CHAPTER 7 Applications of Vectors - Ms. Ma's...

Review of Prerequisite Skills, p. 3501. The velocity relative to the ground has a magnitude equivalent to the hypotenuse of a triangle with sides 800 and 100. So, by thePythagorean theorem we can find the magnitude ofthe velocity.

The velocity of the airplane relative to the ground isabout 806 km h E.2.

The angle between the two displacements is 135°.The magnitude, m, and the angle, of the displacement can be found using the cosine law.

So the displacement is 15.93 units, N.3.

4. a.

b.

8 16.91

5 !286

5 !81 1 9 1 196

5 "(29)2 1 32 1 142

l 5 magnitude

(29, 3, 14)

8 7.87

5 "62

5 "9 1 4 1 49

5 "32 1 (22)2 1 72

l 5 magnitude

(3, 22, 7)

yA(0, 1, 0)

z

x

B(–3, 2, 0)

C(–2, 0, 1)

D(0, 2, –3)

W 32.2°

8 32.2°

u 5 cos21 a134.76

159.3b

cos u 5134.76

159.3

2134.76 5 2159.3 cos u

144 5 253.76 1 25 2 159.3 cos u

122 5 15.932 1 52 2 2(15.93)(5) cos u

8 15.93 units

m 5 "253.85

5 253.85

5 169 1 84.85

5 25 1 144 2 120a2!2

2b

m2 5 52 1 122 2 2(5)(12) cos 135

u,

5

12

displacement

u

N 7.1°>

u 8 7.1°

u 5 tan21 a100

800b

tan u 5100

800

800

100

u

8 806 km>h

v 5 "650 000

5 650 000

5 640 000 1 10 000

v2 5 8002 1 1002

CHAPTER 7Applications of Vectors

7-1Calculus and Vectors Solutions Manual

c. (1, 1, 0)

d.

5. a.In the xy-plane at the point (x, y).b.In the xz-plane at the point (x, z).c.In the yz-plane at the point (y, z).6. a.

b.

c.

d.

7. a.

b.

c.

7.1 Vectors as Forces, pp. 362–3641. a. 10 N is a melon, 50 N is a chair, 100 N is acomputerb. Answers will vary.2. a.

b. 180°3.

The forces should be placed in a line along thesame direction.4. For three forces to be in equilibrium, they mustform a triangle, which is a planar figure.5.

a. The resultant is equivalent in magnitude to thehypotenuse, h, of the triangle with 5 and 12 as sides

and is directed northeast at an angle of

Thus, the resultant is at an angleof E. The equlibrant is equal inmagnitude and opposite in direction of the resultant.Thus, the equilibrant is 13 N at an angle of S 22.6° W.b. The resultant is at an angle of

W. The equilibrant, then, is 15 Nat N 36.9° E.6. For three forces to form equilibrium, they mustbe able to form a triangle or a balanced line, soa. Yes, since these can form a triangle.b. Yes, since these can form a triangle.

c. No, since these cannot form a triangle.d. Yes, since placing the 9 N and 10 Nforce in a line directly opposing the 19 N forceachieves equilibrium.

9 1 10 5 19,

"5 1 6 , 9

9 1 40 . 41

3 1 4 . 7

sin21 1215 5 S 36.9°

"92 1 122 5 15 N

sin21 1213 5 N 22.6°

"52 1 122 5 13 N

sin21 12

h.

resultant

f1

f2equilibrant

10 N20 N

10 N

20 N

30 N

5 12i>

1 j>

2 2k>

5 (6i>

1 4j>

2 2k>

) 1 (6i>

2 3j>

)

5 2(3i>

1 2j>

2 k>

) 2 3(22i>

1 j>

)

2a>

2 3b>

5 5i>

1 j>

2 k>

5 (3i>

1 2j>

2 k>

) 1 (2i>

2 j>

)

5 (3i>

1 2j>

2 k>

) 2 (22i>

1 j>

)

a>

2 b>

5 i>

1 3j>

2 k>

5 (3i>

1 2j>

2 k>

) 1 (22i>

1 j>

)

a>

1 b>

5 4i>

2 6j>

1 8k>

5 (22i>

1 3j>

2 4k>

) 1 (6i>

2 9j>

1 12k>

)

5 21

2(4i

>

2 6j>

1 8k>

) 13

2(4i

>

2 6j>

1 8k>

)

21

2(4, 26, 8) 1

3

2(4, 26, 8)

5 28i>

1 11j>

1 3k>

5 (22i>

1 2j>

1 6k>

) 1 (26i>

1 9j>

2 3k>

)

5 2(2i>

1 j>

1 3k>

) 1 3(22i>

1 3j>

2 k>

)

2(21, 1, 3) 1 3(22, 3, 21)

5 6i>

2 2j>

5 (4i>

2 j>

1 3k>

) 2 (22i>

1 j>

1 3k>

)

(4, 21, 3) 2 (22, 1, 3)

5 i>

2 7j>

5 (26i>

1 0j>

) 1 (7i>

2 7j>

)

5 (26i>

1 0j>

) 1 7(i>

2 j>

)

(26, 0) 1 7(1, 21)

C(0, y, z)

B(x, 0, z)

A(x, y, 0)

8 9.22

5 "85

5 "4 1 0 1 81

5 "22 1 02 1 (29)2

l 5 magnitude

(2, 0, 29)

8 1.41

5 "2

5 "12 1 12 1 02

l 5 magnitude

7-2 Chapter 7: Applications of Vectors

7. Arms 90 cm apart will yield a resultant with asmaller magnitude than at 30 cm apart. A resultantwith a smaller magnitude means less force to counter your weight, hence a harder chin-up.8. Using the cosine law, the resultant has a magnitude,r, of

Using the sine law, the resultant’s angle, can befound by

from the 6 N force toward the 8 Nforce. The equilibrant, then, would be 12.17 N at

from the 6 N force awayfrom the 8 N force.9.

We know that the components of and perpendicular to the 10 N force must be equal, so wecan write

Now we look at and We know

So Substituting then solving for yields

Now we solve for :

So the force from the 10 N force is 9.66 N andthe force perpendicular to it is 2.59 N.10. The force of the block is

98 N. The component of thisforce parallel to the ramp is

directed down theramp. So the force preventing this block from moving would be 49 N directed up the ramp.11. a.

b. Using the cosine law for the angle, we have

5 120

u 5 cos21 21

2

cos u 5256

112

56 5 2112 cos u

169 5 64 1 49 2 112 cos u

132 5 82 1 72 2 2(8)(7) cos u

u,

13 N

7 N

8 N

(98) sin 30° 5 (98)A12B 5 49 N,

(10 kg)(9.8 N>kg) 5

15°

@ f1

>

@ 5 2.59 N

@ f1

>

@ 5 (9.66)(0.268)

5 (9.66) cos 75

cos 15@ f1

>

@

5 @ f2

>

@ cos 75

cos 15@ f1

>

@

f1

>

@ f2

>

@ 5 9.66 N

@ f2

>

@ (1.035) 5 10

@ f2

>

@ acos 75

cos 15 sin 15 1 sin 75b 5 10

@ f2

>

@ sin 75 5 10@ f2

>

@ cos 75

cos 15 sin 15 1

f2

>

@ f2

>

@ sin 75 5 10@ f1

>

@ sin 15 1

x1 1 x2 5 10

x2 5 @ f2

>

@ sin 75

x1 5 @ f1

>

@ sin 15

x2.x1

5 @ f2

>

@ cos 75

cos 15@ f1

>

@

5 @ f2

>

@ cos 75@ f1

>

@ cos 15

f2

>

f1

>

x2 5 component of f2

>

parallel to the 10 N force


>


f2

>

5 force perpendicular to f1

>

f1

>

5 force 15° from the 10 N force

10 N

f1

f2

180° 2 34.7° 5 145.3°

8 34.7°

u 5 sin21 8

"32

12.17

sin u 5 8

"32

12.17

sin u

85

sin 120°

12.17

u,

8 12.17 N

r 5 "148

5 148

5 36 1 64 1 48

5 62 1 82 2 2(6)(8)a21

2b

r2 5 @ f1

>

@ 2 1 @ f2

>

@ 2 2 2 @ f1

>

@ @ f2

>

@ cos 120°


This is the angle between the vectors when placedhead to tail. So the angle between the vectors whenplaced tail to tail is 12. The 10 N force and the 5 N force result in a 5 Nforce east. The 9 N force and the 14 N force resultin a 5 N force south. The resultant of these is nowequivalent to the hypotenuse of the right trianglewith 5 N as both bases and is directed 45° south of

east. So the resultant is 45° south of east.13.

a. Using the Pythagorean theorem,

b. The angle, between and the resultant is given by

So the angle between and the equilibrant is

14. a.

For these three equal forces to be in equilibrium, theymust form an equilateral triangle. Since the resultantwill lie along one of these lines, and since all angles

of an equilateral triangle are 60°, the resultant will beat a 60° angle with the other two vectors.b. Since the equilibrant is directed opposite theresultant, the angle between the equilibrant and theother two vectors is 15. Since and act opposite one another, theynet a 10 N force directed west. Since and actopposite one another, they net a 10 N force directed45° north of east. So using the cosine law to findthe resultant,

Since our net forces are equal at 10 N, the angle ofthe resultant is directed halfway between the two, orat from toward 16.

Let be the tension in the 30° rope and be thetension in the 45° rope.Since this system is in equilibrium, we know thatthe horizontal components of and are equaland opposite and the vertical components add to beopposite the action of the mass. Also, the forceproduced by the mass is So we have a system of two equations: the first,

represents the balance of the horizontal components, and the second,

represents thebalance of the vertical components with the mass. So solving this system of two equations with twovariable gives the desired tensions.

T2aacos 45°

cos 30°b sin 30° 1 sin 45°b 5 196

aT2 cos 45°

cos 30°b sin 30° 1 T2 sin 45° 5 196

T1 sin 30° 1 T2 sin 45° 5 196

T1 5 T2 cos 45°

cos 30°

T1 cos 30° 5 T2 cos 45°

(T1) sin 30° 1 (T2) sin 45° 5 196

(T1) cos 30° 5 (T2) cos 45°

(20 kg)(9.8 N>kg) 5 196 N.

T2T1

T2T1

20 kg

T 2 T 1

f3

>

.f2

>

12 (135°) 5 67.5°

8 7.65 N

@ fr

>

@ 5Å

200 2 200a"2

2b

5 200 2 200a"2

2b

5 200 2 200 cos 45°

@ fr

>

@ 2 5 102 1 102 2 2(10)(10) cos 45°

fr

>

,

f4

>

f3

>

f2

>

f1

>

180° 2 60° 5 120°.

1 N

1 N1 N

60°60°

60°

180° 2 16.3° 5 163.7°.

f1

>

8 16.3°

u 5 sin21 7

25

sin u 57

25

sin u 5@ f2

>

@

25

f1

>

u,

@ f2

>

@ 5 7

5 49

5 252 2 242

@ f2

>

@ 2 5 252 2 @ f1

>

@ 2 @ f1

>

@ 2 1 @ f2

>

@ 2 5 252

= 24 Nf1

f2

resultant = 25 Nequlibrant

"52 1 52 5 "50 8 7.1 N

180° 2 120° 5 60°.



Thus the tension in the 45° rope is 175.73 N and thetension in the 30° rope is 143.48 N.17.

First, use the Cosine Law to find the angles thestrings make at the point of suspension. Let be theangle made by the 32 cm string and be the anglemade by the 24 cm string.

A keen eye could have recognized this triangle as a 3-4-5 right triangle and simply used the Pythagoreantheorem as well. Now we set up the same system ofequations as in problem 16, with being the tensionin the 32 cm string and being the tension in the 24 cm string, and the force of the mass being

8 29.4 N

T1 5 (39.2)cos 53.1°

cos 36.9°

T2 8 39.2 N

T2(1.25) 5 49

T2aacos 53.1°

cos 36.9°b sin 36.9° 1 sin 53.1°b 5 49

aT2 cos 53.1°

cos 36.9°b sin 36.9° 1 T2 sin 53.1° 5 49

T1 sin 36.9° 1 T2 sin 53.1° 5 49

T1 5 T2 cos 53.1°

cos 36.9°

T1 cos 36.9° 5 T2 cos 53.1°

(5 kg)(9.8 N>kg) 5 49 N.

T2

T1

8 53.1°

u2 5 cos21 1152

1920

21152 5 21920 cos u2

322 5 242 1 402 2 2(24)(40) cos u2

8 36.9°

u1 5 cos21 2048

2560

22048 5 22560 cos u1

242 5 322 1 402 2 2(32)(40) cos u1

u2

u1

5 kg

24 cm 32 cm

40 cm

8 143.48 N

T1 5 (175.73)cos 45°

cos 30°

T2 8 175.73 N

T2(1.12) 5 196 Thus the tension in the 24 cm string is 39.2 N andthe tension in the 32 cm string is 29.4 N.18.

(Port means left and starboard means right.) We arelooking for the resultant of these two force vectorsthat are 35° apart. We don’t know the exact valueof the force, so we will call it x. So the small tug is pulling with a force of x and the large tug ispulling with a force of 2x. To find the magnitudeof the resultant, r, in terms of x, we use the cosinelaw.

Now we use the cosine law again to find the angle,made by the resultant.

from the large tug toward thesmall tug, for a net of 8.5° to the starboard side.

8 11.5°

u 5 cos21 a11.2768

11.5076b

cos u 511.2768

11.5076

211.2768x2 5 211.5076x2 cos u

x2 5 12.2768x2 2 11.5076x2 cos u

x2 5 8.2768x2 1 4x2 2 11.5076x2 cos u

x2 5 r2 1 (2x)2 2 2(2.8769x)(2x) cos u

u,

8 2.8769xr 8 "8.2768x2

8 8.2768x2

8 5x2 1 3.2768x2

8 5x2 2 4x2(20.8192)

5 x2 1 4x2 2 4x2 cos 145°

r2 5 x2 1 (2x)2 2 2(x)(2x) cos 145°

N

S

EW

xu

2x

resultant

35°

19.

a. First we will find the resultant of the 5 N and 8 N forces. Use the Pythagorean theorem to find themagnitude, m.

Next we use the Pythagorean theorem again to findthe magnitude, M, of the resultant of this net forceand the 10 N force.

Since the equilibrant is equal in magnitude to theresultant, we have the magnitude of the equilibrantequal to approximately 13.75 N.b. To find each angle, use the definition of cosinewith respect each force as a leg and the resultant asthe hypotenuse. Let be the angle from the 5 N force to the resultant, be the angle from the8 N force to the resultant, and be the anglefrom the 10 N force to the resultant.Let the sign of the resultant be negative, since it isin a direction away from the head of each of thegiven forces.

20. We know that the resultant of these two forces isequal in magnitude and angle to the diagonal line of the parallelogram formed with and as legs and has diagonal length We also knowfrom the cosine law that

where is the supplement to in our parallelogram.Since we know then

Thus we have

7.2 Velocity, pp. 367–3701. a. Both the woman and the train’s velocities arein the same direction, so we add them.

b. The woman’s velocity is directed opposite that oftrain, so we subtract her velocity from the train’s.

. The resultant is inthe same direction as the train’s movement.2. a. The velocity of the wind is directed opposite thatof the airplane, so we subtract the wind’s velocityfrom the airplane’s.

north.b. Both the wind and the airplane’s velocities are inthe same direction, so we add them.

north.3. We use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, made.

The resultant is 304.14 km h, S.W 9.5°> 8 9.5°.

u 5 tan21 50

300

tan u 550

300

8 304.14 km>h

m 5 "92 500

5 92 500

5 90 000 1 2500

m2 5 3002 1 502

u,

600 km>h 1 100 km>h 5 700 km>h

600 km>h 2 100 km>h 5 500 km>h

80 km>h 2 4 km>h 5 76 km>h

80 km>h 1 4 km>h 5 84 km>h

@ f1

>

1 f2

>

@ 5 " @ f1

>

@ 2 1 @ f2

>

@ 2 1 2 @ f1

>

@ @ f2

>

@ cos u

5 @ f1

>

@ 2 1 @ f2

>

@ 2 1 2 @ f1

>

@ @ f2

>

@ cos u

@ f1

>

1 f2

>

@ 2 5 @ f1

>

@ 2 1 @ f2

>

@ 2 2 2 @ f1

>

@ @ f2

>

@ cos f

cos f 5 cos (180 2 u) 5 2cos u.

f 5 180 2 u,

uf

@ f1

>

1 f2

>

@ 2 5 @ f1

>

@ 2 1 @ f2

>

@ 2 2 2 @ f1

>

@ @ f2

>

@ cos f

@ f1

>

1 f2

>

@ .

f2

>

f1

>

8 136.7°

u10N 5 cos21 a10

213.75b

cos u10N 510

213.75

8 125.6°

u8N 5 cos21 a8

213.75b

cos u8N 58

213.75

8 111.3°

u5N 5 cos21 a5

213.75b

cos u5N 55

213.75

u10N

u8N

u5N

M 5 "189 8 13.75

5 189

5 89 1 100

M2 5 m2 1 102

m 5 "89 8 9.4

5 89

5 25 1 64

m2 5 52 1 82

5N

8N 10N



4. Adam must swim at an angle, upstream so asto counter the 1 km h velocity of the stream. This isequivalent to Adam swimming along the hypotenuseof a right traingle with 1 km h leg and a 2 km hhypotenuse. So the angle is found using the definitionof cosine.

upstream5. a. 2 m s forwardb. 20 m s 2 m s 22 m s in the direction of the car6. Since the two velocities are at right angles wecan use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, made.

from the direction of the boat towardthe direction of the current. This results in a net of

, or 7. a. First we find the components of the resultantdirected north and directed west. The componentdirected north is the velocity of the airplane, 800,

since the wind forms a 45°angle south of west. The western component of the resultant is simply So we use thePythagorean theorem to find the magnitude, m, ofthe resultant and the definition of sine to find theangle, of the resultant.

Use the sine law to determine the direction.

The direction is

b. The airplane is travelling at approximately 732.71 km h, so in 1 hour the airplane will travelabout 732.71 km.8. a. First we find the velocity of the airplane. Weuse the Pythagorean theorem to find the magnitude,m, of the resultant.

So in 3 hours, the airplane will travel about

b. To find the angle, the airplane travels, we usethe definition of sine.

east of north.9. a. To find the angle, at which to fly is theequivalent of the angle of a right triangle with 44 asthe opposite leg and 244 as the hypotenuse. So weuse the definition of sine to find this angle.

south of west.b. By the Pythagorean Theorem, the resultant groundspeed of the airplane is Since the duration of theflight is simply 10. a. Since Judy is swimming perpendicular to the flow of the river, her resultant velocity is simplythe hypotenuse of a right triangle with 3 and 4 asbases, which is a 3-4-5 right triangle. Thus, Judy’s resultant velocity is 5 km h. The direction is determined by downstreamb. Judy’s distance traveled down the river would be the“4” leg of the 3-4-5 triangle formed by the vectors, butscaled down so that 1m (the width of the river) isequivalent to the “3” leg. So her distance traveled is

km. This makes her about 0.67 km fromHelen’s cottage.c. While in the river, Judy is swimming at 5 km h for a distance of Since

her time taken is 53 km

5 km>h5 1

3 hours 5 20 minutes.

time 5 distance>rate,

53 km.>

43 8 1.33

u 8 53.1°tan u 5 43.

>

(480 km)> (240 km>h) 5 2 h.


"(2442 2 442) 5 240 km>h.

8 10.4°

u 5 sin21 44

244

sin u 544

244

u,

8 12.5°

u 5 sin21 100

461

sin u 5100

461

u,

(461 km>h)(3 h) 5 1383 km.

8 461 km>h

m 5 "212 500

5 212 500

5 202 500 1 10 000

m2 5 4502 1 1002

>

N 5.5° W.

u 8 5.5°

sin u

1005

sin 45°

732.71

732.71 km>h m 8

8 536 863.8082

8 (729.29)2 1 (71.71)2

m2 5 (800 2 100 sin 45°)2 1 (100 cos 45°)2

u,

100 cos 45°.

minus 100 sin 45°,

N 37.6° W.22.6° 1 15° 5 37.6°

8 22.6°

u 5 sin21 5

13

sin u 55

13

5 13 m>s

m 5 "169

5 169

5 144 1 25

m2 5 122 1 52

u,

>5>1>

>

5 60°

u 5 cos21 1

2

cos u 51

2

>>

>

u,

11.

a. and b. Here, 205 km h directed 30° north of eastis the resultant of 212 km h directed east, and thewind speed, v, directed at some angle. This problemis more easily approached finding the wind speed,v, first. So we will do that using the cosine law.

Now to find the wind’s direction, we simply find theangle supplementary to the lesser angle, formed by the parallelogram of these three velocities. Wecan use the sine law for this.

Thus, the direction of v is the angle supplementaryto in the parallelogram:

west of north.12.

Since her swimming speed is a maximum of 4 km h,this is her maximum resultant magnitude, which isalso the hypotenuse of the triangle formed by her andthe river’s velocity vector. Since one of these legs is 5 km h, we have a triangle with a leg larger than itshypotenuse, which is impossible.13. a. First we need to find Mary’s resultant velocity, v. Since this resultant is the diagonal of theparallelogram formed by hers and the river’s velocity, we can use the cosine law with the angle,of the parallelogram adjacent 30°.

So in 10 seconds, Mary travels about

b. Since Mary is travelling at 3 m s at an angle of 30°,to find the component of her velocity, v, perpendicularto the current, we use the definition of sine.

perpendicular to the current.So since the time taken is

14. a. So we have a 5.5 m s vector and a 4 m s vector with a resultant vector that is directed 45°south of west. Letting be the angle between the 4 km h vector and the resultant, we can construct a parallelogram using these three vectors and a subsequent triangle with opposite the 5.5 m s vector and 45° opposite the 4 m s vector. We nowuse the sine law to find

from the resultant.Since the resultant is 45° west of south, Dave’sdirection is west of south,which is equivalent to about upstream.b. First, we find the magnitude, m, of Dave’s 4 m svelocity in the direction perpendicular to the river. This is done using the definition of sine.

perpendicular to the river.Since time is we have

15. Let b represent the speed of the steamboat and crepresent the speed of the current. On the waydownstream, the effective speed is andupstream is The distance upstream anddownstream is the same, so So, This means that the speed of the boat is 6 times the speed of the current. So, (6c 1 c) ? 5

b 5 6c.

5(b 1 c) 5 7(b 2 c).

b 2 c.

b 1 c,

(200 m)> (3.41 m>s) 8 58.6 s.

distance>rate,

8 3.41 m>s

m 5 4 sin 58.5°

>

180° 2 121.5° 5 58.5°

76.5° 1 45° 5 121.5°

8 76.5°

u 5 sin21 a5.5asin 45°

4bb

sin u 5 5.5asin 45°

4b

sin u

5.55

sin 45°

4

u.

>

>u

>

u

>>

(150 m)> (1.5 m>s) 5 100 s.


5 1.5 m>s

5 3a1

2b

v 5 3 sin 30

>

(6.8 m>s)(10 s) 5 68 m.

8 6.8 m>s

v 5 "45.8

5 45.8

5 25 1 20.8

5 9 1 16 2 24 cos 150°

v2 5 32 1 42 2 2(3)(4) cos 150°

u,

>

>

5

4

180° 2 71.6° 5 108.4° 5 18.4°

u

8 71.6°

u 5 sin21 a205asin 30°

108bb

sin u 5 205asin 30°

108b

sin u

2055

sin 30°

108

u,

8 108 km>h

v 5 "11 694

5 11 694

5 86 969 2 75 275

5 42 025 1 44 944 2 86 920 cos 30°

v2 5 2052 1 2122 2 2(205)(212) cos 30°

>

>

212 km/h

30°v

205 km/h

u



or 35c is the distance. This means that it would takea raft 35 hours moving with the speed of the currentto get from A to B.

7.3 The Dot Product of Two GeometricVectors, pp. 377–378

1. This means or or To be guaranteed that thetwo vectors are perpendicular, the vectors must benonzero.2. is a scalar, and a dot product is only definedfor vectors, so is meaningless.

3. Answers may vary. Let but

4. because 5. Since and are unit vectors, andsince they are pointing in opposite directions then

so Therefore 6. a.

b.

c.

d.

e.

f.

7. a.

b.

c.

d.

e.

f.

8.

Note: is the angle between the two vectors whenthey are tail to tail, so 9. a.

b.

10. so the dot product of any vector with is 0.0

>

@0>

@ 5 0

5 6 0 x>

0 2 2 19x>

? y>

1 3 0 y>

0 25 6 0 x

>

0 2 2 9x>

? y>

2 x>

? y>

2 9x>

? y>

1 3 0 y>

0 21 3 0 y

>

0 21 3 0 x

>

0 2 2 x>

? y>

2 (23y>

? 23x>

)5 3 0 x>

0 2 2 3x>

? 3y>

3x>

? (x>

2 3y>

) 2 (x>

2 3y>

) ? (23x>

1 y>

)

1 7a>

? b>

5 2 0 a>

0 2 2 15 @b>

@ 22 3a

>

? b>

1 10a>

? b>

5 2 0 a>

0 2 2 15 @b>

@ 21 5b

>

? 2a>

2 5b>

? 3b>

(a>

1 5b>

) ? (2a>

2 3b>

) 5 a>

? 2a>

2 a>

? 3b>

u 2 120°.

u

5 22.5

5 (45)a1

2b

5 (7.5)(6) cos (180° 2 120°)

a>

? b>

5 0 a>

0 @b>

@ cos u

u 5 120°

21

25 cos u

250 5 (10)(10) cos u

u>

? v>

5 0 u>

0 0 v>

0cos u

u 5 60°

1

25 cos u

10.5 5 (7)(3) cos u

a>

? b>

5 0 a>

0 @b>

@ cos u

u 8 127°

23

55 cos u

23 5 (5)(1) cos u

p>

? q>

5 0 p>

0 0 q>

0cos u

u 8 53°

3

55 cos u

3 5 (5)(1) cos u

p>

? q>

5 0 p>

0 0 q>

0cos u

u 8 80°

1

65 cos u

(6) 5 (6)(6) cos u

m>

? n>

5 0m>

0 0 n>

0cos u

u 5 30°

"3

25 cos u

12"3 5 (8)(3) cos u

x>

? y>

5 0 x>

0 0 y>

0cos u

8 226.2

5 (4)(8) cos 145°

u>

? v>

5 0 u>

0 0 v>

0cos u

5 0

5 (10)(0)

5 (2)(5) cos (90°)

m>

? n>

5 0m>

0 0 n>

0cos u

5 21

5 (1)(21)

5 (1)(1) cos (180°)

p>

? q>

5 0 p>

0 0 q>

0cos u

5 0

5 (0)(8) cos (100°)

a>

? b>

5 0 a>

0 @b>

@ cos u

8 26.93

5 (8)a2"3

2b

5 (2)(4) cos (150°)

x>

? y>

5 0 x>

0 0 y>

0cos u

5 16

5 (32)(.5)

5 (4)(8) cos (60°)

p>

? q>

5 0 p>

0 0 q>

0cos u

a>

? b>

5 21.cos u 5 21.u 5 180°

0 a>

0 5 @b>

@ 5 1b>

a>

c>

5 a>

a>

? b>

5 b>

? a>

5 b>

? c>

a>

5 2c>

.b>

? c>

5 0,a>

? b>

5 0,

c>

5 2i>

.b>

5 j>

,a>

5 i>

,

(a>

? b>

) ? c>

a>

? b>

cos u 5 0.@b>

@ 5 0,

0 a>

0 5 0,a>

? b>

5 0 a>

0 @b>

@ cos u 5 0.

11.

12. a.

b.

13. a.

b.

Therefore This is just what the Pythagorean theorem says,where and are the legs of the right triangle.14.

15.

16.

17.

18.

because is a scalar

because

7.4 The Dot Product for AlgebraicVectors, pp. 385–387

1.

Any vector of the form (c, c) is perpendicular to Therefore there are infinitely many vectors perpendicular to Answers may vary. For example:(1, 1), (2, 2), (3, 3).2. a.

b.

is acutec.

is obtuse3. Any vector in the xy-plane is of the form

Let

Therefore (0, 0, 1) is perpendicular to every vectorin the xy-plane.Any vector in the xz-plane is of the form

Let

Therefore (0, 1, 0) is perpendicular to every vectorin the xz-plane.

5 0

c>

? d>

5 (0)(c1) 1 (0)(1) 1 (0)(c3)

d>

5 (0, 1, 0).c>

5 (c1, 0, c3).

5 0

a>

? b>

5 (0)(a1) 1 (0)(a2) 1 (0)(1)

b>

5 (0, 0, 1).a>

5 (a1, a2, 0).

u

cos u , 0

5 23 , 0

5 3 1 4 2 10

a>

? b>

5 (1)(3) 1 (22)(22) 1 (5)(22)

u

cos u . 0

5 34 . 0

5 8 1 9 1 17

a>

? b>

5 (2)(4) 1 (3)(3) 1 (21)(217)

u 5 90°

5 0

a>

? b>

5 (22)(1) 1 (1)(2)

a>

.

a>

.

b2 5 b1

(21)b1 1 b2 5 0

a>

? b>

5 0

d>

? a>

5 0

c>

? a>

5 d>

? a>

1 c>

? a>

0 a>

0 5 1c>

? a>

5 (d>

1 c>

) ? a>

c>

? a>

5 (b>

? a>

) 0 a>

0 2b>

? a>

c>

? a>

5 (b>

? a>

)(a>

? a>

)

c>

? a>

5 ((b>

? a>

) a>

) ? a>

b>

5 d>

1 c>

d>

5 b>

2 c>

a>

? b>

1 a>

? c>

1 b>

? c>

5 27

2(a>

? b>

1 a>

? c>

1 b>

? c>

) 5 214

1 1 4 1 9 1 2(a>

? b>

1 a>

? c>

1 b>

? c>

) 5 0

1 c>

? a>

1 c>

? b>

1 0 c>

0 2 5 0

0 a>

0 2 1 a>

? b>

1 a>

? c>

1 b>

? a>

1 @b>

@ 2 1 b>

? c>

1 c>

? (a>

1 b>

1 c>

) 5 0

a>

? (a>

1 b>

1 c>

) 1 b>

? (a>

1 b>

1 c>

)

a>

1 b>

1 c>

5 0>

5 3

5 2 1 2a1

2b 1

1

22

1

2

1 @b>

@ 0 c>

0cos (120°)

2 0 a>

0 @b>

@ cos (60°) 1 0 a>

0 0 c>

0cos (60°) 1 15 1 1

5 0 a>

0 2 1 a>

? b>

1 a>

? c>

1 b>

? a>

1 @b>

@ 2 1 b>

? c>

(a>

1 b>

) ? (a>

1 b>

1 c>

)

5 2 0 u>

0 2 1 2 0 v>

0 25 0 u

>

0 2 1 2u>

? v>

1 0 v>

0 2 1 0 u>

0 2 2 2u>

? v>

1 0 v>

0 25 (u

>

1 v>

) ? (u>

1 v>

) 1 (u>

2 v>

) ? (u>

2 v>

)

0 u>

1 v>

0 2 1 0 u>

2 v>

0 25 14

5 (1)2 1 (2)2 1 (3)2

1 2 0 u>

0 0w>

0cos (90°) 1 2 0 v>

0 0w>

0cos (90°)

5 0 u>

0 2 1 0 v>

0 2 1 0w>

0 2 1 2 0 u>

0 0 v>

0cos (90°)

1 v>

? w>

1 w>

? u>

1 w>

? v>

1 w>

? w>

5 u>

? u>

1 u>

? v>

1 u>

? w>

1 v>

? u>

1 v>

? v>

(u>

1 v>

1 w>

) ? (u>

1 v>

1 w>

)

c>

b>

0 a>

0 2 5 @b>

@ 2 1 0 c>

0 2.

b>

? c>

5 @b>

@ 0 c>

0cos (90°) 5 0

5 @b>

@ 2 1 2b>

? c>

1 0 c>

0 25 (b

>

1 c>

) ? (b>

1 c>

)

0 a>

0 2 5 a>

? a>

5 0 a>

0 2 2 @b>

@ 25 0 a

>

0 2 2 a>

? b>

1 a>

? b>

2 @b>

@ 22 b

>

? b>

(a>

1 b>

) ? (a>

2 b>

) 5 a>

? a>

2 a>

? b>

1 b>

? a>

5 0 a>

0 2 1 2a>

? b>

1 @b>

@ 21 @b

>

@ 25 0 a

>

0 2 1 a>

? b>

1 a>

? b>

1 b>

? a>

1 b>

? b>

(a>

1 b>

) ? (a>

1 b>

) 5 a>

? a>

1 a>

? b>

5 1

a>

? b>

51

6( 0 a

>

0 2 1 5 @b>

@ 2)

0 a>

0 2 1 5 @b>

@ 2 5 6a>

? b>

0 a>

0 2 2 a>

? b>

2 5b>

? a>

1 5 @b>

@ 2 5 0

cos (90°)(a>

2 5b>

) ? (a>

2 b>

) 5 @a>

2 5b>

@ @a>

2 b>

@


Any vector in the yz-plane is of the formLet .

Therefore (1, 0, 0) is perpendicular to every vectorin the yz-plane.4. a.

b. If any of the vectors were collinear then onewould be a scalar multiple of the other. Comparingthe signs of the individual components of each vector eliminates and All ofthe components of have the samesign and the same is true for (4, 3, 10), but (4, 3, 10)is not a scalar multiple of Thereforenone of the vectors are collinear.5. a. Using the strategy of Example 5 yields

and and

Therefore the only result is or (0, 0).This is because and (1, 1) both lie on thexy-plane and are not collinear, so any vector that isperpendicular to both vectors must be in whichdoes not exist in b. If we select any two vectors that are not collinearin then any vector that is perpendicular to bothcannot be in and must be in This is not possible since does not exist in

6. a.

b.

c.

d.

7. a.

b.

8. a.

b.

2

(0, 1)

(1, 0)0

21

–1–2

1–2 –1

x

y

2

(0, 1)

(1, 0)0

21

–1–2

1–2 –1

x

y k $ 0

k 5 0 k 0

k 5 "2 0 k 01

"2

(1)(0) 1 (1)(k) 5 "1 1 1"k2 cos (45°)

a>

? b>

5 0 a>

0 @b>

@ cos u

k 52

3

3k 5 2

6k 2 2 2 3k 5 0

cos (90°)

5 0 a>

0 @b>

@(21)(26k) 1 (2)(21) 1 (23)(k)

a>

? b>

5 0 a>

0 @b>

@ cos u

u 8 154°

5282

91

5282

(7)(13)

5(2)(25) 1 (3)(0) 1 (26)(12)

"4 1 9 1 36"25 1 144

cos u 5a>

? b>

0 a>

0 @b>

@

u 8 64°

54

9

54

(3)(3)

5(2)(2) 1 (2)(1) 1 (1)(22)

"4 1 4 1 1"4 1 1 1 4

cos u 5a>

? b>

0 a>

0 @b>

@

u 8 123°

5214

"680

5(21)(6) 1 (4)(22)

"1 1 16"36 1 4

cos u 5a>

? b>

0 a>

0 @b>

@

u 8 148°

5211

"170

5211

"(34)(5)

5(5)(21) 1 (3)(22)

"25 1 9"1 1 4

cos u 5a>

? b>

0 a>

0 @b>

@

R2.R3

R3.R2

R2,

R2.

R3

(1, 22)

x 5 y 5 0,

3y 5 0

x 1 y 5 0x 2 2y 5 0

(x, y) ? (1, 1) 5 0(x, y) ? (1, 22) 5 0

(24, 25, 26).

(24, 25, 26)(5, 23, 25

6).(1, 2, 21)

5 0

(24, 25, 26) ? a5, 23, 25

6b 5 220 1 15 1 5

5 0

(1, 2, 21) ? (4, 3, 10) 5 4 1 6 2 10

5 0

e>

? f>

5 (1)(0) 1 (0)(e2) 1 (0)(e3)

f>

5 (1, 0, 0)e>

5 (0, e2, e3).


The diagonals are andor

and.

c.

or

9. a.

b.

10. a. i.

ii. Answers may vary. For example:

Let

b. In part a., the values are unique because bothvectors have their third component specified, andthe ratios must be the same for each component In part b. the values are not unique; any value of p could have been chosen, each resulting in adifferent value of q.

11.

12. a.

b.

13. a. Answers may vary. For example:

Let (3, 1, 1) is perpendicular to and

b. Answers may vary. For example:

(x, y, z) ? (21, 22, 3) 5 0

x 5 4z 2 3y x 1 3y 2 4z 5 0

(x, y, z) ? (1, 3, 24) 5 0

(1, 25, 2).

(21, 3, 0)

y 5 1.

y 5 z 22y 1 2z 5 0

x 2 5y 1 2z 5 0

(x, y, z) ? (1, 25, 2) 5 0

x 5 3y 2x 1 3y 5 0

(x, y, z) ? (21, 3, 0) 5 0

u 8 50°

58

905 cos u

49 2 16 1 25 5 90 cos u

3 "49 1 16 1 25 cos u

5 "49 1 16 1 25(27, 4, 5) ? (27, 24, 5)

AE>

? BF>

5 @AE>

@ @BF>

@ cos u

F 5 (0, 0, 5)

E 5 (0, 4, 5),D 5 (7, 0, 5),C 5 (0, 4, 0),

B 5 (7, 4, 0),A 5 (7, 0, 0),O 5 (0, 0, 0),

uC 8 63.4°

uC 5 180° 2 uA 2 uB

uB 8 26.6°

180° 2 uB 8 153.4°

5 2Å

4

5

5240

"(40)(50)

5210 2 30

"4 1 36"25 1 25

cos (180° 2 uB) 5AB

>

? BC>

@AB>

@ @BC>

@

uA 5 90°

180° 2 uA 5 90°

5 0

56 2 6

@AB>

@ @CA>

@

cos (180° 2 uA) 5AB

>

? CA>

@AB>

@ @CA>

@

CA>

5 (3, 21)BC>

5 (25, 25),AB>

5 (2, 6),

b>

.

q 5 250

p 5 1

q 5 22p 2 48

2q 1 4p 1 96 5 0

a>

? b>

5 0

q 5 3

2 52

3q

p 58

3

p 5 4a2

3b

k 52

3

8 5 12ka>

5 kb>

u 5 30°

5"3

2

53"2

"8"3

"2 2 1 1 "2 1 1 1 "2

"(2 2 2"2 1 1) 1 (2 1 2"2 1 1) 1 2"1 1 1 1 15

cos u 5a>

? b>

0 a>

0 @b>

@

u 5 90°

5 0

5(1 2 "2)(1) 1 ("2 2 1)(1)

0 a>

0 @b>

@

cos u 5a>

? b>

0 a>

0 @b>

@

5 0

5 21 1 1

(1, 1) ? (21, 1)

5 0

5 1 2 1

(1, 1) ? (1, 21)

(0, 1) 2 (1, 0) 5 (21, 0)

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

(1, 0) 2 (0, 1) 5 (1, 21)

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



14.

15. a.

b.

16. Answers may vary. For example: Note that, so they are collinear. Therefore any

vector that is perpendicular to is alsoperpendicular to



17.

or

18. a.

Therefore, since the two diagonals are perpendicular,all the sides must be the same length.

b.

c.

19. a.

The coordinates of vertex C are

b.

20. The two vectors representing the body diagonalsare and

Mid-Chapter Review, pp. 388–3891. a.

5 3

5 (6)1

2

a>

? b>

5 (3)(2) cos (60°)

a 8 109.5°

a 5 180° 2 u

u 8 70.5°

1

35 cos u

1 2 1 1 1 5 3 cos u

(21, 1, 1) ? (21, 21, 1) 5 "3"3 cos u

(0 2 1, 0 2 1, 1 2 0) 5 (21, 21, 1)

(0 2 1, 1 2 0, 1 2 0) 5 (21, 1, 1)

u 8 87.4°

16

3545 cos u

7 1 128 2 119 5 354 cos u

3 "1 1 64 1 289 cos u

(7, 16, 27) ? (1, 8, 17) 5 "49 1 256 1 49

AC>

? BD>

5 @AC>

@ @BD>

@ cos u

(6, 18, 24).

z 5 24y 5 18,x 5 6,

(24, 212, 25) 5 (2 2 x, 6 2 y, 29 2 z)

DA>

5 CB>

q 5 14

21 2 q 1 15 5 0

212 1 8 2 4q 1 60 5 0

AB>

? DA>

5 0

DA>

5 (24, 2 2 q, 25)AB>

5 (3, 4, 212),

u2 5 120°

2u1 1 2u2 5 360°

u1 5 60°

1

25 cos u1

2 1 2 2 1 5 6 cos u1

AB>

? BC>

5 @AB>

@ @BC>

@ cos u1

@AB>

@ 5 @BC>

@ 5 "6

5 (2, 1, 1)

BC>

51

2(a>

2 b>

)

5 (1, 2, 21)

AB>

51

2(a>

1 b>

)

5 0

a>

? b>

5 23 1 3

244

65p 5 4

p 5216 6 "(2216)2 2 4(65)(2176)

2(65)

65p2 2 216p 2 176 5 0

9p2 2 24p 1 16 5320

91

16

9p2

9p2 2 24p 1 16 5 49(20 1 p2)a4

21b

2

(3p 2 4)2 5 a7"20 1 p2 cos ub2

8 1 3p 2 12 5 "20 1 p2(7) cos u

5 "16 1 p2 1 4"4 1 9 1 36 cos u

(24, p, 22) ? (22, 3, 6)

x>

? y>

5 0 x>

0 0 y>

0cos u

(22, 24, 2).

(1, 2 2 1)

x 5 y 5 1.

(22, 24, 2).

(1, 2, 21)

x 5 z 5 1.

x 1 2y 2 z 5 0

(x, y, z) ? (1, 2, 21) 5 0

r>

.

s>

s>

5 22r>

p 5 0

3 1 4p 2 3 5 0

3 1 4p 1 q 5 0

23 2 4p 2 q 5 0

(23, p, 21) ? (1, 24, q) 5 0

p 5 3 or 21

p 5 1 6 2

p 52 6 "22 2 4(23)

2

p2 2 2p 2 3 5 0

(p, p, 1) ? (p, 22, 23) 5 0

(21, 22, 3).

(1, 3, 24)

y 5 1.

y 5 z 3y 2 4z 2 2y 1 3z 5 0

2x 2 2y 1 3z 5 0


b.

2.

Let be the tension in the 15 cm cord and bethe tension in the 20 cm cord. Let be the anglethe 15 cm cord makes with the ceiling and be theangle the 20 cm cord makes with the ceiling. By thecosine law:

Horizontal Components:

Vertical Components:

Therefore the tension in the 15 cm cord is 117.60 Nand the tension in the 20 cm cord is 88.20 N.3. The diagonals of a square are perpendicular, sothe dot product is 0.4. a.

By the cosine law:

By the cosine law:

The resultant velocity of the airplane is at S7.06°Eb.

hours5. a.

b.

6.

7. a.

b.

8. a.

b.

c.

d.

e.

5 34

5 15 1 4 1 15

? (5j>

2 4j>

1 3k>

)

1 5k>

)(a>

1 b>

) ? (b>

1 c>

) 5 (3i>

2 j>

5 0

5 5 2 8 1 3

a>

? (b>

1 c>

) 5 (i>

1 2j>

1 k>

) ? (5i>

2 4j>

1 3k>

)

5 5i>

2 4j>

1 3k>

b>

1 c>

5 (2i>

2 3j>

1 4k>

) 1 (3i>

2 j>

2 k>

)

5 55 6 1 3 2 4

b>

? c>

5 (2i>

2 3j>

1 4k>

) ? (3i>

2 j>

2 k>

)

5 05 2 2 6 1 4

a>

? b>

5 (i>

1 2j>

1 k>

) ? (2i>

2 3j>

1 4k>

)

cos (u) 534

63

34 5 "16 1 25 1 400 "1 1 4 1 4 cos (u)

a>

? b>

5 0 a>

0 @b>

@ cos (u)

5 34

a>

? b>

5 (4)(1) 1 (25)(2) 1 (20)(2)

5 4.5

5 (3)(3)(0.5)

@b>

@ cos (60°)a>

? b>

5 0 a>

0

u 5 60°

6u 5 360°

@F>

@ 8 94.49 N

@F>

@ 5 @E>

@ sin (40°)

@E'

>

@ 8 112.61 N

@E'

>

@ 5 (9.8)(15)cos (40°)

@E'

>

@ 5 @E>

@ cos (40°)

E F

t 8 1.74

t 81000

575.1?

km

(km/h)

(distance) 5 (rate)(time)

575.1 km>h

a 8 7.06°

sin (a) 8 0.123

sin (a)

1005

sin (135°)

575.1

0 v>

1 w>

0 8 575.1

2 2(500)(100) cos (135°)

0 v>

1 w>

0 2 5 (500)2 1 (100)2

0w>

0 5 1000 v>

0 5 500,

135°

w

v v w+

a

T2 5 88.2 N

T2 5 (0.75)T1

T1 5 117.6 N

(1.25)T1 5 147

(0.8)T1 1 (0.6)(0.75)T1 5 147

T1 sin (u1) 1 T2 sin (u2) 2 (15)(9.8) 5 0

T2 5 (0.75)T1

(0.8)T2 5 (0.6)T1

2T1 cos (u1) 1 T2 cos (u2) 5 0

sin (u1) 5 0.8

cos (u1) 5 0.6

(20)2 5 (15)2 1 (25)2 2 (2)(15)(25) cos (u1)

sin (u2) 5 0.6

sin (u2) 5 "1 2 cos2 (u2)

cos (u2) 5 0.8

(15)2 5 (20)2 1 (25)2 2 2(20)(25) cos (u2)

u2

u1

T2T1

20 cm

15 cm25 cm

u1

u2

5 81

5 12(3)2 2 3 2 6(2)2

1 8b>

? a>

2 6 @b>

@ 2(3a

>

1 2b>

) ? (4a>

2 3b>

) 5 12 0 a>

0 2 2 9a>

? b>


f.

9. a.

or

b. If and are parallel then one is a scalar multiple of the other.

where n is a constant

by the componentby the component

by the component

Therefore there is no value of x that will make thesetwo vectors parallel.10. a.

b.

c.

or 1.41

d.

e.

11.

12.

N, 25.3° from the 4 N force towards the 3 N force.

N, from the 4 N force away from the 3 N force.13. a. The diagonals are and

b.

u 8 87.9°cos u 8 0.0374

2 2"38 "75 cos u

(9 1 100) 5 (4 1 9 1 25) 1 (1 1 49 1 25)

0m>

2 n>

0 2 5 0m>

0 2 1 0 n>

0 2 2 2 0m>

0 0 n>

0cos u

u 8 109.1°

cos u 8 20.3276

"9 1 100 cos u3 2 40 5 "1 1 16 1 100

(m>

1 n>

) ? (m>

2 n>

) 5 0m>

1 n>

0 0m>

2 n>

0cos u

m>

2 n>

5 (3, 210, 0)

m>

1 n>

5 (1, 4, 10)

m>

2 n>

m>

1 n>

180° 2 25.3° 5 154.7°E>

8 6.08

F>

8 6.08

u 8 25.3°

cos u 544

8"37

(3)2 5 (4)2 1 ("37)22 2(4)("37) cos u

F 8 6.08 N

(F)2 5 37

(F)2 5 25 2 24 cos (120°)

(F)2 5 (3)2 1 (4)2 2 2(3)(4) cos (180° 2 60°)

u 8 126.9° 180° 2 u 8 53.1

0.6 5 cos (180° 2 u)

(4)2 5 (5)2 1 (3)2 2 2(3)(5) cos (180° 2 u)

180° - u

5 N

3 N

4 N

5 212

5 23(1 1 2 1 1)

5 23(i>

2 2j>

2 k) ? (i>

2 j>

2 k>

)

5 23x>

? y>

2x>

? y>

2 5y>

? x>

5 2x>

? y>

2 5x>

? y>

5 245 25 1 6 2 5

? (5i>

2 6j>

2 5k>

)

5 (2i>

2 j>

1 k>

)

1 (4 i>

2 4 j>

2 4k>

)

1 (( i>

2 2j>

2 k>

)

2 (3i>

2 3j>

2 3k>

)) ?

2k>

)(2x>

2 3y>

) ? (x>

1 4y>

) 5 ((2i>

2 4j>

2

5 "2

5 "(2i>

1 k>

) ? (2i>

1 k>

)

5 @2i>

1 k>

@0 x>

2 2y>

0 5 @ (i>

2 2j>

2 k>

) 2 (2i>

2 2j>

2 2k>

) @5 245 6 1 12 1 6

3x>

? 2y>

5 (3i>

2 6j>

2 3k>

) ? (2i>

2 2j>

2 2k>

)

5 i>

2 4j>

2 k>

3x>

2 2y>

5 (3i>

2 6j>

2 3k>

) 2 (2i>

2 2j>

2 2k>

)

1 2 0

j>

1 5 30(0)

x 5 0

i>

x 5 9xk>

n 5 3

xi>

1 j>

1 3k>

5 n(3xi>

1 10xj>

1 k>

)

p>

5 nq>

q>

p>

x 5 21

3x 5 23

x 5210 6 8

6

x 5210 6 "(10)2 2 4(3)(3)

2(3)

3x2 1 10x 1 3 5 0

5 0(xi>

1 j>

1 3k>

) ? (3xi>

1 10xj>

1 k>

)

p>

? q>

5 0

5 9

5 220 1 39 2 10

? (5i>

1 3j>

1 k>

)

5 (24i>

1 13j>

2 10k>

)

1 (3i>

2 j>

1 k>

))

1 2k>

)? ((2i>

1 4j>

2 (6i>

2 9j>

1 12k>

))

(2a>

2 3b>

) ? (2a>

1 c>

) 5 ((2i>

1 4j>

1 2k>

)


14. a.

b.km h

hours15.

choose

or

16. a. m s

mb.

m s

seconds17. a.

when and havethe same length.b. Vectors and determine a parallelogram. Theirsum is one diagonal of the parallelogramformed, with its tail in the same location as the tailsof and Their difference is the otherdiagonal of the parallelogram.18.

N

7.5 Scalar and Vector Projections,pp. 398–400

1. a. Scalar projection of on is where

and is the positive x-axis (X, 0).

The vector projection is the scalar projection

multiplied by where is the x-axis divided by

the magnitude of the x-axis which is equal to The scalar projection of 2 multiplied by equals

b. Scalar projection of on is where

and is now the positive y-axis (0, Y).


multiplied by where is the y-axis divided

by the magnitude of the y-axis which is equal to The scalar projection of 3 multiplied by equals 2. Using the formula would cause a division by 0.Generally the has any direction and 0 magnitude.You can not project onto nothing.3. You are projecting onto the tail of which is a point with magnitude 0. Therefore it is theprojections of onto the tail of are also 0 and 0

>

.

a>

b>

0>

;

b>

a>

0>

3j>

.j>

j>

.

b>

@ b>

@b>

@ b>

@

5 3;

a> @b

>

@@b>

@5

3YY

5 Y @b

>

@ 5 "02 1 Y2

5 3Y 5 0 1 3Y

a>

? b>

5 (2 3 0) 1 (3Y)

b>

a>

5 (2, 3)

a>

?b>

@b>

@b>

a>

2i>

.i>

i>

.

b>

@ b>

@b>

@ b>

@

5 2;

a>

? b>

@b>

@5

2XX

5 X @b

>

@ 5 "X2 1 02

5 2X 5 2X 1 0

a>

? b>

5 (2X) 1 (3 3 0)

b>

a>

5 (2, 3)

a>

?b>

@b>

@b>

a>

8 268.12

@F>

@ 5 350 cos (40°)

a>

2 b>

b>

.a>

a>

1 b>

b>

a>

y>

x>

(x>

1 y>

) ? (x>

2 y>

) 5 0

0 x>

0 2 5 0 y>

0 2 0 x>

0 2 2 x>

? y>

1 y>

? x>

2 0 y>

0 2 5 0

(x>

1 y>

) ? (x>

2 y>

) 5 0

t 8 84.9

t 8180

2.12

> 8 2.12

w 5 3 sin (45°)

8 61.2

d 8 (6.12)(10)

> 8 6.12

v 5 4 1 3 cos (45°)

a21

!6,

2

!6, 2

1

!6bx

>

5 a1

!6,2

2

!6,

1

!6b

x>

51

!6(1, 22, 1)

x1 5 1

x2 5 22

x3 5 1

x2 1 2x3 5 0

2x2 1 5x3 1 3x2 1 5x3 5 0

x1 1 3x2 1 5x3 5 0

b>

? x>

5 0

x1 5 2x2 1 5x3

2x1 1 2x2 1 5x3 5 0

a>

? x>

5 0

t 8 2.17

t 81000

460.5

> 8 460.5

v 5 500 cos (2.6°) 2 45 cos (30°)

u 8 N 2.6° E45 sin (150°) 5 500 sin u


4. Answers may vary. For example:

Scalar projection on Vector projection on Scalar projection on Vector projection on

5. When and then

Therefore the scalar projection is

The vector equation is

Under the same approach, when and then



The same is also true when andthen



Without having to use formulae, a projection of on or is the same as a projection

of on (0, 2, 0) on and (0, 0, 5) on which intuitively yields the same result.6. a.


b. Direction angles for where

include and

Therefore

Therefore

8 74.9°;

b 5 cos21 a6

23b

56

23,

56

"529

56

"9 1 36 1 484

56

"32 1 62 1 (222)2

cos b 5b

"a2 1 b2 1 c2

8 82.5°;

a 5 cos21 a3

23b

53

23,

53

"529

53

"9 1 36 1 484

53

"32 1 62 1 (222)2

cos a 5a

"a2 1 b2 1 c2g.b,a,

p>

5 (a, b, c)p>

5458

441(24, 5, 20).

(24, 5, 220)

21. 5

458

21

The vector equation 5458

213

q>

0 q>

0

p>

? q>

0 q>

05

458

21,

5 21

5 "441

5 "16 1 25 1 400

0 q>

0 5 "(24)2 1 52 1 (220)2

5 4585 212 1 30 1 440

1 (222 3 220)

p>

? q>

5 (3 3 24) 1 (6 3 5)

k>

j>

,i>

,(21, 0, 0)

k>

j>

,i>

,(21, 2, 5)

5 5;

5 3b>

@b>

@5 5 3

(0, 0, 1)

1

5 5,

a>

? b>

@b>

@5

5

1

5 1

@b>

@ 5 "02 1 02 1 12

5 5

a>

? b>

5 (21 3 0 1 2 3 0 1 5 3 1)

b>

5 (0, 0, 1)

a>

5 (21, 2, 5)

5 2;

2 3b>

@b>

@5 2 3

(0, 1, 0)

1

5 2,

a>

? b>

@b>

@5

2

1

5 1

@b>

@ 5 "02 1 1 1 02

5 2

a>

? b>

5 (21 3 0 1 2 3 1 1 5 3 0)

b>

5 (0, 1, 0),

a>

5 (21, 2, 5)

5 21;

21 3b>

@b>

@5 21 3

(1, 0, 0)

1

5 21;

a>

? b>

@b>

@5

21

1

5 1

@b>

@ 5 "12 1 02 1 02

5 21

a>

? b>

5 (21 3 1 1 2 3 0 1 5 3 0)

b>

5 (1, 0, 0)a>

5 (21, 2, 5)

p>

5 AD>

q>

p>

5 @AD>

@ ;q>

q>

5 AC>

;p>

q>

5 @AC>

@ ;p>

A

D

C

E

Bq

p

q>

5 AB>

p>

5 AE>

,



Therefore

7. a.

The scalar projection is

The vector projection is

b.



c.



8. a. The scalar projection of on the x-axis

is

The vector projection of on the x-axis is

The scalar projection of on the y-axis is

The vector projection of on the y-axis is

The scalar projection of on the z-axis is

The vector projection of on the z-axis is

b. The scalar projection of on the x-axisis


5 2mi>

;

2m 3(X, 0, 0)

"X2 1 02 1 025 2m 3

(X, 0, 0)

X

ma>

5 2m

52mX

X

1(4m 3 0)

"X2 1 02 1 02

ma>

? (X, 0, 0)

0 (X, 0, 0) 05

(2m 3 X) 1 (2m 3 0)

"X2 1 02 1 02

(X, 0, 0)

m a>

5 4k>

.

4 3(0, 0, Z)

"02 1 02 1 Z25 4 3

(0, 0, Z)

Z

a>

5 4;

54ZZ

a>

? (0, 0, Z)

0 (0, 0, Z) 05

(21 3 0) 1 (2 3 0) 1 (4 3 Z)

"02 1 02 1 Z2

(0, 0, Z)a>

5 2j>

;

2 3(0, Y, 0)

"02 1 Y2 1 025 2 3

(0, Y, 0)

Y

a>

5 2

52YY

a>

? (0, Y, 0)

0 (0, Y, 0) 05

(21 3 0) 1 (2 3 Y) 1 (4 3 0)

"02 1 Y2 1 02

(0, Y, 0)a>

5 2i>

;

21 3(X, 0, 0)

"X2 1 02 1 025 21 3

(X, 0, 0)

X

a>

5 21;

52XX

a>

? (X, 0, 0)

0 (X, 0, 0) 05

(21 3 X) 1 (2 3 0) 1 (4 3 0)

"X2 1 02 1 02

a>

? (X, 0, 0)

0 (X, 0, 0) 0(X, 0, 0)

a>

550

169(25, 12)

50

133

y>

0 y>

05

50

133

(25, 12)

13

x>

? y>

0 y>

05

50

13.

5 13

5 "169

5 "25 1 144

0 y>

0 5 "(25)2 1 122

5 50

5 210 1 60

x>

? y>

5 (2 3 25) 1 (5 3 12)

5 2i>

2 3y>

0 y>

05 2 3

(1, 0)

1

5 2;

x>

? y>

0 y>

05

2

1

5 1

0 y>

0 5 "12 1 02

5 2

x>

? y>

5 (2 3 1) 1 (2"3 3 0)

0 3y>

0 y>

05 0

>

5 0;

x>

? y>

0 y>

05

0

"2

5 "2

0 y>

0 5 "12 1 (21)2

5 0

5 1 1 (21)

x>

? y>

5 (1 3 1) 1 (1 3 21)

8 163.0°

g 5 cos21 a222

23b

5222

23,

5222

"529

5222

"9 1 36 1 484

5222

"32 1 62 1 (222)2

cos g 5c

"a2 1 b2 1 c2





9. a.

projected onto itself will yield itself. The scalarprojection will be the magnitude of itself.b. Using the formula for the scalar projection


multiplied by

10. a.

b.

So the vector projection is

11. a.

The scalar projection of on the x-axis is






b. The angle made with the y-axis is

51

3,

51

"9

51

"4 1 1 1 4

51

"(22)2 1 12 1 22

cos b 5b

"a2 1 b2 1 c2

b

5 2k>

2 3(0, 0, Z)

"02 1 02 1 Z25 2 3

(0, 0, Z)

Z

AB>

5 2;

52ZZ

a>

? (0, 0, Z)

0 (0, 0, Z) 05

(22 3 0) 1 (1 3 0) 1 (2 3 Z)

"02 1 02 1 Z2

(0, 0, Z)AB>

5 j>

;

1 3(0, Y, 0)

"02 1 Y2 1 025 1 3

(0, Y, 0)

Y

AB>

5 1;

5YY

a>

? (0, Y, 0)

0 (0, Y, 0) 05

(22 3 0) 1 (1 3 Y) 1 (2 3 0)

"02 1 Y2 1 02

(0, Y, 0)AB>

5 22i>

;

22 3(X, 0, 0)

"X2 1 02 1 025 22 3

(X, 0, 0)

X

AB>

5 22;

522X

X

a>

? (X, 0, 0)

0 (X, 0, 0) 05

(22 3 X) 1 (1 3 0) 1 (2 3 0)

"X2 1 02 1 02

(X, 0, 0)AB>

5 (22, 1, 2)

5 (21, 3, 4) 2 (1, 2, 2)

AB>

5 Point B 2 Point A

2 0 a>

0 a0 a>

0

0 a>

0b 5 2a

>

.

5 2 0 a>

0

(2a>

) ? a>

0 a>

0 5

2 0 a>

0 2

0 a>

0

–a aB O A

0 a>

0 3a>

0 a>

05 a

>

.a>

0 a>

0,

5 0 a>

0 .

5 0 a>

0 (1)

0 a>

0cos u 5 0 a>

0cos 0

a>

a

5 4mk>

.

4m 3(0, 0, Z)

"02 1 02 1 Z25 4m 3

(0, 0, Z)

Z

ma>

5 4m;

54mZ

Z

1(4m 3 Z)

"02 1 02 1 Z2

ma

>

? (0, 0, Z)

0 (0, 0, Z) 05

(2m 3 0) 1 (2m 3 0)

"02 1 02 1 Z2

(0, 0, Z)ma>

5 2mj>

;

2m 3(0, Y, 0)

"02 1 Y2 1 025 2m 3

(0, Y, 0)

Y

ma>

5 2m;

52mY

Y

1(4m 3 0)

"02 1 y2 1 02

ma>

? (0, Y, 0)

0 (0, Y, 0) 05

(2m 3 0) 1 (2m 3 Y)

"02 1 Y2 1 02

(0, Y, 0)ma>


Therefore

12. a.

b.

c. In an isosceles triangle, CD is a median and aright bisector of BA. Therefore and have thesame magnitude projected on .d. Yes, not only do they have the same magnitude,but they are in the same direction as well whichmakes them have equivalent vector projections.13. a. Use the formula for the scalar projection of on

And the formula for the scalar projection of on

b.

is the vector projection of on is the vector projection of on

14. a.

The scalar projection of on is

b.



c. Same lengths and both are in the direction of Add to get one vector.15. a.

1c2

a2 1 b2 1 c2

5a2

a2 1 b2 1 c2 1b2

a2 1 b2 1 c2

1 ac

"a2 1 b2 1 c2b

2

5 aa

"a2 1 b2 1 c2b

2

1 ab

"a2 1 b2 1 c2b

2

1 5 cos2 a 1 cos2 b 1 cos2 g

OD>

.

5 6

518

3

518

"9

54 1 12 1 2

"1 1 4 1 4

AC>

? OD>

@OD>

@5

(24 3 21) 1 (6 3 2) 1 (1 3 2)

"(21)2 1 22 1 22

OD>

AC>

5 (24, 6, 1)

5 (26, 7, 5) 2 (22, 1, 4)

AC>

5 Point C 2 Point A5 6

518

3

AB>

? OD>

@OD>

@1

BC>

? OD>

@OD>

@5 2

1

31

19

3

519

3

519

"9

57 1 8 1 4

"1 1 4 1 4

BC

>

? OD>

@OD>

@5

(27 3 21) 1 (4 3 2) 1 (2 3 2)

"(21)2 1 22 1 22

OD>

BC>

5 (27, 4, 2)

5 (26, 7, 5) 2 (1, 3, 3)

BC>

5 Point C 2 Point B

5 21

3

521

"9

5(23) 1 4 1 (22)

"1 1 4 1 4

AB

>

? OD>

@OD>

@5

(3 3 21) 1 (2 3 2) 1 (21 3 2)

"(21)2 1 22 1 22

OD>

AB>

5 (3, 2, 21)

5 (1, 3, 3) 2 (22, 1, 4)

AB>

5 Point B 2 Point A

b>

a>

OP>

a>

b>

OQ>

10

P

Q O

12

a

b

135°

5 28.49

5 12 cos 135°

a>

5 @b>

@ cos u

b>

5 27.07

5 10 cos 135°

b>

5 0 a>

0cos u

a>

c>

b>

a>

C

DBu u

A

abb

c

@BD>

@

C

B u u A

a b

D c

@BD>

@8 70.5°

b 5 cos21 a1

3b


b.

y is a multiple of

z is a multiple of

Answers include etc.c. If two angles add to , then all three will add to

.16. a.

b. For obtuse, use

17.


7.6 The Cross Product of Two Vectors, pp. 407–408

1. a.

is perpendicular to Thus, their dot productmust equal 0. The same applies to the second case.

b. is still in the same plane formed by andthus is perpendicular to making the

dot product 0 again.c. Once again, is still in the same planeformed by and thus is perpendicular to

making the dot product 0 again.2. produces a vector, not a scalar. Thus, theequality is meaningless.3. a. It’s possible because there is a vector crossedwith a vector, then dotted with another vector,producing a scalar.b. This is meaningless because produces ascalar. This results in a scalar crossed with a vector,which is meaningless.

a>

? b>

a>

3 b>

a>

3 b>

a>

2 b>

b>

,a>

a>

2 b>

a>

3 b>

a>

1 b>

b>

,

a>

a>

1 b>

b

y

x

z

a3b a1b

a

a>

.a>

3 b>

b

y

z

x

a x b

a

A (a, b, 0)

B (0, c, d)

y

z

x

sin2 a 1 sin2 b 1 sin2 g 5 2

1 5 3 2 (sin2 a 1 sin2 b 1 sin2 g)

1 5 (1 2 sin2 a) 1 (1 2 sin2 b) 1 (1 2 sin2 g)


cos2 x 5 1 2 sin2 xcos2 x 1 sin2 x 5 1

x 8 125.3°

x 5 cos21 a2Å

1

3b

cos x 5 2Å

1

3.

x 8 54.7°

x 5 cos21 Å

1

3

Å

1

35 cos x

1

35 cos2 x

1 5 3 cos2 x 1 5 cos2 a 1 cos2 b 1 cos2 g

cos2 a 5 cos2 b 5 cos2 g

cos a 5 cos b 5 cos g

a 5 b 5 g

180°

90°

Q0, "32 , 12R, Q0, "3, 1R,

1

2.

51

2,

cos g 5 cos 60°

"32 .

5"3

2,

cos b 5 cos 30°

x 5 0

5 0,

cos a 5 cos 90°

g 5 60°b 5 30°,a 5 90°,

5 1

5a2 1 b2 1 c2

a2 1 b2 1 c2


c. This is possible. produces a vector, andalso produces a vector. The result is a vector

dotted with a vector producing a scalar.d. This is possible. produces a scalar, and

produces a vector. The product of a scalarand vector produces a vector.e. This is possible. produces a vector, and

produces a vector. The cross product of avector and vector produces a vector.f. This is possible. produces a vector. Whenadded to another vector, it produces another vector.4. a.

b.

c.

d.

e.

f.

5.

If we look at the x component, we know that:

6. a.

b. Vectors of the form are in the yz-plane. Thus, the only vectors perpendicular to theyz-plane are those of the form because theyare parallel to the x-axis.7. a.

b.

Using the commutative law of multiplication wecan rearrange this:

8. a.

b.

9. a.

5 k>

5 (0, 0, 1)

5 (0 2 0, 0 2 0, 1 2 0)

i>

3 j>

5 (1, 0, 0) 3 (0, 1, 0)

5 (23, 2, 5)

(0, 28, 4)5 (23, 10, 1) 1

4(1) 2 1(0))2(0) 2 4(2),

4(1) 2 1(3)) 1 (1(2) 2 2(1),

2(3) 2 4(21),p>

3 q>

1 p>

3 r>

5 (1(21) 2 2(1),

5 (23, 2, 5)

4(2) 2 1(3))

5 (1(1) 2 2(2), 3(2) 2 4(1),

5 (4, 1, 2) 3 (3, 2, 1)

5 (4, 1, 2) 3 (3, 1 1 1, 21 1 2)

1 (0, 1, 2)4

p>

3 (q>

1 r>

) 5 (4, 1, 2) 3 3(3, 1, 21)

5 (226, 27, 3)

5 (222, 23, 4) 1 (24, 24, 21)

1(1) 1 2(21))4(21) 2 1(0),

1 (22(0) 2 4(1),

1(2) 1 2(1))4(1) 2 1(7),

p>

3 q>

1 p>

3 r>

5 (22(7) 2 4(2),

5 (226, 27, 3)

1(3) 1 2(0))

5 (22(7) 2 4(3), 4(0) 2 1(7),

5 (1, 22, 4) 3 (0, 3, 7)

5 (1, 22, 4) 3 (1 2 1, 2 1 1, 7 1 0)

1 (21, 1, 0)4

3(1, 2, 7)p>

3 (q>

1 r>

) 5 (1, 22, 4) 3

5 (0, 0, 0)

5 (bck 2 bck, ack 2 ack, abk 2 abk)

a(kb) 2 b(ka))

5 (b(kc) 2 c(kb), c(ka) 2 a(kc),

(a, b, c) 3 (ka, kb, kc)

5 (0, 0, 0)

5 (2(2) 2 1(4), 1(2) 2 1(2), 1(4) 2 2(2))

(1, 2, 1) 3 (2, 4, 2)

(a, 0, 0)

(0, b, c)

5 (24, 0, 0)

0(5) 2 0(1))

a>

3 b>

5 (1(1) 2 1(5), 1(0) 2 0(1),

a 5 1

25(a) 5 25

3(1) 2 5(a) 5 22

21(a) 2 3(0))

5 (3(1) 2 5(a), 5(0) 2 (21)(1),

(21, 3, 5) 3 (0, a, 1)

(21, 2, 4) ? (28, 226, 11) 5 0

(5, 1, 6) ? (28, 226, 11) 5 0

5 (28, 226, 11)

5(2) 2 1(21))

5 (1(4) 2 6(2), 6(21) 2 5(4),

(5, 1, 6) 3 (21, 2, 4)

(1, 21, 0) ? (3, 3, 21) 5 0

(22, 3, 3) ? (3, 3, 21) 5 0

5 (3, 3, 21)

22(21) 2 3(1))

5 (3(0) 2 3(21), 3(1) 2 (22)(0),

(22, 3, 3) 3 (1, 21, 0)

(22, 3, 4) ? (219, 222, 7) 5 0

(1, 2, 9) ? (219, 222, 7) 5 0

5 (219, 222, 7)

1(3) 2 2(22))

5 (2(4) 2 9(3), 9(22) 2 1(4),

(1, 2, 9) 3 (22, 3, 4)

(2, 4, 7) ? (211, 233, 22) 5 0

(5, 21, 1) ? (211, 233, 22) 5 0

5 (211, 233, 22)

5(4) 2 (21)(2))

5 (21(7) 2 1(4), 1(2) 2 5(7),

(5, 21, 1) 3 (2, 4, 7)

(3, 21, 2) ? (1, 5, 1) 5 0

(2, 21, 3) ? (1, 5, 1) 5 0

5 (1, 5, 1)

2(21) 2 (21)(3))

5 (21(2) 2 3(21), 3(3) 2 2(2),

(2, 21, 3) 3 (3, 21, 2)

(0, 21, 4) ? (27, 28, 22) 5 0

(2, 23, 5) ? (27, 28, 22) 5 0

5 (27, 28, 22)

2(21) 2 (23)(0))

5(0) 2 2(4),5 (23(4) 2 5(21),

(2, 23, 5) 3 (0, 21, 4)

a>

3 b>

c>

3 d>

a>

3 b>

c>

3 d>

a>

? b>

c>

1 d>

a>

3 b>


b.

c.

10.

is perpendicular to 11. a.

b.(by part a.)

c. All the vectors are in the xy-plane. Thus, the crossproduct in part b. is between vectors parallel to thez-axis and so parallel to each other. The cross product of parallel vectors is 12. Let

Then

Thus 13.By the distributive property of cross product:

By the distributive property again:

A vector crossed with itself equals thus:

7.7 Applications of the Dot Productand Cross Product, pp. 414–415

1. By pushing as far away from the hinge as possible, is increased making the cross productbigger. By pushing at right angles, sine is its largestvalue, 1, making the cross product larger.2. a.

b. This makes sense because the vectors lie on thesame line. Thus, the parallelogram would just be aline making its area 0.3. a.b.

The axes are tilted to illustrate the force of gravitycan be split up into components to find the part inthe direction of the motion. Let x be the componentof force going in the motion’s direction.

Now we have our force, so:(392) cos 50° N ? 40 m 8 10 078.91 J

x 5 (392) cos (50°)

cos (50°) 5x

392

50° 392 N

40 m

xy

f>

? s>

5 3 ? 150 5 450 J

@a>

3 b>

@ 5 0

5 (0, 0, 0)

2 1(2), 1(4) 2 2(2))

5 (2(2) 2 1(4), 1(2)

a>

3 b>

5 (1, 2, 1) 3 (2, 4, 2)

0 r>

0

5 2a>

3 b>

5 a>

3 b>

2 (2a>

3 b>

)

5 a>

3 b>

2 b>

3 a>

5 2b>

3 a>

1 a>

3 b>

0>

,

5 a>

3 a>

2 b>

3 a>

1 a>

3 b>

2 b>

3 b>

5 (a>

2 b>

) 3 a>

1 (a>

2 b>

) 3 b>

(a>

2 b>

) 3 (a>

1 b>

)

(x>

3 y>

) 3 z>

2 x>

3 (y>

3 z>

).

5 (2, 0, 22)

x>

3 (y>

3 z>

) 5 (0 1 2, 1 2 1, 22 2 0)

5 (1, 22, 1)

y>

3 z>

5 (3 2 2, 1 2 3, 2 2 1)

5 (22, 4, 23)

(x>

3 y>

) 3 z>

5 (0 2 2, 1 2 (23), 23 2 0)

5 (21, 0, 1)

x>

3 y>

5 (0 2 1, 1 2 1, 1 2 0)

z>

5 (1, 2, 3)

y>

5 (1, 1, 1)

x>

5 (1, 0, 1)

0>

.

5 (0, 0, 0)

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

(a>

3 b>

) 3 (c>

3 d>

) 5 (0, 0, 6) 3 (0, 0, 26)

5 (0, 0, 26)

5 (0 2 0, 0 2 0, 6 2 12)

c>

3 d>

5 (2, 3, 0) 3 (4, 3, 0)

5 (0, 0, 6)

5 (0 2 0, 0 2 0, 6 2 0)

a>

3 b>

5 (2, 0, 0) 3 (0, 3, 0)

k(a>

3 b>

).a>

5 0

5 k(0)

1 a3a1b2 2 a3a2b1)

5 k(a1a2b3 2 a1a3b2 1 a2a3b1 2 a2a1b3

? (a1, a2, a3)

k(a2b3 2 a3b2, a3b1 2 a1b3, a1b2 2 a2b1)

5 j>

5 (0, 1, 0)

5 (0 2 0, 0 2 (21), 0 2 0)

2i>

3 k>

5 (21, 0, 0) 3 (0, 0, 1)

5 j>

5 (0, 1, 0)

5 (0 2 0, 1 2 0, 0 2 0)

k>

3 i>

5 (0, 0, 1) 3 (1, 0, 0)

5 i>

5 (1, 0, 0)

5 (0 2 (21), 0 2 0, 0 2 0)

2k>

3 j>

5 (0, 0, 21) 3 (0, 1, 0)

5 i>

5 (1, 0, 0)

5 (1 2 0, 0 2 0, 0 2 0)

j>

3 k>

5 (0, 1, 0) 3 (0, 0, 1)

5 k>

5 (0, 0, 1)

5 (0 2 0, 0 2 0, 0 2 (21))

2j>

3 i>

5 (0, 21, 0) 3 (1, 0, 0)


c.

First find the x component of the force:

Calculate work:

d.

First calculate the x component of the force:

Calculate work:

4. a.The square formed by the 2 vectors has an area of 1.The 2 vectors are on the xy-plane, thus, the crossproduct must be by the right hand rule.b.Once again, the area is 1, making the possible vectorhave a magnitude of 1. Also, the 2 vectors are on thexy-plane again so the cross product must lie on thez axis. However, because of the right hand rule, theproduct must be this time.c.The square has an area of 1, so the magnitude of thevector produced must be 1. The 2 vectors are on thexz-plane. The new vector must be on the y axismaking it – because of the right hand rule.d.The square has an area of 1. The 2 vectors are onthe xz-plane. So the new vector must be becauseof the right hand rule.5. a.

So the area of the parallelogram is square units.b.

So the area of the parallelogram is square units.

6.

7. a.

As we see from the picture, the area of the triangleABC is just half the area of the parallelogram determined by vectors and Thus, we use themagnitude of the cross product to calculate the area.

Since triangle ABC is half the area of the parallelogram, its area is square units.b. This is just a different way of describing the first triangle, thus the area is square units.c. Any two sides of a triangle can be used to calculate its area.8.

9.

We know that the area of a parallelogram is equal toits height multiplied with its base. Its height is BNand its base is as can be seen from the picture. We can calculate the area using the givenvectors, then use the area to find BN.

@ OA>

3 OB>

@ 5 "16 1 16 1 4 5 "36 5 6

5 (4, 24, 22)

OA>

3 OB>

5 (8 2 4, 12 2 16, 4 2 6)

AC>

5 OB>

OA

B

O

A

C

OB

BN

8 0.99 J

5 (0.14) sin (45°) ? 10

@r>

3 f>

@ 5 ( 0 r>

0sin (u)) @ f>

@

5"62

5"62

@AB>

3 AC>

@ 5 "25 1 25 1 100 5 5"6

AB>

3 AC>

5 (1 1 4, 23 1 8, 6 1 4) 5 (5, 5, 10)

AC>

5 (2 1 2, 3 2 1, 2 2 3) 5 (4, 2, 21)

AB>

5 (1 1 2, 0 2 1, 1 2 3) 5 (3, 21, 22)

AC>

.AB>

AB

C

B

A

AC

212

55 2,

522 6 22

10

a 522 6 "22 2 4(5)(224)

2(5)

5a2 1 2a 2 24 5 0

9 1 4a2 1 4a 1 1 1 a2 2 2a 1 1 5 35

9 1 (2a 1 1)2 1 (a 2 1)2 5 35

0 p>

3 q>

0 5 "9 1 (2a 1 1)2 1 (a 2 1)2 5 "35

5 (3, 2a 1 1, a 2 1)

5 (2 1 1, 22a 2 1, a 2 1)

p>

3 q>

5 (a, 1, 21) 3 (1, 1, 2)

"213

@a>

3 b>

@ 5 "196 1 1 1 16 5 "213

5 (214, 21, 4)

5 (28 2 6, 3 2 4, 2 1 2)

a>

3 b>

5 (1, 22, 3) 3 (1, 2, 4)

"3

@a>

3 b>

@ 5 "1 1 1 1 1 5 "3

5 (1, 21, 21)

5 (1 2 0, 0 2 1, 0 2 1)

a>

3 b>

5 (1, 1, 0) 3 (1, 0, 1)

j>

2i>

3 k>

5 2j>

j>

i>

3 k>

5 2j>

2k>

2i>

3 j>

5 2k>

k>

i>

3 j>

5 k>

100 cos 45° ? 500 m 5 35 355.34 J

x 5 (100) cos (45°)

45°

100 N

500 m

140 cos 20° N ? 250 m 8 32 889.24 J

(140) cos (20°) 5 x

20°140 N

250 m


Now we need to calculate to know the lengthof the base.

Substituting these results into the equation for area:

or about 1.18

10. a.

Looking at x components:;

y components:

Substitute in a:

Substitute b back into the x components:;

Check in z components:

b.

Review Exercise, pp. 418–4211. a.

b.

c. 16d. The cross products are parallel, so the originalvectors are in the same plane.2. a.b.c.

d.

e.f.

3. a. If then will be twice thus collinear.b.

4.

5. a.

b. We can use the dot product of the 2 diagonals tocalculate the angle. The diagonals are the vectors

and

u 8 77.9°

u 5 cos21 a

9

3"205b

@OA>

2 OB>

@ 5 "62 1 (23)2 5 3"5

@OA>

1 OB>

@ 5 "42 1 52 5 "41

(OA>

1 OB>

) ? (OA>

2 OB>

) 5 4(6) 1 5(23) 5 9

cos (u) 5(OA

>

1 OB>

) ? (OA>

2 OB>

)

@OA>

1 OB>

@ @OA>

2 OB>

@

OA>

2 OB>

5 (5 1 1, 1 2 4) 5 (6, 23)

OA>

1 OB>

5 (5 2 1, 1 1 4) 5 (4, 5)

OA>

2 OB>

.OA>

1 OB>

40

42

–2–4

2–4 –2

x

y

OB OA

u 8 18.52°

u 5 cos21 a

458

483b

@b>

@ 5 "(23)2 1 62 1 222 5 23

0 a>

0 5 "42 1 52 1 202 5 21

a>

? b>

5 4(23) 1 5(6) 1 20(22) 5 458

cos (u) 5a>

? b>

0 a>

0 @b>

@

a 5254

5

15a 5 2162

3a 1 12a 1 162 5 0

x>

3 y>

5 (3, a, 9) ? (a, 12, 18) 5 0

x>

,y>

a 5 6,

a>

? (a>

2 2b>

) 5 2(210) 2 1(27) 1 2(6) 5 21

5 (210, 27, 6)

a>

2 2b>

5 (2 2 12, 21 2 6, 2 1 4)

a>

? b>

5 2(6) 2 1(3) 1 2(22) 5 5

@a>

1 b>

@ 5 "82 1 22 1 02 5 2"17

5 (8, 2, 0)

a>

1 b>

5 (2 1 6, 21 1 3, 2 2 2)

@a>

2 b>

@ 5 "(24)2 1 (24)2 1 42 5 4"3

5 (24, 24, 4)

a>

2 b>

5 (2 2 6, 21 2 3, 2 1 2)

@b>

@ 5 "62 1 32 1 (22)2 5 7

0 a>

0 5 "22 1 (21)2 1 22 5 3

5 (24, 0, 24)

b>

3 c>

5 (0 2 4, 25 1 5, 24 2 0)

5 (2, 0, 2)

a>

3 b>

5 (2 2 0, 21 1 1, 0 1 2)

5 (25, 5, 212)

5 (22 2 3, 21 1 6, 23 2 9)

5 (2, 21, 23) 2 (3, 26, 9)

(p>

? r>

)q>

2 (q>

? r>

)p>

5 21(2, 1, 3) 2 3(1, 22, 3)

q>

? r>

5 2 1 1 1 0 5 3

p>

? r>

5 1 2 2 1 0 5 21

29 2 3 5 212

3a 1 3b 5 212

a 5 23a 5 25 1 2

b 5 21

5b 5 25

10 1 4b 1 b 5 5

22a 1 b 5 5

a 5 25 2 2ba 1 2b 5 25

a(1, 22, 3) 1 b(2, 1, 3) 5 (25, 5, 212)

5 (25, 5, 212)

(p>

3 q>

) 3 r>

5 (0 2 5, 5 1 0, 29 2 3)

5 (29, 3, 5)

p>

3 q>

5 (26 2 3, 6 2 3, 1 1 4)

BN 56

"26

"26 BN 5 6

@OB>

@ ? BN 5 6

AC>

5 @OB>

@ 5 "9 1 1 1 16 5 "26

@OB>

@


6.

The vertical components of the tensions must equalthe downward force:

The horizontal components:

Substitute in :

Substitute this back in to get :

7.

The resultant velocity is 304.14 km h,8. a.

b.

9.

The cross product is perpendicular to the givenvectors, but its magnitude is

or A unit vectorperpendicular to the given vectors is

10. a.

So is the largest angle.b. The area is half the magnitude of the crossproduct of and

11. The triangle formed by the two strings and theceiling is similar to a 3-4-5 right triangle, with the30 cm and 40 cm strings as legs. So the angleadjacent to the 30 cm leg satisfies

cos u 5 3

5

1

2 AB

>

3 AC>

51

2 0 (63, 231, 20) 0 8 36.50

AC>

.AB>

b 8 77.64°

g 5 180 2 a 2 b 8 36.68°

8 77.64°

b 5 cos21 16

7"114

@BC>

@ 5 "52 1 52 1 (28)2 5 "144

@BA>

@ 5 "22 1 62 1 32 5 7

BA>

? BC>

5 2(5) 1 6(5) 1 3(28) 5 16

BC>

5 (5 2 0, 2 1 3, 24 2 4, ) 5 (5, 5, 28)

BA>

5 2AB>

5 (2, 6, 3)

b 5 cos21 BA

>

? BC>

@BA>

@ @BC>

@

8 65.68°

5 cos21 33

7"131

a 5 cos21 AB

>

? AC>

@AB>

@ @AC>

@

@AC>

@ 5 "32 1 (21)2 1 (211)2 5 "131

@AB>

@ 5 "(22)2 1 (26)2 1 (23)2 5 7

AB>

? AC>

5 22(3) 2 6(21) 2 3(211) 5 33

AC>

5 (5, 2, 24) 2 (2, 3, 7) 5 (3, 21, 211)

AB>

5 (0, 23, 4) 2 (2, 3, 7) 5 (22, 26, 23)

cos (a) 5AB

>

? AC>

@AB>

@ @AC>

@

a9

!115, 2

5

!115, 2

3

!115b.

2"115."182 1 (210)2 1 (26)2,

5 (18, 210, 26)5 (3 1 15, 210 2 0, 0 2 6)

(0, 3, 25) 3 (2, 3, 1)

0 x>

3 y>

0 5 "502 1 202 1 182 5 "3224 8 56.78

5 (250, 220, 18)

x>

3 y>

5 (215 2 35, 25 2 15, 21 2 3)

y

z

x

x

y

W 9.46° N.>

tan21 a

50

300b 8 9.46°

x 5 "502 1 3002 8 304.14

x

300 km/h

50 km/h

T1 8 71.74NT1

T2 8 87.86N

2"6 2 "2

2T2 5 298"3

98"3 2"6

2T2 5 298"3

T1

"3

2T1 2

1

"2T2 5 0

T1 cos (30°) 1 T2 cos (45°) 5 0 N

T1 5 196 2 "2T2

1

2T1 1

1

"2T2 5 98

T1 sin (30°) 1 T2 sin (45°) 5 98 N

x

y

T2T1

98 N

45°30°


The angle adjacent to the 40 cm leg satisfies

cos

Also,

sin and sin

Let T1 be the tension in the 30 cm string, and T2 bethe tension in the 40 cm string. Then

Also,

So the tension in the 30 cm string is 78.4 N and thetension in the 40 cm string is 58.8 N.12. a.

b. The east- and west-pulling forces result in a forceof 5 N west. The north- and south-pulling forcesresult in a force of 12 N north. The 5 N west and 12 N north forces result in a force pulling in thenorth-westerly direction with a force of

by using the Pythagorean theorem. To find the exactdirection of this force, use the definition of sine.

If is the angle west of north, then

So the resultant is 13 N in a direction N22.6°W. The equilibrant is 13 N in a directionS22.6°E.13. a. Let D be the origin, then:

b.

c.By part b.:

14.

(cosine law)

15. a. The angle to the bank, , will satisfy

sin

b. By the Pythagorean theorem, Kayla’s netswimming speed will be

km h.So since distance rate time, it will take her

hmin 3 sec

to swim across. 8 8

8 0.13

t 50.3

"5

35

>"32 2 22 5 "5

u 8 48.2°

90° 2 u 8 41.8°

(90° 2 u) 5 2

3

u

5 7.5

5 213 cos (u) 1 1

5 213a>

? b>

1 1

5 213a>

? b>

1 6a>

? a>

2 5b>

? b>

(2a>

2 5b>

) ? (b>

1 3a>

)

cos (u) 5 21

2

a>

? b>

5 0 a>

0 @b>

@ cos (u) 5 cos (u)

8 3.58

5 (5) cos (44.31°)

Scalar projection 5 @AF>

@ cos (u)

u 8 44.31°

u 5 cos21 a16

10"5b

cos (u) 5AF

>

? AC>

@AF>

@ @AC>

@

@AC>

@ 5 "(22)2 1 42 1 02 5 2"5

@AF>

@ 5 "02 1 42 1 32 5 5

AF>

? AC>

5 0 1 16 1 0 5 16

AC>

5 (22, 4, 0)

AF>

5 (0, 4, 3)

H 5 (0, 0, 3)G 5 (0, 4, 3)

F 5 (2, 4, 3),E 5 (2, 0, 3),D 5 (0, 0, 0),

C 5 (0, 4, 0),B 5 (2, 4, 0),A 5 (2, 0, 0),

u 8 22.6°

sin u 5 5

13

u

"52 1 122 5 13 N,

25 N30 N

42 N

54 N

5 78.4 N

T1 54

3(58.8)

T2 5 58.8 N

5

3T2 5 98

a4

3T2b

4

51 T2

3

55 98

T1

4

52 T2

3

55 98

T1 sin u 1 T2 sin f 5 (10)(9.8) 5 98

T1 54

3T2

T1

3

52 T2

4

55 0

T1 cos u 2 T2 cos f 5 0

f 5 3

5.u 5

4

5

f 5 4

5


c. Such a situation would have resulted in a righttriangle where one of the legs is longer than thehypotenuse, which is impossible.16. a. The diagonals are and

b.

17. a. The z value is double, so if andthe vector will be collinear.

b. If and are perpendicular, then their dot product will equal 0.

c. Let and then we have a vector perpendicular to Now it must be divided by itsmagnitude to make it a unit vector:

So the unit vector is:

18. a.

b.

c. Scalar projection multiplied with the unit vectorin the direction of

d.

19. a. If the dot product is 0, then the vectors areperpendicular:

special

b.

not special

20. a.

b.

c.

d.

21. Since the angle between the two vectors is 60°,the angle formed when they are placed head-to-tailis So the resultant, along with these two vectors, forms an isosceles triangle with top angle

and two equal angles By the cosine law,the two equal forces satisfy

22.5 (2, 28, 210)

a>

3 b>

5 (2 2 0, 25 2 3, 0 2 10)

8 11.55 N

F 520

"3

F 2 5400

3

202 5 2F 2 2 2F 2 cos 120°

30°.120°

120°.

5 (21, 1, 3)

p>

3 q>

5 (22 1 1, 2 2 1, 21 1 4)

(p>

3 r>

) ? r>

5 0 1 2 2 2 5 0

5 (3, 2, 1)

p>

3 r>

5 (4 2 1, 0 1 2, 1 2 0)

5 (22, 2, 6)

(p>

2 q>

) 3 (p>

1 q>

) 5 (22 2 0, 0 1 2, 3 2 (23))

p>

1 q>

5 (3, 23, 2)

p>

2 q>

5 (21, 21, 0)

5 (21, 1, 3)

5 (22(1) 2 1(21), 1(2) 2 1(1), 1(21) 1 2(2))

p>

3 q>

5 0 1 0 1 21

!3 5 2

1

!3

a21

!3,

1

!3,

1

!3b ? (0, 0, 21)

a1

!2,

1

!2, 0b ? (0, 0, 21) 5 0 1 0 1 0 5 0

5 0

5 21

!61

1

!61 0

a1

!2,

1

!2, 0b ? a

21

!3,

1

!3,

1

!3b

(0, 0, 21) ? (0, 1, 0) 5 0 1 0 1 0 5 0

(1, 0, 0) ? (0, 1, 0) 5 0 1 0 1 0 5 0

(1, 0, 0) ? (0, 0, 21) 5 0 1 0 1 0 5 0

8 138.59°

u 5 cos21 a23

4b

m>

? k>

5 23

5 (0.1875)("3, 22, 23)

5 (0.75)("3, 22, 23)

4

5 (0.75)m>

0m>

0

m>

:

8 0.75

5 2"2 cos (74.62°)

Scalar projection 5 0 n>

0cos (u)

8 74.62°

u 5 cos21 a3

8"2b

cos (u) 5m>

? n>

0m>

0 0 n>

0

0 n>

0 5 "4 1 3 1 1 5 2"2

0m>

0 5 "3 1 4 1 9 5 4

m>

? n>

5 2"3 2 2"3 1 3 5 3

a1

!5, 0,

2

!5b

0 p>

0 5 "81 1 0 1 324 5 9"5

p>

.

b 5 0,a 5 9,

p>

? q>

5 2a 2 2b 2 18 5 0

q>

p>

q>

b 5 24,

a 5 4

8 84.36°

u 5 cos21 5 a6

14"19b

cos (u) 5OA

>

? OB>

@OA>

@ @OB>

@

@OB>

@ 5 "(26)2 1 62 1 (22)2 5 2"19

@OA>

@ 5 "32 1 22 1 (26)2 5 7

OA>

? OB>

5 3(26) 1 2(6) 2 6(22) 5 6

5 (9, 24, 24)

OA>

2 OB>

5 (3 1 6, 2 2 6, 26 1 2)

5 (23, 8, 28)

OA>

1 OB>

5 (3 2 6, 2 1 6, 26 2 2)

OA>

2 OB>

.

OA>

1 OB>


23. First we need to determine the dot product of and

By the distributive property:

24.Since the magnitude of the scalar projection is 4,the scalar projection itself has value 4 or If it is 4, we get

If it is we get

So the two possible values for m are 5 and 25.

26. a.b.c.

27.

a. Using properties of parallelograms, we know thatthe other angle is 130 (Angles must add up to 360 , opposite angles are congruent). Using the cosine law,

b. Using the cosine law,

c. is the vector in the opposite direction of, but with the same magnitude. So:

28. a. Scalar:

Vector:

b. Scalar:

Vector:

c. Scalar:

Vector:

29. a. If its magnitude is 1, it’s a unit vector:

not a unit vector

unit vector

unit vector

not a unit vector

b. is. When dotted with it equals 0.30.31. a.b. with the x-axis:

with the y-axis:

with the z-axis:

with the x-axis:

cos (a) 53

"11

b>

@b>

@ 5 "9 1 1 1 1 5 "11

cos (g) 521

"30

a>

cos (b) 55

"30

a>

cos (a) 52

"30

0 a>

0 5 "4 1 25 1 1 5 "30

a>

a>

? b>

5 6 2 5 2 1 5 0

25 ? sin (30°) ? 0.6 5 7.50 J

d>

,a>

@d>

@ 5 "1 1 1 1 1 2 1,

0 c>

0 5 "14 1 1

2 1 14 5 1,

@b>

@ 5 "13 1 1

3 1 13 5 1,

0 a>

0 5 "14 1 1

9 1 136 2 1

1

"2?

(k>

1 j>

)

@k>

1 j>

@5

1

2(k

>

1 j>

)

(i>

1 j>

) ? (k>

1 j>

)

@k>

1 j>

@5

1

"2

1aj>

@ j>

@b 5 j

>

(i>

1 j>

) ? (j>

)

@ j>

@5 1

1ai>

@ i>

@b 5 i

>

(i>

1 j>

) ? (i>

)

@ i>

@5 1

@e>

2 d>

@ 5 @d>

2 e> @ 8 3.84

d>

2 e>

e>

2 d>

@d>

2 e> @ 8 3.84

@d>

2 e> @ 2 5 32 1 52 2 2(3)(5) cos 50°

@d>

1 e> @ 8 7.30

@d>

1 e> @ 2 5 32 1 52 2 2(3)(5) cos 130°

°

°

50°130°d

e

8 111.1°

u 5 cos21 a218

50b

CF>

? OP>

5 29 1 16 2 25 5 218

@OP>

@ 5 "9 1 16 1 25 5 5"2

OP>

5 (3, 4, 5)

@CF>

@ 5 "9 1 16 1 25 5 5"2

CF>

5 (0, 4, 0) 2 (3, 0, 5) 5 (23, 4, 25)

F 5 (0, 4, 0)C 5 (3, 0, 5),

8 103.34°

u 5 cos21 a23

13b

0 a>

0 5 "144 1 9 1 16 5 13

a>

? j>

5 23

27.

m 5 27

2 1 2m 5 212

(1, m, 0) ?(2, 2, 1)

3 5 24

24,

m 5 5

2 1 2m 5 12

(1, m, 0) ?(2, 2, 1)

3 5 4

24.

0 (2, 2, 1) 0 5 "22 1 22 1 12 5 3

5 2141

5 4 1 15 2 10 2 150

5 x>

? x>

1 3x>

? y>

2 2x>

? y>

2 6y>

? y>

(x>

2 2y>

) ? (x>

1 3y>

)

5 5

5 (10) cos (60°)

x>

? y>

5 0 x>

0 0 y>

0cos u

y>

:x>


with the y-axis:

with the z-axis:

c.

32. Need to show that the magnitudes of the diagonals are equal to show that it is a rectangle.

33. a. Direction cosine for x-axis:

We know the identity

Since and we get

So there are two possibilities, depending uponwhether is acute or obtuse.b. If is acute, then

If Á is obtuse, then

34.

35.

36.

37.

But this means that angle , so that thistriangle is degenerate. For completeness, though,notice that and . Thismeans that point C sits at the midpoint of the linesegment joining A and B. So angle

and angle . Socos cos The area of triangle ABC is, of course, 0.

Chapter 7 Test, p. 4221. a. We use the diagram to calculate noting

and

1 1

1

2

1 1So, a

>

3 b>

5 (24, 21, 23)

z 5 21(1) 2 1(2) 5 23

21

y 5 1(2) 2 (21)(23) 5 21

23

x 5 1(23) 2 1(1) 5 24

b>

a>

b3 5 23.

b2 5 1,b1 5 2,a3 5 1a2 5 1,a1 5 21,

a>

3 b>

,

C 5 21.

B 5 1;

B 5 0°C 5 180°

AB>

5 2 AC>

BC>

5 2AC>

A 5 0°

5 1

510

10

cos A 5AB

>

? AC>

@AB>

@ @AC>

@

@BC>

@ 5 "1 1 0 1 4 5 "5

BC>

5 (21, 0, 22)

5 "1 1 0 1 4 5 "5@AC>

@@AC

>

@ 5 (1, 0, 2)

@AB>

@ 5 "4 1 0 1 16 5 2"5

AB>

5 (2, 0, 4)

5 0 a>

0 2 1 @b>

@ 2 2 2 0 a>

0 @b>

@ cos u

5 0 a>

0 2 1 @b>

@ 2 2 2a>

? b>

5 b>

? b>

2 a>

? b>

1 a>

? a>

2 a>

? b>

5 (b>

2 a>

) ? (b>

2 a>

)

0 c>

0 2 5 @b>

2 a>

@ 2c>

5 b>

2 a>

1

4@a>

1 b>

@ 2 21

4@a>

2 b>

@ 2 566

42

98

45 28

@a>

2 b>

@ 5 "1 1 81 1 16 5 "98

a>

2 b>

5 (1, 9, 24)

@a>

1 b>

@ 5 "1 1 1 1 64 5 "66

a>

1 b>

5 (21, 21, 28)

a>

? b>

5 0 2 20 1 12 5 28

m 5 25

21

2m 5

5

2

m 11

22

3

2m 2

6

25 0

ma>

? a>

1 a>

? b>

2 3ma>

? b>

2 3b>

? b>

5 0

(a>

2 3b>

) ? (ma>

1 b>

) 5 0

a>

? b>

5 0 a>

0 @b>

@ cos (u) 51

2

g 8 110.7°

cos g 51

2"2

g 8 69.3°

cos g 51

2"2

g

b 5 g

cos a 5"3

2

cos b 5 cos g 5 61

2"2

2 cos2 b 5 1 23

4

b 5 g,a 5 30g,

cos2 a 1 cos2 b 1 cos2 g 5 1.

cos (30°) 5"3

2

@2i>

1 9j>

2 6k>

@ 5 "1 1 81 1 36 5 "118

@3i>

1 3j>

1 10k>

@ 5 "9 1 9 1 100 5 "118

m>

1 ? m>

2 56

!3302

5

!3302

1

!3305 0

cos (g) 51

"11

b>

cos (b) 521

"11

b>


x

y

z

b. We use the diagram again:

1 1

2 5

1 1So,c.

d. We could use the diagram method again, or, wenote that for any vectors so letting we have from the lastequation. Since from the first twoparts of the problem,2. a. To find the scalar and vector projections of on we need to calculate and

So,The scalar projection of on is

and

the vector projection of on is

b. We find the direction cosines for

c. The area of the parallelogram is the magnitude ofthe cross product.

1 2

So, and thus,

So the area of the parallelogram formed by and is or 5.10 square units.3. We first draw a diagram documenting the situation:

In triangle DEF, we use the cosine law:

We now use the sine law to find

The equilibrant force is equal in magnitude andopposite in direction to the resultant force, so bothforces have a magnitude of 78.10 N. The resultantmakes an angle 33.7 to the 40 N force and 26.3 tothe 50 N force. The equilibrant makes an angle 146.3to the 40 N force and 153.7 to the 50 N force.°

°

°°

/EDF 8 33.7°

sin /EDF 8 0.5544

sin /EDF

508

sin 120°

78.10

sin /EDF

@EF>

@5

sin /DEF

@R>

@

/EDF:

@R>

@ 8 78.10

@R>

@ 5 "402 1 502 2 2(40)(50) cos (120°)

"26

b>

a>

5 "26

@a>

3 b>

@ 5 "32 1 42 1 12

a>

3 b>

5 (3, 4, 1)

2121

z 5 (1)(21) 2 (21)(2) 5 1

y 5 1(2) 2 (1)(22) 5 4

221

x 5 (21)(22) 2 1(21) 5 3

2121

b>

a>

g 8 131.8°.

cos (g) 5b3

@b>

@5

22

3

b 8 109.5°.

cos (b) 5b2

@b>

@5

21

3

a 8 48.2°.

cos (a) 5b1

@b>

@5

2

3

b>

:

aa>

? b>

@b>

@ 2bb

>

5 19(2, 21, 22).

b>

a>

a>

? b>

@b>

@5 1

3,

b>

a>

@b>

@ 5 3

5 3

@b>

@ 5 "22 1 (21)2 1 (22)2

5 1

5 (1)(2) 1 (21)(21) 1 (1)(22)

a>

? b>

5 (1, 21, 1) ? (2, 21, 22)

@b>

@ 5 "b>

? b>

a>

? b>

b>

,

a>

(a>

3 b>

) 3 (b>

3 c>

) 5 0.

a>

3 b>

5 b>

3 c>

x>

3 x>

5 0y>

5 x>

,

x>

3 y>

5 2 (y>

3 x>

),y>

,x>

,

5 0

1 (1)(23)

5 (21)(24) 1 (1)(21)

a>

? (b>

3 c>

) 5 (21, 1, 1) ? (24, 21, 23)

b>

3 c>

5 (24, 21, 23)

z 5 2(1) 2 1(5) 5 23

y 5 23(5) 2 (2)(27) 5 21

2 723

x 5 1(27) 2 (23)(1) 5 24

c>

b>


60°

40 N

50 N

60°

120°

40 N

50 N

R

D G

FE

x

y

z

x

y

z

4. We find the resultant velocity of the airplane.

Since the airplane’s velocity is perpendicular to thewind, the resultant’s magnitude is given by thePythagorean theorem:

The angle is determined using the tangent ratio:

Thus, the resultant velocity is 1004.99 km h,N 5.7 W (or W 84.3 N).5. a. The canoeist will travel 200 m across thestream, so the total time he will paddle is:

The current is flowing 1.2 m s downstream, so thedistance that the canoeist travels downstream is:

So, the canoeist will drift 96 m south.b. In order to arrive directly across stream, thecanoeist must take into account the change in hisvelocity caused by the current. That is, he must initially paddle upstream in a direction such that the resultant velocity is directed straight across the stream. The resultant velocity:

Since the resultant velocity is perpendicular to thecurrent, the direction in which the canoeist shouldhead is determined by the sine ratio.

The canoeist should head upstream.6. The area of the triangle is exactly:

0

1

3

0So, and

So,

The area of the triangle is 3.50 square units.

7.

The system is in equilibrium (i.e. it is not moving),so we know that the horizontal components of and are equal:

@T>

2 @ 5sin (45°)

sin (70°)@T>

1 @

@T>

1 @ sin (45°) 5 @T>

2 @ sin (70°)

T>

2

T>

1

25 kg

458708

T1 T2

ADABC 51

2@AB

>

3 BC>

@ 57

2.

5 7

5 "49

@AB>

3 BC>

@ 5 "(22)2 1 32 1 (26)2

AB>

3 BC>

5 (22, 3, 26)

22

z 5 (3)(0) 2 (22)(23) 5 26

23

y 5 (22)(23) 2 (3)(1) 5 3

22

x 5 (22)(1) 2 (22)(0) 5 22

22

BC>

AB>

5 (23, 0, 1)

BC>

5 (21, 1, 4) 2 (2, 1, 3)

5 (3, 22, 22)

AB>

5 (2, 1, 3) 2 (21, 3, 5)

ADABC 51

2@AB

>

3 BC>

@

28.7°

/EDF 8 28.7°

sin /EDF 51.2

2.5

1.2 m/s

E

FRD

2.5 m/s

d 5 96 m

d 5 (1.2 m>s)(80 s)

d 5 rcurrent 3 t

>

t 5 80 s

t 5200 m

2.5 m>s

t 5d

rcanoeist

°°

>

/EDF 8 5.7°

tan /EDF 5100

1000

@R>

@ 8 1004.99

@R>

@ 5 "10002 1 1002

R

DG

FEF

1000 km/h

Position diagram Vector diagram


x

y

z

Also, the vertical component of must equalthe gravitational force on the block:

Substituting in for , we find that:

So, we can now find

The direction of the tensions are indicated in thediagram.8. a. We explicitly calculate both sides of the equation. The left side is:

We perform a few computations before computingthe right side:

Thus, the right side is

So, the equation holds for these vectors.b. We now verify that the formula holds in general.We will compute the right side of the equation, butwe first perform some intermediary computations:

So, the right side of the equation is:

Thus, the equation holds for arbitrary vectors. 5 x

>

? y>

1

40 x>

1 y>

0 2 21

40 x>

2 y>

0 2 51

4(4(x

>

? y>

))

5 (x>

? x>

) 2 2(x>

? y>

) 1 (y>

? y>

)

1 (2y>

? 2y>

)

5 (x>

? x>

) 1 (x>

? 2y>

) 1 (2y>

? x>

)

0 x>

2 y>

0 2 5 (x>

2 y>

) ? (x>

2 y>

)

5 (x>

? x>

) 1 2(x>

? y>

) 1 (y>

? y>

)

5 (x>

? x>

) 1 (x>

? y>

) 1 (y>

? x>

) 1 (y>

? y>

)

0 x>

1 y>

0 2 5 (x>

1 y>

) ? (x>

1 y>

)

5 0

1

40 x>

1 y>

0 2 21

40 x>

2 y>

0 2 51

4(33) 2

1

4(33)

5 33

5 42 1 12 1 42

0 x>

2 y>

0 2 5 (x>

2 y>

) ? (x>

2 y>

)

5 (4, 1, 4)

x>

2 y>

5 (3, 3, 1) 2 (21, 2, 23)

5 33

5 22 1 52 1 (22)2

0 x>

1 y>

0 2 5 (x>

1 y>

) ? (x>

1 y>

)

5 (2, 5, 22)

x>

1 y>

5 (3, 3, 1) 1 (21, 2, 23)

5 0

5 (3)(21) 1 (3)(2) 1 (1)(23)

x>

? y>

5 (3, 3, 1) ? (21, 2, 23)

@T>

2 @ 8 191.1 N

@T>

2 @ 8sin (45°)

sin (70°)(254.0 N)

@T>

2 @ 5sin (45°)

sin (70°)@T>

1 @

@T>

1 @ 8 254.0 N

@T>

1 @ (0.9645) 8 245 N

@T>

1 @ acos 45° 1sin 45°

sin 70° cos 70°b 5 245 N

(9.8 m>s2)5 (25 kg)sin 45°

sin 70° cos 70°@T

>

1 @

@T>

1 @ cos 45° 1

T>

2

@T>

1 @ cos 45° 1 @T>

2 @ cos 70° 5 (25 kg)(9.8 m>s2)

T>

1 1 T>

2


Review of Prerequisite Skills, p. 3501. The velocity relative to the ground has a magnitude equivalent to the hypotenuse of a triangle with sides 800 and 100. So, by thePythagorean theorem we can find the magnitude ofthe velocity.

The velocity of the airplane relative to the ground isabout 806 km h E.2.

The angle between the two displacements is 135°.The magnitude, m, and the angle, of the displacement can be found using the cosine law.

So the displacement is 15.93 units, N.3.

4. a.

b.

8 16.91

5 !286

5 !81 1 9 1 196

5 "(29)2 1 32 1 142

l 5 magnitude

(29, 3, 14)

8 7.87

5 "62

5 "9 1 4 1 49

5 "32 1 (22)2 1 72

l 5 magnitude

(3, 22, 7)

yA(0, 1, 0)

z

x

B(–3, 2, 0)

C(–2, 0, 1)

D(0, 2, –3)

W 32.2°

8 32.2°

u 5 cos21 a134.76

159.3b

cos u 5134.76

159.3

2134.76 5 2159.3 cos u

144 5 253.76 1 25 2 159.3 cos u

122 5 15.932 1 52 2 2(15.93)(5) cos u

8 15.93 units

m 5 "253.85

5 253.85

5 169 1 84.85

5 25 1 144 2 120a2!2

2b

m2 5 52 1 122 2 2(5)(12) cos 135

u,

5

12

displacement

u

N 7.1°>

u 8 7.1°

u 5 tan21 a100

800b

tan u 5100

800

800

100

u

8 806 km>h

v 5 "650 000

5 650 000

5 640 000 1 10 000

v2 5 8002 1 1002

CHAPTER 7Applications of Vectors


c. (1, 1, 0)

d.

5. a.In the xy-plane at the point (x, y).b.In the xz-plane at the point (x, z).c.In the yz-plane at the point (y, z).6. a.

b.

c.

d.

7. a.

b.

c.

7.1 Vectors as Forces, pp. 362–3641. a. 10 N is a melon, 50 N is a chair, 100 N is acomputerb. Answers will vary.2. a.

b. 180°3.

The forces should be placed in a line along thesame direction.4. For three forces to be in equilibrium, they mustform a triangle, which is a planar figure.5.

a. The resultant is equivalent in magnitude to thehypotenuse, h, of the triangle with 5 and 12 as sides

and is directed northeast at an angle of

Thus, the resultant is at an angleof E. The equlibrant is equal inmagnitude and opposite in direction of the resultant.Thus, the equilibrant is 13 N at an angle of S 22.6° W.b. The resultant is at an angle of

W. The equilibrant, then, is 15 Nat N 36.9° E.6. For three forces to form equilibrium, they mustbe able to form a triangle or a balanced line, soa. Yes, since these can form a triangle.b. Yes, since these can form a triangle.

c. No, since these cannot form a triangle.d. Yes, since placing the 9 N and 10 Nforce in a line directly opposing the 19 N forceachieves equilibrium.

9 1 10 5 19,

"5 1 6 , 9

9 1 40 . 41

3 1 4 . 7

sin21 1215 5 S 36.9°

"92 1 122 5 15 N

sin21 1213 5 N 22.6°

"52 1 122 5 13 N

sin21 12

h.

resultant

f1

f2equilibrant

10 N20 N

10 N

20 N

30 N

5 12i>

1 j>

2 2k>

5 (6i>

1 4j>

2 2k>

) 1 (6i>

2 3j>

)

5 2(3i>

1 2j>

2 k>

) 2 3(22i>

1 j>

)

2a>

2 3b>

5 5i>

1 j>

2 k>

5 (3i>

1 2j>

2 k>

) 1 (2i>

2 j>

)

5 (3i>

1 2j>

2 k>

) 2 (22i>

1 j>

)

a>

2 b>

5 i>

1 3j>

2 k>

5 (3i>

1 2j>

2 k>

) 1 (22i>

1 j>

)

a>

1 b>

5 4i>

2 6j>

1 8k>

5 (22i>

1 3j>

2 4k>

) 1 (6i>

2 9j>

1 12k>

)

5 21

2(4i

>

2 6j>

1 8k>

) 13

2(4i

>

2 6j>

1 8k>

)

21

2(4, 26, 8) 1

3

2(4, 26, 8)

5 28i>

1 11j>

1 3k>

5 (22i>

1 2j>

1 6k>

) 1 (26i>

1 9j>

2 3k>

)

5 2(2i>

1 j>

1 3k>

) 1 3(22i>

1 3j>

2 k>

)

2(21, 1, 3) 1 3(22, 3, 21)

5 6i>

2 2j>

5 (4i>

2 j>

1 3k>

) 2 (22i>

1 j>

1 3k>

)

(4, 21, 3) 2 (22, 1, 3)

5 i>

2 7j>

5 (26i>

1 0j>

) 1 (7i>

2 7j>

)

5 (26i>

1 0j>

) 1 7(i>

2 j>

)

(26, 0) 1 7(1, 21)

C(0, y, z)

B(x, 0, z)

A(x, y, 0)

8 9.22

5 "85

5 "4 1 0 1 81

5 "22 1 02 1 (29)2

l 5 magnitude

(2, 0, 29)

8 1.41

5 "2

5 "12 1 12 1 02

l 5 magnitude


7. Arms 90 cm apart will yield a resultant with asmaller magnitude than at 30 cm apart. A resultantwith a smaller magnitude means less force to counter your weight, hence a harder chin-up.8. Using the cosine law, the resultant has a magnitude,r, of

Using the sine law, the resultant’s angle, can befound by

from the 6 N force toward the 8 Nforce. The equilibrant, then, would be 12.17 N at

from the 6 N force awayfrom the 8 N force.9.

We know that the components of and perpendicular to the 10 N force must be equal, so wecan write

Now we look at and We know

So Substituting then solving for yields

Now we solve for :

So the force from the 10 N force is 9.66 N andthe force perpendicular to it is 2.59 N.10. The force of the block is

98 N. The component of thisforce parallel to the ramp is

directed down theramp. So the force preventing this block from moving would be 49 N directed up the ramp.11. a.

b. Using the cosine law for the angle, we have

5 120

u 5 cos21 21

2

cos u 5256

112

56 5 2112 cos u

169 5 64 1 49 2 112 cos u

132 5 82 1 72 2 2(8)(7) cos u

u,

13 N

7 N

8 N

(98) sin 30° 5 (98)A12B 5 49 N,

(10 kg)(9.8 N>kg) 5

15°

@ f1

>

@ 5 2.59 N

@ f1

>

@ 5 (9.66)(0.268)

5 (9.66) cos 75

cos 15@ f1

>

@

5 @ f2

>

@ cos 75

cos 15@ f1

>

@

f1

>

@ f2

>

@ 5 9.66 N

@ f2

>

@ (1.035) 5 10

@ f2

>

@ acos 75

cos 15 sin 15 1 sin 75b 5 10

@ f2

>

@ sin 75 5 10@ f2

>

@ cos 75

cos 15 sin 15 1

f2

>

@ f2

>

@ sin 75 5 10@ f1

>

@ sin 15 1

x1 1 x2 5 10

x2 5 @ f2

>

@ sin 75

x1 5 @ f1

>

@ sin 15

x2.x1

5 @ f2

>

@ cos 75

cos 15@ f1

>

@

5 @ f2

>

@ cos 75@ f1

>

@ cos 15

f2

>

f1

>


>



>


f2

>

5 force perpendicular to f1

>

f1

>

5 force 15° from the 10 N force

10 N

f1

f2

180° 2 34.7° 5 145.3°

8 34.7°

u 5 sin21 8

"32

12.17

sin u 5 8

"32

12.17

sin u

85

sin 120°

12.17

u,

8 12.17 N

r 5 "148

5 148

5 36 1 64 1 48

5 62 1 82 2 2(6)(8)a21

2b

r2 5 @ f1

>

@ 2 1 @ f2

>

@ 2 2 2 @ f1

>

@ @ f2

>

@ cos 120°


This is the angle between the vectors when placedhead to tail. So the angle between the vectors whenplaced tail to tail is 12. The 10 N force and the 5 N force result in a 5 Nforce east. The 9 N force and the 14 N force resultin a 5 N force south. The resultant of these is nowequivalent to the hypotenuse of the right trianglewith 5 N as both bases and is directed 45° south of

east. So the resultant is 45° south of east.13.

a. Using the Pythagorean theorem,

b. The angle, between and the resultant is given by

So the angle between and the equilibrant is

14. a.

For these three equal forces to be in equilibrium, theymust form an equilateral triangle. Since the resultantwill lie along one of these lines, and since all angles

of an equilateral triangle are 60°, the resultant will beat a 60° angle with the other two vectors.b. Since the equilibrant is directed opposite theresultant, the angle between the equilibrant and theother two vectors is 15. Since and act opposite one another, theynet a 10 N force directed west. Since and actopposite one another, they net a 10 N force directed45° north of east. So using the cosine law to findthe resultant,

Since our net forces are equal at 10 N, the angle ofthe resultant is directed halfway between the two, orat from toward 16.

Let be the tension in the 30° rope and be thetension in the 45° rope.Since this system is in equilibrium, we know thatthe horizontal components of and are equaland opposite and the vertical components add to beopposite the action of the mass. Also, the forceproduced by the mass is So we have a system of two equations: the first,

represents the balance of the horizontal components, and the second,

represents thebalance of the vertical components with the mass. So solving this system of two equations with twovariable gives the desired tensions.

T2aacos 45°

cos 30°b sin 30° 1 sin 45°b 5 196

aT2 cos 45°

cos 30°b sin 30° 1 T2 sin 45° 5 196

T1 sin 30° 1 T2 sin 45° 5 196

T1 5 T2 cos 45°

cos 30°

T1 cos 30° 5 T2 cos 45°

(T1) sin 30° 1 (T2) sin 45° 5 196

(T1) cos 30° 5 (T2) cos 45°

(20 kg)(9.8 N>kg) 5 196 N.

T2T1

T2T1

20 kg

T 2 T 1

f3

>

.f2

>

12 (135°) 5 67.5°

8 7.65 N

@ fr

>

@ 5Å

200 2 200a"2

2b

5 200 2 200a"2

2b

5 200 2 200 cos 45°

@ fr

>

@ 2 5 102 1 102 2 2(10)(10) cos 45°

fr

>

,

f4

>

f3

>

f2

>

f1

>

180° 2 60° 5 120°.

1 N

1 N1 N

60°60°

60°

180° 2 16.3° 5 163.7°.

f1

>

8 16.3°

u 5 sin21 7

25

sin u 57

25

sin u 5@ f2

>

@

25

f1

>

u,

@ f2

>

@ 5 7

5 49

5 252 2 242

@ f2

>

@ 2 5 252 2 @ f1

>

@ 2 @ f1

>

@ 2 1 @ f2

>

@ 2 5 252

= 24 Nf1

f2

resultant = 25 Nequlibrant

"52 1 52 5 "50 8 7.1 N

180° 2 120° 5 60°.



Thus the tension in the 45° rope is 175.73 N and thetension in the 30° rope is 143.48 N.17.

First, use the Cosine Law to find the angles thestrings make at the point of suspension. Let be theangle made by the 32 cm string and be the anglemade by the 24 cm string.

A keen eye could have recognized this triangle as a 3-4-5 right triangle and simply used the Pythagoreantheorem as well. Now we set up the same system ofequations as in problem 16, with being the tensionin the 32 cm string and being the tension in the 24 cm string, and the force of the mass being

8 29.4 N

T1 5 (39.2)cos 53.1°

cos 36.9°

T2 8 39.2 N

T2(1.25) 5 49

T2aacos 53.1°

cos 36.9°b sin 36.9° 1 sin 53.1°b 5 49

aT2 cos 53.1°

cos 36.9°b sin 36.9° 1 T2 sin 53.1° 5 49

T1 sin 36.9° 1 T2 sin 53.1° 5 49

T1 5 T2 cos 53.1°

cos 36.9°

T1 cos 36.9° 5 T2 cos 53.1°

(5 kg)(9.8 N>kg) 5 49 N.

T2

T1

8 53.1°

u2 5 cos21 1152

1920

21152 5 21920 cos u2

322 5 242 1 402 2 2(24)(40) cos u2

8 36.9°

u1 5 cos21 2048

2560

22048 5 22560 cos u1

242 5 322 1 402 2 2(32)(40) cos u1

u2

u1

5 kg

24 cm 32 cm

40 cm

8 143.48 N

T1 5 (175.73)cos 45°

cos 30°

T2 8 175.73 N

T2(1.12) 5 196 Thus the tension in the 24 cm string is 39.2 N andthe tension in the 32 cm string is 29.4 N.18.

(Port means left and starboard means right.) We arelooking for the resultant of these two force vectorsthat are 35° apart. We don’t know the exact valueof the force, so we will call it x. So the small tug is pulling with a force of x and the large tug ispulling with a force of 2x. To find the magnitudeof the resultant, r, in terms of x, we use the cosinelaw.

Now we use the cosine law again to find the angle,made by the resultant.

from the large tug toward thesmall tug, for a net of 8.5° to the starboard side.

8 11.5°

u 5 cos21 a11.2768

11.5076b

cos u 511.2768

11.5076

211.2768x2 5 211.5076x2 cos u

x2 5 12.2768x2 2 11.5076x2 cos u

x2 5 8.2768x2 1 4x2 2 11.5076x2 cos u

x2 5 r2 1 (2x)2 2 2(2.8769x)(2x) cos u

u,

8 2.8769xr 8 "8.2768x2

8 8.2768x2

8 5x2 1 3.2768x2

8 5x2 2 4x2(20.8192)

5 x2 1 4x2 2 4x2 cos 145°

r2 5 x2 1 (2x)2 2 2(x)(2x) cos 145°

N

S

EW

xu

2x

resultant

35°

19.

a. First we will find the resultant of the 5 N and 8 N forces. Use the Pythagorean theorem to find themagnitude, m.

Next we use the Pythagorean theorem again to findthe magnitude, M, of the resultant of this net forceand the 10 N force.

Since the equilibrant is equal in magnitude to theresultant, we have the magnitude of the equilibrantequal to approximately 13.75 N.b. To find each angle, use the definition of cosinewith respect each force as a leg and the resultant asthe hypotenuse. Let be the angle from the 5 N force to the resultant, be the angle from the8 N force to the resultant, and be the anglefrom the 10 N force to the resultant.Let the sign of the resultant be negative, since it isin a direction away from the head of each of thegiven forces.

20. We know that the resultant of these two forces isequal in magnitude and angle to the diagonal line of the parallelogram formed with and as legs and has diagonal length We also knowfrom the cosine law that

where is the supplement to in our parallelogram.Since we know then

Thus we have

7.2 Velocity, pp. 367–3701. a. Both the woman and the train’s velocities arein the same direction, so we add them.

b. The woman’s velocity is directed opposite that oftrain, so we subtract her velocity from the train’s.

. The resultant is inthe same direction as the train’s movement.2. a. The velocity of the wind is directed opposite thatof the airplane, so we subtract the wind’s velocityfrom the airplane’s.

north.b. Both the wind and the airplane’s velocities are inthe same direction, so we add them.

north.3. We use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, made.

The resultant is 304.14 km h, S.W 9.5°> 8 9.5°.

u 5 tan21 50

300

tan u 550

300

8 304.14 km>h

m 5 "92 500

5 92 500

5 90 000 1 2500

m2 5 3002 1 502

u,

600 km>h 1 100 km>h 5 700 km>h

600 km>h 2 100 km>h 5 500 km>h

80 km>h 2 4 km>h 5 76 km>h

80 km>h 1 4 km>h 5 84 km>h

@ f1

>

1 f2

>

@ 5 " @ f1

>

@ 2 1 @ f2

>

@ 2 1 2 @ f1

>

@ @ f2

>

@ cos u

5 @ f1

>

@ 2 1 @ f2

>

@ 2 1 2 @ f1

>

@ @ f2

>

@ cos u

@ f1

>

1 f2

>

@ 2 5 @ f1

>

@ 2 1 @ f2

>

@ 2 2 2 @ f1

>

@ @ f2

>

@ cos f

cos f 5 cos (180 2 u) 5 2cos u.

f 5 180 2 u,

uf

@ f1

>

1 f2

>

@ 2 5 @ f1

>

@ 2 1 @ f2

>

@ 2 2 2 @ f1

>

@ @ f2

>

@ cos f

@ f1

>

1 f2

>

@ .

f2

>

f1

>

8 136.7°

u10N 5 cos21 a10

213.75b

cos u10N 510

213.75

8 125.6°

u8N 5 cos21 a8

213.75b

cos u8N 58

213.75

8 111.3°

u5N 5 cos21 a5

213.75b

cos u5N 55

213.75

u10N

u8N

u5N

M 5 "189 8 13.75

5 189

5 89 1 100

M2 5 m2 1 102

m 5 "89 8 9.4

5 89

5 25 1 64

m2 5 52 1 82

5N

8N 10N



4. Adam must swim at an angle, upstream so asto counter the 1 km h velocity of the stream. This isequivalent to Adam swimming along the hypotenuseof a right traingle with 1 km h leg and a 2 km hhypotenuse. So the angle is found using the definitionof cosine.

upstream5. a. 2 m s forwardb. 20 m s 2 m s 22 m s in the direction of the car6. Since the two velocities are at right angles wecan use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, made.

from the direction of the boat towardthe direction of the current. This results in a net of

, or 7. a. First we find the components of the resultantdirected north and directed west. The componentdirected north is the velocity of the airplane, 800,

since the wind forms a 45°angle south of west. The western component of the resultant is simply So we use thePythagorean theorem to find the magnitude, m, ofthe resultant and the definition of sine to find theangle, of the resultant.

Use the sine law to determine the direction.

The direction is

b. The airplane is travelling at approximately 732.71 km h, so in 1 hour the airplane will travelabout 732.71 km.8. a. First we find the velocity of the airplane. Weuse the Pythagorean theorem to find the magnitude,m, of the resultant.

So in 3 hours, the airplane will travel about

b. To find the angle, the airplane travels, we usethe definition of sine.

east of north.9. a. To find the angle, at which to fly is theequivalent of the angle of a right triangle with 44 asthe opposite leg and 244 as the hypotenuse. So weuse the definition of sine to find this angle.

south of west.b. By the Pythagorean Theorem, the resultant groundspeed of the airplane is Since the duration of theflight is simply 10. a. Since Judy is swimming perpendicular to the flow of the river, her resultant velocity is simplythe hypotenuse of a right triangle with 3 and 4 asbases, which is a 3-4-5 right triangle. Thus, Judy’s resultant velocity is 5 km h. The direction is determined by downstreamb. Judy’s distance traveled down the river would be the“4” leg of the 3-4-5 triangle formed by the vectors, butscaled down so that 1m (the width of the river) isequivalent to the “3” leg. So her distance traveled is

km. This makes her about 0.67 km fromHelen’s cottage.c. While in the river, Judy is swimming at 5 km h for a distance of Since

her time taken is 53 km

5 km>h5 1

3 hours 5 20 minutes.


53 km.>

43 8 1.33

u 8 53.1°tan u 5 43.

>

(480 km)> (240 km>h) 5 2 h.


"(2442 2 442) 5 240 km>h.

8 10.4°

u 5 sin21 44

244

sin u 544

244

u,

8 12.5°

u 5 sin21 100

461

sin u 5100

461

u,

(461 km>h)(3 h) 5 1383 km.

8 461 km>h

m 5 "212 500

5 212 500

5 202 500 1 10 000

m2 5 4502 1 1002

>

N 5.5° W.

u 8 5.5°

sin u

1005

sin 45°

732.71

732.71 km>h m 8

8 536 863.8082

8 (729.29)2 1 (71.71)2

m2 5 (800 2 100 sin 45°)2 1 (100 cos 45°)2

u,

100 cos 45°.

minus 100 sin 45°,

N 37.6° W.22.6° 1 15° 5 37.6°

8 22.6°

u 5 sin21 5

13

sin u 55

13

5 13 m>s

m 5 "169

5 169

5 144 1 25

m2 5 122 1 52

u,

>5>1>

>

5 60°

u 5 cos21 1

2

cos u 51

2

>>

>

u,

11.

a. and b. Here, 205 km h directed 30° north of eastis the resultant of 212 km h directed east, and thewind speed, v, directed at some angle. This problemis more easily approached finding the wind speed,v, first. So we will do that using the cosine law.

Now to find the wind’s direction, we simply find theangle supplementary to the lesser angle, formed by the parallelogram of these three velocities. Wecan use the sine law for this.

Thus, the direction of v is the angle supplementaryto in the parallelogram:

west of north.12.

Since her swimming speed is a maximum of 4 km h,this is her maximum resultant magnitude, which isalso the hypotenuse of the triangle formed by her andthe river’s velocity vector. Since one of these legs is 5 km h, we have a triangle with a leg larger than itshypotenuse, which is impossible.13. a. First we need to find Mary’s resultant velocity, v. Since this resultant is the diagonal of theparallelogram formed by hers and the river’s velocity, we can use the cosine law with the angle,of the parallelogram adjacent 30°.

So in 10 seconds, Mary travels about

b. Since Mary is travelling at 3 m s at an angle of 30°,to find the component of her velocity, v, perpendicularto the current, we use the definition of sine.

perpendicular to the current.So since the time taken is

14. a. So we have a 5.5 m s vector and a 4 m s vector with a resultant vector that is directed 45°south of west. Letting be the angle between the 4 km h vector and the resultant, we can construct a parallelogram using these three vectors and a subsequent triangle with opposite the 5.5 m s vector and 45° opposite the 4 m s vector. We nowuse the sine law to find

from the resultant.Since the resultant is 45° west of south, Dave’sdirection is west of south,which is equivalent to about upstream.b. First, we find the magnitude, m, of Dave’s 4 m svelocity in the direction perpendicular to the river. This is done using the definition of sine.

perpendicular to the river.Since time is we have

15. Let b represent the speed of the steamboat and crepresent the speed of the current. On the waydownstream, the effective speed is andupstream is The distance upstream anddownstream is the same, so So, This means that the speed of the boat is 6 times the speed of the current. So, (6c 1 c) ? 5

b 5 6c.

5(b 1 c) 5 7(b 2 c).

b 2 c.

b 1 c,

(200 m)> (3.41 m>s) 8 58.6 s.

distance>rate,

8 3.41 m>s

m 5 4 sin 58.5°

>

180° 2 121.5° 5 58.5°

76.5° 1 45° 5 121.5°

8 76.5°

u 5 sin21 a5.5asin 45°

4bb

sin u 5 5.5asin 45°

4b

sin u

5.55

sin 45°

4

u.

>

>u

>

u

>>

(150 m)> (1.5 m>s) 5 100 s.


5 1.5 m>s

5 3a1

2b

v 5 3 sin 30

>

(6.8 m>s)(10 s) 5 68 m.

8 6.8 m>s

v 5 "45.8

5 45.8

5 25 1 20.8

5 9 1 16 2 24 cos 150°

v2 5 32 1 42 2 2(3)(4) cos 150°

u,

>

>

5

4

180° 2 71.6° 5 108.4° 5 18.4°

u

8 71.6°

u 5 sin21 a205asin 30°

108bb

sin u 5 205asin 30°

108b

sin u

2055

sin 30°

108

u,

8 108 km>h

v 5 "11 694

5 11 694

5 86 969 2 75 275

5 42 025 1 44 944 2 86 920 cos 30°

v2 5 2052 1 2122 2 2(205)(212) cos 30°

>

>

212 km/h

30°v

205 km/h

u



or 35c is the distance. This means that it would takea raft 35 hours moving with the speed of the currentto get from A to B.

7.3 The Dot Product of Two GeometricVectors, pp. 377–378

1. This means or or To be guaranteed that thetwo vectors are perpendicular, the vectors must benonzero.2. is a scalar, and a dot product is only definedfor vectors, so is meaningless.

3. Answers may vary. Let but

4. because 5. Since and are unit vectors, andsince they are pointing in opposite directions then

so Therefore 6. a.

b.

c.

d.

e.

f.

7. a.

b.

c.

d.

e.

f.

8.

Note: is the angle between the two vectors whenthey are tail to tail, so 9. a.

b.

10. so the dot product of any vector with is 0.0

>

@0>

@ 5 0

5 6 0 x>

0 2 2 19x>

? y>

1 3 0 y>

0 25 6 0 x

>

0 2 2 9x>

? y>

2 x>

? y>

2 9x>

? y>

1 3 0 y>

0 21 3 0 y

>

0 21 3 0 x

>

0 2 2 x>

? y>

2 (23y>

? 23x>

)5 3 0 x>

0 2 2 3x>

? 3y>

3x>

? (x>

2 3y>

) 2 (x>

2 3y>

) ? (23x>

1 y>

)

1 7a>

? b>

5 2 0 a>

0 2 2 15 @b>

@ 22 3a

>

? b>

1 10a>

? b>

5 2 0 a>

0 2 2 15 @b>

@ 21 5b

>

? 2a>

2 5b>

? 3b>

(a>

1 5b>

) ? (2a>

2 3b>

) 5 a>

? 2a>

2 a>

? 3b>

u 2 120°.

u

5 22.5

5 (45)a1

2b

5 (7.5)(6) cos (180° 2 120°)

a>

? b>

5 0 a>

0 @b>

@ cos u

u 5 120°

21

25 cos u

250 5 (10)(10) cos u

u>

? v>

5 0 u>

0 0 v>

0cos u

u 5 60°

1

25 cos u

10.5 5 (7)(3) cos u

a>

? b>

5 0 a>

0 @b>

@ cos u

u 8 127°

23

55 cos u

23 5 (5)(1) cos u

p>

? q>

5 0 p>

0 0 q>

0cos u

u 8 53°

3

55 cos u

3 5 (5)(1) cos u

p>

? q>

5 0 p>

0 0 q>

0cos u

u 8 80°

1

65 cos u

(6) 5 (6)(6) cos u

m>

? n>

5 0m>

0 0 n>

0cos u

u 5 30°

"3

25 cos u

12"3 5 (8)(3) cos u

x>

? y>

5 0 x>

0 0 y>

0cos u

8 226.2

5 (4)(8) cos 145°

u>

? v>

5 0 u>

0 0 v>

0cos u

5 0

5 (10)(0)

5 (2)(5) cos (90°)

m>

? n>

5 0m>

0 0 n>

0cos u

5 21

5 (1)(21)

5 (1)(1) cos (180°)

p>

? q>

5 0 p>

0 0 q>

0cos u

5 0

5 (0)(8) cos (100°)

a>

? b>

5 0 a>

0 @b>

@ cos u

8 26.93

5 (8)a2"3

2b

5 (2)(4) cos (150°)

x>

? y>

5 0 x>

0 0 y>

0cos u

5 16

5 (32)(.5)

5 (4)(8) cos (60°)

p>

? q>

5 0 p>

0 0 q>

0cos u

a>

? b>

5 21.cos u 5 21.u 5 180°

0 a>

0 5 @b>

@ 5 1b>

a>

c>

5 a>

a>

? b>

5 b>

? a>

5 b>

? c>

a>

5 2c>

.b>

? c>

5 0,a>

? b>

5 0,

c>

5 2i>

.b>

5 j>

,a>

5 i>

,

(a>

? b>

) ? c>

a>

? b>

cos u 5 0.@b>

@ 5 0,

0 a>

0 5 0,a>

? b>

5 0 a>

0 @b>

@ cos u 5 0.

11.

12. a.

b.

13. a.

b.

Therefore This is just what the Pythagorean theorem says,where and are the legs of the right triangle.14.

15.

16.

17.

18.

because is a scalar

because

7.4 The Dot Product for AlgebraicVectors, pp. 385–387

1.

Any vector of the form (c, c) is perpendicular to Therefore there are infinitely many vectors perpendicular to Answers may vary. For example:(1, 1), (2, 2), (3, 3).2. a.

b.

is acutec.

is obtuse3. Any vector in the xy-plane is of the form

Let

Therefore (0, 0, 1) is perpendicular to every vectorin the xy-plane.Any vector in the xz-plane is of the form

Let

Therefore (0, 1, 0) is perpendicular to every vectorin the xz-plane.

5 0

c>

? d>

5 (0)(c1) 1 (0)(1) 1 (0)(c3)

d>

5 (0, 1, 0).c>

5 (c1, 0, c3).

5 0

a>

? b>

5 (0)(a1) 1 (0)(a2) 1 (0)(1)

b>

5 (0, 0, 1).a>

5 (a1, a2, 0).

u

cos u , 0

5 23 , 0

5 3 1 4 2 10

a>

? b>

5 (1)(3) 1 (22)(22) 1 (5)(22)

u

cos u . 0

5 34 . 0

5 8 1 9 1 17

a>

? b>

5 (2)(4) 1 (3)(3) 1 (21)(217)

u 5 90°

5 0

a>

? b>

5 (22)(1) 1 (1)(2)

a>

.

a>

.

b2 5 b1

(21)b1 1 b2 5 0

a>

? b>

5 0

d>

? a>

5 0

c>

? a>

5 d>

? a>

1 c>

? a>

0 a>

0 5 1c>

? a>

5 (d>

1 c>

) ? a>

c>

? a>

5 (b>

? a>

) 0 a>

0 2b>

? a>

c>

? a>

5 (b>

? a>

)(a>

? a>

)

c>

? a>

5 ((b>

? a>

) a>

) ? a>

b>

5 d>

1 c>

d>

5 b>

2 c>

a>

? b>

1 a>

? c>

1 b>

? c>

5 27

2(a>

? b>

1 a>

? c>

1 b>

? c>

) 5 214

1 1 4 1 9 1 2(a>

? b>

1 a>

? c>

1 b>

? c>

) 5 0

1 c>

? a>

1 c>

? b>

1 0 c>

0 2 5 0

0 a>

0 2 1 a>

? b>

1 a>

? c>

1 b>

? a>

1 @b>

@ 2 1 b>

? c>

1 c>

? (a>

1 b>

1 c>

) 5 0

a>

? (a>

1 b>

1 c>

) 1 b>

? (a>

1 b>

1 c>

)

a>

1 b>

1 c>

5 0>

5 3

5 2 1 2a1

2b 1

1

22

1

2

1 @b>

@ 0 c>

0cos (120°)

2 0 a>

0 @b>

@ cos (60°) 1 0 a>

0 0 c>

0cos (60°) 1 15 1 1

5 0 a>

0 2 1 a>

? b>

1 a>

? c>

1 b>

? a>

1 @b>

@ 2 1 b>

? c>

(a>

1 b>

) ? (a>

1 b>

1 c>

)

5 2 0 u>

0 2 1 2 0 v>

0 25 0 u

>

0 2 1 2u>

? v>

1 0 v>

0 2 1 0 u>

0 2 2 2u>

? v>

1 0 v>

0 25 (u

>

1 v>

) ? (u>

1 v>

) 1 (u>

2 v>

) ? (u>

2 v>

)

0 u>

1 v>

0 2 1 0 u>

2 v>

0 25 14

5 (1)2 1 (2)2 1 (3)2

1 2 0 u>

0 0w>

0cos (90°) 1 2 0 v>

0 0w>

0cos (90°)

5 0 u>

0 2 1 0 v>

0 2 1 0w>

0 2 1 2 0 u>

0 0 v>

0cos (90°)

1 v>

? w>

1 w>

? u>

1 w>

? v>

1 w>

? w>

5 u>

? u>

1 u>

? v>

1 u>

? w>

1 v>

? u>

1 v>

? v>

(u>

1 v>

1 w>

) ? (u>

1 v>

1 w>

)

c>

b>

0 a>

0 2 5 @b>

@ 2 1 0 c>

0 2.

b>

? c>

5 @b>

@ 0 c>

0cos (90°) 5 0

5 @b>

@ 2 1 2b>

? c>

1 0 c>

0 25 (b

>

1 c>

) ? (b>

1 c>

)

0 a>

0 2 5 a>

? a>

5 0 a>

0 2 2 @b>

@ 25 0 a

>

0 2 2 a>

? b>

1 a>

? b>

2 @b>

@ 22 b

>

? b>

(a>

1 b>

) ? (a>

2 b>

) 5 a>

? a>

2 a>

? b>

1 b>

? a>

5 0 a>

0 2 1 2a>

? b>

1 @b>

@ 21 @b

>

@ 25 0 a

>

0 2 1 a>

? b>

1 a>

? b>

1 b>

? a>

1 b>

? b>

(a>

1 b>

) ? (a>

1 b>

) 5 a>

? a>

1 a>

? b>

5 1

a>

? b>

51

6( 0 a

>

0 2 1 5 @b>

@ 2)

0 a>

0 2 1 5 @b>

@ 2 5 6a>

? b>

0 a>

0 2 2 a>

? b>

2 5b>

? a>

1 5 @b>

@ 2 5 0

cos (90°)(a>

2 5b>

) ? (a>

2 b>

) 5 @a>

2 5b>

@ @a>

2 b>

@


Any vector in the yz-plane is of the formLet .

Therefore (1, 0, 0) is perpendicular to every vectorin the yz-plane.4. a.

b. If any of the vectors were collinear then onewould be a scalar multiple of the other. Comparingthe signs of the individual components of each vector eliminates and All ofthe components of have the samesign and the same is true for (4, 3, 10), but (4, 3, 10)is not a scalar multiple of Thereforenone of the vectors are collinear.5. a. Using the strategy of Example 5 yields

and and

Therefore the only result is or (0, 0).This is because and (1, 1) both lie on thexy-plane and are not collinear, so any vector that isperpendicular to both vectors must be in whichdoes not exist in b. If we select any two vectors that are not collinearin then any vector that is perpendicular to bothcannot be in and must be in This is not possible since does not exist in

6. a.

b.

c.

d.

7. a.

b.

8. a.

b.

2

(0, 1)

(1, 0)0

21

–1–2

1–2 –1

x

y

2

(0, 1)

(1, 0)0

21

–1–2

1–2 –1

x

y k $ 0

k 5 0 k 0

k 5 "2 0 k 01

"2

(1)(0) 1 (1)(k) 5 "1 1 1"k2 cos (45°)

a>

? b>

5 0 a>

0 @b>

@ cos u

k 52

3

3k 5 2

6k 2 2 2 3k 5 0

cos (90°)

5 0 a>

0 @b>

@(21)(26k) 1 (2)(21) 1 (23)(k)

a>

? b>

5 0 a>

0 @b>

@ cos u

u 8 154°

5282

91

5282

(7)(13)

5(2)(25) 1 (3)(0) 1 (26)(12)

"4 1 9 1 36"25 1 144

cos u 5a>

? b>

0 a>

0 @b>

@

u 8 64°

54

9

54

(3)(3)

5(2)(2) 1 (2)(1) 1 (1)(22)

"4 1 4 1 1"4 1 1 1 4

cos u 5a>

? b>

0 a>

0 @b>

@

u 8 123°

5214

"680

5(21)(6) 1 (4)(22)

"1 1 16"36 1 4

cos u 5a>

? b>

0 a>

0 @b>

@

u 8 148°

5211

"170

5211

"(34)(5)

5(5)(21) 1 (3)(22)

"25 1 9"1 1 4

cos u 5a>

? b>

0 a>

0 @b>

@

R2.R3

R3.R2

R2,

R2.

R3

(1, 22)

x 5 y 5 0,

3y 5 0

x 1 y 5 0x 2 2y 5 0

(x, y) ? (1, 1) 5 0(x, y) ? (1, 22) 5 0

(24, 25, 26).

(24, 25, 26)(5, 23, 25

6).(1, 2, 21)

5 0

(24, 25, 26) ? a5, 23, 25

6b 5 220 1 15 1 5

5 0

(1, 2, 21) ? (4, 3, 10) 5 4 1 6 2 10

5 0

e>

? f>

5 (1)(0) 1 (0)(e2) 1 (0)(e3)

f>

5 (1, 0, 0)e>

5 (0, e2, e3).


The diagonals are andor

and.

c.

or

9. a.

b.

10. a. i.

ii. Answers may vary. For example:

Let

b. In part a., the values are unique because bothvectors have their third component specified, andthe ratios must be the same for each component In part b. the values are not unique; any value of p could have been chosen, each resulting in adifferent value of q.

11.

12. a.

b.

13. a. Answers may vary. For example:


b. Answers may vary. For example:

(x, y, z) ? (21, 22, 3) 5 0

x 5 4z 2 3y x 1 3y 2 4z 5 0

(x, y, z) ? (1, 3, 24) 5 0

(1, 25, 2).

(21, 3, 0)

y 5 1.

y 5 z 22y 1 2z 5 0

x 2 5y 1 2z 5 0

(x, y, z) ? (1, 25, 2) 5 0

x 5 3y 2x 1 3y 5 0

(x, y, z) ? (21, 3, 0) 5 0

u 8 50°

58

905 cos u

49 2 16 1 25 5 90 cos u

3 "49 1 16 1 25 cos u

5 "49 1 16 1 25(27, 4, 5) ? (27, 24, 5)

AE>

? BF>

5 @AE>

@ @BF>

@ cos u

F 5 (0, 0, 5)

E 5 (0, 4, 5),D 5 (7, 0, 5),C 5 (0, 4, 0),

B 5 (7, 4, 0),A 5 (7, 0, 0),O 5 (0, 0, 0),

uC 8 63.4°

uC 5 180° 2 uA 2 uB

uB 8 26.6°

180° 2 uB 8 153.4°

5 2Å

4

5

5240

"(40)(50)

5210 2 30

"4 1 36"25 1 25

cos (180° 2 uB) 5AB

>

? BC>

@AB>

@ @BC>

@

uA 5 90°

180° 2 uA 5 90°

5 0

56 2 6

@AB>

@ @CA>

@

cos (180° 2 uA) 5AB

>

? CA>

@AB>

@ @CA>

@

CA>

5 (3, 21)BC>

5 (25, 25),AB>

5 (2, 6),

b>

.

q 5 250

p 5 1

q 5 22p 2 48

2q 1 4p 1 96 5 0

a>

? b>

5 0

q 5 3

2 52

3q

p 58

3

p 5 4a2

3b

k 52

3

8 5 12ka>

5 kb>

u 5 30°

5"3

2

53"2

"8"3

"2 2 1 1 "2 1 1 1 "2

"(2 2 2"2 1 1) 1 (2 1 2"2 1 1) 1 2"1 1 1 1 15

cos u 5a>

? b>

0 a>

0 @b>

@

u 5 90°

5 0

5(1 2 "2)(1) 1 ("2 2 1)(1)

0 a>

0 @b>

@

cos u 5a>

? b>

0 a>

0 @b>

@

5 0

5 21 1 1

(1, 1) ? (21, 1)

5 0

5 1 2 1

(1, 1) ? (1, 21)

(0, 1) 2 (1, 0) 5 (21, 0)

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

(1, 0) 2 (0, 1) 5 (1, 21)

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



14.

15. a.

b.

16. Answers may vary. For example: Note that, so they are collinear. Therefore any

vector that is perpendicular to is alsoperpendicular to



17.

or

18. a.

Therefore, since the two diagonals are perpendicular,all the sides must be the same length.

b.

c.

19. a.

The coordinates of vertex C are

b.

20. The two vectors representing the body diagonalsare and

Mid-Chapter Review, pp. 388–3891. a.

5 3

5 (6)1

2

a>

? b>

5 (3)(2) cos (60°)

a 8 109.5°

a 5 180° 2 u

u 8 70.5°

1

35 cos u

1 2 1 1 1 5 3 cos u

(21, 1, 1) ? (21, 21, 1) 5 "3"3 cos u

(0 2 1, 0 2 1, 1 2 0) 5 (21, 21, 1)

(0 2 1, 1 2 0, 1 2 0) 5 (21, 1, 1)

u 8 87.4°

16

3545 cos u

7 1 128 2 119 5 354 cos u

3 "1 1 64 1 289 cos u

(7, 16, 27) ? (1, 8, 17) 5 "49 1 256 1 49

AC>

? BD>

5 @AC>

@ @BD>

@ cos u

(6, 18, 24).

z 5 24y 5 18,x 5 6,

(24, 212, 25) 5 (2 2 x, 6 2 y, 29 2 z)

DA>

5 CB>

q 5 14

21 2 q 1 15 5 0

212 1 8 2 4q 1 60 5 0

AB>

? DA>

5 0

DA>

5 (24, 2 2 q, 25)AB>

5 (3, 4, 212),

u2 5 120°

2u1 1 2u2 5 360°

u1 5 60°

1

25 cos u1

2 1 2 2 1 5 6 cos u1

AB>

? BC>

5 @AB>

@ @BC>

@ cos u1

@AB>

@ 5 @BC>

@ 5 "6

5 (2, 1, 1)

BC>

51

2(a>

2 b>

)

5 (1, 2, 21)

AB>

51

2(a>

1 b>

)

5 0

a>

? b>

5 23 1 3

244

65p 5 4

p 5216 6 "(2216)2 2 4(65)(2176)

2(65)

65p2 2 216p 2 176 5 0

9p2 2 24p 1 16 5320

91

16

9p2

9p2 2 24p 1 16 5 49(20 1 p2)a4

21b

2

(3p 2 4)2 5 a7"20 1 p2 cos ub2

8 1 3p 2 12 5 "20 1 p2(7) cos u

5 "16 1 p2 1 4"4 1 9 1 36 cos u

(24, p, 22) ? (22, 3, 6)

x>

? y>

5 0 x>

0 0 y>

0cos u

(22, 24, 2).

(1, 2 2 1)

x 5 y 5 1.

(22, 24, 2).

(1, 2, 21)

x 5 z 5 1.

x 1 2y 2 z 5 0

(x, y, z) ? (1, 2, 21) 5 0

r>

.

s>

s>

5 22r>

p 5 0

3 1 4p 2 3 5 0

3 1 4p 1 q 5 0

23 2 4p 2 q 5 0

(23, p, 21) ? (1, 24, q) 5 0

p 5 3 or 21

p 5 1 6 2

p 52 6 "22 2 4(23)

2

p2 2 2p 2 3 5 0

(p, p, 1) ? (p, 22, 23) 5 0

(21, 22, 3).

(1, 3, 24)

y 5 1.

y 5 z 3y 2 4z 2 2y 1 3z 5 0

2x 2 2y 1 3z 5 0


b.

2.

Let be the tension in the 15 cm cord and bethe tension in the 20 cm cord. Let be the anglethe 15 cm cord makes with the ceiling and be theangle the 20 cm cord makes with the ceiling. By thecosine law:

Horizontal Components:

Vertical Components:

Therefore the tension in the 15 cm cord is 117.60 Nand the tension in the 20 cm cord is 88.20 N.3. The diagonals of a square are perpendicular, sothe dot product is 0.4. a.

By the cosine law:

By the cosine law:

The resultant velocity of the airplane is at S7.06°Eb.

hours5. a.

b.

6.

7. a.

b.

8. a.

b.

c.

d.

e.

5 34

5 15 1 4 1 15

? (5j>

2 4j>

1 3k>

)

1 5k>

)(a>

1 b>

) ? (b>

1 c>

) 5 (3i>

2 j>

5 0

5 5 2 8 1 3

a>

? (b>

1 c>

) 5 (i>

1 2j>

1 k>

) ? (5i>

2 4j>

1 3k>

)

5 5i>

2 4j>

1 3k>

b>

1 c>

5 (2i>

2 3j>

1 4k>

) 1 (3i>

2 j>

2 k>

)

5 55 6 1 3 2 4

b>

? c>

5 (2i>

2 3j>

1 4k>

) ? (3i>

2 j>

2 k>

)

5 05 2 2 6 1 4

a>

? b>

5 (i>

1 2j>

1 k>

) ? (2i>

2 3j>

1 4k>

)

cos (u) 534

63

34 5 "16 1 25 1 400 "1 1 4 1 4 cos (u)

a>

? b>

5 0 a>

0 @b>

@ cos (u)

5 34

a>

? b>

5 (4)(1) 1 (25)(2) 1 (20)(2)

5 4.5

5 (3)(3)(0.5)

@b>

@ cos (60°)a>

? b>

5 0 a>

0

u 5 60°

6u 5 360°

@F>

@ 8 94.49 N

@F>

@ 5 @E>

@ sin (40°)

@E'

>

@ 8 112.61 N

@E'

>

@ 5 (9.8)(15)cos (40°)

@E'

>

@ 5 @E>

@ cos (40°)

E F

t 8 1.74

t 81000

575.1?

km

(km/h)

(distance) 5 (rate)(time)

575.1 km>h

a 8 7.06°

sin (a) 8 0.123

sin (a)

1005

sin (135°)

575.1

0 v>

1 w>

0 8 575.1

2 2(500)(100) cos (135°)

0 v>

1 w>

0 2 5 (500)2 1 (100)2

0w>

0 5 1000 v>

0 5 500,

135°

w

v v w+

a

T2 5 88.2 N

T2 5 (0.75)T1

T1 5 117.6 N

(1.25)T1 5 147

(0.8)T1 1 (0.6)(0.75)T1 5 147

T1 sin (u1) 1 T2 sin (u2) 2 (15)(9.8) 5 0

T2 5 (0.75)T1

(0.8)T2 5 (0.6)T1

2T1 cos (u1) 1 T2 cos (u2) 5 0

sin (u1) 5 0.8

cos (u1) 5 0.6

(20)2 5 (15)2 1 (25)2 2 (2)(15)(25) cos (u1)

sin (u2) 5 0.6

sin (u2) 5 "1 2 cos2 (u2)

cos (u2) 5 0.8

(15)2 5 (20)2 1 (25)2 2 2(20)(25) cos (u2)

u2

u1

T2T1

20 cm

15 cm25 cm

u1

u2

5 81

5 12(3)2 2 3 2 6(2)2

1 8b>

? a>

2 6 @b>

@ 2(3a

>

1 2b>

) ? (4a>

2 3b>

) 5 12 0 a>

0 2 2 9a>

? b>


f.

9. a.

or

b. If and are parallel then one is a scalar multiple of the other.

where n is a constant

by the componentby the component

by the component

Therefore there is no value of x that will make thesetwo vectors parallel.10. a.

b.

c.

or 1.41

d.

e.

11.

12.

N, 25.3° from the 4 N force towards the 3 N force.

N, from the 4 N force away from the 3 N force.13. a. The diagonals are and

b.

u 8 87.9°cos u 8 0.0374

2 2"38 "75 cos u

(9 1 100) 5 (4 1 9 1 25) 1 (1 1 49 1 25)

0m>

2 n>

0 2 5 0m>

0 2 1 0 n>

0 2 2 2 0m>

0 0 n>

0cos u

u 8 109.1°

cos u 8 20.3276

"9 1 100 cos u3 2 40 5 "1 1 16 1 100

(m>

1 n>

) ? (m>

2 n>

) 5 0m>

1 n>

0 0m>

2 n>

0cos u

m>

2 n>

5 (3, 210, 0)

m>

1 n>

5 (1, 4, 10)

m>

2 n>

m>

1 n>

180° 2 25.3° 5 154.7°E>

8 6.08

F>

8 6.08

u 8 25.3°

cos u 544

8"37

(3)2 5 (4)2 1 ("37)22 2(4)("37) cos u

F 8 6.08 N

(F)2 5 37

(F)2 5 25 2 24 cos (120°)

(F)2 5 (3)2 1 (4)2 2 2(3)(4) cos (180° 2 60°)

u 8 126.9° 180° 2 u 8 53.1

0.6 5 cos (180° 2 u)

(4)2 5 (5)2 1 (3)2 2 2(3)(5) cos (180° 2 u)

180° - u

5 N

3 N

4 N

5 212

5 23(1 1 2 1 1)

5 23(i>

2 2j>

2 k) ? (i>

2 j>

2 k>

)

5 23x>

? y>

2x>

? y>

2 5y>

? x>

5 2x>

? y>

2 5x>

? y>

5 245 25 1 6 2 5

? (5i>

2 6j>

2 5k>

)

5 (2i>

2 j>

1 k>

)

1 (4 i>

2 4 j>

2 4k>

)

1 (( i>

2 2j>

2 k>

)

2 (3i>

2 3j>

2 3k>

)) ?

2k>

)(2x>

2 3y>

) ? (x>

1 4y>

) 5 ((2i>

2 4j>

2

5 "2

5 "(2i>

1 k>

) ? (2i>

1 k>

)

5 @2i>

1 k>

@0 x>

2 2y>

0 5 @ (i>

2 2j>

2 k>

) 2 (2i>

2 2j>

2 2k>

) @5 245 6 1 12 1 6

3x>

? 2y>

5 (3i>

2 6j>

2 3k>

) ? (2i>

2 2j>

2 2k>

)

5 i>

2 4j>

2 k>

3x>

2 2y>

5 (3i>

2 6j>

2 3k>

) 2 (2i>

2 2j>

2 2k>

)

1 2 0

j>

1 5 30(0)

x 5 0

i>

x 5 9xk>

n 5 3

xi>

1 j>

1 3k>

5 n(3xi>

1 10xj>

1 k>

)

p>

5 nq>

q>

p>

x 5 21

3x 5 23

x 5210 6 8

6

x 5210 6 "(10)2 2 4(3)(3)

2(3)

3x2 1 10x 1 3 5 0

5 0(xi>

1 j>

1 3k>

) ? (3xi>

1 10xj>

1 k>

)

p>

? q>

5 0

5 9

5 220 1 39 2 10

? (5i>

1 3j>

1 k>

)

5 (24i>

1 13j>

2 10k>

)

1 (3i>

2 j>

1 k>

))

1 2k>

)? ((2i>

1 4j>

2 (6i>

2 9j>

1 12k>

))

(2a>

2 3b>

) ? (2a>

1 c>

) 5 ((2i>

1 4j>

1 2k>

)


14. a.

b.km h

hours15.

choose

or

16. a. m s

mb.

m s

seconds17. a.

when and havethe same length.b. Vectors and determine a parallelogram. Theirsum is one diagonal of the parallelogramformed, with its tail in the same location as the tailsof and Their difference is the otherdiagonal of the parallelogram.18.

N

7.5 Scalar and Vector Projections,pp. 398–400

1. a. Scalar projection of on is where

and is the positive x-axis (X, 0).


multiplied by where is the x-axis divided by

the magnitude of the x-axis which is equal to The scalar projection of 2 multiplied by equals

b. Scalar projection of on is where

and is now the positive y-axis (0, Y).


multiplied by where is the y-axis divided

by the magnitude of the y-axis which is equal to The scalar projection of 3 multiplied by equals 2. Using the formula would cause a division by 0.Generally the has any direction and 0 magnitude.You can not project onto nothing.3. You are projecting onto the tail of which is a point with magnitude 0. Therefore it is theprojections of onto the tail of are also 0 and 0

>

.

a>

b>

0>

;

b>

a>

0>

3j>

.j>

j>

.

b>

@ b>

@b>

@ b>

@

5 3;

a> @b

>

@@b>

@5

3YY

5 Y @b

>

@ 5 "02 1 Y2

5 3Y 5 0 1 3Y

a>

? b>

5 (2 3 0) 1 (3Y)

b>

a>

5 (2, 3)

a>

?b>

@b>

@b>

a>

2i>

.i>

i>

.

b>

@ b>

@b>

@ b>

@

5 2;

a>

? b>

@b>

@5

2XX

5 X @b

>

@ 5 "X2 1 02

5 2X 5 2X 1 0

a>

? b>

5 (2X) 1 (3 3 0)

b>

a>

5 (2, 3)

a>

?b>

@b>

@b>

a>

8 268.12

@F>

@ 5 350 cos (40°)

a>

2 b>

b>

.a>

a>

1 b>

b>

a>

y>

x>

(x>

1 y>

) ? (x>

2 y>

) 5 0

0 x>

0 2 5 0 y>

0 2 0 x>

0 2 2 x>

? y>

1 y>

? x>

2 0 y>

0 2 5 0

(x>

1 y>

) ? (x>

2 y>

) 5 0

t 8 84.9

t 8180

2.12

> 8 2.12

w 5 3 sin (45°)

8 61.2

d 8 (6.12)(10)

> 8 6.12

v 5 4 1 3 cos (45°)

a21

!6,

2

!6, 2

1

!6bx

>

5 a1

!6,2

2

!6,

1

!6b

x>

51

!6(1, 22, 1)

x1 5 1

x2 5 22

x3 5 1

x2 1 2x3 5 0

2x2 1 5x3 1 3x2 1 5x3 5 0

x1 1 3x2 1 5x3 5 0

b>

? x>

5 0

x1 5 2x2 1 5x3

2x1 1 2x2 1 5x3 5 0

a>

? x>

5 0

t 8 2.17

t 81000

460.5

> 8 460.5

v 5 500 cos (2.6°) 2 45 cos (30°)

u 8 N 2.6° E45 sin (150°) 5 500 sin u



Scalar projection on Vector projection on Scalar projection on Vector projection on

5. When and then



Under the same approach, when and then



The same is also true when andthen



Without having to use formulae, a projection of on or is the same as a projection

of on (0, 2, 0) on and (0, 0, 5) on which intuitively yields the same result.6. a.


b. Direction angles for where

include and

Therefore

Therefore

8 74.9°;

b 5 cos21 a6

23b

56

23,

56

"529

56

"9 1 36 1 484

56

"32 1 62 1 (222)2

cos b 5b

"a2 1 b2 1 c2

8 82.5°;

a 5 cos21 a3

23b

53

23,

53

"529

53

"9 1 36 1 484

53

"32 1 62 1 (222)2

cos a 5a

"a2 1 b2 1 c2g.b,a,

p>

5 (a, b, c)p>

5458

441(24, 5, 20).

(24, 5, 220)

21. 5

458

21

The vector equation 5458

213

q>

0 q>

0

p>

? q>

0 q>

05

458

21,

5 21

5 "441

5 "16 1 25 1 400

0 q>

0 5 "(24)2 1 52 1 (220)2

5 4585 212 1 30 1 440

1 (222 3 220)

p>

? q>

5 (3 3 24) 1 (6 3 5)

k>

j>

,i>

,(21, 0, 0)

k>

j>

,i>

,(21, 2, 5)

5 5;

5 3b>

@b>

@5 5 3

(0, 0, 1)

1

5 5,

a>

? b>

@b>

@5

5

1

5 1

@b>

@ 5 "02 1 02 1 12

5 5

a>

? b>

5 (21 3 0 1 2 3 0 1 5 3 1)

b>

5 (0, 0, 1)

a>

5 (21, 2, 5)

5 2;

2 3b>

@b>

@5 2 3

(0, 1, 0)

1

5 2,

a>

? b>

@b>

@5

2

1

5 1

@b>

@ 5 "02 1 1 1 02

5 2

a>

? b>

5 (21 3 0 1 2 3 1 1 5 3 0)

b>

5 (0, 1, 0),

a>

5 (21, 2, 5)

5 21;

21 3b>

@b>

@5 21 3

(1, 0, 0)

1

5 21;

a>

? b>

@b>

@5

21

1

5 1

@b>

@ 5 "12 1 02 1 02

5 21

a>

? b>

5 (21 3 1 1 2 3 0 1 5 3 0)

b>

5 (1, 0, 0)a>

5 (21, 2, 5)

p>

5 AD>

q>

p>

5 @AD>

@ ;q>

q>

5 AC>

;p>

q>

5 @AC>

@ ;p>

A

D

C

E

Bq

p

q>

5 AB>

p>

5 AE>

,



Therefore

7. a.



b.



c.



8. a. The scalar projection of on the x-axis

is






b. The scalar projection of on the x-axisis


5 2mi>

;

2m 3(X, 0, 0)

"X2 1 02 1 025 2m 3

(X, 0, 0)

X

ma>

5 2m

52mX

X

1(4m 3 0)

"X2 1 02 1 02

ma>

? (X, 0, 0)

0 (X, 0, 0) 05

(2m 3 X) 1 (2m 3 0)

"X2 1 02 1 02

(X, 0, 0)

m a>

5 4k>

.

4 3(0, 0, Z)

"02 1 02 1 Z25 4 3

(0, 0, Z)

Z

a>

5 4;

54ZZ

a>

? (0, 0, Z)

0 (0, 0, Z) 05

(21 3 0) 1 (2 3 0) 1 (4 3 Z)

"02 1 02 1 Z2

(0, 0, Z)a>

5 2j>

;

2 3(0, Y, 0)

"02 1 Y2 1 025 2 3

(0, Y, 0)

Y

a>

5 2

52YY

a>

? (0, Y, 0)

0 (0, Y, 0) 05

(21 3 0) 1 (2 3 Y) 1 (4 3 0)

"02 1 Y2 1 02

(0, Y, 0)a>

5 2i>

;

21 3(X, 0, 0)

"X2 1 02 1 025 21 3

(X, 0, 0)

X

a>

5 21;

52XX

a>

? (X, 0, 0)

0 (X, 0, 0) 05

(21 3 X) 1 (2 3 0) 1 (4 3 0)

"X2 1 02 1 02

a>

? (X, 0, 0)

0 (X, 0, 0) 0(X, 0, 0)

a>

550

169(25, 12)

50

133

y>

0 y>

05

50

133

(25, 12)

13

x>

? y>

0 y>

05

50

13.

5 13

5 "169

5 "25 1 144

0 y>

0 5 "(25)2 1 122

5 50

5 210 1 60

x>

? y>

5 (2 3 25) 1 (5 3 12)

5 2i>

2 3y>

0 y>

05 2 3

(1, 0)

1

5 2;

x>

? y>

0 y>

05

2

1

5 1

0 y>

0 5 "12 1 02

5 2

x>

? y>

5 (2 3 1) 1 (2"3 3 0)

0 3y>

0 y>

05 0

>

5 0;

x>

? y>

0 y>

05

0

"2

5 "2

0 y>

0 5 "12 1 (21)2

5 0

5 1 1 (21)

x>

? y>

5 (1 3 1) 1 (1 3 21)

8 163.0°

g 5 cos21 a222

23b

5222

23,

5222

"529

5222

"9 1 36 1 484

5222

"32 1 62 1 (222)2

cos g 5c

"a2 1 b2 1 c2





9. a.

projected onto itself will yield itself. The scalarprojection will be the magnitude of itself.b. Using the formula for the scalar projection


multiplied by

10. a.

b.

So the vector projection is

11. a.

The scalar projection of on the x-axis is






b. The angle made with the y-axis is

51

3,

51

"9

51

"4 1 1 1 4

51

"(22)2 1 12 1 22

cos b 5b

"a2 1 b2 1 c2

b

5 2k>

2 3(0, 0, Z)

"02 1 02 1 Z25 2 3

(0, 0, Z)

Z

AB>

5 2;

52ZZ

a>

? (0, 0, Z)

0 (0, 0, Z) 05

(22 3 0) 1 (1 3 0) 1 (2 3 Z)

"02 1 02 1 Z2

(0, 0, Z)AB>

5 j>

;

1 3(0, Y, 0)

"02 1 Y2 1 025 1 3

(0, Y, 0)

Y

AB>

5 1;

5YY

a>

? (0, Y, 0)

0 (0, Y, 0) 05

(22 3 0) 1 (1 3 Y) 1 (2 3 0)

"02 1 Y2 1 02

(0, Y, 0)AB>

5 22i>

;

22 3(X, 0, 0)

"X2 1 02 1 025 22 3

(X, 0, 0)

X

AB>

5 22;

522X

X

a>

? (X, 0, 0)

0 (X, 0, 0) 05

(22 3 X) 1 (1 3 0) 1 (2 3 0)

"X2 1 02 1 02

(X, 0, 0)AB>

5 (22, 1, 2)

5 (21, 3, 4) 2 (1, 2, 2)

AB>

5 Point B 2 Point A

2 0 a>

0 a0 a>

0

0 a>

0b 5 2a

>

.

5 2 0 a>

0

(2a>

) ? a>

0 a>

0 5

2 0 a>

0 2

0 a>

0

–a aB O A

0 a>

0 3a>

0 a>

05 a

>

.a>

0 a>

0,

5 0 a>

0 .

5 0 a>

0 (1)

0 a>

0cos u 5 0 a>

0cos 0

a>

a

5 4mk>

.

4m 3(0, 0, Z)

"02 1 02 1 Z25 4m 3

(0, 0, Z)

Z

ma>

5 4m;

54mZ

Z

1(4m 3 Z)

"02 1 02 1 Z2

ma

>

? (0, 0, Z)

0 (0, 0, Z) 05

(2m 3 0) 1 (2m 3 0)

"02 1 02 1 Z2

(0, 0, Z)ma>

5 2mj>

;

2m 3(0, Y, 0)

"02 1 Y2 1 025 2m 3

(0, Y, 0)

Y

ma>

5 2m;

52mY

Y

1(4m 3 0)

"02 1 y2 1 02

ma>

? (0, Y, 0)

0 (0, Y, 0) 05

(2m 3 0) 1 (2m 3 Y)

"02 1 Y2 1 02

(0, Y, 0)ma>


Therefore

12. a.

b.

c. In an isosceles triangle, CD is a median and aright bisector of BA. Therefore and have thesame magnitude projected on .d. Yes, not only do they have the same magnitude,but they are in the same direction as well whichmakes them have equivalent vector projections.13. a. Use the formula for the scalar projection of on

And the formula for the scalar projection of on

b.

is the vector projection of on is the vector projection of on

14. a.


b.



c. Same lengths and both are in the direction of Add to get one vector.15. a.

1c2

a2 1 b2 1 c2

5a2

a2 1 b2 1 c2 1b2

a2 1 b2 1 c2

1 ac

"a2 1 b2 1 c2b

2

5 aa

"a2 1 b2 1 c2b

2

1 ab

"a2 1 b2 1 c2b

2


OD>

.

5 6

518

3

518

"9

54 1 12 1 2

"1 1 4 1 4

AC>

? OD>

@OD>

@5

(24 3 21) 1 (6 3 2) 1 (1 3 2)

"(21)2 1 22 1 22

OD>

AC>

5 (24, 6, 1)

5 (26, 7, 5) 2 (22, 1, 4)

AC>

5 Point C 2 Point A5 6

518

3

AB>

? OD>

@OD>

@1

BC>

? OD>

@OD>

@5 2

1

31

19

3

519

3

519

"9

57 1 8 1 4

"1 1 4 1 4

BC

>

? OD>

@OD>

@5

(27 3 21) 1 (4 3 2) 1 (2 3 2)

"(21)2 1 22 1 22

OD>

BC>

5 (27, 4, 2)

5 (26, 7, 5) 2 (1, 3, 3)

BC>

5 Point C 2 Point B

5 21

3

521

"9

5(23) 1 4 1 (22)

"1 1 4 1 4

AB

>

? OD>

@OD>

@5

(3 3 21) 1 (2 3 2) 1 (21 3 2)

"(21)2 1 22 1 22

OD>

AB>

5 (3, 2, 21)

5 (1, 3, 3) 2 (22, 1, 4)

AB>

5 Point B 2 Point A

b>

a>

OP>

a>

b>

OQ>

10

P

Q O

12

a

b

135°

5 28.49

5 12 cos 135°

a>

5 @b>

@ cos u

b>

5 27.07

5 10 cos 135°

b>

5 0 a>

0cos u

a>

c>

b>

a>

C

DBu u

A

abb

c

@BD>

@

C

B u u A

a b

D c

@BD>

@8 70.5°

b 5 cos21 a1

3b
