
1

L=∫

1+ 4t 2 +9t 4


dt

By the Parabolic Rule (n = 10):
i
0
1
2
3
4
5
6
7
8
9
10

xi
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2

4sin 2 t + cos 2 t + 1 dt =


3sin 2 t + 2 dt

∫0

1601 dt = ⎡⎣ 401 1601 t ⎤⎦ =
0

1601 ≈ 25.14

L = ∫1

dx
dy
dz
= −2sin t ,
= cos t , = 1
dt
dt
dt
L = ∫0

1
4sin 2 t + 4 cos 2 t + 400
dt =

8π

8π 1


∫0

35.

ci c i ⋅ f ( x i )
1
3.7417
4 17.4433
2 10.0841
4 23.1395
2 13.1779
4 29.8161
2 16.7598
4 37.4655
2 20.8268
4 46.0832
1 12.6886
approximation 7.7075

f ( xi )
3.7417
4.3608
5.0421
5.7849
6.5890
7.4540
8.3799
9.3664
10.4134
11.5208

12.6886

36.

xi
0
1.88
3.77
5.65
7.54
9.42
11.3
13.2
15.1
17
18.8

ci c i ⋅ f ( x i )
1
1.4142
4
8.6843
2
3.4851
4
6.9702
2
4.3421
4
5.6569

2
4.3421
4
6.9702
2
3.4851
4
8.6843
1
1.4142
approximation 34.8394

f ( xi )
1.4142
2.1711
1.7425
1.7425
2.1711
1.4142
2.1711
1.7425
1.7425
2.1711
1.4142

dx
dy
dz
= cos t ,
= − sin t , = cos t

dt
dt
dt
2π

L = ∫0
2π

∫0

cos 2 t + sin 2 t + cos 2 t dt =

cos 2 t + 1 dt

By the Parabolic Rule (n = 10):
i
0
1
2
3
4
5
6
7
8
9
10

xi
0

0.63
1.26
1.88
2.51
3.14
3.77
4.4
5.03
5.65
6.28

ci c i ⋅ f ( x i )
1
1.4142
4
5.1451
2
2.0933
4
4.1866
2
2.5726
4
5.6569
2
2.5726
4
4.1866
2
2.0933

4
5.1451
1
1.4142
approximation 7.6405

f ( xi )
1.4142
1.2863
1.0467
1.0467
1.2863
1.4142
1.2863
1.0467
1.0467
1.2863
1.4142

37. The center of the sphere is the midpoint of the
diameter, so it is
11 ⎞
⎛ –2 + 4 3 –1 6 + 5 ⎞ ⎛
,
,
⎜
⎟ = ⎜ 1, 1, ⎟ . The radius is
2
2
2

2⎠
⎝
⎠ ⎝
1
53
. The
(–2 – 4)2 + (3 + 1) 2 + (6 – 5) 2 =
2
2
2

53
⎛ 11 ⎞
equation is ( x − 1)2 + ( y –1) 2 + ⎜ z – ⎟ = .
2⎠
4
⎝

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Section 11.1

671

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


38. Since the spheres are tangent and have equal
1

radii, the radius of each sphere is of the
2
distance between the centers.
1
r=
(–3 – 5)2 + (1 + 3)2 + (2 – 6)2 = 2 6. The
2

41. a.

b. Plane perpendicular to the xy-plane whose
trace in the xy-plane is the line x = y.
c.

spheres are ( x + 3) + ( y –1) + ( z – 2) = 24 and
2

2

2

( x – 5)2 + ( y + 3)2 + ( z – 6) 2 = 24.

Plane parallel to and two units above the
xy-plane

Union of the yz-plane (x = 0) and the
xz-plane (y = 0)

d. Union of the three coordinate planes


39. The center must be 6 units from each coordinate
plane. Since it is in the first octant, the center is
(6, 6, 6). The equation is

e.

Cylinder of radius 2, parallel to the z-axis

f.

Top half of the sphere with center (0, 0, 0)
and radius 3

( x − 6)2 + ( y – 6) 2 + ( z – 6)2 = 36.

42. The points of the intersection satisfy both
40. x + y = 12 is parallel to the z-axis. The distance
from (1, 1, 4) to the plane x + y = 12 is the same
as the distance in the xy-plane of (1, 1, 0) to the
line x + y – 12 = 0. That distance is
1 + 1 –12
= 5 2. The equation of the sphere is
(1 + 1)1/ 2

( x –1)2 + ( y + 2)2 + ( z + 1)2 = 10 and z = 2, so
( x –1)2 + ( y + 2)2 + (2 + 1) 2 = 10 . This simplifies

to ( x –1)2 + ( y + 2)2 = 1, the equation of a circle
of radius 1. The center is (1, –2, 2).


( x –1)2 + ( y –1)2 + ( z – 4)2 = 50.

43. If P(x, y, z) denotes the moving point,

( x –1) 2 + ( y – 2)2 + ( z + 3)2 = 2 ( x –1)2 + ( y – 2)2 + ( z – 3)2 ,

which simplifies to ( x –1)2 + ( y – 2) 2 + ( z – 5) 2 = 16, is a sphere with radius 4 and center (1, 2, 5).
44. If P(x, y, z) denotes the moving point, ( x –1) 2 + ( y – 2)2 + ( z + 3)2 = ( x – 2)2 + ( y – 3)2 + ( z – 2) 2 ,
which simplifies to x + y + 5z = 3/2, a plane.
⎡ ⎛ h ⎞⎤
45. Note that the volume of a segment of height h in a hemisphere of radius r is πh 2 ⎢ r − ⎜ ⎟ ⎥ .
⎣ ⎝ 3 ⎠⎦
The resulting solid is the union of two segments, one for each sphere. Since the two spheres have the same radius,
each segment will have the same value for h. h is the radius minus half the distance between the centers of the two
spheres.
1
3 1
h = 2−
(2 − 1)2 + (4 − 2) 2 + (3 − 1)2 = 2 − =
2
2 2
⎡ ⎛ 1 ⎞2 ⎛
⎤
1 ⎞ 11π
V = 2 ⎢π ⎜ ⎟ ⎜ 2 − ⎟ ⎥ =
2
6 ⎠ ⎥ 12
⎢⎣ ⎝ ⎠ ⎝
⎦


46. As in Problem 45, the resulting solid is the union of two segments. Since the radii are not the same, the segments
will have different heights. Let h1 be the height of the segment from the first sphere and let h2 be the height from
the second sphere. r1 = 2 is the radius of the first sphere and r2 = 3 is the radius of the second sphere .

Solving for the equation of the plane containing the intersection of the spheres ( x − 1) 2 + ( y − 2) 2 + ( z − 1)2 − 4 = 0
and ( x − 2) 2 + ( y − 4)2 + ( z − 3)2 − 9 = 0, we get x + 2y + 2z – 9 = 0.
The distance from (1, 2, 1) to the plane is
h1 = 2 −

2
7
, and the distance from (2, 4, 3) to the plane is .
3
3

2 4
7 2
= ; h2 = 3 − =
3 3
3 3
2

2

4⎞
2⎞
⎛4⎞ ⎛
⎛2⎞ ⎛
V = π ⎜ ⎟ ⎜ 2 − ⎟ + π ⎜ ⎟ ⎜ 3 − ⎟ = 4π

3
9
3
9
⎝ ⎠ ⎝
⎠
⎝ ⎠ ⎝
⎠

672

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Instructor’s Resource Manual

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


47. Plots will vary. We first note that the sign of c will influence the vertical direction an object moves (along the
helix) with increasing time; if c is negative the object will spiral downward, whereas if c is positive it will spiral
upward. The smaller c is the “tighter” the spiral will be; that is the space between successive “coils” of the helix

decreases as c decreases.
48. Plots will vary. We first note that the sign of a will influence the rotational direction that an object moves (along
the helix) with increasing time; if a is negative the object will rotate in a clockwise direction, whereas if a is
positive rotation will be counterclockwise. The smaller a is the narrower the spiral will be; that is the circles

traced out will be of smaller radius as a decreases


11.2 Concepts Review

7.

w = u cos 60° + v cos 60° =

8.

w = u cos 45° + v cos 45° =

1. magnitude; direction

1 1
+ =1
2 2

2. they have the same magnitude and direction.

2
2
+
= 2
2
2

9. u + v = −1 + 3, 0 + 4 = 2, 4
3. the tail of u; the head of v

u − v = −1 − 3, 0 − 4 = −4, −4
u = (−1) 2 + (0) 2 = 1 = 1


4. 3

v = (3) 2 + (4) 2 = 25 = 5

Problem Set 11.2
10. u + v = 0 + (−3), 0 + 4 = −3, 4
1.

u − v = 0 − (−3), 0 − 4 = 3, −4
u = (0) 2 + (0) 2 = 0 = 0
v = (−3) 2 + (4) 2 = 25 = 5

11. u + v = 12 + (−2),12 + 2 = 10,14
2.

u − v = 12 − (−2),12 − 2 = 14,10
u = (12) 2 + (12) 2 = 288 = 12 2
v = (−2) 2 + (2) 2 = 8 = 2 2

12. u + v = (−0.2) + (−2.1), 0.8 + 1.3 = −2.3, 2.1

3.

u − v = (−0.2) − (−2.1), 0.8 − 1.3 = 1.9, −0.5
u = (−0.2) 2 + (0.8) 2 = 0.68 ≈ 0.825
v = (−2.1) 2 + (1.3) 2 = 6.10 ≈ 2.47

4. 0
5. w =


1
1
1
(u + v) = u + v
2
2
2

1
1
1
6. n = ( v − u) = v − u
2
2
2
1
1
1
m = v – n = v – ( v − u) = v + u
2
2
2

Instructor’s Resource Manual

13. u + v = −1 + 3, 0 + 4, 0 + 0 = 2, 4, 0
u − v = −1 − 3, 0 − 4, 0 − 0 = −4, −4, 0
u = (−1) 2 + (0) 2 + (0) 2 = 1 = 1
v = (3) 2 + (4) 2 + (0) 2 = 25 = 5


Section 11.2

673

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


14. u + v = 0 + (−3), 0 + 3, 0 + 1 = −3,3,1
u − v = 0 − (−3), 0 − 3, 0 − 1 = 3, −3, −1
u = (0) 2 + (0) 2 + (0) 2 = 0 = 0
2

2

15. u + v = 1 + (−5), 0 + 0,1 + 0 = −4, 0,1
u − v = 1 − (−5), 0 − 0,1 − 0 = 6, 0,1

)

v sin θ = 80sin 60° – 60sin 30° = 40 3 – 30

(

= 10 4 3 – 3
2

)


2

2

v = v cos 2 θ + v sin 2 θ

(

)

(

u = (1) 2 + (0) 2 + (1) 2 = 2 ≈ 1.414

= 100 3 3 + 4

v = (−5) 2 + (0) 2 + (0) 2 = 25 = 5

= 100(100) = 10,000
v = 10, 000 = 100

16. u + v = 0.3 + 2.2, 0.3 + 1.3, 0.5 + (−0.9) =

tan θ =

2.5,1.6, −0.4
u − v = 0.3 − 2.2, 0.3 − 1.3, 0.5 − (−0.9) =

−1.9, −1.0,1.4
u = (0.3) 2 + (0.3) 2 + (0.5) 2 = 0.43 ≈ 0.656

v = (2.2) 2 + (1.3)2 + (−0.9)2 = 7.34 ≈ 2.709

17. Let θ be the angle of w measured clockwise from
south.
w cosθ = u cos 30° + v cos 45° = 25 3 + 25 2
= 25

(

3+ 2

)

w sin θ = v sin 45° – u sin 30° = 25 2 – 25

= 25

(

2

)

2 –1
2

2

w = w cos 2 θ + w sin 2 θ


( 3 + 2 ) + 625 ( 2 –1)
= 625 ( 8 − 2 2 + 2 6 )
w = 625 ( 8 − 2 2 + 2 6 ) = 25
2

= 625

2

8−2 2 + 2 6

≈ 79.34
tan θ =

w sin θ
w cos θ

⎛

=

2 –1
3+ 2

2 –1 ⎞
⎟⎟ = 7.5°
⎝ 3+ 2⎠
w has magnitude 79.34 lb in the direction
S 7.5° W.


θ = tan –1 ⎜⎜

674

(

= 10 3 3 + 4

v = (−3) + (3) + (1) = 19 ≈ 4.359
2

18. Let v be the resulting force. Let θ be the angle of
v measured clockwise from south.
v cos θ = 60 cos 30° + 80 cos 60° = 30 3 + 40

Section 11.2

v sin θ
v cos θ

2

+ 100 4 3 – 3

=

)

2


4 3 –3
3 3+4

⎛ 4 3 –3⎞
⎟⎟ ≈ 23.13°
⎝3 3+4⎠
The resultant force has magnitude 100 lb in the
direction S 23.13° W.

θ = tan –1 ⎜⎜

19. The force of 300 N parallel to the plane has
magnitude 300 sin 30° = 150 N. Thus, a force of
150 N parallel to the plane will just keep the
weight from sliding.
20. Let a be the magnitude of the rope that makes an
angle of 27.34°. Let b be the magnitude of the
rope that makes an angle of 39.22°.
1. a sin 27.34° = b sin 39.22°
2. a cos 27.34° + b cos 39.22° = 258.5
Solve 1 for b and substitute in 2.
sin 27.34°
a cos 27.34° + a
cos 39.22° = 258.5
sin 39.22°
258.5
a=
≈ 178.15
cos 27.34° + sin 27.34° cot 39.22°
a sin 27.34°

b=
≈ 129.40
sin 39.22°
The magnitudes of the forces exerted by the
ropes making angles of 27.34° and 39.22° are
178.15 lb and 129.40 lb, respectively.
21. Let θ be the angle the plane makes from north,
measured clockwise.
425 sin θ = 45 sin 20°
9
sin θ = sin 20°
85
⎞
–1 ⎛ 9
θ = sin ⎜ sin 20° ⎟ ≈ 2.08°
⎝ 85
⎠
Let x be the speed of airplane with respect to the
ground.
x = 45 cos 20° + 425 cos θ ≈ 467
The plane flies in the direction N 2.08° E, flying
467 mi/h with respect to the ground.
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22. Let v be his velocity relative to the surface. Let θ
be the angle that his velocity relative to the

surface makes with south, measured clockwise.
v cos θ = 20, v sin θ = 3
2

2

2

v = v cos θ + v sin θ = 400 + 9 = 409
2

e. a(bu) = a ( bu1 , bu2

v sin θ
v cos θ

=

a (bu1 ), a(bu2 ) =

(ab)u1 , (ab)u2 = (ab)u

f. a(u + v ) = a u1 + v1 , u2 + v2 =

2

a (u1 + v1 ), a(u2 + v2 ) =

v = 409 ≈ 20.22


tan θ =

)=

au1 + av1 , au2 + av2 =
au1 , au2 + av1 , av2 = au + av

3
20

3
≈ 8.53°
20
His velocity has magnitude 20.22 mi/h in the
direction S 8.53° W.

θ = tan –1

23. Let x be the air speed.
x cos 60° = 40
40
x=
= 80
cos 60°
The air speed of the plane is 80 mi/hr

2

x sin θ
63sin11.5°

=
x cosθ 837 + 63cos11.5°
⎛ 63sin11.5° ⎞
θ = tan –1 ⎜
⎟ ≈ 0.80°
⎝ 837 + 63cos11.5° ⎠
The plane should fly in the direction N 0.80° W
at an air speed of 898.82 mi/h.
tan θ =

25. Let u = u1 , u2 , v = v1 , v2 , and w = w1 , w2
a. u + v = u1 + v1 , u2 + v2 =
v1 + u1 , v2 + u2 = v + u

b. (u + v) + w = u1 + v1 , u2 + v2 + w1 , w2 =
(u1 + v1 ) + w1 , (u2 + v2 ) + w2 =
u1 + (v1 + w1 ), u2 + (v2 + w2 ) =
u1 , u2 v1 + w1 , v2 + w2 = u + ( v + w )

c. u + 0 = u1 + 0, u2 + 0 = u1 , u2 = u

u1 + (−u1 ), u2 + (−u2 ) = 0, 0 = 0

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h. 1u = 1 ⋅ u1 ,1 ⋅ u2 = u1 , u2 = u
au = au1 , au2

= (au1 ) 2 + (au2 ) 2 =


a 2 u12 + u2 2 = a ⋅ u

= (837 + 63cos11.5°) + (63sin11.5°)
= 704,538 + 105,462 cos 11.5°
x = 704,538 + 105, 462 cos11.5° ≈ 898.82

d. u + (-u) = u1 , u2 + −u1 , −u2 =

au1 , au2 + bu1 , bu2 = au + bu

a 2u12 + a 2u2 2 = a 2 (u12 + u2 2 ) =

x 2 = x 2 cos 2 θ + x 2 sin 2 θ

u + 0 = 0 + u by part a.

au1 + bu1 , au2 + bu2 =

i.

24. Let x be the air speed. Let θ be the angle that the
plane makes with north measured counterclockwise.
x cos θ = 837 + 63 cos 11.5°
x sin θ = 63 sin 11.5°
2

g. (a + b)u = (a + b)u1 , (a + b)u2 =

26. Let u = u1 , u2 , u3 , v = v1 , v2 , v3 , and
w = w1 , w2 , w3


a. u + v = u1 + v1 , u2 + v2 , u3 + v3 =
v1 + u1 , v2 + u2 , v3 + u3 = v + u
b. (u + v) + w = u1 + v1 , u2 + v2 , u3 + v3
+ w1 , w2 , w3 =
(u1 + v1 ) + w1 , (u2 + v2 ) + w2 , (u3 + v3 ) + w3 =
u1 + (v1 + w1 ), u2 + (v2 + w2 ), u3 + (v3 + w3 ) =
u1 , u2 , u3 v1 + w1 , v2 + w2 , v3 + w3 =
u + (v + w)

c. u + 0 = u1 + 0, u2 + 0, u3 + 0 = u1 , u2 , u3 = u
u + 0 = 0 + u by part a.

d. u + (-u) = u1 , u2 , u3 + −u1 , −u2 , −u3 =
u1 + (−u1 ), u2 + (−u2 ), u3 + (−u3 ) =

0, 0, 0 = 0

e. a(bu) = a ( bu1 , bu2 , bu3

)=

a (bu1 ), a(bu2 ), a (bu3 ) =
(ab)u1 , (ab)u2 , (ab)u3 = (ab)u

g. (a + b)u = (a + b)u1 , (a + b)u2 , (a + b)u3 =
au1 + bu1 , au2 + bu2 , au3 + bu3 =
au1 , au2 , au3 + bu1 , bu2 , bu3 = au + bu

h. 1u = 1 ⋅ u1 ,1 ⋅ u2 ,1 ⋅ u3 = u1 , u2 , u3 = u


Section 11.2

675

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27. Given triangle ABC, let D be the midpoint of AB
and E be the midpoint of BC. u = AB , v = BC ,
w = AC , z = DE

u+v=w
1
1
1
1
z = u + v = (u + v) = w
2
2
2
2
Thus, DE is parallel to AC.
28. Given quadrilateral ABCD, let E be the midpoint
of AB, F the midpoint of BC, G the midpoint of
CD, and H the midpoint of AD.
ABC and ACD are triangles. From Problem 17,
EF and HG are parallel to AC. Thus, EF is
parallel to HG. By similar reasoning using

triangles ABD and BCD, EH is parallel to FG.
Therefore, EFGH is parallelogram.

2π
. Thus the vectors
n
(placed head to tail from v1 to v n ) form a regular
n-gon. From Problem 19, the sum of the vectors
is 0.

a regular n-gon is π –

31. The components of the forces along the lines
containing AP, BP, and CP are in equilibrium;
that is,
W = W cos α + W cos β
W = W cos β + W cos γ
W = W cos α + W cos γ
Thus, cos α + cos β = 1, cos β + cos γ = 1, and
cos α + cos γ = 1. Solving this system of
1
equations results in cos α = cos β = cos γ = .
2
Hence α = β = γ = 60°.
Therefore, α + β = α + γ = β + γ = 120°.
32. Let A′, B ′, C ′ be the points where the weights are
attached. The center of gravity is located
AA′ + BB ′ + CC ′
units below the plane of the
3

triangle. Then, using the hint, the system is in
equilibrium when AA′ + BB ′ + CC ′ is
maximum. Hence, it is in equilibrium when
AP + BP + CP is minimum, because the total

29. Let Pi be the tail of vi . Then
v1 + v 2 + …+ v n
⎯⎯
→

⎯⎯
→

= P1P2 + P2 P3 +

⎯⎯
→

+ Pn P1

⎯⎯
→

= P1 P1 = 0.

30. Consider the following figure of the circle.

2π
n
The vectors have the same length. Consider the

following figure for adding vectors vi and vi +1.

α=

Then β = π –

676

2π
. Note that the interior angle of
n

Section 11.2

length of the string is
AP + AA′ + BP + BB ′ + CP + CC ′ .
33. The components of the forces along the lines
containing AP, BP, and CP are in equilibrium;
that is,
5w cos α + 4w cos β = 3w
3w cos β + 5w cos γ = 4w
3w cos α + 4w cos γ = 5w
Thus,
5 cos α + 4 cos β = 3
3 cos β + 5 cos γ = 4
3 cos α + 4 cos γ = 5.
Solving this system of equations results in
3
4
cos α = , cos β = 0, cos γ = , from which it

5
5
4
3
follows that sin α = , sin β = 1, sin γ = .
5
5
4
Therefore, cos(α + β ) = – , cos(α + γ ) = 0,
5
3
cos( β + γ ) = – , so
5
⎛ 4⎞
α + β = cos –1 ⎜ – ⎟ ≈ 143.13°, α + γ = 90°,
⎝ 5⎠
⎛ 3⎞
β + γ = cos –1 ⎜ – ⎟ ≈ 126.87°.
⎝ 5⎠

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This problem can be modeled with three strings
going through A, four strings through B, and five
strings through C, with equal weights attached to
the twelve strings. Then the quantity to be

minimized is 3 AP + 4 BP + 5 CP .

f.

35. By symmetry, the tension on each wire will be
the same; denote it by T where T can be the
tension vector along any of the wires. The
chandelier exerts a force of 100 lbs. vertically
downward. Each wire exerts a vertical tension of
T sin 45 upward. Since a state of equilibrium
exists,
100

36. By symmetry, the tension on each wire will be
the same; denote it by T where T can be the
tension vector along any of the wires. The
chandelier exerts a force of 100 lbs. vertically
downward. Each wire exerts a vertical tension of
T sin 45 upward. Since a state of equilibrium

b ⋅ b − b = (2)(2) + (−3)(−3) − 4 + 9

–4a + 3b = −12, 4 + 3, −3 = −9, 1

b.

b ⋅ c = (1)(0) + (–1)(5) = –5

c.


(a + b ) ⋅ c =

4, −2 ⋅ 0,5

= (4)(0) + (–2)(5) = –10

d.

2c ⋅ (3a + 4b) = 2 0, 5 ⋅ ( 9, – 3 + 4, – 4

)

= 2 0, 5 ⋅ 13, – 7 = 2[(0)(13) + (5)(–7)]

≈ 35.36

2 2
The tension in each wire is approximately 35.36
lbs.

a c ⋅ a = 4 + 9[(0)(−2) + (−5)(3)] = −15 13

= 13 − 13

2. a.

34. Written response.

4 ⋅ T sin 45 = 100 or T =


e.

= –70
e.

b b ⋅ a = 1 + 1[(1)(3) + (−1)(−1)] = 4 2

f.

c

2

− c⋅c =

(

0 + 25

)

2

− [(0)(0) + (5)(5)]

=0

3. a.

200

≈ 47.14
3 2
The tension in each wire is approximately 47.14
lbs.

exists, 3 ⋅ T sin 45 = 100 or T =

a ⋅b
(1)(−1) + (−3)(2)
7
=
=−
a b
50
10
5

cos θ =
=–

( )( )

7
5 2

≈ –0.9899

11.3 Concepts Review
1. u1v1 + u2 v2 + u3v3 ; u v cos θ
2. 0

3. F • D
4.

b.

cos θ =

A, B, C

=–

Problem Set 11.3
1. a.

1
5

a ⋅b
(−1)(6) + (−2)(0)
6
=
=−
a b
6 5
5 (6)

( )

≈ −0.4472


2a – 4b = (–4i + 6j) + (–8i + 12j)
= –12i + 18j

b.

a ⋅ b = (–2)(2) + (3)(–3) = –13

c.

a ⋅ (b + c) = (–2i + 3j) ⋅ (2i – 8 j)
= (–2)(2) + (3)(–8) = –28

d.

(–2a + 3b) ⋅ 5c = 5[(10i – 15 j) ⋅ (–5 j)]
= 5[(10)(0) + (–15)(–5)] = 375

Instructor’s Resource Manual

Section 11.3

677

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


c.

cos θ =


a ⋅b
(2)(−2) + (−1)(−4) 0
=
=
=0
a b
10
5 2 5

( )(

)

c. cos θ =

a ⋅b
(−1)(2) + (3)(−6) −20
=
=
= −1
a b
20
10 2 10

( )(

)

−1


θ = cos − 1 = 180

d.

cos θ =
=

a ⋅b
(4)(−8) + (−7)(10)
=
a b
65 2 41

(

–102
2 2665

=–

51
2665

)(

)

d. cos θ =


≈ –0.9879

=

a ⋅b
(12)(−5) + (0)(0)
=
a b
(12)(5)

–60
= −1
60

θ = cos −1 − 1 = 180

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

3
3

=1

θ = cos −1 1 = 0

4. a.
cos θ =


a ⋅b
=
a b

5. a.
b.

aib = (1)(0) + (2)(1) + (−1)(1) = 1
(a + c)ib = (3j + k )i(j + k)

= (0)(0) + (3)(1) + (1)(1) = 4

c.

a
1
(i + 2j − k )
=
2
2
a
1 + 2 + (−1)2
= 6 i + 6 j− 6 k
6
3
6

d.

(b − c)ia = (i − k )i(i + 2 j − k)


= (1)(1) + (0)(2) + (−1)(−1) = 2

b. cos θ =

a ⋅b
(4)(−8) + (3)(−6) −50
=
=
= −1
a b
(5)(10)
50

e.

aib
(1)(0) + (2)(1) + (−1)(1)
=
a b
12 + 22 + (−1) 2 02 + 12 + 12
=

−1

θ = cos − 1 = 180
f.
6. a.

1

6 2

=

3
6

By Theorem A (5), bib − b

2

=0

aic = ( 2)(−2) + ( 2)(2) + (0)(1) = 0

b. (a − c)ib =

2 − (−2), 2 − 2, 0 − 1 i 1, −1,1 =

(2 + 2)(1) + ( 2 − 2)(−1) + (−1)(1) = 3

678

Section 11.3

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.



1
a
=
2
a
( 2) + ( 2) 2 + 02

c.

1
2

2, 2, 0 =

⎛ −2 + 2 + 0 ⎞
⎟=
⎝ 2⋅ 9 ⎠

θb,c = cos −1 ⎜
cos −1 0 = 90

2 2
,
,0
2 2

(b − c)ia = 3, −3, 0 i

d.


2, 2, 0 =

2, 2, 0 =

(3)( 2) + (−3)( 2) + (0)(0) = 0

9. The basic formulae are
u
u
u
cos α u = 1 cos βu = 2 cos γ u = 3
u
u
u
a.

(1)(−2) + (−1)(2) + (1)(1)
bic
=
=
2
b c
1 + (−1) 2 + 12 (−2) 2 + 22 + 12

e.

−3
3 9


=−

3
3

By Theorem A (5), aia − a

f.

7. The basic formula is cos θ =

2

=0

ui v
u v

b.

⎛ 2 − 2 +0⎞
⎟=
4 ⋅ 3 ⎟⎠
⎝

θa,b = cos −1 ⎜⎜
cos −1 0 = 90

⎛ −2 2 + 2 2 + 0 ⎞
⎟⎟ =

4⋅ 9
⎝
⎠

θa,c = cos −1 ⎜⎜

c.

cos −1 0 = 90
⎛ −2 − 2 + 1 ⎞
⎟=
⎝ 3⋅ 9 ⎠

θb,c = cos −1 ⎜

⎛
3⎞
cos −1 ⎜⎜ −
⎟⎟ ≈ 125.26
3
⎝
⎠

8. The basic formula is cos θu, v =

θ a ,b

ui v
u v


⎛ 3
⎞
3
−
+0⎟
⎜
3
3
⎟=
= cos ⎜
⎜
1⋅ 2 ⎟
⎜
⎟
⎝
⎠

2, 2, 0

a =2

cos α a =

a1
2
=
, α a = 45
a
2


cos βa =

a2
2
=
, βa = 45
a
2

cos γ a =

a3 0
= = 0, γ a = 90
2
a

b = 1, −1,1

b = 3

cos α b =

b1
1
=
≈ 0.577, α b ≈ 54.74
b
3

cos βb =


b2
−1
=
≈ −0.577, βb ≈ 125.26
b
3

cos γ b =

b3
1
=
≈ 0.577, γ b ≈ 54.74
b
3

c = −2, 2,1

c =3

cos α c =

c1
2
= − , α c ≈ 131.81
c
3

cos βc =


c2 2
= , βc ≈ 48.19
c
3

cos γ c =

c3 1
= , γ c ≈ 70.53
c 3

10. The basic formulae are
u
u
u
cos α u = 1 cos βu = 2 cos γ u = 3
u
u
u

−1

cos −1 0 = 90

θ a ,c

a=

⎛ −2 3 −2 3

3⎞
+
+
⎜
⎟
3
3 ⎟=
= cos ⎜ 3
⎜
⎟
1⋅ 9
⎜
⎟
⎝
⎠
−1

3
≈ 125.26
cos −
3
−1

Instructor’s Resource Manual

a.

a=

3 3 3

,
,
3 3 3

a =1

cos α a =

a1
3
=
≈ 0.577, α a ≈ 54.74
a
3

cos βa =

a2
3
=
≈ 0.577, βa ≈ 54.74
a
3

cos γ a =

a3
3
=
≈ 0.577, γ a ≈ 54.74

a
3

Section 11.3

679

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


b = 1, −1, 0

b.

16. If x, y , z is perpendicular to both 1, -2, -3 and
-3, 2, 0 , then x – 2y – 3z = 0 and –3x + 2y = 0.

cos α b =

b1
2
=
, α b = 45
b
2

cos βb =

b2

2
=−
, βb ≈ 135
b
2

cos γ b =

b3
0
=
= 0, γ b = 90
b
2

c = −2, −2,1

c.

11.

b = 2

Solving these equations for y and z in terms of x
3
2
yields y = x, z = – x. Thus, all the vectors
2
3
3

2
have the form x, x, – x where x is a real
2
3
number.

c =3

cos α c =

c1
2
= − , α c ≈ 131.81
c
3

cos βc =

c2
2
= − , βc ≈ 131.81
3
c

cos γ c =

c3 1
= , γ c ≈ 70.53
c 3


17. A vector equivalent to BA is
u = 1 + 4, 2 − 5,3 − 6 = 5, −3, −3 .

A vector equivalent to BC is
v = 1 + 4, 0 - 5, 1 - 6 = 5, -5, -5 .
cos θ =
=

6,3 i −1, 2 = (6)(−1) + (3)(2) = 0

Therefore the vectors are orthogonal
12. aib = (1)(1) + (1)(−1) + (1)(0) = 0
aic = (1)(−1) + (1)(−1) + (1)(2) = 0
bic = (1)(−1) + (−1)(−1) + (0)(2) = 0
Therefore the vectors are mutually orthogonal.

u⋅v
5 ⋅ 5 + (–3)(–5) + (–3)(–5)
=
u v
25 + 9 + 9 25 + 25 + 25

55
43 75

=

11
129


, so θ = cos –1

11
129

18. A vector equivalent to BA is
u = 6 - 3, 3 - 1, 3 + 1 = 3, 2, 4 .

A vector equivalent to BC is
v = -1 - 3, 10 - 1, -2.5 + 1 = -4, 9, -1.5 .
u ⋅ v = 3(–4) + 2 ⋅ 9 + 4 ⋅ (–1.5) = 0 so u is
perpendicular to v. Thus the angle at B is a right
angle.

13. aib = (1)(1) + (−1)(1) + (0)(0) = 0
aic = (1)(0) + (−1)(0) + (0)(2) = 0
bic = (1)(0) + (1)(0) + (0)(2) = 0
Therefore the vectors are mutually orthogonal.

19.

14. (u + v ) ⋅ (u – v ) = u ⋅ u – u ⋅ v + v ⋅ u – v ⋅ v

20. (2ci − 8 j)i(3i + cj) = 0 ⇒ 6c − 8c = 0 ⇒

−2c = 0 ⇒ c = 0
21. (ci + j + k )i(0i + 2 j + dk ) = 0 ⇒ 0c + 2 + d = 0 ⇒
c is any number, d = −2

Thus, u 2 = v 2 or u = v

15. If xi + yj + zk is perpendicular to –4i + 5j + k and
4i + j, then –4x + 5y + z = 0 and 4x + y = 0 since
the dot product of perpendicular vectors is 0.
Solving these equations yields y = –4x and
z = 24x. Hence, for any x, xi – 4xj + 24xk is
perpendicular to the given vectors.
xi − 4 xj + 24 xk = x + 16 x + 576 x
2

10

40

j+

240

10
593

k and
593
593
593
10
40
240
–
i+
j–

k.
593
593
593

are

680

i–

Section 11.3

22.

a, 0,1 i 0, 2, b = 0 ⇒ b = 0
a, 0,1 i 1, c,1 = 0 ⇒ a + 1 = 0

0, 2, b i 1, c,1 = 0 ⇒ 2c + b = 0

Thus: a = −1 and c = b = 0

2

= x 593

This length is 10 when x = ±

c, 6 i c, −4 = 0 ⇒ c 2 − 24 = 0 ⇒
c 2 = 24 ⇒ c = ±2 6


= u 2− v 2 =0

2

≈ 14.4°.

. The vectors

For problems 23-34, the formula to use is
⎛ bia ⎞
proja b = ⎜ 2 ⎟ a
⎜ a ⎟
⎝
⎠
23. u = 1, 2 , v = 2, −1 , w = 1,5

projv u =

(1)(2) + (2)(−1)
22 + (−1) 2

2, −1 = 0

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.



24. u = 1, 2 , v = 2, −1 , w = 1,5
proju v =

(2)(1) + (−1)(2)
12 + 22

34. u = 3, 2,1 , v = 2, 0, −1 , w = 1,5, −3

1, 2 = 0

proji u =

25. u = 1, 2 , v = 2, −1 , w = 1,5
proju w =

(1)(1) + (5)(2)
12 + 22

11 22
1, 2 =
,
5 5

35. a.

26. u = 1, 2 , v = 2, −1 , w = 1,5 , w − v = −1, 6
proju (w − v ) =

(−1)(1) + (6)(2)
1 +2

2

2

1, 2 =

11 22
,
5 5

27. u = 1, 2 , v = 2, −1 , w = 1,5

projju =

(1)(0) + (2)(1)

b.

36. a.

0,1 = 0, 2

02 + (1) 2

28. u = 1, 2 , v = 2, −1 , w = 1,5
proji u =

(1)(1) + (2)(0)
12 + (0) 2


b.

1, 0 = 1, 0

29. u = 3, 2,1 , v = 2, 0, −1 , w = 1, 5, −3
projv u =

(3)(2) + (2)(0) + (1)(−1)
22 + 02 + (−1) 2

2, 0, −1 =

(2)(3) + (0)(2) + (−1)(1)
32 + 22 + 12

3, 2,1 =

⎛ u ⋅ (−u) ⎞
⎟ ( −u ) = ⎛⎜ u ⋅ u ⎞⎟ u = u
proj−u u = ⎜
⎜ −u 2 ⎟
⎝ u⋅u ⎠
⎝
⎠
⎛ ( −u ) ⋅ u ⎞
⎟ ( −u )
proju ( −u ) = ⎜
⎜ u2 ⎟
⎝
⎠

−
⋅
u
u
⎛
⎞
=⎜
⎟ ( −u ) = u
⎝ u ⋅u ⎠
⎛ ( −u ) ⋅ ( −u ) ⎞
⎟ ( −u )
proj−u ( −u ) = ⎜
2
⎜
⎟
−
u
⎝
⎠
⎛ u⋅u ⎞
=⎜
⎟ ( −u ) = −u
⎝ u⋅u ⎠

(

)

38. u ⋅ v = 5 – 5 + 5( 5) + 2(1) = 2
v = 5 + 5 + 1 = 11


u⋅v
2
=
v
11

31. u = 3, 2,1 , v = 2, 0, −1 , w = 1,5, −3
(1)(3) + (5)(2) + (−3)(1)
32 + 22 + 12

3, 2,1 =

39.

u = (4 + 9 + z 2 )1/ 2 = 5 and z > 0, so
z = 2 3 ≈ 3.4641.

15 10 5
, ,
7 7 7

40. cos 2 (46°) + cos 2 (108°) + cos 2 γ = 1
⇒ cos γ ≈ ±0.6496

u = 3, 2,1 , v = 2, 0, −1 , w = 1,5, −3

⇒ γ ≈ 49.49° or γ ≈ 130.51°

w + v = 3,5, −4

proju (w + v) =

(3)(3) + (5)(2) + (−4)(1)
32 + 22 + 12

3, 2,1 =

45 30 15
, ,
14 14 14

41. There are infinitely many such pairs. Note that
–4, 2, 5 ⋅ 1, 2, 0 = –4 + 4 + 0 = 0, so
u = 1, 2, 0 is perpendicular to −4, 2,5 . For

any c, –2, 1, c ⋅ 1, 2, 0 = –2 + 2 + 0 = 0 so

33. u = 3, 2,1 , v = 2, 0, −1 , w = 1, 5, −3

projk u =

⎛ u ⋅u ⎞
⎟ u = ⎜⎛ u ⋅ u ⎟⎞ u = u
proju u = ⎜
⎜ u2⎟
⎝ u⋅u ⎠
⎝
⎠

u⋅v

3
=
= 3
v
3

15 10 5
, ,
14 14 14

32.

1, 0, 0 = 3, 0, 0

v = 1+1+1 = 3

30. u = 3, 2,1 , v = 2, 0, −1 , w = 1,5, −3

proju w =

12 + 02 + 02

37. u ⋅ v = (–1)(–1) + 5 ⋅1 + 3(–1) = 3

2, 0, −1

proju v =

(3)(1) + (2)(0) + (1)(0)


(3)(0) + (2)(0) + (1)(1)
02 + 02 + (1) 2

Instructor’s Resource Manual

v = −2, 1, c is a candidate.
–4, 2, 5 ⋅ –2, 1, c = 8 + 2 + 5c

0, 0,1 = 0, 0,1

8 + 2 + 5c = 0 ⇒ c = –2, so one pair is
u = 1, 2, 0 , v = −2,1, −2 .
Section 11.3

681

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


42. The midpoint is
5 1⎞
⎛ 3 + 5 2 – 7 –1 + 2 ⎞ ⎛
,
,
⎜
⎟ = ⎜ 4, – , ⎟ , so the
2
2
2

2 2⎠
⎝
⎠ ⎝
5 1
vector is 4, – , .
2 2

52. r = ka + mb ⇒ 6 = k(–4) + m(2) and
–7 = k(3) + m(–1)
–4k + 2m = 6
3k – m = –7
Solve the system of equations to get
k = –4, m = –5.

43. The following do not make sense.

53. a and b cannot both be zero. If a = 0, then the line
ax + by = c is horizontal and n = bj is vertical, so
n is perpendicular to the line. Use a similar
argument if b = 0. If a ≠ 0 and b ≠ 0 , then
⎛c ⎞
⎛ c⎞
P1 ⎜ , 0 ⎟ and P2 ⎜ 0, ⎟ are points on the line.
a
⎝
⎠
⎝ b⎠

a.


v ⋅ w is not a vector.

b.

u ⋅ w is not a vector.

44. The following do not make sense.
c.

u is not a vector.

d.

u + v is not a scalar.

⎯⎯
→
c ⎞
⎛ c
n ⋅ P1 P2 = (ai + bj) ⋅ ⎜ – i + j ⎟ = – c + c = 0
⎝ a b ⎠

54.
45. au + bu = a u1 , u2 + b u1 , u2

u+v

2

+ u−v


2

= (u + v ) ⋅ (u + v )

+(u − v ) ⋅ (u − v ) = [u ⋅ u + 2(u ⋅ v) + v ⋅ v]
+[u ⋅ u − 2(u ⋅ v) + v ⋅ v] = 2(u ⋅ u) + 2( v ⋅ v)

= au1 , au2 + bu1 , bu2 = au1 + bu1 , au2 + bu2

(a + b)u1 , (a + b)u2 = (a + b) u1 , u2

=2 u

2

+2 v

2

u+v

2

− u−v

= (a + b)u

55.


46. ui v = u1v1 + u2 v2 = v1u1 + v2u2 = viu
47. c(u ⋅ v ) = c ( u1 , u2 ⋅ v1 , v2

= c (u1v1 + u2 v2 ) = c(u1v1 ) + c(u2 v2 )
= (cu1 )v1 + (cu2 )v2 = cu1 , cu2 ⋅ v1 , v2

= ( c u1 , u2 ) ⋅ v1 , v2 = (cu) ⋅ v

)

= u1 , u2 ⋅ v1 + w1 , v2 + w2
= u1 (v1 + w1 ) + u2 (v2 + w2 )

= (u1v1 + u2 v2 ) + (u1w1 + u2 w2 ) = u ⋅ v + u ⋅ w

49. 0iu = 0u1 + 0u2 = 0
50. uiu = u12 + u2 2 =

(

u12 + u2 2

)

2

= u

2


51. r = ka + mb ⇒ 7 = k(3) + m(–3) and
–8 = k(–2) + m(4)
3k – 3m = 7
–2k + 4m = –8
Solve the system of equations to get
2
5
k = ,m = – .
3
3

= (u + v ) ⋅ (u + v )

−(u − v) ⋅ (u − v ) = [u ⋅ u + 2(u ⋅ v) + v ⋅ v ]
−[u ⋅ u − 2(u ⋅ v ) + v ⋅ v] = 4(u ⋅ v )
1
2 1
2
so u ⋅ v = u + v − u − v
4
4

)

48. u ⋅ ( v + w ) = u1 , u2 ⋅ ( v1 , v2 + w1 , w2

2

56. Place the cube so that its corners are at the points
(0, 0, 0), (1, 0, 0), (0,1, 0), (1, 1, 0), (0, 0, 1),

(1, 0, 1), (0,1, 1), and (1, 1, 1). The main
diagonals are (0, 0, 0) to (1, 1, 1), (1, 0, 0) to
(0, 1, 1), (0, 1, 0) to (1, 0, 1), and (0, 0, 1) to
(1, 1, 0). The corresponding vectors are
1, 1, 1 , −1,1,1 , 1, −1,1 , and 1,1, −1 .

Because of symmetry, we need only address one
situation; let’s choose the diagonal from
(0, 0, 0) to (1,1,1) and the face that lies in the
xy − plane. A vector in the direction of the
diagonal is d = 1,1,1 and a vector normal to the
chosen face is n = 0, 0,1 . The angle between
the diagonal and the face is the complement of
the angle between d and n ; that is
⎛ din ⎞
90 − θ = 90 − cos −1 ⎜
=
⎜ d n ⎟⎟
⎝
⎠
⎛ 1 ⎞
−1 ⎛ 3 ⎞
90 − cos −1 ⎜⎜
⎟⎟ = 90 − cos ⎜⎜
⎟⎟
⎝ 3 1⎠
⎝ 3 ⎠
≈ 90 − 54.74 = 35.26

682


Section 11.3

Instructor’s Resource Manual

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


57. Place the box so that its corners are at the points
(0, 0, 0), (4, 0, 0), (0, 6, 0), (4, 6, 0), (0, 0, 10),
(4, 0, 10), (0, 6, 10), and (4, 6, 10). The main
diagonals are (0, 0, 0) to (4, 6, 10), (4, 0, 0) to
(0, 6, 10), (0, 6, 0) to (4, 0, 10), and (0, 0, 10) to
(4, 6, 0). The corresponding vectors are
4, 6, 10 , −4, 6,10 , 4, −6,10 , and
4, 6, −10 .

62. D = 12j
Work = F· D = (–5)(0) + (8)(12) = 96 joules
63. D = (4 – 0)i + (4 – 0)j + (0– 8)k = 4 +4j – 8 k
Thus, W = F ⋅ D = 0(4) + 0(4) − 4(−8) = 32 joules.
64. D = (9 – 2)i + (4 – 1)j + (6 – 3)k = 7i + 3j + 3k
Thus, W = F ⋅ D = 3(7) – 6(3) + 7(3) = 24 ft-lbs.

All of these vectors have length
16 + 36 + 100 = 152. Thus, the smallest angle
θ between any pair, u and v of the diagonals is
found from the largest value of u ⋅ v , since
u⋅v

u⋅v
=
cos θ =
.
152
u v

65. 2( x − 1) − 4( y − 2) + 3( z + 3) = 0
2 x − 4 y + 3 z = −15

There are six ways of pairing the four vectors.
The largest value of u ⋅ v is 120 which occurs
with
u = 4, 6, 10 and v = −4, 6,10 . Thus,

67. ( x − 1) + 4( y − 2) + 4( z − 1) = 0
x + 4 y + 4 z = 13

120 15
15
cos θ =
=
so θ = cos –1 ≈ 37.86°.
152 19
19

58. Place the box so that its corners are at the points
(0, 0, 0), (1, 0, 0), (0,1, 0), (1, 1, 0), (0, 0, 1),
(1, 0, 1), (0,1, 1), and (1, 1, 1). The main
diagonals are (0, 0, 0) to (1, 1, 1), (1, 0, 0) to

(0, 1, 1), (0, 1, 0) to (1, 0, 1), and (0, 0, 1) to
(1, 1, 0). The corresponding vectors are
1, 1, 1 , −1,1,1 , 1, −1,1 , and 1,1, −1 .
All of these vectors have length 1 + 1 + 1 = 3.
Thus, the angle θ between any pair, u and v of
the diagonals is found from
u⋅v
u⋅v
=
cos θ =
.
u v
3
There are six ways of pairing the four vectors,
but due to symmetry, there are only two cases we
need to consider. In these cases, u ⋅ v = 1 or
1
u ⋅ v = −1 . Thus we get that cosθ = or
3
1
cos θ = − . Solving for θ gives
3

θ ≈ 70.53 or θ ≈ 109.47 .
59. Work = F ⋅ D = (3i + 10 j) ⋅ (10 j)
= 0 + 100 = 100 joules
60. F = 100 sin 70°i – 100 cos 70°j
D = 30i
Work = F ⋅ D
= (100sin 70°)(30) + (−100 cos 70°)(0)

= 3000 sin 70° ≈ 2819 joules
61. D = 5i + 8j
Work = F · D = (6)(5) + (8)(8) = 94 ft-lb
Instructor’s Resource Manual

66. 3( x + 2) − 2( y + 3) − 1( z − 4) = 0
3x − 2 y − z = −4

68. z + 3 = 0
z = −3
69. The planes are 2x – 4y + 3z = –15 and
3x – 2y – z = –4. The normals to the planes are
u = 2, −4,3 and v = 3, −2, −1 . If θ is the angle
between the planes,
u⋅v
6+8−3
11
cos θ =
=
=
, so
u v
29 14
406

θ = cos –1

11
406


≈ 56.91°.

70. An equation of the plane has the form
2x + 4y – z = D. And this equation must satisfy
2(–1) + 4(2) – (–3) = D, so D = 9. Thus an
equation of the plane is 2x + 4y – z = 9.
71. a.

Planes parallel to the xy-plane may be
expressed as z = D, so z = 2 is an equation of
the plane.

b. An equation of the plane is
2(x + 4) – 3(y + 1) – 4(z – 2) = 0 or
2x – 3y – 4z = –13.
72. a.

Planes parallel to the xy-plane may be
expressed as z = D, so z = 0 is an equation of
the plane.

b. An equation of the plane is
x + y + z = D ; since the origin is in the
plane, 0 + 0 + 0 = D . Thus an equation is
x+ y+z =0.
73. Distance =

(1) + 3(–1) + (2) – 7
1+ 9 +1


=

7
11

≈ 2.1106

Section 11.3

683

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


74. The distance is 0 since the point is in the plane.
( (–3)(2) + 2(6) + (3) – 9 = 0 )
75. (0, 0, 9) is on –3x + 2y + z = 9. The distance from
(0, 0, 9) to 6x – 4y – 2z = 19 is
6(0) – 4(0) – 2(9) –19
37
=
≈ 4.9443 is the
36 + 16 + 4
56
distance between the planes.
76. (1, 0, 0) is on 5x – 3y – 2z = 5. The distance from
(1, 0, 0) to –5x + 3y + 2z = 7 is
–5(1) + 3(0) + 2(0) – 7
12

=
≈ 1.9467.
25 + 9 + 4
38
77. The equation of the sphere in standard form is
( x + 1)2 + ( y + 3) 2 + ( z – 4) 2 = 26, so its center is

(–1, –3, 4) and radius is 26. The distance from
the sphere to the plane is the distance from the
center to the plane minus the radius of the sphere
or
3(–1) + 4(–3) + 1(4) –15
– 26 = 26 – 26 = 0,
9 + 16 + 1
so the sphere is tangent to the plane.
78. The line segment between the points is
perpendicular to the plane and its midpoint,
(2, 1, 1), is in the plane. Then
6 − (−2),1 − 1, −2 − 4 = 8, 0, −6 is
perpendicular to the plane. The equation of the
plane is
8(x – 2) + 0(y – 1) – 6(z – 1) = 0 or 8x – 6z = 10.
79. u ⋅ v = cos θ u v ≤ u v since cos θ ≤ 1.
80.

u+v

2

= u


2

+ 2u ⋅ v + v

≤ u

2

+ 2 u⋅v + v

≤ u

2

+2 u v + v

2

2

2

=( u + v

)2

Therefore, u + v ≤ u + v .
81. The 3 wires must offset the weight of the object,
thus

(3i + 4j + 15k) + (–8i – 2j + 10k) + (ai + bj + ck)
= 0i + 0j + 30k
Thus, 3 – 8 + a = 0, so a = 5;
4 – 2 + b = 0, so b = –2;
15 + 10 + c = 30, so c = 5.

83. Let x = x, y, z , so
( x − a) ⋅ ( x − b) =
x − a1 , y − a2 , z − a3 ⋅ x − b1 , y − b2 , z − b3

= x 2 − (a1 + b1 ) x + a1b1 + y 2 − (a2 + b2 ) y + a2b2
+ z 2 − (a3 + b3 ) z + a3b3
Setting this equal to 0 and completing the squares
yields
2

2

2

a3 + b3 ⎞
a1 + b1 ⎞ ⎛
a2 + b2 ⎞ ⎛
⎛
⎜x− 2 ⎟ +⎜ y− 2 ⎟ +⎜z − 2 ⎟
⎝
⎠ ⎝
⎠ ⎝
⎠ .
1⎡

= (a1 − b1 )2 + (a2 − b2 )2 + (a3 − b3 )2 ⎤
⎦
4⎣
⎛ a +b a +b a +b ⎞
A sphere with center ⎜ 1 1 , 2 2 , 3 3 ⎟
2
2 ⎠
⎝ 2
and radius
1⎡
1
2
(a1 – b1 ) 2 + (a2 – b2 ) 2 + (a3 – b3 )2 ⎤ = a – b
⎦ 4
4⎣

84. Let P ( x0 , y0 , z0 ) be any point on
Ax + By + Cz = D, so Ax0 + By0 + Cz0 = D . The
distance between the planes is the distance from
P ( x0 , y0 , z0 ) to Ax + By + Cz = E, which is
Ax0 + By0 + Cz0 – E
A2 + B 2 + C 2

=

D–E
A2 + B 2 + C 2

.


85. If a, b, and c are the position vectors of the
vertices labeled A, B, and C, respectively, then
the side BC is represented by the vector c – b.
The position vector of the midpoint of BC is
1
1
b + (c – b) = (b + c). The segment from A to
2
2
1
the midpoint of BC is (b + c) – a. Thus, the
2
position vector of P is
2 ⎡1
⎤ a+b+c
a + ⎢ (b + c) – a ⎥ =
3 ⎣2
3
⎦
If the vertices are (2, 6, 5), (4, –1, 2), and
(6, 1, 2), the corresponding position vectors are
2, 6, 5 , 4, −1, 2 , and 6, 1, 2 . The position
vector of P is
1
1
2 + 4 + 6, 6 – 1 + 1, 5 + 2 + 2 = 12, 6, 9 =
3
3
4, 2, 3 . Thus P is (4, 2, 3).


82. Let v1 , v 2 , …, v n represent the sides of the
polygon connected tail to head in succession
around the polygon.
Then v1 + v 2 +…+ v n = 0 since the polygon is
closed, so
F ⋅ v1 + F ⋅ v 2 +…+ F ⋅ v n = F ⋅ ( v1 + v 2 +…+ v n )
= F · 0 = 0.
684

Section 11.3

Instructor’s Resource Manual

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


86. Let A, B, C, and D be the vertices of the
tetrahedron with corresponding position vectors
a, b, c, and d. The vector representing the
segment from A to the centroid of the opposite
1
face, triangle BCD, is (b + c + d) – a by
3
Problem 85. Similarly, the segments from B, C,
and D to the opposite faces are
1
1
(a + c + d) – b, (a + b + d) – c, and
3

3
1
(a + b + c) – d, respectively. If these segments
3
meet in one point which has a nice formulation as
some fraction of the way from a vertex to the
centroid of the opposite face, then there is some
number k, such that
⎡1
⎤
⎡1
⎤
a + k ⎢ (b + c + d) – a ⎥ = b + k ⎢ (a + c + d) – b ⎥
3
3
⎣
⎦
⎣
⎦
⎡1
⎤
⎡1
⎤
= c + k ⎢ (a + b + d) – c ⎥ = d + k ⎢ (a + b + c) – d ⎥ .
⎣3
⎦
⎣3
⎦
k
3

Thus, a(1 – k ) = a, so k = . Hence the
3
4
segments joining the vertices to the centroids of
the opposite faces meet in a common point which
3
is of the way from a vertex to the
4
3
corresponding centroid. With k = , all of the
4
1
above formulas simplify to (a + b + c + d), the
4
position vector of the point.

Problem Set 11.4
i

1. a.

i

1.

j

k

2


1

2 –2
–3 –2
–3 2
j+
k
i–
2 –4
–1 –4
–1 2
=(–8 + 4)i – (12 – 2)j + (–6 + 2)k
= –4i – 10j – 4k
b. b + c = 6i + 5j – 8k, so
i j k
a × (b + c) = –3 2 –2
6 5 –8
=

2 –2

c.

i–

1

j+


–3 2

a ⋅ (b + c) = –3(6) + 2(5) – 2(–8) = 8
i

d.

j

k

b × c = –1 2 –4
7 3 –4
=

2 –4

i–

–1 –4

j+

–1 2

k
3 –4
7 –4
7 3
= (–8 + 12)i – (4 + 28)j + (–3 – 14)k

= 4i – 32j – 17k
i
j
k
a × (b × c) = –3

2

–2

4 –32 –17
=

2

–2

i–

–3

–2

j+

–3

2

–32 –17

4 –17
4 –32
= (–34 – 64)i – (51 + 8)j + (96 – 8)k
= –98i – 59j + 88k

a×b = 3

k

j k
3

1

–2 –1 0

–1 2

–1 2 1 =
i–
j+
k
1 –1
3 –1
3 1
3 1 –1

–3 –2

k

5 –8
6 –8
6 5
= (–16 + 10)i – (24 + 12)j + (–15 – 12)k
= –6i – 36j – 27k

2. a.

= (–2 – 1)i – (1 – 3)j + (–1 – 6)k
= –3i + 2j – 7k = -3, 2, -7
2.

a × b = –3 2 –2
–1 2 –4

i

–1

k

=

87. After reflecting from the xy-plane, the ray has
direction ai + bj – ck. After reflecting from the
xz-plane, the ray now has direction ai – bj – ck.
After reflecting from the yz-plane, the ray now
has direction –ai – bj – ck, the opposite of its
original direction. ( a < 0 , b < 0 , c < 0 )


11.4 Concepts Review

j

=

3 1

i–

3 1

j+

3

3

k
–1 0
–2 0
–2 –1
= (0 + 1)i – (0 + 2)j + (–3 + 6)k = i – 2j + 3k
= 1, − 2, 3

u v sin θ

3. –(v × u)
4. parallel


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5. u = 3 − 1, −1 − 3, 2 − 5 = 2, −4, −3 and

b. b + c = -2 - 2, -1 - 3, 0 - 1

v = 4 − 1, 0 − 3,1 − 5 = 3, −3, −4 are in the

= -4, -4, -1
i

j

plane.

k

i

a × (b + c) = 3 3 1
–4 –4 –1


3 1
3 1
3 3
j+
k
i–
–4 –1
–4 –1
–4 –4
= (–3 + 4)i – (–3 + 4)j + (–12 + 12)k
= i – j = 1, - 1, 0
=

c.

i

j

k

b × c = –2

–1

0

–1

0


–3 –1

–2

i–

0

–2 –1

j+

–2

–1

–2 –3

k

a ⋅ (b × c) = 3(1) + 3(−2) + 1(4) = 1
i

=

3 1

i–


3

1

1 –2

4

3 1

3. a × b =

j

k

1 2

3

j+

3

3

–2 2 –4
3
1 3
1 2

i–
j+
k
2 –4
–2 –4
–2 2
= (–8 – 6)i – (–4 + 6)j + (2 + 4)k = –14i – 2j + 6k
is perpendicular to both a and b. All
perpendicular vectors will have the form
c(–14i – 2j + 6k) where c is a real number.
=

2

i

j

k

4. a × b = –2 5 –2
3 –2 4
=

5 –2

i–

–2 –2


3 –3 –4
=

–4 –3

2 –3

i–

j+

2 –4

k
–3 –4
3 –4
3 –3
= (16 – 9)i – (–8 + 9)j + (–6 + 12)k
= 7, −1, 6 is perpendicular to the plane.
1

7, – 1, 6 = ±

7

,–

1

,


6

j+

–2

6. u = 5 + 1,1 − 3, 2 − 0 = 6, −2, 2 and
v = 4 + 1, −3 − 3, −1 − 0 = 5, −6, −1 are in the
plane.
i

j

k

–2 2
6 2
6 –2
j+
k
i–
–6 –1
5 –1
5 –6
= (2 + 12)i – (–6 – 10)j + (–36 + 10)k
= 14,16, −26
=

k

–2 4
1 4
1 –2
= (12 + 2)i – (12 – 1)j + (–6 – 3)k
= 14, −11, −9
i

–3

u × v = 6 –2 2
5 –6 –1

j k

a × (b × c) = 3

u × v = 2 –4

49 + 1 + 36
86
86
86
are the unit vectors perpendicular to the plane.

= (1 – 0)i – (2 – 0)j + (6 – 2)k = 1, -2, 4

d.

k


±

–2 –3 –1
=

j

5

k

is perpendicular to the plane.
1
±
14, 16, – 26
196 + 256 + 676
7
8
13
=±
,
,–
282
282
282
are the unit vectors perpendicular to the plane.
i j k
7. a × b = –1 1 –3
4 2 –4


1 –3
–1 –3
–1 1
j+
k
i–
2 –4
4 –4
4 2
= (–4 + 6)i – (4 + 12)j + (–2 – 4)k
= 2i – 16j – 6k
Area of parallelogram = a × b
=

= 4 + 256 + 36 = 2 74

–2 4
3 4
3 –2
= (20 – 4)i – (–8 + 6)j + (4 – 15)k
= 16i + 2j – 11k
All vectors perpendicular to both a and b will
have the form c(16i + 2j – 11k) where c is a real
number.

686

Section 11.4

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i

j

11. u = 0 − 1,3 − 3, 0 − 2 = −1, 0, −2 and

k

8. a × b = 2 2 –1
–1 1 –4
=

2

–1

2

v = 2 − 1, 4 − 3,3 − 2 = 1, 1, 1 are in the plane.
i

–1

2 2


i–
j+
k
1 –4
–1 –4
–1 1
= (–8 + 1)i – (–8 – 1)j + (2 + 2)k
= –7i + 9j + 4k
Area of parallelogram = a × b = 49 + 81 + 16
= 146

b = −1 − 3, 2 − 2,5 − 1 = −4, 0, 4 are adjacent
sides of the triangle. The area of the triangle is
half the area of the parallelogram with a and b as
adjacent sides.
i j k
a × b = –1 2 5
–4 0 4
2 5
–1 5
–1 2
j+
k
=
i–
0 4
–4 4
–4 0

(


1
1
64 + 256 + 64 = 8 6
2
2

)

10. a = 3 − 1,1 − 2,5 − 3 = 2, −1, 2 and
b = 4 − 1,5 − 2, 6 − 3 = 3, 3, 3 are adjacent
sides of the triangle.
i
j k
a × b = 2 –1 2
3 3 3
3 3

i–

2 2
3 3

j+

0 –2
–1 –2
–1 0
j+
k

i–
1 1
1 1
1 1

= (0 + 2)i – (–1 + 2)j + (–1 – 0)k = 2, −1, −1

2(x – 1) – 1(y – 3) – 1(z – 2) = 0 or
2x – y – z = –3.
12. u = 0 − 1, 0 − 1,1 − 2 = −1, −1, −1 and
v = −2 − 1, −3 − 1, 0 − 2 = −3, −4, −2 are in the
plane.
i

j

k

u × v = –1 –1 –1

=

=4 6

–1 2

=

–3 –4 –2


= (8 – 0)i – (–4 + 20)j + (0 + 8)k = 8, -16, 8

=

k

The plane through (1, 3, 2) with normal
2, −1, −1 has equation

9. a = 2 – 3, 4 – 2, 6 – 1 = –1, 2, 5 and

Area of triangle =

j

u × v = –1 0 –2
1 1 1

–1 –1
–4 –2

i–

–1 –1
–3 –2

j+

–1 –1
–3 –4


k

= (2 – 4)i – (2 – 3)j + (4 – 3)k = −2,1,1
The plane through (0, 0, 1) with normal −2,1,1
has equation –2(x – 0) + 1(y – 0) + 1(z – 1) = 0 or
–2x + y + z = 1.
13. u = 0 − 7,3 − 0, 0 − 0 = −7,3, 0 and
v = 0 − 7, 0 − 0,5 − 0 = −7, 0,5 are in the
plane.
i

j k

u × v = −7 3 0 =
2 –1
3

3

−7 0 5

k

= (–3 – 6)i – (6 – 6)j + (6 + 3)k = -9, 0, 9
1
1
9 2
Area of triangle = a × b =
81 + 81 =

2
2
2

3 0
0 5

i−

−7 0
−7 5

j+

−7 3
−7 0

k

= (15 – 0)i – (–35 – 0)j + (0 + 21)k = 15,35, 21
The plane through (7, 0, 0) with normal
15,35, 21 has equation 15(x – 7) +35(y – 0) +
21(z – 0) = 0 or 15x + 35y +21 z = 105.

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14. u = 0 − a, b − 0, 0 − 0 = − a, b, 0 and
v = 0 − a, 0 − 0, c − 0 = − a, 0, c are in the
plane.
i

j k

u × v = −a b 0 =

20. k = 0, 0,1 is normal to the xy-plane and

−a 0 c

b 0
0 c

i−

−a 0
−a c

j+

−a b
−a 0


v = 3, −2,1 is normal to the plane
k

= bci – (–ac)j + abk = bc, ac, ab
The plane through (a, 0, 0) with normal
bc, ac, ab has equation bc(x – a) + ac(y – 0) +

16. An equation of the plane is
1(x –0) +1(y – 0) +1(z – 2) = 0 or
x + y + z = 2.
17. The plane’s normals will be perpendicular to the
normals of the other two planes. Then a normal is
1, −3, 2 × 2, −2, −1 = 7, 5, 4 . An equation
of the plane is 7(x + 1) + 5(y + 2) + 4(z – 3) = 0
or 7x + 5y + 4z = –5.
18. u = 1,1,1 is normal to the plane x + y + z = 2
and v = 1, −1, −1 is normal to the plane
x − y − z = 4 . A normal vector to the required
plane must be perpendicular to both the other
normals; thus one possibility is
i
j k

1 = 0, 2, −2

k×v = 0

0

1 = 2,3, 0


The plane has an

1

equation of the form: 2 x + 3 y = D. Since the
point (0, 0, 0) is in the plane, 2(0) + 3(0) = D;
thus D = 0 and an equation for the plane is
2x + 3y = 0 .

15. An equation of the plane is
1(x –2) – 1(y – 5) +2(z – 1) = 0 or
x –y +2z = –1.

1

3 x − 2 y + z = 4 . A normal vector to the required
plane must be perpendicular to both the other
normals; thus one possibility is:
i
j k

3 −2

ab(z – 1) = 0 or bcx +ac y +abz = abc.

u× v = 1

19. (4i + 3j – k) × (2i – 5j + 6k) = 13i – 26j – 26k
= 13(i – 2j – 2k)

is normal to the plane. An equation of the plane is
1(x – 2) – 2(y + 3) – 2(z – 2) = 0 or
x – 2y – 2z = 4.

21. Each vector normal to the plane we seek is
parallel to the line of intersection of the given
planes. Also, the cross product of vectors normal
to the given planes will be parallel to both planes,
hence parallel to the line of intersection. Thus,
4, −3, 2 × 3, 2, −1 = −1,10,17 is normal to
the plane we seek. An equation of the plane is
–1(x – 6) + 10(y – 2) + 17(z + 1) = 0 or
–x + 10y + 17z = –3.
22. a × b is perpendicular to the plane containing a
and b, hence a × b is perpendicular to both a and
b. (a × b) × c is perpendicular to a × b hence it
is parallel to the plane containing a and b.

The plane has an

1 −1 −1

equation of the form: 2 y − 2 z = D. Since the
point (2, −1, 4) is in the plane, 2(−1) − 2(4) = D;
thus D = - 10 and an equation for the plane is
2 y − 2 z = −10 or y − z = −5.

23. Volume = 2, 3, 4 ⋅ ( 0, 4, –1 × 5, 1, 3

)


= 2, 3, 4 ⋅ 13, – 5, – 20 = –69 = 69

24. Volume = (3i – 4 j + 2k ) ⋅ [(– i + 2 j + k ) × (3i – 2 j + 5k ) = (3i – 4 j + 2k ) ⋅ (12i + 8 j – 4k ) = –4 = 4
25. a.
b.

688

Volume = u ⋅ ( v × w ) = 3, 2, 1 ⋅ –3, –1, 2 = –9 = 9

Area = u × v = 3, − 5, 1

Section 11.4

= 9 + 25 + 1 = 35

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c.

Let θ be the angle. Then θ is the complement of the smaller angle between u and v × w.
u ⋅ ( v × w)
9
9
⎛π

⎞
=
= , θ ≈ 40.01°
cos ⎜ − θ ⎟ = sin θ =
u v×w
9 + 4 + 1 9 + 1 + 4 14
⎝2
⎠

26. From Theorem C, a ⋅ (b × c) = (a × b) ⋅ c , which leads to a ⋅ (b × c) = c ⋅ (a × b) . Again from Theorem C,
c ⋅ (a × b) = (c × a) ⋅ b = −(a × c) ⋅ b , which leads to c ⋅ (a × b) = b ⋅ (a × c) . Therefore, we have that

a ⋅ (b × c) = b ⋅ (a × c) = c ⋅ (a × b) .

27. Choice (c) does not make sense because (a ⋅ b) is a scalar and can't be crossed with a vector. Choice (d) does not
make sense because (a × b) is a vector and can't be added to a constant.
28. a × b and c × d will both be perpendicular to the common plane. Hence a × b and c × d are parallel so
(a × b) × (c × d) = 0.
29. Let b and c determine the (triangular) base of the tetrahedron. Then the area of the base is

1
b × c which is half of
2

the area of the parallelogram determined by b and c. Thus,
1
1 ⎡1
⎤
(area of base)(height) = ⎢ (area of corresponding parallelogram)(height) ⎥
3

3 ⎣2
⎦
1
1
= (area of corresponding parallelpiped) = a ⋅ (b × c)
6
6
30. a = 4 + 1, −1 − 2, 2 − 3 = 5, −3, −1 ,
b = 5 + 1, 6 − 2,3 − 3 = 6, 4, 0 ,
c = 1 + 1,1 − 2, −2 − 3 = 2, −1, −5
Volume =

1
1
a ⋅ (b × c) =
5, – 3, –1 ⋅ –20, 30, –14
6
6

=

1
88
–176 =
6
3

31. Let u = u1 , u2 , u3 and v = v1 , v2 , v3 then
u × v = u2 v3 – u3v2 , u3v1 – u1v3 , u1v2 – u2 v1


u× v

2

= u22 v32 − 2u2 u3v2 v3 + u32 v22 + u32 v12 − 2u1u3v1v3 + u12 v32 + u12 v22 − 2u1u2 v1v2 + u22 v12

= u12 (v12 + v2 2 + v32 ) – u12 v12 + u22 (v12 + v2 2 + v32 ) – u22 v2 2 + u32 (v12 + v22 + v32 )
– u32 v32 – 2u2 u3v2 v3 – 2u1u3v1v3 – 2u1u2 v1v2
= (u12 + u22 + u32 )(v12 + v22 + v32 ) – (u12 v12 + u22 v22 + u32 v32 + 2u2 u3v2 v3 + 2u1u3v1v3 + 2u1u2 v1v2 )
= (u12 + u22 + u32 )(v12 + v2 2 + v32 ) – (u1v1 + u2 v2 + u3v3 )2 = u

2

v

2

− (u ⋅ v )2

32. u = u1 , u2 , u3 , v = v1 , v2 , v3 , w = w1 , w2 , w3
u × ( v × w ) = (u2 v3 – u3v2 ) + (u2 w3 – u3 w2 ), (u3v1 – u1v3 ) + (u3 w1 – u1w3 ), (u1v2 – u2 v1 ) + (u1w2 – u2 w1 )

= (u × v) + (u × w)
33. (v + w) × u = –[u × (v + w)]
= –[(u × v) + (u × w)] = –(u × v) – (u × w)
= (v × u) + (w × u)
34. u × v = 0 ⇒ u and v are parallel. u ⋅ v = 0 ⇒ u and v are perpendicular. Thus, either u or v is 0.

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Section 11.4

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35.

PQ = – a, b, 0 , PR = – a, 0, c ,

The area of the triangle is half the area of the
parallelogram with PQ and PR as adjacent
sides, so area
1
− a, b, 0 × − a, 0, c
(ΔPQR ) =
2
1
1 2 2
=
bc, ac, ab =
b c + a 2 c2 + a 2b2 .
2
2
36. The area of the triangle is
1
x2 − x1 , y2 − y1 , 0 × x3 − x1 , y3 − y1 , 0 =
2

1
0, 0, ( x2 − x1 )( y3 − y1 ) − ( y2 − y1 )( x3 − x1 )
2
1
= ( x2 y3 – x3 y2 ) – ( x1 y3 – x3 y1 ) + ( x1 y2 – x2 y1 )
2
which is half of the absolute value of the
determinant given. (Expand the determinant
along the third column to see the equality.)
37. From Problem 35, D 2 =
2

2

1 2 2
(b c + a 2 c 2 + a 2b 2 )
4
2

⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞
= ⎜ bc ⎟ + ⎜ ac ⎟ + ⎜ ab ⎟ = A2 + B 2 + C 2 .
⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠

38. Note that the area of the face determined by a
1
and b is a × b .
2
1
Label the tetrahedron so that m = (a × b),
2

1
1
n = (b × c), and p = (c × a) point outward.
2
2
The fourth face is determined by a – c and b – c,
so
1
q = [(b – c) × (a – c)]
2
1
= [(b × a) – (b × c) – (c × a) + (c × c)]
2
1
= [–(a × b) – (b × c) − (c × a)].
2
1
m + n + p + q = [(a × b) + (b × c) + (c × a)
2
− (a × b) − (b × c) − (c × a)] = 0

690

Section 11.4

39. The area of the triangle is A =

(

1

a × b . Thus,
2

)

1
1
2
2
2
a×b =
a b − (a ⋅ b)2
4
4
1⎡ 2 2 1
2
2
2 2⎤
a + b − a−b
= ⎢a b −
⎥
4⎣
4
⎦
1
= ⎡ 4a 2 b 2 – ( a 2 + b 2 – c 2 ) 2 ⎤
⎦
16 ⎣

A2 =


(

)

1
(2a 2 b 2 – a 4 + 2a 2 c 2 – b 4 + 2b 2 c 2 – c 4 ).
16
1
Note that s – a = (b + c – a ),
2
1
1
s – b = (a + c – b), and s – c = (a + b – c).
2
2
s ( s − a )( s − b)( s − c)
=

1
(a + b + c)(b + c − a)(a + c − b)(a + b − c)
16
1
= (2a 2 b 2 – a 4 + 2a 2 c 2 – b 4 + 2b 2 c 2 – c 4 )
16
which is the same as was obtained above.
=

40. u × v = (u1i + u2 j + u3k ) × (v1i + v2 j + v3k )
= (u1v1 )(i × i ) + (u1v2 )(i × j) + (u1v3 )(i × k ) +

(u2 v1 )( j × i ) + (u2 v2 )( j × j) + (u2 v3 )( j × k ) +
(u3v1 )(k × i ) + (u3v2 )(k × j) + (u3v3 )(k × k )
= (u1v1 )(0) + (u1v2 )(k ) + (u1v3 )(− j) +
(u2 v1 )(−k ) + (u2 v2 )(0) + (u2 v3 )(i ) +
(u3v1 )( j) + (u3v2 )(−i ) + (u3v3 )(0)
= (u2 v3 – u3v2 )i + (u3v1 – u1v3 ) j + (u1v2 – u2 v1 )k

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7.

11.5 Concepts Review

lim ln(t 3 ), t 2 ln t , t does not exist because

t →0 +

lim ln(t 3 ) = – ∞.

1. a vector-valued function of a real variable

t →0 +

2. f and g are continuous at c; f ′(t )i + g ′(t ) j
8.


3. position
4. r ′(t ); r ′′(t ); tangent; concave

2

t →0

−

=

lim e −1/ t , lim

Problem Set 11.5
9. a.
1. lim[2ti – t 2 j] = lim(2t )i – lim(t 2 ) j = 2i – j
t →1

t →1

t →1

t →3

⎡ sin t cos t
⎤
7t 3
t
i−
j+

k⎥
5. lim ⎢
t
t
t + 1 ⎦⎥
t →0 ⎢
e
⎣

⎡ t sin t
7t 3
sin t ⎤
i−
j−
k⎥
6. lim ⎢
t
t →∞ ⎢ t 2
t 3 − 3t
⎣
⎦⎥
⎛ 7t 3 ⎞
⎛ t sin t ⎞
⎛ sin t ⎞
= lim ⎜
i − lim ⎜
⎟ j − lim ⎜
⎟
⎟k
t →∞ ⎝ t 2 ⎠

t →∞ ⎜ t 3 − 3t ⎟
t →∞ ⎝ t ⎠
⎝
⎠

= 0, − 1, 0

2
is
t–4
(−∞, 4) ∪ (4, ∞) . The domain of

The domain of f (t ) =

: t ≤ 3}.

is (−∞, ∞) . The

domain of g (t ) = 20 – t is (– ∞ , 20]. The
domain of h(t ) = 3 is (−∞, ∞ ) . Thus, the
domain of r is (– ∞ , 20] or {t ∈ : t ≤ 20}.
c.

The domain of f (t ) = cos t is (−∞, ∞) . The
domain of g (t ) = sin t is also (−∞, ∞) . The
domain of h(t ) = 9 − t 2 is [ −3,3] . Thus,

the domain of r is [ −3,3] , or
{t ∈


10. a.

: − 3 ≤ t ≤ 3}.

The domain of f(t) = ln(t – 1) is (1, ∞ ). The
domain of g (t ) = 20 – t is (– ∞ , 20]. Thus,
the domain of r is (1, 20] or
{t ∈ :1 < t ≤ 20}.

b. The domain of f (t ) = ln(t –1 ) is (0, ∞) ).
The domain of g (t ) = tan –1 t is (– ∞ , ∞ ).
The domain of h(t ) = t is (−∞, ∞) . Thus,
the domain of r is (0, ∞ ) or {t ∈ : t > 0}.

⎛ 7t 3 ⎞
⎛ sin t cos t ⎞
lim
= lim ⎜
i
−
⎜
⎟j
⎟
t
t →0 ⎝
⎠ t →0 ⎜⎝ et ⎟⎠
⎛ t ⎞
+ lim ⎜
⎟k = i
t →0 ⎝ t + 1 ⎠


t
, lim t
t t →0−

b. The domain of f (t ) = t 2

⎡ 2t 2 − 10t − 28
7t 3 ⎤
4. lim ⎢
i−
j⎥
t+2
t − 3 ⎦⎥
t →−2 ⎢
⎣
⎛ 2t 2 − 10t − 28 ⎞
⎛ 7t 3 ⎞
= lim ⎜
⎟ i − lim ⎜
⎟j
⎟ t →−2 ⎜ t − 3 ⎟
t+2
t →−2 ⎜⎝
⎠
⎝
⎠
56
56
= lim (2t − 14)i − j = −18i − j

5
5
t →−2

t →0−

domain of r is (– ∞ , 3] or {t ∈

= lim[2(t − 3)2 ]i − lim(7t 3 ) j = −189 j

⎛
⎞
⎛ ( t − 1)( t + 3) ⎞
t −1
= lim ⎜⎜
⎟ i − lim ⎜
⎟j
t −1
t →1 ⎝ ( t − 1)( t + 1) ⎠⎟
t →1 ⎝
⎠
1
⎛ 1 ⎞
= lim ⎜
⎟ i − lim(t + 3) j = i − 4 j
2
t →1 ⎝ t + 1 ⎠
t →1

t →0


h(t ) = ln 4 − t is (−∞, 4) ∪ (4, ∞) . Thus, the

t →3

⎡ t −1
t 2 + 2t − 3 ⎤
i−
j⎥
3. lim ⎢
t −1
t →1 ⎢⎣ t 2 − 1
⎥⎦

2

−

g (t ) = 3 – t is (– ∞ , 3]. The domain of

2. lim[2(t − 3)2 i − 7t 3 j]
t →3

t
,t
t

lim e−1/ t ,

c.


The domain of g (t ) = 1/ 1 − t 2 is (−1,1) .
The domain of h(t ) = 1/ 9 − t 2 is ( −3,3) .
(The function f is f ( x) = 0 which has
domain (−∞, ∞) .) Thus, domain of r is

( −1,1) .

⎛ sin t ⎞
⎛ sin t ⎞
= lim ⎜
⎟ i − 7 j − lim ⎜
⎟ k = −7 j
t
t →∞ ⎝
t →∞ ⎝ t ⎠
⎠

Instructor’s Resource Manual

Section 11.5

691

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


11. a.


2
is continuous on
t–4

f (t ) =

c.

h(t ) = 1/ 9 − t 2 is continuous on ( −3,3) . (The

(−∞, 4) ∪ (4, ∞) . g (t ) = 3 – t is

function f is f ( x) = 0 which is continuous on

continuous on (– ∞ , 3]. h(t ) = ln 4 − t
is continuous on (−∞, 4) and on
(4, ∞) . Thus, r is continuous on
(– ∞ , 3] or {t ∈ : t ≤ 3}.

g (t ) = 1/ 1 − t 2 is continuous on (−1,1) .

(−∞, ∞) .) Thus, r is continuous on

13. a.

( −1,1) .

2

Dt r (t ) = 9(3t + 4) 2 i + 2tet j + 0k

2

b.

f (t ) = t 2

(–

Dt2r (t ) = 54(3t + 4)i + 2(2t 2 + 1)et j

is continuous on

) (

n + 1, – n ∪

)

n , n + 1 where

n is a non-negative integer.
g (t ) = 20 – t is continuous on

( −∞, 20 ) or {t ∈

: t < 20}. h(t ) = 3

b.

Dt2r (t ) = 2 cos 2ti − 9 cos 3tj + 2k


14. a.

is continuous on (−∞, ∞) . Thus, r is
continuous on

(–

) (

n + 1, – n ∪

)

{t ∈

c.

: t < 20, t 2 not an integer}.

f (t ) = cos t and f (t ) = sin t are

continuous on ( −∞, ∞ ) .
h(t ) = 9 − t 2 is continuous on

[ −3,3] .
[ −3,3] .
12. a.

Thus, r is continuous on


f(t) = ln(t – 1) is continuous on (1, ∞ ) .
g (t ) = 20 – t is continuous on (– ∞ ,
20). Thus, r is continuous on (1, 20) or
{t ∈ :1 < t < 20}.

b.

f (t ) = ln(t –1 ) is continuous on

( 0, ∞ ) .

g (t ) = tan –1 t is continuous on

(– ∞ , ∞ ). h(t ) = t is continuous on
(−∞, ∞ ) .Thus, r is continuous on (0,
∞ ) or {t ∈ : t > 0}.

(
) i + (ln 2)2 j + k
r ′′(t ) = ( e + 4t e – 2e ) i + (ln 2) 2 j

r ′(t ) = et − 2te −t

b.

2

t


2 –t 2

t

k , k + 1 where

n and k are non-negative integers and
k < 400 or

Dt r (t ) = sin 2ti − 3sin 3tj + 2tk

r ′(t ) = 2sec 2 2ti +

–t 2

1
1+ t2

r ′′(t ) = 8 tan 2t sec2 2ti –

2 t

j

2t
(1 + t 2 ) 2

j

2

2
j ; r ''(t ) = e −t i + j
t
t2
2
r (t ) ⋅ r ''(t ) = e −2t − ln t 2
t2
⎡2 1
4
⎤
Dt [r (t ) ⋅ r ''(t )] = −2e−2t − ⎢ ⋅ ⋅ 2t − ln t 2 ⎥
2 2
3
t
⎣t t
⎦

15. r ′(t ) = – e – t i –

( )

( )

= −2e

−2t

−

4

t

3

+

( )

4 ln t 2
t

3

16. r ′(t ) = 3cos 3ti + 3sin 3tj
r (t ) ⋅ r '(t ) = 0
Dt [r (t ) ⋅ r '(t )] = 0
17. h(t )r (t ) = e –3t t – 1i + e –3t ln(2t 2 ) j
e –3t ⎛ 6t – 7 ⎞
⎜
⎟i
2 ⎝ t –1 ⎠
⎛2
⎞
+ e –3t ⎜ – 3ln(2t 2 ) ⎟ j
⎝t
⎠

Dt [h(t )r (t )] = –

18. h(t)r(t) = ln(3t – 2)sin 2ti + ln(3t – 2) coshtj

3sin 2t ⎤
⎡
Dt [h(t )r (t )] = ⎢ 2 ln(3t – 2) cos 2t +
i
3t – 2 ⎥⎦
⎣
3cosh t ⎤
⎡
j
+ ⎢ ln(3t – 2) sinh t +
3t – 2 ⎥⎦
⎣

692

Section 11.5

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.