13.1 SOLUTIONS 907 CHAPTER THIRTEEN - Valenciafd.valenciacollege.edu/file/tsmith143/ch13.pdf13.1 SOLUTIONS 909 30. (a) True, since vectors ~cand f~point in the same direction and have

13.1 SOLUTIONS 907

CHAPTER THIRTEEN

Solutions for Section 13.1

Exercises

1. 4~i + 2~j − 3~i +~j =~i + 3~j

2. ~i + 2~j − 6~i − 3~j = −5~i −~j3. −4~i + 8~j − 0.5~i + 0.5~k = −4.5~i + 8~j + 0.5~k

4. (0.9~i − 1.8~j − 0.02~k )− (0.6~i − 0.05~k ) = 0.3~i − 1.8~j + 0.03~k

5. ‖~v ‖ =√

12 + (−1)2 + 32 =√

11.

6. ‖~v ‖ =√

12 + (−1)2 + 22 =√

6.

7. ‖~v ‖ =√

1.22 + (−3.6)2 + 4.12 =√

31.21 ≈ 5.6.

8. ‖~v ‖ =√

7.22 + (−1.5)2 + 2.12 =√

58.5 ≈ 7.6.

9. 5~b = 5(−3~i + 5~j + 4~k ) = −15~i + 25~j + 20~k .

10. ~a + ~z = (2~j + ~k ) + (~i − 3~j − ~k ) = (0 + 1)~i + (2− 3)~j + (1− 1)~k =~i −~j11. 2~c + ~x = 2(~i + 6~j ) + (−2~i + 9~j ) = (2~i + 12~j ) + (−2~i + 9~j ) = (2− 2)~i + (12 + 9)~j = 21~j .

12. ‖~z ‖ =√

(1)2 + (−3)2 + (−1)2 =√

1 + 9 + 1 =√

11.

13. ‖~y ‖ =√

(4)2 + (−7)2 =√

16 + 49 =√

65.

14.

2~a + 7~b − 5~z = 2(2~j + ~k ) + 7(−3~i + 5~j + 4~k )− 5(~i − 3~j − ~k )

= (4~j + 2~k ) + (−21~i + 35~j + 28~k )− (5~i − 15~j − 5~k )

= (−21− 5)~i + (4 + 35 + 15)~j + (2 + 28 + 5)~k = −26~i + 54~j + 35~k .

15. (a) See Figure 13.1.(b) ‖~v ‖ =

√52 + 72 =

√74 = 8.602.

(c) We see in Figure 13.2 that tan θ = 75

and so θ = 54.46◦.

~v

5

−7

x

y

Figure 13.1

θ

7

5x

y

Figure 13.2

16. Resolving ~v into components gives ~v = 8 cos(40◦)~i − 8 sin(40◦)~j = 6.13~i − 5.14~j . Notice that the component inthe ~j direction must be negative.

17. ~a = −2~j ,~b = 3~i , ~c =~i +~j , ~d = 2~j , ~e =~i − 2~j , ~f = −3~i −~j .

908 Chapter Thirteen /SOLUTIONS

18. The vector we want is the displacement from Q to P , which is given by~QP = (1− 4)~i + (2− 6)~j = −3~i − 4~j

19. ~a = ~b = ~c = 3~k , ~d = 2~i + 3~k , ~e = ~j , ~f = −2~i

20. ~u =~i +~j + 2~k and ~v = −~i + 2~k .

Problems

21. To determine if two vectors are parallel, we need to see if one vector is a scalar multiple of the other one. Since ~u = −2~w ,and ~v = 1

4~q and no other pairs have this property, only ~u and ~w , and ~v and ~q are parallel.

22. The length of the vector~i −~j +2~k is√

12 + (−1)2 + 22 =√

6. We can scale the vector down to length 2 by multiplyingit by 2√

6. So the answer is 2√

6~i − 2√

6~j + 4√

6~k .

23. (a) The displacement from P to Q is given by−−→PQ = (4~i + 6~j )− (~i + 2~j ) = 3~i + 4~j .

Since‖−−→PQ‖ =

√32 + 42 = 5,

a unit vector ~u in the direction of −−→PQ is given by

~u =1

5

−−→PQ =

1

5(3~i + 4~j ) =

3

5~i +

4

5~j .

(b) A vector of length 10 pointing in the same direction is given by

10~u = 10(3

5~i +

4

5~j ) = 6~i + 8~j .

24. Since the component of ~v in the~i -direction is 3, we have ~v = 3~i +b~j for some b. Since ‖~v ‖ = 5, we have√

32 + b2 =5, so b = 4 or b = −4. There are two vectors satisfying the properties given: ~v = 3~i + 4~j and ~v = 3~i − 4~j .

25. (a) The displacement vectors are:

From the submarine to the ship = −2~i + 3~j + 6~k

From the helicopter to the ship = 2~i + 2~j − 10~k

From the submarine to the helicopter = −4~i +~j + 16~k

(b) The distance between the submarine and the ship is√

49 = 7. The distance between the helicopter and the ship is√108 = 10.39. The distance between the submarine and the helicopter is

√273 = 16.52. The submarine and the

ship are the closest.

26. We get displacement by subtracting the coordinates of the origin (0, 0, 0) from the coordinates of the cat (1, 4, 0), givingDisplacement = (1− 0)~i + (4− 0)~j + (0− 0)~k =~i + 4~j .

27. We get displacement by subtracting the coordinates of the bottom of the tree, (2, 4, 0), from the coordinates of the squirrel,(2, 4, 1), giving:

Displacement = (2− 2)~i + (4− 4)~j + (1− 0)~k = ~k .

28.

Displacement = Cat’s coordinates − Bottom of the tree’s coordinates

= (1− 2)~i + (4− 4)~j + (0− 0)~k = −~i .

29.

Displacement = Squirrel’s coordinates − Cat’s coordinates

= (2− 1)~i + (4− 4)~j + (1− 0)~k =~i + ~k .

13.1 SOLUTIONS 909

30. (a) True, since vectors ~c and ~f point in the same direction and have the same length.(b) False, since vectors ~a and ~d point in opposite directions. We have ~a = −~d .(c) False, since −~b points in the opposite direction to~b , the vectors −~b and ~a are perpendicular.(d) True. The vector ~f can be ”moved” to point directly up the z-axis.(e) True. We move in the positive x-direction following vector ~a and then in the positive y-direction following vector−~b . The resulting sum is the vector ~e .

(f) False, vector ~d is the negative of the vector ~g − ~c . It is true that ~d = ~c − ~g .

31.

30 km/hrtruck

6

~r

40 km/hrPolice car�

x East

Northy

C Police carP

Truck

O

Figure 13.3

Since both vehicles reach the crossroad in exactly one hour, at the present the truck is at O in Figure 13.3; the policecar is at P and the crossroads is at C. If ~r is the vector representing the line of sight of the truck with respect to the policecar.

~r = −40~i − 30~j

32. In Figure 13.4 let O be the origin, points A, B, and C be the vertices of the triangle, point D be the midpoint of BC, andQ be the point in the line segment DA that is 1

3|DA| away from D.

A

C

B

DQ

Ox

y

Figure 13.4

From Figure 13.4 we see that

−−→OQ =

−−→OD +

−−→DQ =

−−→OD +

1

3−−→DA

=−−→OD +

1

3(−→OA−−−→OD)

=−−→OD +

1

3

−→OA− 1

3

−−→OD

=1

3

−→OA+

2

3

−−→OD.


Because the diagonals of a parallelogram meet at their midpoint, and 2−−→OD is a diagonal of the parallelogram formed by−−→

OB and −−→OC, we have:−−→OD =

1

2(−−→OB +

−−→OC),

so we can write:

−−→OQ =

1

3

−→OA+

2

3

(1

2

)(−−→OB +

−−→OC) =

1

3(−→OA+

−−→OB +

−−→OC).

Thus a vector from the origin to a point 13

of the way along medianAD fromD, the midpoint, is given by 13(−→OA+

−−→OB+

−−→OC).

In a similar manner we can show that the vector from the origin to the point 13

of the way along any median from themidpoint of the side it bisects is also 1

3(−→OA+

−−→OB +

−−→OC). See Figure 13.5 and 13.6.

A

C

B

Q

x

y

Figure 13.5

A

C

B

Q

x

y

Figure 13.6

Thus the medians of a triangle intersect at a point 13

of the way along each median from the side that each bisects.

33. We want to find an expression for a vector from the origin to a point that is 14

of the way from a centroid to its oppositevertex.

In Figure 13.7 let O be the origin, A, B, C, and D be the vertices of a tetrahedron, P be the centroid of face BCD,and Q be the point on PA that is | 1

4PA| away from P .

−−→OQ =

−−→OP +

−−→PQ

=−−→OP +

1

4

−→PA

=−−→OP +

1

4(−→OA−−−→OP )

=−−→OP +

1

4

−→OA− 1

4

−−→OP

=1

4

−→OA+

3

4

−−→OP.

In Problem 32 we showed that a vector from the origin to P , the centroid of a triangle, is

−−→OP =

1

3(−−→OB +

−−→OC +

−−→OD).

Substituting this into our expression for OQ gives

−−→OQ =

1

4

−→OA+

3

4

(1

3

)(−−→OB +

−−→OC +

−−→OD)

=1

4−→OA+

1

4(−−→OB +

−−→OC +

−−→OD)

=1

4(−→OA+

−−→OB +

−−→OC +

−−→OD).

In a similar manner we can show that a vector from the origin to a point 14

of the way from the centroid of any faceto its opposite vertex is 1

4(−→OA +

−−→OB +

−−→OC +

−−→OD). Thus, lines joining the centroid of each face to its opposite vertex

all meet at a single point which is 14

of the way from any centroid to its opposite face.

13.2 SOLUTIONS 911

z

x

y

A

B

CD

0

P

Q

Figure 13.7

34. We must check that all the points are the same distance apart, i.e., the magnitude of the displacement vectors −→OA, −−→OB,−−→OC, −→BA, −−→CB and −→CA is the same. Here goes:

‖−→OA‖ = ‖(2~i + 0~j + 0~k )− (0~i + 0~j + 0~k )‖ =√

22 + 02 + 02 = 2

‖−−→OB‖ = ‖(1~i +√

3~j + 0~k )− (0~i + 0~j + 0~k )‖ =

√12 + (

√3)2 + 02 = 2

‖−−→OC‖ = ‖(1~i + 1/√

3~j + 2√

2/3~k )− (0~i + 0~j + 0~k )‖ =√

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

‖−→BA‖ = ‖(2~i + 0~j + 0~k )− (1~i +√

3~j + 0~k )‖ =√

1 + 3 + 0 = 2

‖−−→CB‖ = ‖(1~i +√

3~j + 0~k )− (1~i + 1/√

3~j + 2√

2/3~k )‖

=

√02 + (

√3− 1/

√3)2 + 4(2/3) =

√3− 2 + 1/3 + 8/3 = 2

‖−→CA‖ = ‖(2~i + 0~j + 0~k )− (1~i + 1/√

3~j + 2√

2/3~k )‖ =√

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


Exercises

1. Scalar

2. Scalar

3. The magnetic field is a vector because it has both a magnitude (the strength of the field) and a direction (the direction ofthe compass).

4. Temperature is measured by a single number, and so is a scalar.

5. Writing ~P = (P1, P2, · · · , P50) where Pi is the population of the i-th state, shows that ~P can be thought of as a vectorwith 50 components.

6. We need to calculate the length of each vector.

‖21~i + 35~j ‖ =√

212 + 352 =√

1666 ≈ 40.8,

‖40~i ‖ =√

402 = 40.

So the first car is faster.


7. In components, we have ~v = −40 cos(20◦)~i − 40 sin(20◦)~j = −37.59~i − 13.68~j . Notice that both coefficients arenegative. The components are −37.59~i and −13.68~j .

8. In components, we have ~v = 10 cos(45◦)~i − 10 sin(45◦)~j = (5√

2)~i − (5√

2)~j = 7.07~i − 7.07~j . Notice that thecoefficient in the ~j -direction must be negative. The components are 5

√2~i and −5

√2~j .

9. (a) If the car is going east, it is going solely in the positive x direction, so its velocity vector is 50~i .(b) If the car is going south, it is going solely in the negative y direction, so its velocity vector is −50~j .(c) If the car is going southeast, the angle between the x-axis and the velocity vector is −45◦. Therefore

velocity vector = 50 cos(−45◦)~i + 50 sin(−45◦)~j

= 25√

2~i − 25√

2~j .

(d) If the car is going northwest, the velocity vector is at a 45◦ angle to the y-axis, which is 135◦ from the x-axis.Therefore:

velocity vector = 50(cos 135◦)~i + 50(sin 135◦)~j = −25√

2~i + 25√

2~j .

10. See Figure 13.8. Since

tan θ =18

15,

we haveθ = arctan

(18

15

)= 50.194◦.

15~i

18~j~F

θ

x

y

Figure 13.8

Problems

11. Let the velocity vector of the airplane be ~V = x~i +y~j +z~k in km/hr. We know that x = −y because the plane is travelingnorthwest. Also, ‖~V ‖ =

√x2 + y2 + z2 = 200 km/hr and z = 300 m/min = 18 km/hr. We have

√x2 + y2 + z2 =√

x2 + x2 + 182 = 200, so x = −140.8, y = 140.8, z = 18. (The value of x is negative and y is positive because theplane is heading northwest.) Thus,

~v = −140.8~i + 140.8~j + 18~k .

12. (a) The velocity vector for the boat is ~b = 25~i and the velocity vector for the current is

~c = −10 cos(45◦)~i − 10 sin(45◦)~j = −7.07~i − 7.07~j .

The actual velocity of the boat is~b + ~c = 17.93~i − 7.07~j .

(b) ‖~b + ~c ‖ = 19.27 km/hr.

(c) We see in Figure 13.9 that tan θ =7.07

17.93, so θ = 21.52◦ south of east.

6

?7.07

-� 17.93

θ

~b

~c

~b + ~c

Figure 13.9

13.2 SOLUTIONS 913

13. The velocity vector for the wind is ~w = 60 cos(45◦)~i − 60 sin(45◦)~j = 42.43~i − 42.43~j . If the airplane is to headdue east, then the component in the ~j -direction of ~p + ~w must be zero (where ~p represents the velocity vector for theairspeed of the plane.) Thus, we have ~p = A~i + 42.43~j , for some value of A. Since the airplane is flying at an airspeedof 500 km/hr, we have

‖~p ‖ = 500√A2 + 42.432 = 500

A = 498.20.

We have~p = 498.20~i + 42.43~j .

This is the direction the plane should head in order to go due east. We use tan θ =42.43

498.20, so θ = 4.87◦. The plane

should head 4.87◦ north of east. Since~p + ~w = 540.63~i ,

the airplane’s speed relative to the ground is ‖~p + ~w ‖ = 540.63 km/hr.

14. The velocity vector of the plane with respect to the air has the form

~v = a~i + 80~k where ‖~v ‖ = 480.

(See Figure 13.10.) Therefore√a2 + 802 = 480 so a =

√4802 − 802 ≈ 473.3 km/hr. We conclude that ~v ≈ 473.3~i +

80~k .The wind vector is

~w = 100(cos 45◦)~i + 100(sin 45◦)~j

≈ 70.7~i + 70.7~j

The velocity vector of the plane with respect to the ground is then

~v + ~w = (473.3~i + 80~k ) + (70.7~i + 70.7~j )

= 544~i + 70.7~j + 80~k

From Figure 13.11, we see that the velocity relative to the ground is

544~i + 70.7~j .

The ground speed is therefore√

5442 + 70.72 ≈ 548.6 km/hr.

a~i

50~k~v

Figure 13.10: Side view

a~i

Velocity relativeto ground

~w

Figure 13.11: Top view

15. Let the x-axis point east and the y-axis point north. Since the wind is blowing from the northeast at a speed of 50 km/hr,the velocity of the wind is

~w = −50 cos 45◦~i − 50 sin 45◦~j ≈ −35.4~i − 35.4~j .

Let ~a be the velocity of the airplane, relative to the air, and let φ be the angle from the x-axis to ~a ; since ‖~a ‖ =600 km/hr, we have ~a = 600 cosφ~i + 600 sinφ~j . (See Figure 13.12.)


~w

~a

~v

?

φ

x

y

Figure 13.12

Now the resultant velocity, ~v , is given by

~v = ~a + ~w = (600 cosφ~i + 600 sinφ~j ) + (−35.4~i − 35.4~j )

= (600 cosφ− 35.4)~i + (600 sinφ− 35.4)~j .

Since the airplane is to fly due east, i.e., in the x direction, then the y-component of the velocity must be 0, so we musthave

600 sinφ− 35.4 = 0

sinφ =35.4

600.

Thus φ = arcsin(35.4/600) ≈ 3.4◦.

16. (a) See Figure 13.13. Notice that the velocity vectors are tangent to the curve, they point in the direction of motion, andthey are longer when the rocket is moving faster.

(b) If the rocket has a parachute, it comes down more slowly. The velocity vectors on the downward part of the graph areshorter for this rocket.

>> > >

>

>

>

Figure 13.13

P R

Q

Figure 13.14

17. See Figure 13.14.

18. At the point P , the velocity of the car is changing the quickest; not in magnitude, but in direction only. The accelerationvector is therefore the longest at this point. The direction of the vector is directed in toward the center of the track becausethe difference in velocity vectors at nearby points is a vector pointing toward the center.

19. The total scores are out of 300 and are given by the total score vector ~v + 2~w :

~v + 2~w = (73, 80, 91, 65, 84) + 2(82, 79, 88, 70, 92)

= (73, 80, 91, 65, 84) + (164, 158, 176, 140, 184)

= (237, 238, 267, 205, 268).

To get the scores as a percentage, we divide by 3, giving

1

3(237, 238, 267, 205, 268) ≈ (79.00, 79.33, 89.00, 68.33, 89.33).

20. Since there are 16 ounces in a pound, we multiply the vector by 1/16 to get 0.01875~i + 0.0125~j + 0.03125~k in dollarsper ounce.

21. We want the total force on the object to be zero. We must choose the third force ~F 3 so that ~F 1 + ~F 2 + ~F 3 = 0. Since~F 1 + ~F 2 = 11~i − 4~j , we need ~F 3 = −11~i + 4~j .

13.2 SOLUTIONS 915

22. Let the x-axis point east and the y-axis point north. We resolve the forces into components. Since the first force points50◦ south of east with a force of 25 newtons, we have

~F1 = 25 cos(50◦)~i − 25 sin 50◦~j = 16.070~i − 19.151~j .

Since ~F1 lies in the fourth quadrant, the coefficient of~i is positive and the coefficient of ~j is negative.The second force points 70◦ north of west with a force of 60 newtons, so we have

~F2 = −60 cos(70◦)~i + 60 sin 70◦~j = −20.521~i + 56.382~j .

Since ~F2 lies in the second quadrant, the coefficient of~i is negative and the coefficient of ~j is positive.The third force must make the total force equal to zero, so we have

~F1 + ~F2 + ~F3 = ~0

~F3 = −( ~F1 + ~F2 )

= −((16.070~i − 19.151~j ) + (−20.521~i + 56.382~j ))

= −(−4.451~i + 37.231~j )

= 4.451~i − 37.231~j .

The magnitude of this force is ‖ ~F3 ‖ =√

4.4512 + 37.2312 = 37.50 newtons. The direction is arctan(37.231/4.451) =83.20◦ south of east.

23. The force exerted on the object from the first rope ~F 1 = 100 cos(30◦)~i + 100 sin(30◦)~j = 86.60~i + 50~j and the forceexerted from the second rope is ~F 2 = 70 cos(80◦)~i − 70 sin(80◦)~j = 12.16~i − 68.94~j . The sum of these two forcesis ~F 1 + ~F 2 = 98.76~i −18.94~j . See Figure 13.15. In order for the object to move vertically, the total force on the objectmust be in the form ~F = 0~i + 0~j + b~k for some b. Thus the force vector for the crane is

~F c = −98.76~i + 18.94~j + b~k

for some b. To find b, we use the fact that ‖ ~F c‖ = 3000. Thus,

‖~F c‖ = 3000√(98.76)2 + (18.94)2 + b2 = 3000

b = ±2998.31

We use the positive value of b since we want the object to go up rather than down. The force exerted by the crane is

~F c = −98.76~i + 18.94~j + 2998.31~k .

The total force acting on the object is 2998.31~k , or 2998.31 newtons straight up.

~F c

~F 1 = 100 newtons

~F 2 = 70 newtons

30◦

80◦

N

S

EW

Figure 13.15: Horizontal forces on object

O

A

B

C

~v

~w~w

~v

Figure 13.16

24. The vector ~v + ~w is equivalent to putting the vectors −→OA and−→AB end-to-end as shown in Figure 13.16; the vector ~w +~vis equivalent to putting the vectors −−→OC and−−→CB end-to-end. Since they form a parallelogram, ~v + ~w and ~w +~v are bothequal to the vector −−→OB, we have ~v + ~w = ~w + ~v .


25.

β~v

α~v

~v

β~v

Figure 13.17

The vectors ~v , α~v and β~v are all parallel. Figure 13.17 shows them with α, β > 0, so all the vectors are in the samedirection. Notice that α~v is a vector α times as long as ~v and β~v is β times as long as ~v . Therefore α~v +β~v is a vector(α+ β) times as long as ~v , and in the same direction. Thus,

α~v + β~v = (α+ β)~v .

26.

~v + ~w~w

~v

?

scalingby α

α(~v + ~w )

α~v

α~w

Figure 13.18

The effect of scaling the left-hand picture in Figure 13.18 is to stretch each vector by a factor of α (shown withα > 1). Since, after scaling up, the three vectors α~v , α~w , and α(~v + ~w ) form a similar triangle, we know that α(~v + ~w )is the sum of the other two: that is

α(~v + ~w ) = α~v + α~w .

27. Assume α, β > 0. The vector β~v is in the same direction and β times as long as ~v . The vector α(β~v ) is in the samedirection and α times as long as β~v , and so is αβ times as long as ~v and in the same direction as ~v . Thus,

α(β~v ) = (αβ)~v .

28. Since the zero vector has zero length, adding it to ~v has no effect.

29. According to the definition of scalar multiplication, 1 · ~v has the same direction and magnitude as ~v , so it is the same as~v .

30. By Figure 13.19, the vectors ~v + (−1)~w and ~v − ~w are equal.

-

~v

~w ~v − ~w

(−1)~w

(−1)~w~v + (−1)~w

Figure 13.19

O A

B

C

~u

~v

~w

Figure 13.20

13.3 SOLUTIONS 917

31. The vector ~u + ~v is represented by −−→OB. The vector (~u + ~v ) + ~w is represented by −−→OB followed by −−→BC, which istherefore −−→OC. Now ~v + ~w is represented by −→AC. So ~u + (~v + ~w ) is −→OA followed by −→AC, which is −−→OC. Since we getthe vector −−→OC by both methods, we know

(~u + ~v ) + ~w = ~u + (~v + ~w )


Exercises

1. ~a · ~y = (2~j + ~k ) · (4~i − 7~j ) = −14.

2. ~c · ~y = (~i + 6~j ) · (4~i − 7~j ) = (1)(4) + (6)(−7) = 4− 42 = −38.

3. ~a ·~b = (2~j + ~k ) · (−3~i + 5~j + 4~k ) = (0)(−3) + (2)(5) + (1)(4) = 0 + 10 + 4 = 14.

4. ~a · ~z = (2~j + ~k ) · (~i − 3~j − ~k ) = (0)(1) + (2)(−3) + (1)(−1) = 0− 6− 1 = −7.

5. ~c + ~y = (~i + 6~j ) + (4~i − 7~j ) = 5~i −~j , so

~a · (~c + ~y ) = (2~j + ~k ) · (5~i −~j ) = −2.

6. ~c · ~a + ~a · ~y = (~i + 6~j ) · (2~j + ~k ) + (2~j + ~k ) · (4~i − 7~j ) = 12− 14 = −2.

7. Since ~a ·~b is a scalar and ~a is a vector, the answer to this equation is a vector parallel to ~a . We have

~a ·~b = (2~j + ~k ) · (−3~i + 5~j + 4~k ) = 0(−3) + 2(5) + 1(4) = 14.

Thus,(~a ·~b ) · ~a = 14~a = 14(2~j + ~k ) = 28~j + 14~k

8. Since ~a · ~y and ~c · ~z are both scalars, the answer to this equation is the product of two numbers and therefore a number.We have

~a · ~y = (2~j + ~k ) · (4~i − 7~j ) = 0(4) + 2(−7) + 1(0) = −14

~c · ~z = (~i + 6~j ) · (~i − 3~j − ~k ) = 1(1) + 6(−3) + 0(−1) = −17

Thus,(~a · ~y )(~c · ~z ) = 238

9. Since ~c ·~c is a scalar and (~c ·~c )~a is a vector, the answer to this equation is another scalar. We could calculate ~c ·~c , then(~c · ~c )~a , and then take the dot product ((~c · ~c )~a ) · ~a . Alternatively, we can use the fact that

((~c · ~c )~a ) · ~a = (~c · ~c )(~a · ~a ).

Since~c · ~c = (~i + 6~j ) · (~i + 6~j ) = 12 + 62 = 37

~a · ~a = (2~j + ~k ) · (2~j + ~k ) = 22 + 12 = 5,

we have,(~c · ~c )(~a · ~a ) = 37(5) = 185

10. A normal vector can be obtained from the coefficients: ~n = 2~i +~j − ~k .

11. Writing the equation in the form3x+ 4y − z = 7

shows that a normal vector is~n = 3~i + 4~j − ~k


12. The equation can be rewritten as

z − 5x+ 10 = 15− 3y

−5x+ 3y + z = 5

so ~n = −5~i + 3~j + ~k .

13. Rewriting the equation as2x− 2z = 3x+ 3y

orx+ 3y + 2z = 0

tells us that a normal vector is~n =~i + 3~j + 2~k .

Problems

14. (a) Increasing ||~v || increases ~v · ~w because ~v · ~w = ||~v || ||~w || cos θ, and cos θ is positive.(b) Increasing θ decreases ~v · ~w because cos θ is a decreasing function.

15. (a) Any multiple of ~v will work, for example, 8~i + 6~j .(b) Any vector ~w such that ~v · ~w = 0 will work, such as −3~i + 4~j .

16.

cos θ =(~i +~j + ~k ) · (~i −~j − ~k )

‖~i +~j + ~k ‖‖~i −~j − ~k ‖=

(1)(1) + (1)(−1) + (1)(−1)√11 + 12 + 12

√12 + (−1)2 + (−1)2

= −1

3.

So, θ = arccos(− 13) ≈ 1.91 radians, or ≈ 109.5◦.

17. Since 3~i +√

3~j =√

3(√

3~i + ~j ), we know that 3~i +√

3~j and√

3~i + ~j are scalar multiples of one another, andtherefore parallel.

Since (√

3~i +~j ) · (~i −√

3~j ) =√

3−√

3 = 0, we know that√

3~i +~j and~i −√

3~j are perpendicular.Since 3~i +

√3~j and

√3~i +~j are parallel, 3~i +

√3~j and~i −

√3~j are perpendicular, too.

18. In general, ~u and ~v are perpendicular when ~u · ~v = 0.In this case, ~u · ~v = (t~i −~j + ~k ) · (t~i + t~j − 2~k ) = t2 − t− 2.This is zero when t2 − t− 2 = 0, i.e. when (t− 2)(t+ 1) = 0, so t = 2 or −1.In general, ~u and ~v are parallel if and only if ~v = α~u for some real number α.Thus we need αt~i − α~j + α~k = t~i + t~j − 2~k , so we need αt = t, and −α = t, and α = −2. But if α = −2, wecan’t have αt = t unless t = 0, and if t = 0, we can’t have −α = t, so there are no values of t for which ~u and ~v areparallel.

19. Vectors ~v 1, ~v 4, and ~v 8 are all parallel to each other. Vectors ~v 3, ~v 5, and ~v 7 are all parallel to each other, and are allperpendicular to the vectors in the previous sentence. Vectors ~v 2 and ~v 9 are perpendicular.

20. (a) Perpendicular vectors have a dot product of 0. Since ~a · ~c = 1(−2) − 3(−1) − 1 · 1 = 0, and ~b · ~d = 1(−1) +

1(−1) + 2 · 1 = 0, the pairs we want are ~a ,~c and ~b , ~d .(b) Parallel vectors are multiples of one another, so there are no parallel vectors in this set.(c) Since ~v · ~w = ||~v ||||~w || cos θ, the dot product of the vectors we want is positive. We have

~a ·~b = 1 · 1− 3 · 1− 1 · 2 = −4

~a · ~d = 1(−1)− 3(−1)− 1 · 1 = 1

~b · ~c = 1(−2) + 3(−1) + 2 · 1 = −1

~c · ~d = −2(−1)− 1(−1) + 1 · 1 = 4,

and we already know ~a · ~c = ~b · ~d = 0. Thus, the pairs of vectors with an angle of less than π/2 between them are~a , ~d and ~c , ~d .

(d) Vectors with an angle of more than π/2 between them have a negative dot product, so pairs are ~a ,~b and~b ,~c .

13.3 SOLUTIONS 919

21. Let~a = ~a parallel + ~a perp

where ~a parallel is parallel to ~d , and ~a perp is perpendicular to ~d . Then ~a parallel is the projection of ~a in the direction of~d :

~a parallel =

(~a ·

~d

‖~d ‖

)~d

‖~d ‖

=

((3~i + 2~j − 6~k ) · (2~i − 4~j + ~k )√

22 + 42 + 12

)(2~i − 4~j + ~k )√

22 + 42 + 12

= − 8

21(2~i − 4~j + ~k )

= − 8

21~d

Since we now know ~a and ~a parallel, we can solve for ~a perp:

~a perp = ~a − ~a parallel

= (3~i + 2~j − 6~k )−(− 8

21

)(2~i − 4~j + ~k )

=79

21~i +

10

21~j − 118

21~k .

Thus we can now write ~a as the sum of two vectors, one parallel to ~d , the other perpendicular to ~d :

~a = − 8

21~d +

(79

21~i +

10

21~j − 118

21~k)

22. Since a normal vector of the plane is ~n = −~i + 2~j + ~k , an equation for the plane is

−x+ 2y + z = −1 + 2 · 0 + 2 = 1

−x+ 2y + z = 1.

23. Since the plane is normal to the vector 5~i +~j − 2~k and passes through the point (0, 1,−1), an equation for the plane is

5x+ y − 2z = 5 · 0 + 1 · 1 + (−2) · (−1) = 3

5x+ y − 2z = 3.

24. Since the plane is normal to the vector 2~i − 3~j + 7~k and passes through the point (1,−1, 2), an equation for the plane is

2x− 3y + 7z = 2 · 1− 3 · (−1) + 7 · 2 = 19

2x− 3y + 7z = 19.

25. Two planes are parallel if their normal vectors are parallel. Since the plane 2x + 4y − 3z = 1 has normal vector~n = 2~i + 4~j − 3~k , the plane we are looking for has the same normal vector and passes through the point (1, 0,−1).Thus the plane we want has equation:

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

26. Two planes are parallel if their normal vectors are parallel. Since the plane 3x + y + z = 4 has normal vector ~n =3~i + ~j + ~k , the plane we are looking for has the same normal vector and passes through the point (−2, 3, 2). Thus, ithas the equation

3x+ y + z = 3 · (−2) + 3 + 2 = −1.


27. (a) The plane can be written as 5x − 2y − z + 7 = 0, so the vector 5~i − 2~j − ~k is normal to the plane. The vectorλ~i +~j + 0.5~k is parallel to 5~i − 2~j − ~k if one is a scalar multiple of the other. This occurs if the coefficients arein proportion:

λ

5=

1

−2=

0.5

−1.

Solving gives λ = −2.5.(b) Substituting x = a+ 1, y = a, z = a− 1 into the equation of the plane gives

a− 1 = 5(a+ 1)− 2a+ 7

a− 1 = 5a+ 5− 2a+ 7

−13 = 2a

a = −6.5.

28. The plane cuts the x-axis where y = z = 0, so x = −3/5 = −0.6, giving the point

P = (−0.6, 0, 0)

Similarly, the plane cuts the y-axis where x = z = 0, so y = 3/4 = 0.75, so

Q = (0, 0.75, 0)

The plane cuts the z-axis at x = y = 0, so that z = 3, so

R = (0, 0, 3)

Now we have the three vertices of the triangle, P , Q, and R. The vectors along the three sides of the triangle are−−→QP = −0.6~i − 0.75~j−→RP = −0.6~i − 3~k−→QR = −0.75~j + 3~k

The lengths of the sides of the triangle are

‖−−→QP‖ =√

(−0.6)2 + (−0.75)2 ≈ 0.96

‖−→RP‖ =√

(−0.6)2 + (−3)2 ≈ 3.059

‖−→QR‖ =√

(−0.75)2 + (3)2 ≈ 3.092

The angle between the vectors ~v and ~w is given by

cos θ =~v · ~w‖~v ‖‖~w ‖ so θ = arccos

(~v · ~w‖~v ‖‖~w ‖

).

Thus,

Angle at P = arccos

( −−→QP · −→RP‖−−→QP‖‖−→RP‖

)

≈ arccos

((−0.6~i − 0.75~j ) · (−0.6~i − 3~k )

0.96 · 3.059

)≈ arccos

(0.36

0.96 · 3.059

)

≈ arccos(0.123) ≈ 83.0◦.

Angle at Q = arccos

( −−→QP · −→QR‖−−→QP‖‖−→QR‖

)

≈ arccos

((−0.6~i − 0.75~j ) · (−0.75~j + 3~k )

0.96 · 3.092

)≈ arccos

(0.5625

0.96 · 3.092

)

≈ arccos(0.19) ≈ 79.1◦.

Now we use the fact that the angles of the triangle add up to 180◦. Thus

Angle at R ≈ 180◦ − (83.0◦ + 79.1◦) ≈ 17.9◦.

13.3 SOLUTIONS 921

29. The angle between two planes is equal to the angle between the normal vectors of the two planes. A normal vector to theplane 5(x− 1) + 3(y + 2) + 2z = 0 is

~n 1 = 5~i + 3~j + 2~k ,

and a normal vector to the plane x+ 3(y − 1) + 2(z + 4) = 0 is

~n 2 =~i + 3~j + 2~k .

Since ~n 1 · ~n 2 = ‖~n 1‖‖~n 2‖ cos θ, then

cos θ =~n 1 · ~n 2

‖~n 1‖‖~n 2‖=

(5~i + 3~j + 2~k ) · (~i + 3~j + 2~k )√52 + 32 + 22

√12 + 32 + 22

=18√532

= 0.78

Hence, θ ≈ 38.7◦.

30. We first find displacement vectors −→AB = (4− 2)~i + (2− 2)~j + (1− 2)~k = 2~i −~k and−→AC = (2− 2)~i + (3− 2)~j +

(1− 2)~k = ~j − ~k . Then

cos( 6 BAC) =−→AB · −→AC‖−→AB‖‖−→AC‖

=1√5√

2= 0.3162.

Thus angle BAC is 71.57◦ (or 1.25 radians.)

31. See Figure 13.21. One way to find the angle atA is to find the angle between vectors −→AB and−→AC. Since−→AB = −1~i −7~jand −→AC = −5~i − 3~j , we have

cos( 6 BAC) =−→AB · −→AC‖−→AB‖‖−→AC‖

=(−1)(−5) + (−7)(−3)√

50√

34= 0.6306.

Thus the angle at vertex A is 50.91◦. Similarly, we see that the angle at vertex B is 53.13◦ and (since the angles of atriangle add up to 180◦) the angle at vertex C is 75.96◦.

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

A

B

C

x

y

Figure 13.21


32. (a) The points A, B and C are shown in Figure 13.22.

xy

z

B

C

A

Figure 13.22

First, we calculate the vectors which form the sides of this triangle:−→AB = (4~i + 2~j + ~k )− (2~i + 2~j + 2~k ) = 2~i − ~k−−→BC = (2~i + 3~j + ~k )− (4~i + 2~j + ~k ) = −2~i +~j−→AC = (2~i + 3~j + ~k )− (2~i + 2~j + 2~k ) = ~j − ~k

Now we calculate the lengths of each of the sides of the triangles:‖−→AB‖ =

√22 + (−1)2 =

√5

‖−−→BC‖ =√

(−2)2 + 12 =√

5

‖−→AC‖ =√

12 + (−1)2 =√

2

Thus the length of the shortest side of S is√

2.

(b) cos 6 BAC =−→AB·−→AC

‖−→AB‖·‖

−→AC‖

= 2·0+0·1+(−1)·(−1)√5·√

2≈ 0.32

33. We need to find the speed of the wind in the direction of the track. Looking at Figure 13.23, we see that we want thecomponent of ~w in the direction of ~v . We calculate

‖~w parallel‖ = ‖~w ‖ cos θ =~w · ~v‖~v ‖ =

(5~i +~j ) · (2~i + 6~j )

‖2~i + 6~j ‖

=16√40≈ 2.53

< 5

Therefore, the race results will not be disqualified.

x

y Track~v = 2~i + 6~j

θ

Wind~w = 5~i +~j

0

Figure 13.23

13.3 SOLUTIONS 923

34. Let the room be put in the coordinate system as shown in Figure 13.24.

0 = (0, 0, 0)

A = (0, 0, 25)

B = (0, 80, 25)

C = (200, 0, 0)

D = (200, 80, 0)

θ�

�

�

� �

�

80

25

200

Figure 13.24

Then the vectors of the two strings are given by:−−→AD = (200~i + 80~j + 0~k )− (0~i + 0~j + 25~k ) = 200~i + 80~j − 25~k−−→BC = (200~i + 0~j + 0~k )− (0~i + 80~j + 25~k ) = 200~i − 80~j − 25~k .Let the angle between −−→AD and −−→BC be θ. Then we have

cos θ =−−→AD · −−→BC‖−−→AD‖ ‖−−→BC‖

=200(200) + (80)(−80) + (−25)(−25)√

2002 + 802 + (−25)2√

(200)2 + (−80)2 + (−25)2

=34225

47025= 0.727804

35. The vector ~a represents the averages of the exams, written as decimals. The vector ~w represents the weightings.

~w · ~a = 0.1 · 0.75 + 0.15 · 0.91 + 0.25 · 0.84 + 0.5 · 0.87 = 0.8565 = 85.65%

The dot product, 86.65%, represents the class average of the four exams in the course.

36. We have

~p · ~q = (1.00)(43) + (3.50)(57) + (4.00)(12) + (2.75)(78) + (5.00)(20) + (3.00)(35)

= 710 dollars.

The vendor took in $710 in from sales. The quantity ~p · ~q represents the total revenue earned.

37. If ~x and ~y are two consumption vectors corresponding to points satisfying the same budget constraint, then

~p · ~x = k = ~p · ~y .

Therefore we have~p · (~x − ~y ) = ~p · ~x − ~p · ~y = 0.

Thus ~p and ~x − ~y are perpendicular; that is, the difference between two consumption vectors on the same budgetconstraint is perpendicular to the price vector.


38. (a) The geometric definition of the dot product says that

~n · −−→P0P = ||~n ||||−−→P0P || cos θ,

where θ is the angle between ~n and −−→P0P with 0 ≤ θ ≤ π. To say that the dot product ~n · −−→P0P is positive means thatthe angle between ~n and −−→P0P is between 0 and π/2, and strictly less than π/2. Hence ~n and −−→P0P are both pointingto the same side of the plane. Thus, all the points satisfying ~n · −−→P0P > 0 are on the same side of the plane, the sidewhich ~n points to. To say that the dot product is negative is to say that π/2 < θ ≤ π, and this means that −−→P0P and~n are pointing to opposite sides of the plane. Thus, all points satisfying ~n · −−→P0P < 0 are on the side of the planeopposite to ~n .

(b) Suppose the normal vector is ~n = a~i + b~j + c~k , let P0 = (x0, y0, z0) be a point in the plane and let P = (x, y, z)

be a variable point. Then −−→P0P = (x− x0)~i + (y − y0)~j + (z − z0)~k . Then ~n · −−→P0P > 0 means

a(x− x0) + b(y − y0) + c(z − z0) > 0

and ~n · −−→P0P < 0 meansa(x− x0) + b(y − y0) + c(z − z0) < 0

If the equation of the plane is written ax+ by + cz = d (with d = ax0 + by0 + cz0) then the inequalities become

ax+ by + cz > d and ax+ by + cz < d.

(c) We test each of the points P = (−1,−1, 1), Q = (−1,−1,−1) and R = (1, 1, 1), using the coordinate version ofthe inequalities in part (b):

P : 2 · (−1)− 3 · (−1) + 4 · 1 = 5 > 4

Q : 2 · (−1)− 3 · (−1) + 4 · (−1) = −3 < 4

R : 2 · 1− 3 · 1 + 4 · 1 = 3 < 4

Therefore Q and R are on the same side of the plane as each other; P is on the other side.

39. Suppose ~v = v1~i + v2

~j + v3~k and ~w = w1

~i + w2~j + w3

~k .

• Property 1:We calculate both ~v · ~w and ~w · ~v using the algebraic definition of the dot product:

~v · ~w = v1w1 + v2w2 + v3w3

~w · ~v = w1v1 + w2v2 + w3v3

But since ordinary multiplication of scalars is commutative, v1w1 = w1v1 and so on. Therefore

~v · ~w = ~w · ~v .

• Property 2:First we observe that

λ~w = λ(w1~i + w2

~j + w3~k ) = (λw1)~i + (λw2)~j + (λw3)~k

λ~v = λ(v1~i + v2

~j + v3~k ) = (λv1)~i + (λv2)~j + (λv3)~k .

Now we calculate the three quantities ~v · (λ~w ) and λ(~v · ~w ) and (λ~v ) · ~w

~v · (λ~w ) = v1(λw1) + v2(λw2) + v3(λw3)

λ(~v · ~w ) = λ(v1w1 + v2w2 + v3w3)

(λ~v ) · ~w = (λv1)w1 + (λv2)w2 + (λv3)w3

Since ordinary multiplication is associative and commutative, we know thatv1(λw1) = λv1w1 = (λv1)w1 and so on. Thus, we have ~v · (λ~w ) = (λ~v ) · ~w .

In addition, the distributive property of ordinary multiplication tells us that

λ(v1w1 + v2w2 + v3w3) = λv1w1 + λv2w2 + λv3w3

Thus, we know that all three quantities are equal

~v · (λ~w ) = λ(~v · ~w ) = (λ~v ) · ~w

13.3 SOLUTIONS 925

• Property 3:First we observe that

~v + ~w = (v1 + w1)~i + (v2 + w2)~j + (v3 + w3)~k .

Next we calculate the quantities ((~v + ~w ) · ~u ) and (~v · ~u + ~w · ~u ).

(~v + ~w ) · ~u = (v1 + w1)u1 + (v2 + w2)u2 + (v3 + w3)u3

~v · ~u + ~w · ~u = (v1u1 + v2u2 + v3u3) + (w1u1 + w2u2 + w3u3).

The distributive law of ordinary multiplication shows that (v1 + w1)u1 = v1u1 + w1u1, and so on. Thus, the dotproduct is distributive also:

(~v + ~w ) · ~u = ~v · ~u + ~w · ~u40. Property 2 says that multiplying one of the vectors by a scalar simply multiplies the dot product by the same scalar. If

λ > 0, then when one vector is multiplied by λ, the angle between the vectors does not change, but the length of onevector, and hence the dot product, is multiplied by λ. The result remains true when λ < 0. For a justification in the casewhen λ < 0, see Problem 44 on page 926.

41. We want to show that (~b · ~c )~a − (~a · ~c )~b and ~c are perpendicular. We do this by taking their dot product:

((~b · ~c )~a − (~a · ~c )~b ) · ~c = (~b · ~c )(~a · ~c )− (~a · ~c )(~b · ~c ) = 0.

Since the dot product is 0, the vectors (~b · ~c )~a − (~a · ~c )~b and ~c are perpendicular.

42. Since ~u · ~w = ~v · ~w , (~u − ~v ) · ~w = 0. This equality holds for any ~w , so we can take ~w = ~u − ~v . This gives

‖~u − ~v ‖2 = (~u − ~v ) · (~u − ~v ) = 0,

that is,‖~u − ~v ‖ = 0.

This implies ~u − ~v = 0, that is, ~u = ~v .

43. If ~u = ~0 , then both sides of the equation are zero. If ~u 6= ~0 , write ~v parallel, ~w parallel, and (~v + ~w )parallel for thecomponents of ~v , ~w , and ~v + ~w in the direction of ~u . Then Figure 13.36 shows that

~v parallel + ~w parallel = (~v + ~w )parallel.

So (~v · ~u‖~u ‖2

)~u +

(~w · ~u‖~u ‖2

)~u =

((~v + ~w ) · ~u‖~u ‖2

)~u .

Thus, since ~u 6= ~0 , we deduce that~v · ~u‖~u ‖2 +

~w · ~u‖~u ‖2 −

(~v + ~w ) · ~u‖~u ‖2 = 0,

so~v · ~u + ~w · ~u = (~v + ~w ) · ~u .

44. Suppose θ is the angle between ~u and ~v .

(a) By the definition of scalar multiplication, we know that−~v is in the opposite direction of ~v , so the angle between ~uand −~v is π − θ. (See Figure 13.25.) Hence,

~u · (−~v ) = ‖~u ‖‖ − ~v ‖ cos(π − θ)= ‖~u ‖‖~v ‖(− cos θ)

= −(~u · ~v )

−~v ~v

θπ − θ

~u

Figure 13.25

λ~v

λ~u

λ < 0

~u

~v

θ

π − θ

π − θ

Figure 13.26


(b) If λ < 0, the angle between ~u and λ~v is π − θ, and so is the angle between λ~u and ~v . (See Figure 13.26.) So wehave,

~u · (λ~v ) = ‖~u ‖‖λ~v ‖ cos(π − θ)= |λ|‖~u ‖‖~v ‖(− cos θ)

= −λ‖~u ‖‖~v ‖(− cos θ) since |λ| = −λ= λ‖~u ‖‖~v ‖ cos θ

= λ(~u · ~v )

By a similar argument, we have

(λ~u ) · ~v = ‖λ~u ‖‖~v ‖ cos(π − θ)= −λ‖~u ‖‖~v ‖(− cos θ)

= λ(~u · ~v )

45. Let ~u and ~v be the displacement vectors from C to the other two vertices. Then

c2 = ‖~u − ~v ‖2

= (~u − ~v ) · (~u − ~v )

= ~u · ~u − ~v · ~u − ~u · ~v + ~v · ~v= ‖~u ‖2 − 2‖u‖‖v‖ cosC + ‖~v ‖2

= a2 − 2ab cosC + b2

46. We substitute ~u = u1~i + u2

~j + u3~k and by the result of Problem 43, we expand as follows:

(~u · ~v )geom = (u1~i + u2

~j + u3~k ) · ~v

= (u1~i ) · ~v + (u2

~j ) · ~v + (u3~k ) · ~v

where all the dot products are defined geometrically By the result of Problem 44 we can write

(~u · ~v )geom = u1(~i · ~v )geom + u2(~j · ~v )geom + u3(~k · ~v )geom.

Now substitute ~v = v1~i + v2

~j + v3~k and expand, again using Problem 43 and the geometric definition of the dot

product:

(~u · ~v )geom = u1

(~i · (v1

~i + v2~j + v3

~k ))

geom

+u2

(~j · (v1

~i + v2~j + v3

~k ))

geom

+u3

(~k · (v1

~i + v2~j + v3

~k ))

geom

= u1v1(~i ·~i )geom + u1v2(~i ·~j )geom + u1v3(~i · ~k )geom

+u2v1(~i ·~i )geom + u2v2(~i ·~j )geom + u2v3(~i · ~k )geom

+u3v1(~i ·~i )geom + u3v2(~i ·~j )geom + u3v3(~i · ~k )geom

The geometric definition of the dot product shows that

~i ·~i = ‖~i ‖ ‖~i ‖ cos 0 = 1

~i ·~j = ‖~i ‖ ‖~j ‖ cosπ

2= 0.

Similarly ~j ·~j = ~k · ~k = 1 and~i · ~k = ~j · ~k = 0. Thus, the expression for (~u · ~v )geom becomes

(~u · ~v )geom = u1v1(1) + u1v2(0) + u1v3(0)

+u2v1(0) + u2v2(1) + u2v3(0)

+u3v1(0) + u3v2(0) + u3v3(1)

= u1v1 + u2v2 + u3v3.

13.4 SOLUTIONS 927

47. (a) Since q(t) = (~v + t ~w ) · (~v + t ~w ) = ‖~v + t ~w ‖2 and since the length of any vector is nonnegative, we must have

q(t) = ‖~v + t ~w ‖2 ≥ 0

for all real t.(b) Using the distributive law

q(t) = (~v + t ~w ) · (~v + t ~w ) = ~v · ~v + t ~w · ~v + ~v · t ~w + t2 ~w · ~w= ‖~v ‖2 + 2(~v · ~w )t+ ‖~w ‖2t2.

If ~w 6= 0, then ‖~w ‖ 6= 0 and q(t) is quadratic in t.(c) Since q(t) ≥ 0, the quadratic has one repeated root or no roots, so the discriminant must be less than or equal to zero.

Thus,(2~v · ~w )2 − 4‖~v ‖2‖~w ‖2 ≤ 0.

Taking square roots, we have|~v · ~w | ≤ ‖~v ‖‖~w ‖.

If ~w = 0, then q(t) is no longer a quadratic. However, in that case,

|~v · ~w | = 0 = ‖~v ‖‖~w ‖

so the inequality still holds.


Exercises

1. ~v × ~w = ~k ×~j = −~i (remember~i ,~j ,~k are unit vectors along the axes, and you must use the right hand rule.)

2. ~v = −~i , and ~w = ~j + ~k

~v × ~w =

∣∣∣∣∣∣

~i ~j ~k

−1 0 0

0 1 1

∣∣∣∣∣∣= ~j − ~k

3. ~v =~i + ~k , and ~w =~i +~j

~v × ~w =

∣∣∣∣∣∣

~i ~j ~k

1 0 1

1 1 0

∣∣∣∣∣∣= −~i +~j + ~k

4. ~v =~i +~j + ~k , and ~w =~i +~j − ~k

~v × ~w =

∣∣∣∣∣∣

~i ~j ~k

1 1 1

1 1 −1

∣∣∣∣∣∣= −2~i + 2~j

5. ~v = 2~i − 3~j + ~k , and ~w =~i + 2~j − ~k

~v × ~w =

∣∣∣∣∣∣

~i ~j ~k

2 −3 1

1 2 −1

∣∣∣∣∣∣=~i + 3~j + 7~k

6. ~v = 2~i −~j − ~k , and ~w = −6~i + 3~j + 3~k

~v × ~w =

∣∣∣∣∣∣

~i ~j ~k

2 −1 −1

−6 3 3

∣∣∣∣∣∣= 0~i + 0~j + 0~k = ~0 .


7.

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

8.

(~i +~j )× (~i ×~j ) = (~i +~j )× ~k= (~i × ~k ) + (~j × ~k )

= −~j +~i =~i −~j .

9. By the definition of cross product, 2~i × (~i + ~j ) is in the direction of ~k . The magnitude of it equals to the area of theparallelogram which is

‖2~i ‖ · ‖~i +~j ‖ sinπ

4= 2√

2 sinπ

4= 2√

2 ·√

2

2= 2.

So 2~i × (~i +~j ) = 2~k . See Figure 13.27.

x

y

z

2~i

~i

~k

~j

~i +~j

Figure 13.27

x

z

y

~i

~i −~j

~k

~j

~i +~j

Figure 13.28

10. By definition, (~i +~j )× (~i −~j ) is in the direction of −~k . The magnitude is

‖~i +~j ‖ · ‖~i −~j ‖ sinπ

4=√

2 ·√

2 = 2.

So (~i +~j )× (~i −~j ) = −2~k . See Figure 13.28.

11. We find that ~v × ~w = −6~i + 7~j + 8~k and ~w × ~v = 6~i − 7~j − 8~k . Notice that

~v × ~w = −(~w × ~v ).

12.

~a ×~b =

∣∣∣∣∣∣

~i ~j ~k

3 1 −1

1 −4 2

∣∣∣∣∣∣

=

∣∣∣∣1 −1

−4 2

∣∣∣∣~i −∣∣∣∣3 −1

1 2

∣∣∣∣~j +

∣∣∣∣3 1

1 −4

∣∣∣∣~k

= −2~i − 7~j − 13~k .

13.4 SOLUTIONS 929

Since~a · (~a ×~b ) = 3(−2) + (−7)− (−13) = 0

and~b · (~a ×~b ) = 1(−2)− 4(−7) + 2(−13) = 0,

~a ×~b is perpendicular to both ~a and~b .

13. We can form the displacement vectors ~a = −~i +~j + 0~k from (1, 0, 0) to (0, 1, 0) and~b = −~i + 0~j +~k from (1, 0, 0)

to (0, 0, 1). A normal vector to the plane is ~a × ~b = ~i + ~j + ~k . Using the point (1, 0, 0), the plane can be written as(x− 1) + y + z = 0 or x+ y + z = 1.

14. The displacement vector from (3, 4, 2) to (−2, 1, 0) is:

~a = −5~i − 3~j − 2~k .

The displacement vector from (3, 4, 2) to (0, 2, 1) is:

~b = −3~i − 2~j − ~k .Therefore the vector normal to the plane is:

~n = ~a ×~b = −~i +~j + ~k .

Using the first point, the equation of the plane can be written as:

−(x− 3) + (y − 4) + (z − 2) = 0.

The equation of the plane is thus:−x+ y + z = 3.

Problems

15. (a) Since ~v · ~w = ||~v || ||~w || cos θ and ||~v × ~w || = ||~v || ||~w || sin θ,

tan θ =sin θ

cos θ=||~v × ~w ||~v · ~w =

3

5= 0.6.

(b) Then θ = tan−1(0.6) = 0.540.

16. Since‖~v × ~w ‖ = ‖~v ‖ · ‖~w ‖ sin θ,

and~v · ~w = ‖~v ‖ · ‖~w ‖ cos θ,

so‖~v × ~w ‖~v · ~w =

‖~v ‖ · ‖~w ‖ sin θ

‖~v ‖ · ‖~w ‖ cos θ= tan θ,

so

tan θ =‖2~i − 3~j + 5~k ‖

3

√38

3≈ 2.05.

17. Since ~v × ~w is perpendicular to both ~v and ~w , we can conclude that ~v × ~w is parallel to the z-axis.

18. (a) We first find two displacement vectors: −→AB = (3 − (−1))~i + (2 − 3)~j + (4 − 0)~k = 4~i − ~j + 4~k and−→AC = 2~i − 4~j + 5~k . The normal vector, ~n , to the plane is perpendicular to these two vectors, so we have

~n =−→AB ×−→AC = 11~i − 12~j − 14~k .

Using the normal vector, we see that the equation of the plane is 11x−12y−14z = d for some number d. Substitutingone of the points gives d = −47. Therefore, an equation for the plane is

11x− 12y − 14z = −47.

(b) The area of the triangle is given by

Area =1

2‖−→AB ×−→AC‖ =

1

2

√461 = 10.74.


19. (a) If we let −−→PQ in Figure 13.29 be the vector from point P to point Q and −→PR be the vector from P to R, then−−→PQ = −~i + 2~k

−→PR = 2~i − ~k ,

then the area of the parallelogram determined by −−→PQ and −→PR is:

Area ofparallelogram = ‖−−→PQ×−→PR‖ =

∣∣∣∣∣∣

∣∣∣∣∣∣

∣∣∣∣∣∣

~i ~j ~k

−1 0 2

2 0 −1

∣∣∣∣∣∣

∣∣∣∣∣∣

∣∣∣∣∣∣= ‖3~j ‖ = 3.

Thus, the area of the triangle PQR is(

Area oftriangle

)=

1

2

(Area of

parallelogram

)=

3

2= 1.5.

x

y

z

P = (0, 1, 0)

Q = (−1, 1, 2)

(2, 1,−1) = R

Figure 13.29

(b) Since ~n =−−→PQ × −→PR is perpendicular to the plane PQR, and from above, we have ~n = 3~j , the equation of the

plane has the form 3y = C. At the point (0, 1, 0) we get 3 = C, therefore 3y = 3, i.e., y = 1.

20. The normal vectors to the two planes are ~n 1 = 4~i − 3~j + 2~k and ~n 2 = ~i + 5~j − ~k . A vector parallel to the line ofintersection of the two planes is perpendicular to both these normal vectors, so

Vector parallel to line = ~n 1 × ~n 2 = −7~i + 6~j + 23~k .

21. The normal vectors to the planes are ~n1 = 2~i − 3~j + 5~k and ~n2 = 4~i + ~j − 3~k . The line of intersection isperpendicular to both normal vectors (picture the pages in a partially open book). Hence the vector we need is ~n1 × ~n2 =4~i + 26~j + 14~k .

22. The vector parallel to the line of intersection is 4~i + 26~j + 14~k and this is normal to the desired plane. Therefore,4x+ 26y + 14z = 0 is the equation of the plane.

23. We use the same normal vector ~n = 4~i +26~j +14~k and the point (4, 5, 6) to get 4(x−4)+26(y−5)+14(z−6) = 0.

24. Normal vectors to the planes are

~n1 =~i −~j + ~k and ~n2 = 2~i +~j − 2~k .

The vector ~n1 × ~n2 is perpendicular to both planes and is normal to the plane we want:

~n1 × ~n2 =

∣∣∣∣∣∣

~i ~j ~k

1 −1 1

2 1 −2

∣∣∣∣∣∣=~i + 4~j + 3~k .

The plane through the origin with normal ~n1 × ~n2 is

x+ 4y + 3z = 0.

13.4 SOLUTIONS 931

25. First let~a = a1

~i + a2~j + a3

~k ~b = b1~i + b2~j + b3~k ~c = c1~i + c2~j + c3~k

so~b + ~c = (b1 + c1)~i + (b2 + c2)~j + (b3 + c3)~k . Now, using the general formula for cross products, we have:

~a × (~b + ~c )

= [a2(b3 + c3)− a3(b2 + c2)]~i + [a3(b1 + c1)− a1(b3 + c3)]~j + [a1(b2 + c2)− a2(b1 + c1)]~k

= (a2b3 + a2c3 − a3b2 − a3c2)~i + (a3b1 + a3c1 − a1b3 − a1c3)~j

+(a1b2 + a1c2 − a2b1 − a2c1)~k

= (a2b3 − a3b2)~i + (a2c3 − a3c2)~i + (a3b1 − a1b3)~j + (a3c1 − a1c3)~j

+(a1b2 − a2b1)~k + (a1c2 − a2c1)~k

= (a2b3 − a3b2)~i + (a3b1 − a1b3)~j + (a1b2 − a2b1)~k + (a2c3 − a3c2)~i + (a3c1 − a1c3)~j

+(a1c2 − a2c1)~k

= (~a ×~b ) + (~a × ~c )

Thus, ~a × (~b + ~c ) = ~a ×~b + ~a × ~c .

26. Any vector ~v that is perpendicular to both ~a and ~b will have the property that its dot product with ~a and ~b is 0, that is

~a · ~v = a1x+ a2y + a3z = 0,

~b · ~v = b1x+ b2y + b3z = 0.

Multiply the first equation by b1 and the second by a1 and subtract to get

(b1a2 − a1b2)y + (b1a3 − a1b3)z = 0 or y =−(b1a3 − a1b3)z

(b1a2 − a1b2)(for b1a2 6= a1b2)

Multiply the second equation by a2 and the first by b2 and subtract to get

(b2a1 − a2b1)x+ (b2a3 − a2b3)z = 0 or x =−(b2a3 − a2b3)z

(b2a1 − a2b1).

So

~v =−(b2a3 − a2b3)z

(b2a1 − a2b1)~i − (b1a3 − a1b3)z

(b1a2 − a1b2)~j + z~k .

Pick z = b2a1 − b1a2 and multiply out, and we see that the algebraic method of finding a cross product yields the sameresult as our standard method.

27. (a) Figure 13.30 shows the vectors ~a , ~b , and ~c satisfying the conditions 0 < a2 < a1 and 0 < b1 < b2.

y

x

~a = a1~i + a2~j

~b = b1~i + b2~j

~c = −a2~i + a1~j

Height ofparallelogram

θ�

�

Figure 13.30

(b) ~c ·~a = −a2a1 + a1a2 = 0 and ~c ·~c = a22 + a1

2 = ‖~a ‖2. Thus ~c is orthogonal (perpendicular) to ~a and has thesame length as ~a .

(c) ~c ·~b = −a2b1 + a1b2. Since a1 > a2 > 0, and b2 > b1 > 0, we know that ~c ·~b is positive.


(d) If θ is the angle between ~a and~b and α is the angle between ~c and~b , then α = (π2− θ). Thus cosα = sin θ, so

~c ·~b = ‖~c ‖‖~b ‖ cosα = ‖~a ‖‖~b ‖ sin θ.

Since ‖~a ‖ = Base of the parallelogram and‖~b ‖ sin θ = Height of the parallelogram, we have

~c ·~b = Base · Height = Area of the parallelogram formed by ~a and ~b .

(e) By the right-hand rule, ~a ×~b is in the direction of the positive z-axis. See Figure 13.31. Since we know that

Area of the parallelogram = ~c ·~b = a1b2 − a2b1.

the definition of ~a ×~b tells us that

~a ×~b = (Area of Parallelogram)~k = (~c ·~b )~k = (a1b2 − a2b1)~k .

Thus,~a ×~b = (a1b2 − a2b1)~k .

x

y

z

~a

~b

~a ×~b (parallel to~k )

Figure 13.31: Cross product of two vectors in the xy-plane

28. If λ = 0, then all three cross products are ~0 , since the cross product of the zero vector with any other vector is always 0.If λ > 0, then λ~v and ~v are in the same direction and ~w and λ~w are in the same direction. Therefore the unit

normal vector ~n is the same in all three cases. In addition, the angles between λ~v and ~w , and between ~v and ~w , andbetween ~v and λ~w are all θ. Thus,

(λ~v )× ~w = ‖λ~v ‖‖~w ‖ sin θ~n

= λ‖~v ‖‖~w ‖ sin θ~n

= λ(~v × ~w )

= ‖~v ‖‖λ~w ‖ sin θ~n

= ~v × (λ~w )

If λ < 0, then λ~v and ~v are in opposite directions, as are ~w and λ~w in opposite directions. Therefore if ~n is thenormal vector in the definition of ~v × ~w , then the right-hand rule gives −~n for (λ~v )× ~w and ~v × (λ~w ). In addition,if the angle between ~v and ~w is θ, then the angle between λ~v and ~w and between ~v and λ~w is (π− θ). Since if λ < 0,we have |λ| = −λ, so

(λ~v )× ~w = ‖λ~v ‖‖~w ‖ sin(π − θ)(−~n )

= |λ| ‖~v ‖‖~w ‖ sin(π − θ)(−~n )

= −λ‖~v ‖‖~w ‖ sin θ(−~n )

= λ‖~v ‖‖~w ‖ sin θ~n

= λ(~v × ~w ).

Similarly,

~v × (λ~w ) = ‖~v ‖‖λ~w ‖ sin(π − θ)(−~n )

= −λ‖~v ‖‖~w ‖ sin θ(−~n )

= λ(~v × ~w ).

13.4 SOLUTIONS 933

29. The quantities∣∣~a · (~b × ~c )

∣∣ and∣∣(~a ×~b ) · ~c

∣∣ both represent the volume of the same parallelepiped, namely that defined

by the three vectors ~a ,~b , and ~c , and therefore must be equal. Thus, the two triple products ~a · (~b × ~c ) and (~a ×~b ) · ~cmust be equal except perhaps for their sign. In fact, both are positive if ~a , ~b , ~c are right-handed and negative if ~a , ~b , ~care left-handed. This can be shown by drawing a picture:

~a

~c

~b

right-handed

~a

~b

~c

left-handed

Figure 13.32

30. If θ is the angle between ~a and~b , then

‖~a ×~b ‖2 = (‖~a ‖‖~b ‖ sin θ)2

= ‖~a ‖2‖~b ‖2 sin2 θ

= ‖~a ‖2‖~b ‖2(1− cos2 θ)

= ‖~a ‖2‖~b ‖2 − ‖~a ‖2‖~b ‖2 cos2 θ

= ‖~a ‖2‖~b ‖2 − (~a ·~b )2.

31. Write ~v and ~w in components and expand using the distributive property of the cross product.

~v × ~w = (v1~i + v2

~j + v3~k )× (w1

~i + w2~j + w3

~k )

= v1w1~i ×~i + v1w2

~i ×~j + v1w3~i × ~k

+v2w1~j ×~i + v2w2

~j ×~j + v2w3~j × ~k

+v3w1~k ×~i + v3w2

~k ×~j + v3w3~k × ~k

Now we use the fact that ~i ×~i = ~0 ,~i × ~j = ~k ,~i × ~k = −~j ,~j ×~i = −~k ,~j × ~j = ~0 ,~j × ~k = ~i ,~k ×~i =~j ,~k ×~j = −~i ,~k × ~k = ~0 . Thus we have

~v × ~w = ~0 + v1w2~k + v1w3(−~j ) + v2w1(−~k ) + ~0 + v2w3

~i + v3w1~j + v3w2(−~i ) +~0

= (v2w3 − v3w2)~i + (v3w1 − v1w3)~j + (v1w2 − v2w1)~k .

32. (a) Since ~c is perpendicular to ~a ×~b , and since ~a ×~b is normal to the plane containing ~a and~b , it follows that ~c mustbe in the plane containing ~a and~b .

(b) Using the expression given in the problem for ~c , we get

~a · ~c = ~a · (~a × (~b × ~a ))

= (~a × ~a ) · (~b × ~a )

= ~0 · (~b × ~a ) = 0.

and~b · ~c = ~b · (~a × (~b × ~a ))

= (~b × ~a ) · (~b × ~a )

= ‖~b × ~a ‖2

= ‖~a ‖2‖~b ‖2 − (~a ·~b )2.


(c) Since ~c lies in the plane containing ~a and ~b , it is of the form ~c = x~a + y~b for some scalars x and y. Thus, usingthe fact that ~a · ~c = 0 from part (b), we have

~a · ~c = ~a · (x~a + y~b ) = x‖~a ‖2 + y(~a ·~b ) = 0.

Similarly, using the fact that~b · ~c = ‖~a ‖2‖~b ‖2 − (~a ·~b )2 from part (b), we have

~b · ~c = ~b · (x~a + y~b ) = x(~a ·~b ) + y‖~b ‖2 = ‖~a ‖2‖~b ‖2 − (~a ·~b )2.

Solving these two linear equations in x and y, we find x = −~a ·~b and y = ‖~a ‖2.

33. Problem 29 tells us that (~u × ~v ) · ~w = ~u · (~v × ~w ). Using this result on the triple product of (~a + ~b ) × ~c with anyvector ~d together with the fact that the dot product distributes over addition gives us:

[(~a +~b )× ~c ] · ~d = (~a +~b ) · (~c × ~d )

= ~a · (~c × ~d ) +~b · (~c × ~d ) (dot product is distributive)

= (~a × ~c ) · ~d + (~b × ~c ) · ~d (using Problem 29 again)

= [(~a × ~c ) + (~b × ~c )] · ~d . (dot product is distributive)

So, since [(~a +~b )× ~c ] · ~d = [(~a × ~c ) + (~b × ~c )] · ~d , then

[(~a +~b )× ~c ] · ~d − [(~a × ~c ) + (~b × ~c )] · ~d = 0,

Since the dot product is distributive, we have

[((~a +~b )× ~c )− (~a × ~c )− (~b × ~c )] · ~d = 0.

Since this equation is true for all vectors ~d , by letting

~d = ((~a +~b )× ~c )− (~a × ~c )− (~b × ~c ),

we get‖(~a +~b )× ~c − ~a × ~c −~b × ~c ‖2 = 0

and hence(~a +~b )× ~c − (~a × ~c )− (~b × ~c ) = ~0 .

Thus(~a +~b )× ~c = (~a × ~c ) + (~b × ~c ).

34. The area vector for face OAB = 12~b × ~a .

The area vector for face OBC = 12~a × ~c .

The area vector for face OAC = 12~b × ~c .

The area vector for face ABC = 12(~b − ~a )× (~c − ~a ).

1

2~b × ~a +

1

2~c ×~b +

1

2~a × ~c +

1

2(~b − ~a )× (~c − ~a ) =

1

2~b × ~a +

1

2~c ×~b +

1

2~a × ~c +

1

2(~b × ~c −~b × ~a − ~a × ~c − ~a × ~a ) = 0.

Solutions for Chapter 13 Review

Exercises

1. ~v + 2~w = 2~i + 3~j − ~k + 2(~i −~j + 2~k ) = 4~i +~j + 3~k .

2. 3~v − ~w − ~v = 2~v − ~w = 2(2~i + 3~j − ~k )− (~i −~j + 2~k ) = 3~i + 7~j − 4~k .

SOLUTIONS to Review Problems for Chapter Thirteen 935

3. ~v · ~w = (2~i + 3~j − ~k ) · (~i −~j + 2~k ) = 2 · 1 + 3 · (−1) + (−1) · 2 = −3.

4. ~v × ~w =

∣∣∣∣∣∣

~i ~j ~k

2 3 −1

1 −1 2

∣∣∣∣∣∣= (6− 1)~i − (4 + 1)~j + (−2− 3)~k = 5~i − 5~j − 5~k .

5. For any vector ~v , we have ~v × ~v = ~0 .

6. ||~v + ~w || = ||3~i + 2~j + ~k || =√

32 + 22 + 12 =√

14.

7. Since ~v · ~w = 2 · 1 + 3(−1) + (−1)2 = −3, we have (~v · ~w )~v = −6~i − 9~j + 3~k .

8. We have ~v × ~w = 5~i − 5~j − 5~k , so

(~v × ~w )× ~w =

∣∣∣∣∣∣

~i ~j ~k

5 −5 −5

1 −1 2

∣∣∣∣∣∣= (−10− 5)~i − (10 + 5)~j + (−5 + 5)~k = −15~i − 15~j .

9. Since ~v × ~w is perpendicular to ~w , we have (~v × ~w ) · ~w = 0.

10. The cross product of two parallel vectors is ~0 , so the cross product of any vector with itself is ~0 .

11. (a) We have ~v · ~w = 3 · 4 + 2 · (−3) + (−2) · 1 = 4.(b) We have ~v × ~w = −4~i − 11~j − 17~k .(c) A vector of length 5 parallel to ~v is

5

‖~v ‖~v =5√17

(3~i + 2~j − 2~k ) = 3.64~i + 2.43~j − 2.43~k .

(d) The angle between vectors ~v and ~w is found using

cos θ =~v · ~w‖~v ‖‖~w ‖ =

4√17√

26= 0.190,

so θ = 79.0◦.(e) The component of vector ~v in the direction of vector ~w is

~v · ~w‖~w ‖ =

4√26

= 0.784.

(f) The answer is any vector ~a such that ~a · ~v = 0. One possible answer is 2~i − 2~j + ~k .(g) A vector perpendicular to both is the cross product:

~v × ~w = −4~i − 11~j − 17~k .

12. Since ||2~i + 3~j − ~k || =√

22 + 32 + (−1)2 =√

14, vectors of length 10 are

± 10√14

(2~i + 3~j − ~k ).

13. We take the cross product of~i +~j and~i −~j − ~k and then make a unit vector parallel to the cross product.

(~i +~j )× (~i −~j − ~k ) =

∣∣∣∣∣∣

~i ~j ~k

1 1 0

1 −1 −1

∣∣∣∣∣∣= −~i +~j − 2~k .

Since || −~i +~j − 2~k || =√

(−1)2 + 12 + (−2)2 = 6, unit vectors are

±−~i +~j − 2~k√

6.


14. We want a unit vector of the form a~i + b~j such that

(a~i + b~j ) · (3~i − 2~j ) = 3a− 2b = 0.

Let’s take a = 2 and b = 3. Then the vector 2~i + 3~j is perpendicular to 3~i − 2~j , but 2~i + 3~j is not a unit vector. Since||2~i + 3~j || =

√13, unit vectors are

±2~i + 3~j√13

.

15. The vector ~w we want is shown in Figure 13.33, where the given vector is ~v = 4~i + 3~j . The vectors ~v and ~w are thesame length and the two angles marked α are equal, so the two right triangles shown are congruent. Thus

a = −3 and b = 4.

Therefore~w = −3~i + 4~j .

3 4

34

(a, b)

(4, 3)

~v

~w

α

αx

y

Figure 13.33

Problems

16. (a) We need 6~i + 8~j + 3~k = λ(2~i + (t2 + 23t+ 1)~j + t~k ) for some λ. This gives

6 = 2λ

8 = (t2 +2

3t+ 1)λ

3 = tλ

From the first equation, we have λ = 3. Substituting λ = 3 into the third equation gives t = 1. Check the secondequation, it says 8 = 8, if t = 1 and λ = 3. So for t = 1, the two vectors are parallel to each other.

(b) Similar to part (a), we need to solve

2 = tλ

−4 = λ

1 = λ(t− 1)

From the first two equations we have λ = −4 and t = − 12

. Substituting this into the third equation gives 1 = 6.Thus this system of equations has no solution, so the pair of vectors is not parallel to each other for any value of t.

(c) 2t~i + t~j + t~k = t3(6~i + 3~j + 3~k ). For any t, the two vectors are parallel to each other.

17. (a) Since ~v · ~w = ||~v || ||~w || cos θ and ||~v × ~w || = ||~v || ||~w || sin θ, we find

||~v × ~w || = ||12~i − 3~j + 4~k || =√

122 + (−3)2 + 42 = 13.

Then

tan θ =sin θ

cos θ=||~v × ~w ||~v · ~w =

13

8= 1.625.

(b) Then θ = tan−1(1.625) = 1.019.


18. ~n = 4~i + 6~k (the coefficients of x, y, z are the same as the coefficients of~i , ~j , and ~k .)

19. If the planes are parallel, they have a common normal vector ~n . Rewrite the equation of the plane as 4x− 3y − z = −8so that ~n = 4~i − 3~j − ~k and the desired plane is 4(x− 0)− 3(y − 0)− (z − 0) = 0 or 4x− 3y − z = 0.

20. (a) On the x-axis, y = z = 0, so 5x = 21, giving x = 215

. So the only such point is ( 215, 0, 0).

(b) Other points are (0,−21, 0), and (0, 0, 3). There are many other possible answers.(c) ~n = 5~i −~j + 7~k . It is the normal vector.(d) The vector between two points in the plane is parallel to the plane. Using the points from part (b), the vector 3~k −

(−21~j ) = 21~j + 3~k is parallel to the plane.

21. Let ~r 1 be the displacement vector −−→PQ and let ~r 2 be the displacement vector −→PR. Then

~r 1 = (1 + 2)~i + (3− 2)~j + (−1− 0)~k = 3~i +~j − ~k ,~r 2 = (−4 + 2)~i + (2− 2)~j + (1− 0)~k = −2~i + ~k ,

~r 1 × ~r 2 =

∣∣∣∣∣∣

~i ~j ~k

3 1 −1

−2 0 1

∣∣∣∣∣∣=~i − (3− 2)~j + 2~k =~i −~j + 2~k .

The area of the triangle = 12‖~r 1 × ~r 2‖ = 1

2

√12 + 12 + 22 =

√6

2.

22. (a) The displacement vector −→AB lies in the plane and is given by

−→AB = (0− 2)~i + (1− 1)~j + (3− 0)~k = −2~i + 3~k .

Similarly, the displacement vector −→AC also lies in the plane,

−→AC = (1− 2)~i + (0− 1)~j + (1− 0)~k = −~i −~j + ~k .

(b) The vector ~n =−→AB ×−→AC is perpendicular to both −→AB and −→AC and is therefore perpendicular to the plane.

−→AB ×−→AC = (−2~i + 3~k )× (−~i −~j + ~k ) = 3~i −~j + 2~k .

(c) The normal vector to the plane is ~n = 3~i −~j + 2~k , so the equation is of the form

3x− y + 2z = d.

Substituting, for example, x = 1, y = 0, z = 1 gives d = 5:

3x− y + 2z = 5.

23. (a) Since −−→PQ = (3~i + 5~j + 7~k )− (~i + 2~j + 3~k ) = 2~i + 3~j + 4~k ,

and

−→PR = (2~i + 5~j + 3~k )− (~i + 2~j + 3~k ) =~i + 3~j ,

−−→PQ×−→PR =

∣∣∣∣∣∣

~i ~j ~k

2 3 4

1 3 0

∣∣∣∣∣∣= −12~i + 4~j + 3~k ,

which is a vector perpendicular to the plane containing P , Q and R. Since

‖−−→PQ×−→PR‖ =√

(−12)2 + 42 + 32 = 13,

the unit vectors which are perpendicular to a plane containing P , Q, and R are

−12

13~i +

4

13~j +

3

13~k ,

or the unit vector pointing to the opposite direction,

12

13~i − 4

13~j − 3

13~k .


(b) The angle between PQ and PR is θ for which

cos θ =−−→PQ · −→PR

‖−−→PQ‖ · ‖−→PR‖=

2 · 1 + 3 · 3 + 4 · 0√22 + 32 + 42 ·

√12 + 32 + 02

=11√290

,

soθ = cos−1 (

11√290

) ≈ 49.76◦.

(c) The area of triangle PQR = 12‖−−→PQ×−→PR‖ = 13

2.

(d) Let d be the distance from R to the line through P and Q (see Figure 13.34), then

1

2d · ‖−−→PQ‖ = the area of 4 PQR =

13

2.

Therefore,

d =13

‖−−→PQ‖=

13√22 + 32 + 42

=13√29.

P

R

Q

d

Figure 13.34

24. Find an arbitrary point on the plane 2x + 4y − z = −1, say A = (0, 0, 1). The normal ~n to the plane at B is ~n =

2~i + 4~j − ~k and −→PA = −2~i +~j − 2~k . See Figure 13.35.

θ

BA

~n

P

Figure 13.35

So the distance d from the point P to the plane is

d = ‖−−→PB‖ = ‖−→PA‖ cos θ

=−→PA · ~n‖~n ‖ since −→PA · ~n = ‖−→PA‖‖~n ‖ cos θ)

=(−2~i +~j − 2~k ) · (2~i + 4~j − ~k )√

22 + 42 + (−1)2

=2√21.


25. The displacement from (1, 1, 1) to (1, 4, 5) is

~r1 = (1− 1)~i + (4− 1)~j + (5− 1)~k = 3~j + 4~k .

The displacement from (−3,−2, 0) to (1, 4, 5) is

~r2 = (1 + 3)~i + (4 + 2)~j + (5− 0)~k = 4~i + 6~j + 5~k .

A normal vector is

~n = ~r1 × ~r2 =

∣∣∣∣∣∣

~i ~j ~k

0 3 4

4 6 5

∣∣∣∣∣∣= (15− 24)~i − (−16)~j + (−12)~k = −9~i + 16~j − 12~k .

The equation of the plane is

−9x+ 16y − 12z = −9 · 1 + 16 · 1− 12 · 1 = −5

9x− 16y + 12z = 5.

We pick a point A on the plane, A = ( 59, 0, 0) and let P = (0, 0, 0). (See Figure 13.36.) Then ~PA = (5/9)~i .

θ

BA

~n

P

Figure 13.36

So the distance d from the point P to the plane is

d = ‖−−→PB‖ = ‖−→PA‖ cos θ

=−→PA · ~n‖~n ‖ since −→PA · ~n = ‖−→PA‖‖~n ‖ cos θ)

=

∣∣∣∣( 5

9~i ) · (−9~i + 16~j − 12~k )√

92 + 162 + 122

∣∣∣∣

=5√481

= 0.23.

26. Suppose ~u represents the velocity of the plane relative to the air and ~w represents the velocity of the wind. We can addthese two vectors by adding their components. Suppose north is in the y-direction and east is the x-direction. The vectorrepresenting the airplane’s velocity makes an angle of 45◦ with north; the components of ~u are

~u = 700 sin 45◦~i + 700 cos 45◦~j ≈ 495~i + 495~j .

Since the wind is blowing from the west, ~w = 60~i . By adding these we get a resultant vector ~v = 555~i + 495~j . Thedirection relative to the north is the angle θ shown in Figure 13.37 given by

θ = tan−1 x

y= tan−1 555

495

≈ 48.3◦


The magnitude of the velocity is

‖~v ‖ =√

4952 + 5552 =√

553,050

= 744 km/hr.

~w

θ

~u

~v

E

N

45◦

0 -� x

6

?

y

Figure 13.37: Note that θ is the angle between north and the vector ~v

27. (a) Let x-axis be the East direction and y-axis be the North direction. From Figure 13.38,

θ = sin−1(4/5) = 53.1◦.

That is, he should steer at 53.1◦ east of south.

*

*

*

-

66

x

yN

θ 5 km/hr

Steering direction

4 km/hr

Figure 13.38

(b)

-66

x

yN

θ

4 km/hr

π4

10 km/hr

~R

5 km/hr

~R

Figure 13.39

Let ~R be the resultant of the wind and river velocities, that is

~R = −4~i + (10 cos(π

4)~i + 10 cos(

π

4)~j )

= (−4 + 5√

2)~i + 5√

2~j .


From Figure 13.39, we see that to get the the x-component of his rowing velocity and the x-component of ~R tocancel each other, we must have

5 sin θ = −4 + 5√

2

θ = sin−1

(−4 + 5

√2

5

)= 37.9◦.

However for this value of θ, the y-component of the velocity is

5√

2− 5 cos(37.9◦) = 3.1.

Since the y-component is positive, the man will not move across the river in a southward direction.

28. The speed of the particle before impact is v, so the speed after impact is 0.8v. If we consider the barrier as being alongthe x-axis (see Figure 13.40), then the~i -component is 0.8v cos 60◦ = 0.8v(0.5) = 0.4v.

Similarly, the ~j -component is 0.8v sin 60◦ = 0.8v(0.8660) ≈ 0.7v. Thus

~v after = 0.4v~i + 0.7v~j .

~v before

~v after

v sin 60◦

0.8v sin 60◦

60◦ 60◦

-�v cos 60◦

-�0.8v cos 60◦

x

y

Figure 13.40

29. (a) 500 km/hr in the west direction, so ~v = −500~i .(b) While traveling at constant altitude, the plane travels 250 km westward. Thus the coordinates of the point where the

plane begins to descend are (550, 60, 4)− (250, 0, 0) = (300, 60, 4).(c) The vector from the plane to the airport at the time it begins its descent is (200~i + 10~j ) − (300~i + 60~j + 4~k ) =

−100~i − 50~j − 4~k . Velocity is a vector of length 200 km/hr in the direction of −100~i − 50~j − 4~k . Since√(−100)2 + (−50)2 + (−4)2 ≈ 111.9, a unit vector in the direction of descent is − 100

111.9~i − 50

111.9~j − 4

111.9~k .

ThusVelocity vector = 200

(− 100

111.9~i − 50

111.9~j − 4

111.9~k)

= −178.7~i − 89.4~j − 7.2~k .

30. (a) The displacement vector of the moon relative to the earth is

~r = 384~i .

The displacement vector of the spaceship relative to the earth is

~rE = 280~i + 90~j .

The displacement vector of the spaceship relative to the moon is

~rL = ~rE − ~r = −104~i + 90~j .

See Figure 13.41.(b) Distance of spaceship from Earth = || ~rE || =

√2802 + 902 =

√86500 = 294.109 thousand km.

Distance of spaceship from the moon = || ~rL || =√

(−104)2 + 902 =√

18916 = 137.535 thousand km.


(c) See Figure 13.41. The gravitational force of the earth, −→FE , is parallel to ~rE but of length 461 and in the oppositedirection:

−→FE = − 461√

186500(280~i + 90~j ) = −438.885~i − 141.070~j .

The gravitational force of the moon, −→FL, is parallel to ~rL but of length 26 and in the opposite direction:

−→FL = − 26√

18916(−104~i + 90~j ) = 19.660~i − 17.041~j .

The resulting force, ~F is~F =

−→FE +

−→FL = 419.225~i − 158.084~j .

Earth

~r E=

280~i+ 90~j

Spacecraft

~r L

Moon

~r = 384~i

Figure 13.41

31. Let ~R be the resultant force, and let ~F 1 and ~F 2 be the forces exerted by the larger and smaller tugs. See Figure 13.42.Then ‖ ~F1 ‖ = 5

4‖ ~F2 ‖. The y components of the vectors ~F 1 and ~F 2 must cancel each other in order to ensure that the

ship travels due east, hence‖ ~F1 ‖ sin 30◦ = ‖ ~F2 ‖ sin θ,

so5

4‖ ~F2 ‖ sin 30◦ = ‖ ~F2 ‖ sin θ,

giving sin θ = 58, and hence θ = sin−1 5

8= 38.7◦.

30◦~R

~F 1

~F 2

θ

6

-

6

x

yN

Figure 13.42

32. Let the x-axis point east and the y-axis point north. We use ~C , ~W , and ~E to represent the current, wind, and enginevectors, respectively. We resolve the current and wind velocity vectors into components. Since the current points 25◦

north of east with a speed of 12, we have

~C = 12 cos(25◦)~i + 12 sin (25◦)~j = 10.876~i + 5.071~j .

Since ~C lies in the first quadrant, both coefficients are positive.The wind points 80◦ south of east with a speed of 7 km/hr, so we have

~W = 7 cos(80◦)~i − 7 sin (80◦)~j = 1.216~i − 6.894~j .


Since ~W lies in the fourth quadrant, the coefficient of~i is positive and the coefficient of ~j is negative.The combined velocity on the boat is due east at a speed of 40 km/hr, so we want

~C + ~W + ~E = 40~i .

We solve for ~E :

~E = 40~i − ( ~C + ~W )

= 40~i − ((10.876~i + 5.071~j ) + (1.216~i − 6.894~j ))

= 40~i − (12.092~i − 1.823~j )

= 27.908~i + 1.823~j .

The engine should push the boat with a speed of ‖ ~E ‖ =√

27.9082 + 1.8232 = 27.97 km/hr, and in directionarctan(1.823/27.908) = 3.74◦ north of east.

33. Let the x-axis point east and the y-axis point north. Denote the forces exerted by Charlie, Sam and Alice by ~F C , ~F S and~F A (see Figure 13.43).

y

x~FC

~FS

~FA

62◦43◦

φ

Figure 13.43

Since ‖ ~F C‖ = 175 newtons and the angle θ from the x-axis to ~F C is 90◦ + 62◦ = 152◦, we have

~F C = 175 cos 152◦~i + 175 sin 152◦~j ≈ −154.52~i + 82.16~j .

Similarly,~F S = 200 cos 47◦~i + 200 sin 47◦~j ≈ 136.4~i + 146.27~j .

Now Alice is to counterbalance Sam and Charlie, so the resultant force of the three forces ~F C , ~F S and ~F A must be 0,that is,

~F C + ~F S + ~F A = 0.

Thus, we have

~F A = −~F C − ~F S

≈ −(−154.52~i + 82.16~j )− (136.4~i + 146.27~j )

= 18.12~i − 228.43~j

and, ‖ ~F A‖ =√

18.122 + (−228.43)2 ≈ 229.15 newtons.If φ is the angle from the x-axis to ~F A, then

φ = arctan−228.43

18.12≈ −85.5◦.


34. (a) Since the radius of the circle is 1 meter, the circumference is 2π meters. Thus, the object is moving at 2π me-ters/minute, or π/30 meters/second ≈ 0.11 meters/second.

(b) 30 seconds after passing the point (0, 1), the object is at the point (−1, 0). (Since it completes 1 revolution eachminute, it will move π radians in 30 seconds.) This is true regardless of whether the point is moving clockwiseor counterclockwise. However, since the velocity vector, ~v , is tangential to the curve in the direction of motion, itwill have an opposite sign if the motion is in the opposite direction. So, moving clockwise ~v = 2π~j , and movingcounterclockwise ~v = −2π~j , if the speed is measured in meters/minute.

35. The speed is a scalar which equals 30 times the circumference of the circle per minute. So it is a constant. The velocity is avector. Since the direction of the motion changes all the time, the velocity is not constant. This implies that the accelerationis nonzero.

36. Let ~v = vx~i + vy~j + vz~k be the vector. We will use the properties given in the problem to find vx, vy , and vz . If ~v hasmagnitude 10, then ‖~v ‖ = 10.

If ~v makes an angle of 45◦ with the x-axis, then its x-component, vx, is given by:

vx = ~v ·~i = ‖~v ‖ cos 45◦ = 10

(√2

2

)= 7.0710.

Similarly, if ~v makes a 75◦ angle with the y-axis, then its y-component, vy , is given by:

vy = ~v ·~j = ‖~v ‖ cos 75◦ = 10(0.25882) = 2.5882.

We now have two components of ~v :

~v = 7.0710~i + 2.5882~j + vz~k .

We only need to find vz . To do this we use the fact that√~v · ~v = ‖~v ‖ = 10.

~v · ~v = 100

v2x + v2

y + v2z = 100

v2z = 100− v2

x − v2y

v2z = ±

√100− v2

x − v2y

vz = ±6.580

Since the problem tells us that the ~k -component is positive, vz = +6.580. Thus

~v = 7.0710~i + 2.5882~j + 6.580~k .

37. (a)

x

y

z

TS

Q

αβ

γ

Figure 13.44

Suppose ~v =−−→OP as in Figure 13.44. The~i component of −−→OP is the projection of −−→OP on the x-axis:

−→OT = v cosα~i .


Similarly, the ~j and ~k components of −−→OP are the projections of −−→OP on the y-axis and the z-axis respectively. So:

−→OS = v cosβ~j−−→OQ = v cos γ~k

Since ~v =−→OT +

−→OS +

−−→OQ, we have

~v = v cosα~i + v cosβ~j + v cos γ~k .

(b) Since

v2 = ~v · ~v = (v cosα~i + v cosβ~j + v cos γ~k ) ·(v cosα~i + v cosβ~j + v cos γ~k )

= v2(cos2 α+ cos2 β + cos2 γ)

socos2 α+ cos2 β + cos2 γ = 1.

38. In Figure 13.45, let l1 be a line with direction vector ~v1 passing through P1. Let l2 be a line with direction vector ~v2

passing through P2. Lines l1 and l2 are skew if they are not parallel and do not intersect.

P1

P2

~v1

~v2

l1

l2

Figure 13.45

If we draw a line l3 parallel to l1, i.e., with a direction vector v1, passing through P2, as shown in Figure 13.46, thenl2 and l3 determine a plane that is parallel to l1:

P1

P2

~n

~v1

~v2

~v1

l1

l2

l3

Figure 13.46


The minimum distance between l1 and l2 is equal to the distance of l1 to the plane. So it is equal to the projection of−−−→P1P2 in the direction of the normal vector of the plane. The unit normal vector is given by:

~n =~v 1 × ~v 2

||~v 1 × ~v 2||,

so the component of −−−→P1P2 in the direction of ~n is:

−−−→P1P2 · ~n =

−−−→P1P2 · ~v 1 × ~v 2

||~v 1 × ~v 2||.

Thus the minimum distance between l1 and l2 is:

|−−−→P1P2 · (~v 1 × ~v 2)|||~v 1 × ~v 2||

.

CAS Challenge Problems

39. (~a ×~b ) · ~c = 0 , (~a ×~b )× (~a × ~c ) = ~0

Since ~c is the sum of a scalar multiple of ~a and a scalar multiple of ~b , it lies in the plane containing ~a and ~b . Onthe other hand, ~a ×~b is perpendicular to this plane, so ~a ×~b is perpendicular to ~c . Therefore, (~a ×~b ) · ~c = 0. Also,~a × ~c is also perpendicular to the plane, thus parallel to ~a ×~b , and thus (~a ×~b )× (~a × ~c ) = ~0 .

40. The first parallelepiped has volume

|(~a ×~b ) · ~c | = |ywr − vzr + zus− xws+ xvt− yut|.

The second has volume |(~a × ~b ) · (2~a −~b + ~c )|, which also simplifies to |ywr − vzr + zus − xws + xvt − yut|.Both parallelepipeds have base with edges ~a and ~b . The third edge of the first one is ~c and the third edge of the secondone is ~c + 2~a − ~b . Thus the top face of the second parallelepiped is obtained by shifting the top face of the first by2~a −~b . Since this is parallel to the base, the second parallelepiped has the same altitude as the first. Since the volume ofa parallelepiped is product of the area of its base with its height, the two parallelepipeds have the same volume.

41. (a) From the geometric definition of the dot product, we have

cos θ =|~a ·~b |‖~a ‖‖~b ‖

=10√14√

9.

Using sin2 θ = 1− cos2 θ, we get

x+ 2y + 3z = 0

2x+ y + 2z = 0

x2 + y2 + z2 = ‖~a ‖2‖~b ‖2(1− cos2 θ) = (14)(9)

(1− 100

(14)(9)

)

Solving these equations we get x = −1, y = −4, z = 3 or x = 1, y = 4, and z = −3. Thus ~c = −~i − 4~j + 3~kor ~c =~i + 4~j − 3~k .

(b) ~a ×~b = ~i + 4~j − 3~k . This is the same as one of the answers in part (a). The conditions in part (a) ensured that~c is perpendicular to ~a and ~b and that it has magnitude ‖~a ‖‖~b ‖| sin θ|. The cross product is the solution that, inaddition, satisfies the right-hand rule.

42. (a) We have

‖−→AB‖ = ‖2~i ‖ = 2

‖−→AC‖ = ‖~i +√

3~j ‖ =√

1 + 3 = 2

‖−−→AD‖ = ‖~i + (1/√

3)~j + 2√

2/3~k ‖ =√

1 + (1/3) + (8/3) =√

4 = 2

‖−−→BC‖ = ‖ −~i +√

3~j ‖ =√

1 + 3 = 2

‖−−→BD‖ = ‖ −~i + (1/√

3)~j + 2√

2/3~k ‖ =√

1 + (1/3) + (8/3) =√

4 = 2

‖−−→CD‖ = ‖(1/√

3−√

3)~j + 2√

2/3~k )‖ =√

(1/3− 2 + 3) + 8/3 =√

4 = 2

Thus all the points are 2 units apart.

CHECK YOUR UNDERSTANDING 947

(b) By solving the equations

x2 + y2 + z2 = (x− 2)2 + y2 + z2

x2 + y2 + z2 = (x− 1)2 + (y −√

3)2 + z2

x2 + y2 + z2 = (x− 1)2 + (y − 1/√

3)2 + (z − 2√

2/3)2

we get P = (1, 1/√

3,√

6/6).(c) The cosine of the angle APB is 1/3 and the angle is 109.471◦.

43. (a) −−→PQ×−→PR is perpendicular to the plane containing P ,Q,R, and therefore parallel to the normal vector a~i +b~j +c~k .(b)

−−→PQ×−→PR = (tv − sw − ty + wy + sz − vz)~i+

(−tu+ rw + tx− wx− rz + uz)~j + (su− rv − sx+ vx+ ry − uy)~k

(c) After substituting z = (d− ax− by)/c, w = (d− au− bv)/c, t = (d− ar − bs)/c into the result of part (a), andsimplifying the expression, we obtain:

−−→PQ×−→PR =

a(s(u− x) + vx− uy + r(−v + y))

c~i+

b(s(u− x) + vx− uy + r(−v + y))

c~j + (s(u− x) + vx− uy + r(−v + y))~k

=(s(u− x) + vx− uy + r(−v + y))

c(a~i + b~j + c~k ).

Thus −−→PQ×−→PR is a scalar multiple of a~i + b~j + c~k , and hence parallel to it.

CHECK YOUR UNDERSTANDING

1. False. There are exactly two unit vectors: one in the same direction as ~v and the other in the opposite direction. Explicitly,

the unit vectors parallel to ~v are ± 1

‖~v ‖~v .

2. False. The length of this vector is√

1/3 + 1/3 + 4/3 =√

2, not 1.

3. True. Multiplying by a scalar greater than one stretches the length of the vector by the scalar.

4. False. If ~v and ~w are not parallel, the three vectors ~v , ~w and ~v + ~w can be thought of as three sides of a triangle. (Ifthe tail of ~w is placed at the head of ~v , then ~v + ~w is a vector from the tail of ~v to the head of ~w .) The length of oneside of a triangle is less than the sum of the lengths of the other two sides. Alternatively, a counterexample is ~v = ~i and~w = ~j . Then ‖~i +~j ‖ =

√2 but ‖~i ‖+ ‖~j ‖ = 2.

5. False. If ~v and ~w are not parallel, the three vectors ~v , ~w and ~v − ~w can be thought of as three sides of a triangle. (Ifthe tails of ~v and ~w are placed together, then ~v − ~w is a vector from the head of ~w to the head of ~v .) The length of oneside of a triangle is less than the sum of the lengths of the other two sides. Alternatively, a counterexample is ~v = ~i and~w = ~j . Then ‖~i −~j ‖ =

√2 but ‖~i ‖ − ‖~j ‖ = 0.

6. False. Two vectors are parallel if and only if one is a nonzero scalar multiple of the other. If c(~i −2~j +~k ) = 2~i −~j +~k ,then c = 2 so that the~i components are equal, but multiplication by 2 does not make the ~j or ~k components equal. Thus,there is no scalar multiple of~i − 2~j + ~k that is equal to 2~i −~j + ~k .

7. False. As a counterexample, take 3~i and −~i . Then the sum is 2~i which has magnitude 2 (smaller than ‖3~i ‖ = 3).

8. False. Since magnitudes are nonnegative this cannot be true when c < 0. The correct statement is ‖c~v ‖ = |c|‖~v ‖.9. False. To find the displacement vector from (1, 1, 1) to (1, 2, 3) we subtract ~i + ~j + ~k from ~i + 2~j + 3~k to get

(1− 1)~i + (2− 1)~j + (3− 1)~k = ~j + 2~k .

10. False. The displacement vector from (a, b) to (c, d) has the same magnitude but opposite direction as the displacementvector from (c, d) to (a, b).

11. False. The dot product is a scalar.

12. True. Components of a normal vector can be read directly from coefficients of x, y and z in the equation for a plane.

13. True. The cosine of the angle between the vectors is negative when the angle is between π/2 and π.


14. False. The equation z = x + y has normal~i + ~j − ~k , which is not parallel to~i + ~j + ~k . An equation satisfying thegiven conditions is x+ y + z = 6.

15. True. The vector from (0, 1, 0) to (1, 1, 0) is~i , while the vector from (0, 1, 0) to (0, 1, 1) is ~k , and~i · ~k = 0.

16. True. ~v · ~v = ‖~v ‖2, which cannot be negative.

17. False. If the vectors are nonzero and perpendicular, the dot product will be zero (e.g.~i ·~j = 0).

18. False. If ~v and ~w are different vectors, but both are perpendicular to ~u , then both ~u · ~v and ~u · ~w are zero, yet ~v 6= ~w .For example, take ~u =~i , ~v = ~j and ~w = ~k .

19. True. Using the distributive property, and the fact that ~u · (−~v ) = −~u · ~v , we have

(~u + ~v ) · (~u − ~v ) = ~u · ~u + ~u · (−~v ) + ~v · ~u − ~v · ~v = ‖~u ‖2 − ‖~v ‖2.

20. True. This vector is ~vPerp

, the component of ~v perpendicular to the unit vector ~u . To check, calculate the dot product

~u · (~v − (~v · ~u )~u ) = ~u · ~v − (~v · ~u )(~u · ~u ) = ~u · ~v − ~v · ~u = 0,

since ~u · ~u = ‖~u ‖2 = 1.

21. True. The cross product yields a vector.

22. False. ~u × ~v has direction perpendicular to both ~u and ~v .

23. False. This is only true when ~u and ~v are perpendicular. In general, ‖~u × ~v ‖ = ‖~u ‖‖~v ‖ sin θ, where θ is the anglebetween ~u and ~v . The value of ‖~u × ~v ‖ is the area of the parallelogram with sides ~u and ~v .

24. True. The left-hand side evaluates to ~k · ~k = 1, while the right-hand side evaluates to~i ·~i = 1.

25. False. If ~u and ~w are two different vectors both of which are parallel to ~v , then ~v × ~u = ~v × ~w = ~0 , but ~u 6= ~w . Acounterexample is ~v =~i , ~u = 2~i and ~w = 3~i .

26. True. Since (~v × ~w ) is perpendicular to ~v , the dot product with ~v is zero.

27. True. The cross product is a vector in 3-space, while the dot product is a scalar, so they cannot be equal.

28. True. The cross product (~i +~j )× (~j + 2~k ) = 2~i − 2~j +~k , which has magnitude√

22 + (−2)2 + 12 = 3. Since thetriangle has area of 1/2 the parallelogram with the given vectors as sides, the triangle has area 3/2.

29. True. Any vector ~w that is parallel to ~v will give ~v × ~w = ~0 .

30. False. It is not true in general, but there are special cases when ~v × ~w = ~w × ~v . For example, when ~v is parallel to ~w ,or when one of the vectors is ~0 . In either case the cross products ~v × ~w and ~w × ~v are both the zero vector.

PROJECTS FOR CHAPTER THIRTEEN

1. (a) Let r = ‖~a ‖ and s = ‖~b ‖, and let α, β, be the angles between ~a ,~b , and the x-axis as shown in the figure.Suppose θ is the angle between ~a and ~b . We drew the figure with α < β and thus β − α = θ. If α > β,then α− β = θ. In both cases we know that

Area of parallelogram = ‖~a ‖‖~b ‖ sin θ.

Using the formulasin(β − α) = sinβ cosα− cosβ sinα,

and the fact that a1 = r cosα, a2 = r sinα, b1 = s cosβ, and b2 = s sinβ, we get

a1b2 − a2b1 = (r cosα)(s sinβ)− (r sinα)(s cosβ)

= rs(cosα sinβ − sinα cosβ)

= rs sin(β − α) (from sin(β − α) = sinβ cosα− cos β sinα)

= ‖~a ‖‖~b ‖ sin(β − α)

If β > α, we have β − α = θ, so

a1b2 − a2b1 = ‖~a ‖‖~b ‖ sin θ = Area of parallelogram.

PROJECTS FOR CHAPTER THIRTEEN 949

If β < α, we have α− β = θ, so

|a1b2 − a2b1| = ‖~a ‖‖~b ‖| sin(β − α)| = ‖~a ‖‖~b ‖ sin θ = Area of parallelogram.

(b) The sign of a1b2 − a2b1 is the same as the sign of β − α, so the sign of a1b2 − a2b1 tells us whether therotation from ~a to~b is counterclockwise (then a1b2 − a2b1 is positive) or clockwise (then a1b2 − a2b1 isnegative).

(c) Part (a) tells us thatArea of the parallelogram = |a1b2 − a2b1|.

The algebraic definition of the cross product is

~a ×~b = (a1b2 − a2b1)~k .

The geometric definition has magnitude given by ‖~a × ~b ‖ = Area of parallelogram. So the magnitudeof the algebraic definition agrees with the magnitude of the geometric definition. To check agreement ofthe direction of ~a × ~b for the two definitions, we notice that (a1b2 − a2b1)~k is perpendicular to ~a and~b since ~a and ~b are in the~i ,~j -plane. Also, part (b) says (a1b2 − a2b1)~k will point up (down) when therotation from ~a to ~b is counterclockwise (clockwise). So the direction of the algebraic definition obeysthe right-hand rule.

2. (a) We have

‖~a 2‖ =√

0.102 + 0.082 + 0.122 + 0.692 = 0.7120

‖~a 3‖ =√

0.202 + 0.062 + 0.062 + 0.662 = 0.6948

‖~a 4‖ =√

0.222 + 0.002 + 0.202 + 0.572 = 0.6429

~a 2 · ~a 3 = 0.10 · 0.20 + 0.08 · 0.06 + 0.12 · 0.06 + 0.69 · 0.66 = 0.4874

~a 3 · ~a 4 = 0.20 · 0.22 + 0.06 · 0.00 + 0.06 · 0.20 + 0.66 · 0.57 = 0.4322

The distance between the English and the Bantus is given by θ where

cos θ =~a 2 · ~a 3

‖~a 2‖‖~a 3‖=

0.4874

(0.7120)(0.6948)≈ 0.9852

so θ ≈ 9.9o.The distance between the English and the Koreans is given by φ where

cosφ =~a 3 · ~a 4

‖~a 3‖‖~a 4‖=

0.4322

(0.6948)(0.6429)≈ 0.9676

so φ ≈ 14.6o. Hence the English are genetically closer to the Bantus than to the Koreans.(b) Let ~a 5 be the 4-vector for the half Eskimo, half Bantu population. So

~a 5 =1

2~a 1 +

1

2~a 2 = (0.195, 0.04, 0.075, 0.68).

Then

‖~a 5‖ =√

0.1952 + 0.042 + 0.0752 + 0.682 = 0.7125,

~a 3 · ~a 5 = 0.20 · 0.195 + 0.06 · 0.04 + 0.06 · 0.075 + 0.66 · 0.68 = 0.4947.

So the distance between the English population and the half Eskimo, half Bantu population is

θ = arccos~a 3 · ~a 5

‖~a 3‖‖~a 5‖= arccos

0.4947

(0.6948)(0.7125)

= arccos 0.9993 ≈ 2.1o.

Since 2.1 < 9.9, the English are closer to the Bantu/Eskimo mix than to the Bantu alone.


(c) Suppose that x is the fraction of the population that is Eskimo, where 0 ≤ x ≤ 1. Then (1 − x) is thefraction that is Bantu. (For example, x = 0.5, in part (b).) Let ~a 6 be the 4-vector for a population that is xEskimo and (1− x) Bantu. We have

~a 6 = x~a 1 + (1− x)~a 2 = ~a 2 + x(~a 1 − ~a 2)

= (0.10 + 0.19x, 0.08− 0.08x, 0.12− 0.09x, 0.69− 0.02x).

Then

‖~a 6‖ =√

(0.10 + 0.19x)2 + (0.08− 0.08x)2 + (0.12− 0.09x)2 + (0.69− 0.02x)2

=√

0.5069− 0.024x+ 0.051x2

and

~a 3 · ~a 6 = 0.20 · (0.10 + 0.19x) + 0.06 · (0.08− 0.08x)

+0.06 · (0.12− 0.09x) + 0.66 · (0.69− 0.02x)

= 0.4874 + 0.0146x.

Since cos θ is a decreasing function of θ for 0 ≤ θ ≤ π, to minimize the angle θ = arccos~a 3 · ~a 6

‖~a 3‖‖~a 6‖, we

must maximizef(x) =

~a 3 · ~a 6

‖~a 3‖‖~a 6‖=

0.4874 + 0.0146x

0.6948√

0.5069− 0.024x+ 0.051x2.

Using a calculator or computer, we find that the maximum of this function for 0 ≤ x ≤ 1 is

f(0.5293) = 0.9994.

So the minimum distance of θ = arccos(0.9994) = 2.0o occurs at a mix of about 52.93% Eskimo and47.07% Bantu.

Documents

13.1 SOLUTIONS 907 CHAPTER THIRTEEN - Valenciafd.valenciacollege.edu/file/tsmith143/ch13.pdf13.1 SOLUTIONS 909 30. (a) True, since vectors ~cand f~point in the same direction and have