Vector Primer Problem Set Solutions - Pima Community College

Vector PrimerProblem Set Solutions

For physics students everywhere!

Wayne HackerCopyright c©Wayne Hacker 2009. All rights reserved.

vector primer problem set solns Copyright c©Wayne Hacker 2009. All rights reserved. 1

Contents

1 Introduction to Vectors 2

1.1 Identifying Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Given vector, identify on figure . . . . . . . . . . . . . . . . . . . 2

1.1.2 Given vector on figure, identify formula . . . . . . . . . . . . . . . 7

1.1.3 Given direction, identify on figure . . . . . . . . . . . . . . . . . . 13

1.1.4 Given vectors on figure, name quadrant of sum/difference . . . . . 18

1.1.5 Mixing it up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.2 Geometric Vector Addition and Subtraction . . . . . . . . . . . . . . . . 25

1.3 Position vs. Displacement Vector Problems . . . . . . . . . . . . . . . . . 30

1.4 Finding Components of Vectors . . . . . . . . . . . . . . . . . . . . . . . 34

1.5 Algebraic Vector Addition and Subtraction . . . . . . . . . . . . . . . . . 36

1.6 Concept vector and scalar questions . . . . . . . . . . . . . . . . . . . . . 39

1.7 Applications of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

1.7.1 Breaking vectors into components . . . . . . . . . . . . . . . . . . 44

1.7.2 Introduction to force vectors . . . . . . . . . . . . . . . . . . . . . 51

1.8 Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

1.8.1 Using the algebraic definition of dot product . . . . . . . . . . . . 56

1.8.2 Using the geometric form of dot product . . . . . . . . . . . . . . 59

1.8.3 Orthogonal vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 60

1.8.4 Direction cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

1.8.5 Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

1.8.6 Mixing it up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

1.8.7 Application Problems . . . . . . . . . . . . . . . . . . . . . . . . . 65

1.9 Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

1.10 Advanced-Level Problems (Dot and Cross Product) . . . . . . . . . . . . 70

1.10.1 Dot product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

1.10.2 Cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71


1 Introduction to Vectors

Unless stated otherwise, the direction of a vector in xy-space is the counterclockwiseangle θ that it makes with the x-axis; 0 ≤ θ < 360◦ or 0 ≤ θ < 2π rad., and angle mea-surements in your answers should be in the same units as in the statement of the problem.

1.1 Identifying Vectors

1.1.1 Given vector, identify on figure

Problem 1. Which of the vectors in thegraph corresponds to −10ı− 5?

(a) ~A (b) ~B

(c) ~C *(d) ~D(e) None of these

Solution: Only ~D has a negative x-component and a negative y-component.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

Problem 2. Which of the vectors in thegraph corresponds to −6ı + 2?

(a) ~A (b) ~B

*(c) ~C (d) ~D(e) None of these

Solution: Only ~B and ~C have anegative x-component and a positive y-component. In −6ı + 2, the magnitudeof the x-component (6) is greater than themagnitude of the y-component (2). This

is true of ~C but not of ~B.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B


Problem 3. Which of the vectors in thegraph corresponds to 4ı + 4?

*(a) ~A (b) ~B

(c) ~C (d) ~D(e) None of these

Solution: Only ~A has a positive x-component and a positive y-component.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

Problem 4. Which of the vectors in thegraph corresponds to −ı + 2?

(a) ~A *(b) ~B


Solution: Only ~B and ~C have anegative x-component and a positive y-component. In −ı + 2, the magnitude ofthe x-component (1) is smaller than themagnitude of the y-component (2). This

is true of ~B but not of ~C.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

Problem 5. Which of the vectors in thegraph corresponds to −3ı + ?

(a) ~A (b) ~B


Solution: Only ~C has a negative x-component and a positive y-component.

-

6

x

y

��

~A

PPPP

PPi~C

SSSSSSSSw ~D

��~B


Problem 6. Which of the vectors in thegraph corresponds to 2ı + ?

*(a) ~A (b) ~B


Solution: Only ~A and ~B have apositive x-component and a positive y-component. In 2ı + , the magnitude ofthe x-component (2) is greater than themagnitude of the y-component (1). This

is true of ~A but not of ~B.

-

6

x

y

��

�* ~APP

PPPPi~C

SSSSSSSSw ~D

��~B

Problem 7. Which of the vectors in thegraph corresponds to ı + 2?

(a) ~A *(b) ~B


Solution: Only ~A and ~B have apositive x-component and a positive y-component. In ı + 2, the magnitude ofthe x-component (1) is smaller than themagnitude of the y-component (2). This

is true of ~B but not of ~A.

-

6

x

y

��

�* ~A@

@@@

@@I

~C

SSSSSSSSw ~D

��~B

Problem 8. Which of the vectors in thegraph corresponds to 3ı− 4?

(a) ~A (b) ~B


Solution: Only ~D has a positive x-component and a negative y-component.

-

6

x

y

��

~A

PPPP

PPi~C

SSSSSSSSw ~D

AAAAAAAAK

~B


Problem 9. Which of the vectors in thegraph corresponds to 〈−3, 1〉?(a) ~A (b) ~B


Solution: Only ~B and ~C have anegative x-component and a positive y-component. In 〈−3, 1〉, the magnitude ofthe x-component (3) is greater than themagnitude of the y-component (1). This

is true of ~C but not of ~B.

-

6

x

y

��

~A

AAAAAAAAK

~B

PPPP

PPi~C

��~D

Problem 10. Which of the vectors in thegraph corresponds to 〈2, 2〉?*(a) ~A (b) ~B



-

6

x

y

��

~A

AAAAAAAAK

~B

PPPP

PPi~C

��

��~D

Problem 11. Which of the vectors in thegraph corresponds to 〈−1, 2〉?(a) ~A *(b) ~B


Solution: Only ~B and ~C have anegative x-component and a positive y-component. In 〈−1, 2〉, the magnitude ofthe x-component (1) is smaller than themagnitude of the y-component (2). This

is true of ~B but not of ~C.

-

6

x

y

HHHHj ~A

AAAAAAAAK

~B

PPPP

PPi~C

��~D


Problem 12. Which of the vectors in thegraph corresponds to 〈−2,−1〉?(a) ~A (b) ~B


Solution: Only ~D has a negative x-component and a negative y-component.

-

6

x

y

AAAAAAAAU ~A

AAAAAAAAK

~B

PPPP

PPi~C

��~D

Problem 13. Which of the vectors in thegraph corresponds to 〈1, 3〉?*(a) ~A (b) ~B



-

6

x

y

��~A

AAAAAAAAK

~B

PPPP

PPi~C

HHHHj ~D

Problem 14. Which of the vectors in thegraph corresponds to 〈−4,−2〉?(a) ~A (b) ~B


Solution: Only ~C has a negative x-component and a negative y-component.

-

6

x

y

AAAAAAAAU ~D

AAAAAAAAK

~B

��

��~C

HHHHj ~A


Problem 15. Which of the vectors in thegraph corresponds to 〈10,−5〉?(a) ~A *(b) ~B


Solution: Only ~A and ~B have apositive x-component and a negative y-component. In 〈10,−5〉, the magnitudeof the x-component (10) is greater thanthe magnitude of the y-component (5).

This is true of ~B but not of ~A.

-

6

x

y

AAAAAAAAU ~A

HHHHHHj ~B

PPPP

PPi~C

��~D

Problem 16. Which of the vectors in thegraph corresponds to 〈3,−3〉?*(a) ~A (b) ~B


Solution: Only ~A has a positive x-component and a negative y-component.

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D

1.1.2 Given vector on figure, identify formula

Problem 17. Which of the vectorsbelow corresponds to ~A on the graph atright?

(a) ı + (b) −ı + *(c) ı− (d) −ı− (e) None of these

Solution: ~A has a positive x-component and a negative y-component;so (c).

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D


Problem 18. Which of the vectorsbelow corresponds to ~B on the graph atright?

*(a) ı + (b) −ı + (c) ı− (d) −ı− (e) None of these

Solution: ~B has a positive x-component and a positive y-component;so (a).

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D

Problem 19. Which of the vectorsbelow corresponds to ~C on the graph atright?

(a) ı + *(b) −ı + (c) ı− (d) −ı− (e) None of these

Solution: ~C has a negative x-component and a positive y-component;so (b).

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D

Problem 20. Which of the vectorsbelow corresponds to ~D on the graph atright?

(a) ı + (b) −ı + (c) ı− *(d) −ı− (e) None of these

Solution: ~D has a negative x-component and a negative y-component;so (d).

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D



(a) 3ı + (b) −3ı + *(c) 2ı− (d) −ı− 3(e) None of these

Solution: ~A has a positive x-component and a negative y-component;so (c).

-

6

x

y

HHHHH

Hj ~A


*(a) 2ı + 6 (b) 3ı + (c) 2ı− (d) −ı− 3(e) None of these

Solution: ~A has a positive x-component and a positive y-component;so (a) or (b). ~A has a small x-componentand a large y-component; so (a).

-

6

x

y

��~A


(a) 2ı− (b) ı− 2(c) −2ı + *(d) −ı + 2(e) None of these

Solution: ~A has a negative x-component and a positive y-component;so (c) or (d). ~A has a small x-componentand a large y-component; so (d).

-

6

x

y

AAAAAAAAK

~A



(a) 6ı− 3 (b) −3ı + 6*(c) −6ı + 3 (d) 3ı− 6(e) None of these

Solution: ~A has a negative x-component and a positive y-component;so (b) or (c). ~A has a large x-componentand a small y-component; so (c).

-

6

x

y

HHH

HHHY

~A


(a) 〈−1, 2〉 (b) 〈−2, 1〉(c) 〈−1, 2〉 *(d) 〈−2,−1〉(e) None of these

Solution: ~A has a negative x-component and a negative y-component;so (d).

-

6

x

y

��

��~A


(a) 〈−1, 2〉 *(b) 〈1,−2〉(c) 〈−2, 1〉 (d) 〈2,−1〉(e) None of these

Solution: ~A has a positive x-component and a negative y-component;so (b) or (d). ~A has a small x-componentand a large y-component; so (b).

-

6

x

y

AAAAAAU ~A



(a) 〈−6,−3〉 *(b) 〈−3,−6〉(c) 〈6,−3〉 (d) 〈3,−6〉(e) None of these

Solution: ~A has a negative x-component and a negative y-component;so (a) or (b). ~A has a small x-componentand a large y-component; so (b).

-

6

x

y

��~A


(a) 〈−5,−10〉 (b) 〈5,−10〉*(c) 〈−10, 5〉 (d) 〈10,−5〉(e) None of these

Solution: ~A has a negative x-component and a positive y-component;so (c).

-

6

x

y

HHH

HHHY

~A


(a) 〈−2, 4〉 *(b) 〈2, 4〉(c) 〈−4, 2〉 (d) 〈4, 2〉(e) None of these

Solution: ~A has a positive x-component and a positive y-component;so (b) or (d). ~A has a small x-componentand a large y-component; so (b).

-

6

x

y

��~A



*(a) 〈−1, 2〉 (b) 〈−2, 1〉(c) 〈1,−2〉 (d) 〈2,−1〉(e) None of these

Solution: ~A has a negative x-component and a positive y-component;so (a) or (b). ~A has a small x-componentand a large y-component; so (a).

-

6

x

y

AAAAAAK

~A


*(a) 〈2, 1〉 (b) 〈1, 2〉(c) 〈−2, 1〉 (d) 〈−1, 2〉(e) None of these

Solution: ~A has a positive x-component and a positive y-component;so (a) or (b). ~A has a large x-componentand a small y-component; so (a).

-

6

x

y

��

��*

~A


(a) 〈−2,−1〉 (b) 〈−1,−2〉*(c) 〈2,−1〉 (d) 〈1,−2〉(e) None of these

Solution: ~A has a positive x-component and a negative y-component;so (c) or (d). ~A has a large x-componentand a small y-component; so (c).

-

6

x

y

HHHHH

Hj ~A


1.1.3 Given direction, identify on figure

Problem 33. Which of the vectors in thegraph corresponds to the direction 20◦

east of north?

(a) ~A (b) ~B


Solution: Since we want “east ofnorth”, we take the north direction as thereference line. Going 20◦ east of northputs us in the first quadrant; so ~D.

-

6

E

N

�� ~D

AAAAAAK~A

��~B

AAAAAAU ~C


east of south?

(a) ~A (b) ~B


Solution: Since we want “east ofsouth”, we take the south direction as thereference line. Going 20◦ east of southputs us in the fourth quadrant; so ~C.

-

6

E

N

�� ~D

AAAAAAK~A

��~B

AAAAAAU ~C


west of north?

*(a) ~A (b) ~B


Solution: Since we want “west ofnorth”, we take the north direction as thereference line. Going 20◦ west of northputs us in the second quadrant; so ~A.

-

6

E

N

�� ~D

AAAAAAK~A

��~B

AAAAAAU ~C



west of south?

(a) ~A *(b) ~B


Solution: Since we want “west ofsouth”, we take the south direction as thereference line. Going 20◦ west of southputs us in the third quadrant; so ~B.

-

6

E

N

�� ~D

AAAAAAK~A

��~B

AAAAAAU ~C


south of west?

(a) ~A (b) ~B


Solution: Since we want “south ofwest”, we take the west direction as thereference line. Going 30◦ south of westputs us in the third quadrant; so ~C or~D. Since 30◦ < 45◦, our vector should becloser to the west direction than to thesouth direction; so ~C. ( ~D would be morelike 60◦ south of west.)

-

6

E

N

AAAAAAK

~A

HHHH

HHY~B

��~C

��~D


north of west?

(a) ~A *(b) ~B


Solution: Since we want “north ofwest”, we take the west direction as thereference line. Going 30◦ north of westputs us in the second quadrant; so ~A or~B. Since 30◦ < 45◦, our vector should becloser to the west direction than to thenorth direction; so ~B. ( ~A would be morelike 60◦ north of west.)

-

6

E

N

AAAAAAK

~A

HHHHH

HY~B

��~C

��~D



south of west?

(a) ~A (b) ~B


Solution: Since we want “south ofwest”, we take the west direction as thereference line. Going 60◦ south of westputs us in the third quadrant; so ~C or~D. Since 60◦ > 45◦, our vector should becloser to the south direction than to thewest direction; so ~D. (~C would be morelike 30◦ south of west.)

-

6

E

N

AAAAAAK

~A

HHHHH

HY~B

��~C

��~D


north of west?

*(a) ~A (b) ~B


Solution: Since we want “north ofwest”, we take the west direction as thereference line. Going 60◦ north of westputs us in the second quadrant; so ~A or~B. Since 60◦ > 45◦, our vector should becloser to the north direction than to thewest direction; so ~A. ( ~B would be morelike 30◦ north of west.)

-

6

E

N

AAAAAAK

~A

HHHHH

HY~B

��~C

��~D

Problem 41. What is the direction ofvector ~A in the graph at right?

(a) 30◦ north of east*(b) 30◦ south of east(c) 30◦ north of west(d) 30◦ south of west(e) None of these

Solution: Of the solutions, only 30◦

south of east is in the fourth quadrant.

-

6

E

N

HHHHHHj ~A



*(a) 30◦ north of east(b) 30◦ south of east(c) 30◦ north of west(d) 30◦ south of west(e) None of these


north of east is in the first quadrant.

-

6

E

N

��

��*

~A


(a) 30◦ north of east(b) 30◦ south of east(c) 30◦ north of west*(d) 30◦ south of west(e) None of these


south of west is in the third quadrant.

-

6

E

N

��

��~A


(a) 30◦ north of east(b) 30◦ south of east*(c) 30◦ north of west(d) 30◦ south of west(e) None of these


north of west is in the second quadrant.

-

6

E

N

HHH

HHHY

~A



(a) 30◦ east of north(b) 60◦ east of north(c) 30◦ west of north*(d) 60◦ west of north(e) None of these

Solution: 30◦ west of north and 60◦

west of north are both in the second quad-rant. However, 30◦ west of north wouldbe closer to the north direction than tothe west direction. Hence: 60◦ west ofnorth.

-

6

E

N

HHHHH

HY~A


(a) 30◦ east of north(b) 60◦ east of north*(c) 30◦ west of north(d) 60◦ west of north(e) None of these

Solution: 30◦ west of north and 60◦

west of north are both in the second quad-rant. However, 60◦ west of north wouldbe closer to the west direction than tothe north direction. Hence: 30◦ west ofnorth.

-

6

E

N

AAAAAAK

~A


(a) 30◦ east of north*(b) 60◦ east of north(c) 30◦ west of north(d) 60◦ west of north(e) None of these

Solution: 30◦ east of north and 60◦

east of north are both in the first quad-rant. However, 30◦ east of north would becloser to the north direction than to theeast direction. Hence: 60◦ east of north.

-

6

E

N

��

�* ~A



*(a) 30◦ east of north(b) 60◦ east of north(c) 30◦ west of north(d) 60◦ west of north(e) None of these

Solution: 30◦ east of north and 60◦

east of north are both in the first quad-rant. However, 60◦ east of north wouldbe closer to the east direction than to thenorth direction. Hence: 30◦ east of north.

-

6

E

N

��~A

1.1.4 Given vectors on figure, name quadrant of sum/difference

Problem 49. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~A − ~B. What quadrantdoes ~A− ~B lie in?

(a) Quadrant I(b) Quadrant II*(c) Quadrant III(d) Quadrant IV(e) It lies on the x- or y-axis

Solution: Sketch a copy of − ~B, la-belled − ~B′, with its tail at the head of ~A.Then ~A − ~B has its tail at the tail of ~Aand its head at the head of − ~B′.

-

6

x

y

��9~A

��

~B

��

��

− ~B′~A− ~B

Problem 50. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~A + ~B. What quadrantdoes ~A+ ~B lie in?

(a) Quadrant I*(b) Quadrant II(c) Quadrant III(d) Quadrant IV(e) It lies on the x- or y-axis

Solution: Sketch a copy of ~B, labelled~B′, with its tail at the head of ~A. Then~A+ ~B has its tail at the tail of ~A and itshead at the head of ~B′.

-

6

x

y

��)~A

��~B

��

~B′

~A+ ~B



(a) Quadrant I(b) Quadrant II*(c) Quadrant III(d) Quadrant IV(e) It lies on the x- or y-axis


-

6

x

y

��)~A

��~B

��

− ~B′ ~A− ~B

Problem 52. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~B − ~A. What quadrantdoes ~B − ~A lie in?


Solution: Sketch a copy of − ~A, la-belled − ~A′, with its tail at the head of ~B.Then ~B − ~A has its tail at the tail of ~Band its head at the head of − ~A′.

-

6

x

y

��)~A

��~B��)

− ~A′

~B − ~A


(a) Quadrant I(b) Quadrant II(c) Quadrant III(d) Quadrant IV*(e) It lies on the x- or y-axis

Solution: Sketch a copy of ~B, labelled~B′, with its tail at the head of ~A. Then~A+ ~B has its tail at the tail of ~A and itshead at the head of ~B′: on the negativex-axis

-

6

x

y

��)

~A

PPPP

PPi~B

PPPP

PPi

~B′

~A+ ~B



*(a) Quadrant I(b) Quadrant II(c) Quadrant III(d) Quadrant IV(e) It lies on the x- or y-axis


-

6

x

y

��

��1 ~A��~B

��

~B′

~A+ ~B




-

6

x

y

��

��1 ~A��~B��)

− ~A′

~B − ~A


(a) Quadrant I(b) Quadrant II(c) Quadrant III*(d) Quadrant IV(e) It lies on the x- or y-axis


-

6

x

y

��

��1 ~A��~B

��

− ~B′

~A− ~B





-

6

x

y

��

��1 ~A

BBBBBBBN ~B

BBBBBBBM

− ~B′

~A− ~B




-

6

x

y

��

��1 ~A

BBBBBBBN ~B

BBBBBBBN

~B′

~A+ ~B




-

6

x

y

PPPPPPq ~A

BBBBBBBM

~B

BBBBBBBN

− ~B′~A− ~B





-

6

x

y

PPPPPPq ~A

BBBBBBBM

~B

BBBBBBBM

~B′

~A+ ~B




-

6

x

y

PPPPPPq ~A

BBBBBBBM

~BPPPP

PPi − ~A′

~B − ~A

1.1.5 Mixing it up

Problem 62. Five vectors are graphedon the diagram at right and described be-low. Write the letter of each vector besideits description.~E 3ı− ~B −4ı + 2~D 2ı− 3~A ı + 4~C −3ı− 3

-

6

x

y

PPPPPPq ~E

HHHH

HHHHY

~B

JJJJJJJ ~D

��~A

��

��

��~C


Problem 63. Five vectors are graphedon the diagram at right and described be-low. Write the letter of each vector besideits description.~A 〈2, 2〉~C 〈−3, 1〉~E 〈3,−4〉~D 〈−2,−1〉~B 〈−2, 4〉

-

6

x

y

��

~A

PPPP

PPi~C

SSSSSSSSw ~E

��~D

AAAAAAAAK

~B

Problem 64. Five velocity vectors aregraphed on the diagram at right and de-scribed below. Write the letter of eachvector beside its magnitude and direction.~E 14 m/s; 0.79 rad south of east

~D 45 m/s; 1.11 rad south of east

~C 41 m/s; 1.33 rad south of west

~A 45 m/s; 0.46 rad north of west

~B 36 m/s; 0.59 rad south of west

-

6

E

N

@@R ~E

AAAAAAAAU ~D

��~C

HHH

HHH

HHY

~A

��

��

��+~B

Problem 65. Five velocity vectors aregraphed on the diagram at right and de-scribed below. Write the letter of eachvector beside its magnitude and direction.~E 22 mph; 64◦ north of east~C 50 mph; 37◦ south of east~A 36 mph; 34◦ south of west~D 32 mph; 18◦ north of east~B 28 mph; 45◦ south of east

-

6

E

N

��

~E

ZZZZZZZZ~~C

��

��

��+~A

��

��1 ~D

@@@@R~B


Problem 66. The vector ~A is labelled onthe figure at right. Sketch and label thevector − ~A.

Solution: − ~A should be the samelength as ~A, in exactly the opposite di-rection. -

6

x

y

�

~A

�− ~A

Problem 67. The vector ~A is labelled onthe figure at right. Sketch and label thevector − ~A.

Solution: − ~A should be the samelength as ~A, in exactly the opposite di-rection. -

6

x

y

PPPP

Pi~A

PPPPPq

− ~A

Problem 68. Which of the vectors in thegraph corresponds to ı− 3?

(a) ~A (b) ~B


Solution: Since the tail is at the ori-gin, the tip must be at the point (1,−3)(the 3rd quadrant). The magnitude ofthe y-component is greater than the x-component, so the answer must be C.

-

6

x

y

BBBBBBBN ~C

PPPPPPq

~APP

PPPPi

~D

�� ~B


Problem 69. Five vectors are graphedon the diagram at right and described be-low. Write the letter of each vector besideits description.~A = 〈2, 2〉~C = 〈−3, 1〉~E = 〈3,−4〉~D = 〈−2,−1〉~B = 〈−2, 4〉

-

6

x

y

��

~A

PPPP

PPi~C

SSSSSSSSw ~E

��~D

AAAAAAAAK

~B

Problem 70. Which of the vectors in thegraph corresponds to −6ı− 3?

(a) ~A (b) ~B


Solution: Only ~D has negative x- andy-components.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

1.2 Geometric Vector Addition and Subtraction

Problem 71. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~A+ ~B.


-

6

x

y

-

~A

6~B

~B′

�

~A+ ~B


Problem 72. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~A− ~B.


-

6

x

y

-~A

6~B

− ~B′JJJJJJ~A− ~B


-

6

x

y

�~A

6

~B

− ~B′�

~A− ~B

Problem 74. The vectors ~X, ~Y , and ~Zare labelled on the figure at right. Whichof the following equations is true?

(a) ~Z = ~X − ~Y *(b) ~Z = ~Y − ~X

(c) ~Z = ~X + ~Y (d) ~Z = − ~X − ~Y(e) None of these

Solution: Notice that ~Y +(− ~X) = ~Z.

-

6

x

y

�~Y

?~X

JJJJJJ]

~Z


Problem 75. The vectors ~X, ~Y , and ~Zare labelled on the figure at right. Con-sider the vector −~Z. From the figure it isclear that

−~Z = ± ~X ± ~Y .

Which of the following equations is true?

(a) −~Z = − ~X + ~Y (b) −~Z = ~X + ~Y

(c) −~Z = − ~X − ~Y (d) −~Z = ~X − ~Y(e) None of these

Solution: ~X + (−~Y ) = −~Z.

-

6

x

y

�~Y

?~X

JJJJJJ]

~Z



-

6

x

y

��*

~A

HHHH

HY~B

~B′6

~A+ ~B



-

6

x

y

��

~A

AAAAAK

~B

− ~B′

-

~A− ~B




-

6

x

y

��~A

��)

~B

~B′

@@

@@I~A+ ~B



-

6

x

y

PPPPPPq ~A

SSSSSSSSw ~B

− ~B′

6~A− ~B



-

6

x

y

BBBBBBBBN~A

��

~B

~B′

@@@@@R~A+ ~B




-

6

x

y

��9~A

@@

@@@I

~B

− ~B′

��

~A− ~B

Problem 82. The vectors ~A, ~B, and ~Care labelled on the figure at right. Sketchand label the vector ~A+ ~B + ~C.

Solution: Sketch a copy of ~B, labelled~B′, with its tail at the head of ~A. Sketcha copy of ~C, labelled ~C ′, with its tail atthe head of ~B′. Then ~A+ ~B+ ~C will haveits tail at the tail of ~A and its head at thehead of ~C ′.

-

6

x

y

SSSSSSSw

~A

6

~B

BBBBBBBBBM

~C

~B′6

~C ′

��

~A+ ~B + ~C

Problem 83. The vectors ~A, ~B, and ~Care labelled on the figure at right. Sketchand label the vector ~A+ ~B − ~C.

Solution: Sketch a copy of ~B, labelled~B′, with its tail at the head of ~A. Sketcha copy of −~C, labelled −~C ′, with its tailat the head of ~B′. Then ~A + ~B − ~C willhave its tail at the tail of ~A and its headat the head of −~C ′.

-

6

x

y

�� ~A

@@

@@I

~B

HHH

HHH

HHH

HHY~C

@I

~B′−~C ′

��

��1

~A+ ~B − ~C


1.3 Position vs. Displacement Vector Problems

Problem 84. On the graph below, draw and label the position vectors ~r1 and ~r2 corre-sponding to the points P1 and P2 on the graph, and the displacement vector ∆~r = ~r2−~r1 .

-

6

x

y

rP1

rP2

��

~r1BBBBBBM

~r2

XXXXXX

Xy ∆~r

Problem 85. At exactly 12:00, a spiderclimbs onto the tip of a clock’s minutehand, where it remains for the next hour.On the axes at right, sketch and label thespider’s displacement vectors:~A at 12:15~B at 12:30~C at 12:45~D at 1:00

Solution: Don’t forget that you’remeasuring displacement from the 12:00position, not from the center of the clock.

-

6

x

y

@@@@R ~A

?~B

��

��~C

~D = ~0r


Problem 86. At exactly 12:00, a spiderclimbs onto the tip of a clock’s minutehand, where it remains for the next hour.Which of the vectors on the figure atright is in the direction of the spider’sdisplacement vector at 12:15?

(a) ~C *(b) ~D

(c) ~G (d) ~H(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F

Problem 87. At exactly 12:00, aspider climbs onto the tip of a clock’sminute hand, where it remains for thenext hour. Which of the vectors onthe figure at right is in the direction ofthe spider’s displacement vector at 12:30?

(a) The zero vector ~0 (b) ~A

(c) ~B *(d) ~E(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F


(a) ~B *(b) ~F


-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F



*(a) The zero vector ~0 (b) ~A

(c) ~B (d) ~E(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F

Problem 90. At exactly 12:00, a spiderclimbs onto the tip of a clock’s minutehand, where it remains for the next hour.If the origin is at the center of the clock,which of the vectors on the figure at rightis the spider’s position vector at 12:15?

(a) ~B *(b) ~C


-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F


(a) The zero vector ~0 (b) ~A

*(c) ~E (d) ~F(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F



(a) ~B (b) ~F

*(c) ~G (d) ~H(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F


(a) The zero vector ~0 *(b) ~A

(c) ~B (d) ~E(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F


1.4 Finding Components of Vectors

Problem 94. The vector ~A has magnitude 5.9 and direction θ = 0.34 rad. Find the x-and y-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = (5.9)(cos 0.34 rad) = 5.6

Ay = ‖ ~A‖ sin θ = (5.9)(sin 0.34 rad) = 2.0

Problem 95. The vector ~A has magnitude 83 and direction θ = 1.98 rad. Find the x-and y-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = (83)(cos 1.98 rad) = −33

Ay = ‖ ~A‖ sin θ = (83)(sin 1.98 rad) = 76

Problem 96. The vector ~A has magnitude 41 and direction θ = 3.51 rad. Find the x-and y-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = (41)(cos 3.51 rad) = −38

Ay = ‖ ~A‖ sin θ = (41)(sin 3.51 rad) = −15

Problem 97. The vector ~A has magnitude 7.7 and direction θ = 5.66 rad. Find the x-and y-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = (7.7)(cos 5.66 rad) = 6.3

Ay = ‖ ~A‖ sin θ = (7.7)(sin 5.66 rad) = −4.5

Problem 98. The vector ~A has magnitude 0.88 and direction 37◦. Find the x- andy-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = 0.88 cos 37◦ = 0.70

Ay = ‖ ~A‖ sin θ = 0.88 sin 37◦ = 0.53

Problem 99. The vector ~A has magnitude 1.28 and direction 98◦. Find the x- andy-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = 1.28 cos 98◦ = −0.18

Ay = ‖ ~A‖ sin θ = 1.28 sin 98◦ = 1.27

Problem 100. The vector ~A has magnitude 110 and direction 219◦. Find the x- andy-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = 110 cos 219◦ = −85

Ay = ‖ ~A‖ sin θ = 110 sin 219◦ = −69

Problem 101. The vector ~A has magnitude 37 and direction 304◦. Find the x- andy-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = 37 cos 304◦ = 21

Ay = ‖ ~A‖ sin θ = 37 sin 304◦ = −31

Problem 102. ~A = 〈5.1, 8.8〉. Find the magnitude and direction of ~A. Give the directionin radians.

Solution: ‖ ~A‖ =√Ax

2 + Ay2 =√

5.12 + 8.82 = 10.2

Since Ax > 0 and Ay > 0, θ is in the first quadrant. Hence:

θ = tan−1

∣∣∣∣AyAx∣∣∣∣ = tan−1 8.8

5.1= 1.05 rad


Problem 103. ~A = 〈16,−41〉. Find the magnitude and direction of ~A. Give thedirection in radians.


2 + Ay2 =

√162 + (−41)2 = 44

Since Ax > 0 and Ay < 0, θ is in the fourth quadrant. Hence:

θ = 2π − tan−1

∣∣∣∣AyAx∣∣∣∣ = 2π − tan−1 41

16= 5.08 rad

Problem 104. ~A = 〈−0.81,−1.13〉. Find the magnitude and direction of ~A. Give thedirection in degrees.


2 + Ay2 =

√(−0.81)2 + (−1.13)2 = 1.39

Since Ax < 0 and Ay < 0, θ is in the third quadrant. Hence:

θ = 180◦ + tan−1

∣∣∣∣AyAx∣∣∣∣ = 180◦ + tan−1 1.13

0.81= 234◦

Problem 105. ~A = 〈−216, 82〉. Find the magnitude and direction of ~A. Give thedirection in degrees.


2 + Ay2 =

√(−216)2 + 822 = 231

Since Ax < 0 and Ay > 0, θ is in the second quadrant. Hence:

θ = 180◦ − tan−1

∣∣∣∣AyAx∣∣∣∣ = 180◦ − tan−1 82

216= 159◦

Problem 106. ~A = 4.8ı + 7.0. Find the magnitude and direction of ~A. Give thedirection in radians.


2 + Ay2 =√

4.82 + 7.02 = 8.5

Since Ax > 0 and Ay > 0, θ is in the first quadrant. Hence:

θ = tan−1

∣∣∣∣AyAx∣∣∣∣ = tan−1 7.0

4.8= 0.97 rad

Problem 107. ~A = −120ı − 220. Find the magnitude and direction of ~A. Give thedirection in radians.


2 + Ay2 =

√(−120)2 + (−220)2 = 251

Since Ax < 0 and Ay < 0, θ is in the third quadrant. Hence:

θ = π + tan−1

∣∣∣∣AyAx∣∣∣∣ = π + tan−1 220

120= 4.21 rad

Problem 108. ~A = −29ı + 12. Find the magnitude and direction of ~A. Give thedirection in degrees.


2 + Ay2 =

√(−29)2 + 122 = 31

Since Ax < 0 and Ay > 0, θ is in the second quadrant. Hence:

θ = 180◦ − tan−1

∣∣∣∣AyAx∣∣∣∣ = 180◦ − tan−1 12

29= 158◦


Problem 109. ~A = 0.16ı − 0.23. Find the magnitude and direction of ~A. Give thedirection in degrees.


2 + Ay2 =

√0.162 + (−0.23)2 = 0.28

Since Ax > 0 and Ay < 0, θ is in the fourth quadrant. Hence:

θ = 360◦ − tan−1

∣∣∣∣AyAx∣∣∣∣ = 360◦ − tan−1 0.23

0.16= 305◦

1.5 Algebraic Vector Addition and Subtraction

Problem 110. ~A = 3ı− 2 and ~B = 4ı + 5. Find ~A+ ~B.

Solution: ~A+ ~B = (3 + 4)ı + (−2 + 5) = 7ı + 3

Problem 111. ~A = −ı + 7 and ~B = 2ı + 3. Find ~A− ~B.

Solution: ~A− ~B = (−1− 2)ı + (7− 3) = −3ı + 4

Problem 112. ~A = 〈8, 3〉 and ~B = 〈1, 2〉. Find ~A+ ~B.

Solution: ~A+ ~B = 〈8 + 1, 3 + 2〉 = 〈9, 5〉

Problem 113. ~A = 〈2, 6〉 and ~B = 〈−4, 5〉. Find ~A− ~B.

Solution: ~A− ~B = 〈2− (−4), 6− 5〉 = 〈6, 1〉

Problem 114. ~A = 〈4,−1〉 and ~B = 〈−5,−3〉. Find ~A+ ~B.

(a) 〈−9,−4〉 (b) 〈1, 2〉(c) 〈3,−8〉 *(d) 〈−1,−4〉(e) None of these

Solution: 〈4,−1〉+ 〈−5,−3〉 = 〈4 + (−5),−1 + (−3)〉 = 〈−1,−4〉

Problem 115. ~A = 〈−2, 5〉 and ~B = 〈3,−3〉. Find ~A+ ~B.

(a) 〈5, 8〉 *(b) 〈1, 2〉(c) 〈−5, 8〉 (d) 〈−6,−15〉(e) None of these

Solution: Add like components: ~A+ ~B = 〈−2, 5〉+〈3,−3〉 = 〈−2 + 3, 5− 3〉 = 〈1, 2〉.

Problem 116. ~A = 〈−2, 5〉 and ~B = 〈7,−3〉. Find ~A− ~B.

(a) 〈5, 2〉 (b) 〈−9, 2〉*(c) 〈−9, 8〉 (d) 〈5,−2〉(e) None of these

Solution: Add like components: ~A − ~B = 〈−2, 5〉 − 〈7,−3〉 = 〈−2− 7, 5 + 3〉 =〈−9, 8〉.


Problem 117. ~A has magnitude 7.2 and direction 1.1 rad. ~B has magnitude 5.5 anddirection 2.9 rad. Find ~A+ ~B. Give your answer in component form: 〈Cx, Cy〉.Solution: We begin by putting ~A and ~B in component form.

~A =⟨‖ ~A‖ cos θA, ‖ ~A‖ sin θA,

⟩= 〈(7.2)(cos 1.1 rad), (7.2)(sin 1.1 rad)〉

~B =⟨‖ ~B‖ cos θB, ‖ ~B‖ sin θB,

⟩= 〈(5.5)(cos 2.9 rad), (5.5)(sin 2.9 rad)〉

~A+ ~B = 〈(7.2)(cos 1.1 rad) + (5.5)(cos 2.9 rad), (7.2)(sin 1.1 rad) + (5.5)(sin 2.9 rad)〉= 〈−2.1, 7.7〉

Problem 118. ~A has magnitude 0.83 and direction 22◦. ~B has magnitude 1.42 anddirection 157◦. Find ~A− ~B. Give your answer in terms of unit vectors: Cxı + Cy .

Solution: We begin by putting ~A and ~B in unit-vector form:

~A = ‖ ~A‖ cos θA ı + ‖ ~A‖ sin θA = 0.83 cos 22◦ ı + 0.83 sin 22◦

~B = ‖ ~B‖ cos θB ı + ‖ ~B‖ sin θB = 1.42 cos 157◦ ı + 1.42 sin 157◦

~A− ~B = (0.83 cos 22◦ − 1.42 cos 157◦) ı + (0.83 sin 22◦ − 1.42 sin 157◦)

= 2.08ı− 0.24

Problem 119. ~A has magnitude 115 and direction 3.8 rad. ~B has magnitude 303 anddirection 5.1 rad. Find the magnitude and direction of ~A+ ~B.

Solution: This will require three steps: putting ~A and ~B into component form; addingthe two vectors; and then finding the magnitude and direction of the resultant.

~A =⟨‖ ~A‖ cos θA, ‖ ~A‖ sin θA,

⟩= 〈(115)(cos 3.8 rad), (115)(sin 3.8 rad)〉

~B =⟨‖ ~B‖ cos θB, ‖ ~B‖ sin θB,

⟩= 〈(303)(cos 5.1 rad), (303)(sin 5.1 rad)〉

~A+ ~B = 〈(115)(cos 3.8 rad) + (303)(cos 5.1 rad), (115)(sin 3.8 rad) + (303)(sin 5.1 rad)〉= 〈Cx, Cy〉= 〈24,−351〉

You should not use these rounded component values in calculating magnitude and direc-tion. However, you’ll need to know their signs in order to make sure your angle is in theright quadrant. Since the x-component is positive and the y-component is negative, θ isin the fourth quadrant.

‖ ~A+ ~B‖ =√Cx

2 + Cy2 = 352

θ = 2π − tan−1

∣∣∣∣CyCx∣∣∣∣ = 4.78 rad


Problem 120. ~A has magnitude 57 and direction 43◦. ~B has magnitude 22 and direction117◦. Find the magnitude and direction of ~A− ~B.

This will require three steps: putting ~A and ~B in component form; finding ~C = ~A− ~B;and calculating the magnitude and direction of ~C.

~A = ‖ ~A‖ cos θA ı + ‖ ~A‖ sin θA = 57 cos 43◦ ı + 57 sin 43◦

~B = ‖ ~B‖ cos θB ı + ‖ ~B‖ sin θB = 22 cos 117◦ ı + 22 sin 117◦

~A− ~B = (57 cos 43◦ − 22 cos 117◦) ı + (57 sin 43◦ − 22 sin 117◦)

= Cxı + Cy

= 52ı + 19

You should not use these rounded component values in calculating magnitude and direc-tion of ~C. However, you’ll need their signs to make sure you have θ in the right quadrant.Since both components are positive, θ is in the first quadrant.

‖ ~A− ~B‖ =√Cx

2 + Cy2 = 55

θ = tan−1 CyCx

= 20◦

Problem 121. If ~A = 〈3.3, 5.7〉, what is the magnitude of ~A? Round your answer tothe nearest 0.1.

(a) 5.3 (b) 5.9*(c) 6.6 (d) 7.2(e) None of these

Solution: ‖A‖ =√A2x + A2

y =√

(3.3)2 + (5.7)2 = 6.6

Problem 122. If ~A = 〈3, 4〉, what is the magnitude of ~A?

*(a) 5 (b) 6(c) 25 (d) 7(e) None of these

Solution: ‖A‖ =√A2x + A2

y =√

(3)2 + (4)2 = 6.6

Problem 123. If ~A = 3ı− 4, what is the magnitude of ~A?

(a) -1 (b) 2(c) 4 *(d) 5(e) None of these

Solution: ‖ ~A‖ =√

(3)2 + (−4)2 = 5


1.6 Concept vector and scalar questions

The first three problems deal with trigonometric functions.

Problem 124. cos(5 m/s) = ?

Solution: We can only take the cosine of a pure number, so this question doesn’tmake sense.

Problem 125. Consider the expression: sin(ω∗ t∗) = ?, where t∗ has units of time. Whatmust the units of ω∗ be in order for the argument of the sine function to make sense?

Solution: [ω∗] = 1T

.

Problem 126. Why is an angle a dimensionless number when measured in radians?

Solution: The angle in radians is the ratio of the arc intercepted by the angle to theradius of the circle:

angle in radians =arc length

radius

The arc length and the radius are both measured in units of length; so length in thenumerator and denominator cancel, leaving no units.

Problem 127. True or false: If we add the scalars 1 m + 1 m, we get 2 m.

Solution: True. We add scalars as we do regular numbers: 1 + 1 = 2.

Problem 128. True or false: If we add two vectors, each with length 1 m, we get avector with length 2 m.

Solution: False. Suppose the first vector is 1 m long in the positive x-direction, andthe second is 1 m long in the negative x-direction. Then the sum of the two vectors iszero. Unless the two vectors are pointing in exactly the same direction, we can’t find themagnitude of their sum by adding the magnitudes of the two vectors.

Problem 129. Can you find a vector ~v = 〈v1, v2〉 such that ~v 6= 0, but ‖~v‖ = 0?

Solution: No. The magnitude of ~v is

‖~v‖ =√v2

1 + v22

Since it has to be the case that v21 ≥ 0 and v2

2 ≥ 0, the only way ‖~v‖ can be zero is ifv2

1 = v22 = 0. But that means that v1 = v2 = 0, so ~v = 0.

Problem 130. Is it possible for the magnitude of a vector to be smaller than any of itscomponents?


Solution: No. For simplicity’s sake, let’s look at a vector with two components:~v = 〈v1, v2〉; and let’s suppose that v1 < v2. The magnitude of ~v is

‖~v‖ =√v2

1 + v22 ≥

√v2

1 + 0 = |v1| ≥ v1

Hence the magnitude of ~v is no smaller than its smallest component.Technically, we should prove this for an arbitrary vector ~v = 〈v1, . . . , vn〉, where vk is thesmallest component. But the notation would get cumbersome, and the essentials of theproof are the same.

Problem 131. True or false: Let ~v = 〈a, b〉. If ‖~v‖ > 0, then a > 0 and b > 0.

Solution: False. Suppose a = −3 and b = −4. Then

‖~v‖ =√

(−3)2 + (−4)2 =√

25 = 5

In this example, ‖~v‖ > 0, and a < 0 and b < 0.

Problem 132. If ~A = Ax ı+Ay and makes an angle θ with the positive x-axis, expresssin θ in terms of Ax and Ay.

Solution: sin θ =Ay√

A2x + A2

y

Problem 133. If ~A = Ax ı+Ay and makes an angle θ with the positive x-axis, expresscos θ in terms of Ax and Ay.

Solution: cos θ =Ax√

A2x + A2

y

Problem 134. True or false: If ~A = Ax ı + Ay , then the equation for the magnitudesatisfies A = Ax + Ay?

Solution: False. A =√Ax

2 + Ay2 6= Ax + Ay. For example, let Ax = Ay = −1.

Problem 135. If ~A = Ax ı+Ay and makes an angle θ with the positive x-axis, expressθ in terms of Ax and Ay. Assume that Ax > 0 and Ay > 0.

Solution: θ = tan−1

(AyAx

). [Since Ax > 0 and Ay > 0, θ is in the first quadrant; so

we don’t have to worry about reference angles.]

Problem 136. True or false: If ~A = Ax ı + Ay , then the equation for the magnitude

satisfies ‖ ~A‖ ≥ |Ax|. Justify your answer.

Solution: True. ‖ ~A‖ =√Ax

2 + Ay2 ≥

√Ax

2 = |Ax|


Problem 137. True or false: Let ~r = 〈x, y〉 = x ı + y . If ‖~r‖ > 0, then x > 0 andy > 0. Justify your answer.

Solution: False. Let ~r = 〈−1,−1〉. Then ‖~r‖ =√

2 > 0. [There are many othercounterexamples you could use.]

Problem 138. Can you find a situation where the average velocity vector is equal tothe zero vector, but the average speed is not zero? In symbols:

Can we have ~vave =∆~r

∆t= ~0 and speed =

total distance travelled

∆t6= 0 ?

Give an example, or explain why one doesn’t exist.

Solution: A car drives 100 km straight north at 100 km/h, then quickly turns aroundand drives back to its starting point at 100 km/h. The total time it takes is ∆t = 2 h.Since it has returned to its starting point, ∆~r = ~0. The total distance travelled is 200km, so the average speed is 100 km/h. [There are many other examples you could use.]

Problem 139. Consider the statements (i) and (ii). Are the statements true or false?

(i) If k is a scalar, then k + k = 2k

(ii) Let ~v = 〈a, b〉. If ‖v‖ > 0, then a > 0 and b > 0.

Choose the correct answer from below.

*(a) (i) is true; (ii) is false (b) (i) is false; (ii) is true

(c) Both statements are true (d) Both statements are false

Solution: (i) is true. (ii) is false: every nonzero vector has a positive magnitude,including those with negative components. For example, if ~v = 〈−3, 4〉, then ‖~v‖ =√

(−3)2 + 42 = 5.


(i) If ~v is a vector, then ‖2~v‖ = 2‖~v‖.

(ii) Let ~v = 〈a, b〉. If ‖v‖ = 0, then a = 0 and b = 0.


(a) (i) is true; (ii) is false (b) (i) is false; (ii) is true

*(c) Both statements are true (d) Both statements are false


(i) If ~v is a vector, then ‖ − ~v‖ = −‖~v‖.

(ii) Let ~v = 〈a, b〉. Then 2~v = 〈2a, 2b〉.


(a) (i) is true; (ii) is false *(b) (i) is false; (ii) is true


Solution: (ii) is true. (i) is false: the magnitude of a vector is always positive, unlessthe vector is ~0, in which case the magnitude is zero. Hence ‖ − ~v‖ = ‖~v‖.



(i) If ~v is a vector, then ‖ − ~v‖ = ‖~v‖.

(ii) Let ~v = 〈a, b〉. Then −2~v = −2a− 2b.




Solution: (i) is true. (ii) is false: if ~v = 〈a, b〉, then −2~v = 〈−2a,−2b〉.


(i) If ~u and ~v are vectors, then ‖~u+ ~v‖ = ‖~u‖+ ‖~v‖.

(ii) Let ~v = 〈a, b〉. Then ‖~v‖ =√a2 + b2.




Solution: (ii) is true. (i) is false; for example, let ~u = 〈1, 0〉 and ~v = 〈−1, 0〉. Then~u+ ~v = ~0; so ‖~u‖ = ‖~v‖ = 1, but ‖~u+ ~v‖ = 0 6= 1 + 1.


(i) If ~u is a vector and k is a scalar, then ‖k~u‖ = k‖~u‖.

(ii) Let ~v = 〈a, b〉. Then ‖~v‖ =√a+ b.



(c) Both statements are true *(d) Both statements are false

Solution: (i) is false. Let ~u = 〈1, 0〉, and let k = −2. Then ‖~u‖ = 1; k~u = 〈−2, 0〉, so‖k~u‖ = 2 6= k‖~u‖ = (−2)(1). (A true version of (i) would be: ‖k~u‖ = |k| · ‖~u‖.)(ii) is also false. ‖~v‖ =

√a2 + b2.


(i) If ~u is a vector, k is a scalar, and k > 0, then ‖k~u‖ = k‖~u‖.

(ii) Let ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉. Then v1 + v2 = 〈a1 + a2, b1 + b2〉.



*(c) Both statements are true (d) Both statements are false

Solution: Both statements are true. Statement (i) is a specific case of the generalstatement: If ~u is a vector and k is a scalar, then ‖k~u‖ = |k| · ‖~u‖. If k > 0, then |k| = k.



(i) If ~u is a vector, then ‖ − 2~u‖ = 2‖~u‖.

(ii) Let ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉. Then v1 + v2 = 〈a1 + b1, a2 + b2〉.




Statement (i) is a specific case of the general statement: If ~u is a vector and k is a scalar,then ‖k~u‖ = |k| · ‖~u‖. If k = −2, then |k| = 2.

Statement (ii) is a garbled version of the true statement: if ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉,then ~v1 + ~v2 = 〈a1 + a2, b1 + b2〉.


(i) If ~u is a vector, then ~u+ (−~u) = ~0.

(ii) Let ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉. Then v1 − v2 = 〈a1 − b1, a2 − b2〉.




Solution: (i) is true. (ii) is a garbled version of the true statement: if ~v1 = 〈a1, b1〉and ~v2 = 〈a2, b2〉, then v1 − v2 = 〈a1 − a2, b1 − b2〉.


(i) If ~u is a vector, then ~u− 2~u = ~u.

(ii) Let ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉. Then v1 − v2 = 〈a1 − a2, b1 − b2〉.




Solution: (i) is false; a true version would be: ~u− 2~u = −~u. (ii) is true.


1.7 Applications of Vectors

1.7.1 Breaking vectors into components

Problem 149. An airplane is flying at a speed of V in a direction 22◦ north of east.Which expression gives the northward component of the plane’s velocity?

(a) V cos 22◦ (b) V/ cos 22◦

*(c) V sin 22◦ (d) V/ sin 22◦

(e) None of these

Problem 150. An airplane is flying at a speed of V in a direction 22◦ north of east.Which expression gives the eastward component of the plane’s velocity?

*(a) V cos 22◦ (b) V/ cos 22◦

(c) V sin 22◦ (d) V/ sin 22◦

(e) None of these

Problem 151. A car is driving at a speed of V up a hill that makes an angle of 22◦

with the horizontal. What is the vertical component of the car’s velocity?

(a) V cos 22◦ (b) V/ cos 22◦

*(c) V sin 22◦ (d) V/ sin 22◦

(e) None of these

Problem 152. An airplane is flying at a speed of 190 mph in a direction 24◦ north ofeast. What is the northward component of the plane’s velocity? Round your answer tothe nearest mph.

*(a) 77 mph (b) 101 mph(c) 161 mph (d) 175 mph(e) None of these

Solution: vy = v sin θ = (190 mph) sin 24◦ = 77 mph

Problem 153. An airplane is flying at a speed of 310 mph in a direction 77◦ north ofeast. What is the eastward component of the plane’s velocity? Round your answer tothe nearest mph.

*(a) 70 mph (b) 155 mph(c) 180 mph (d) 294 mph(e) None of these

Solution: vx = v cos θ = (310 mph) cos 77◦ = 70 mph

Problem 154. An airplane is flying at 214 m/s in a direction 37◦ north of east. Whatis the eastward component of its velocity? Round your answer to the nearest m/s.

(a) 129 m/s (b) 134 m/s(c) 161 m/s *(d) 171 m/s(e) None of these

Solution: vE = (214) cos 37◦ = 171


Problem 155. An airplane is flying at 220 mph in a direction 30◦ north of east. Whatis the northward component of its velocity?

Solution: You don’t need a calculator, since every good physics student knows thatsin 30◦ = 1/2. Then

vN = v sin θ = (220 mph) sin 30◦ = 110 mph

Problem 156. An airplane is flying at 220 mph in a direction 30◦ east of north. Whatis the northward component of its velocity? Round your answer to the nearest mph.Solution: vN = v cos θ = (220 mph) cos 30◦ = 191 mph

Problem 157. A hill has a slope of 0.30 radians above the horizontal. If you are drivingup that hill at 22 km/hr, how fast are you gaining elevation?

Solution: You want the vertical component of your velocity:

vy = v sin θ = (22 km/hr)(sin 0.30 rad) = 6.5 km/hr

Problem 158. A car is driving at 35 mph up a hill with a slope of 8◦ above the horizontal.What is the horizontal component of the car’s velocity? Round your answer to the nearest0.1 mph.

(a) 4.9 mph (b) 5.1 mph(c) 7.5 mph *(d) 34.7 mph(e) None of these

Solution: vx = v cos θ = (35 mph) cos 8◦ = 34.7 mph

Problem 159. A gun is fired at an elevation of 39◦ above the horizontal. The shellemerges from the muzzle at 320 m/s. What is the vertical component of the shell’svelocity? Round your answer to the nearest m/s.

(a) 85 m/s *(b) 201 m/s(c) 249 m/s (d) 308 m/s(e) None of these

Solution: vy = v sin θ = (320 m/s) sin 39◦ = 201 m/s

Problem 160. (Lab Problem) A spring cannon is fired at an elevation of 34◦ abovethe horizontal. The ball emerges from the cannon at 14.4 m/s. What is the horizontalcomponent of the ball’s velocity? Round your answer to the nearest 0.1 m/s.

(a) 8.1 m/s (b) 9.7 m/s*(c) 11.9 m/s (d) 12.2 m/s(e) None of these

Solution: vx = v cos θ = (14.4 m/s) cos 34◦ = 11.9 m/s


Problem 161. You are on an airplane flying upward at an angle of θ above the horizontal,at 110 m/s. You have a cold, and if you gain altitude faster than 25 m/s, your eardrumswill explode. What is the maximum safe value for θ? Round your answer to the nearestdegree.

(a) 12◦ *(b) 13◦

(c) 14◦ (d) 16◦

(e) None of these

Solution: You know v and vy; you want to know θ.

vy = v sin θ ⇒ sin θ =vyv

⇒ θ = sin−1[vyv

]= sin−1

[25 m/s

110 m/s

]= 13◦

Problem 162. (Lab Problem) A spring cannon is fired at an elevation of 34◦ abovethe horizontal. The ball emerges from the cannon at 14.4 m/s. What is the horizontalcomponent of the ball’s velocity? Round your answer to the nearest 0.1 m/s.

(a) 8.1 m/s (b) 9.7 m/s*(c) 11.9 m/s (d) 12.2 m/s(e) None of these

Solution: vx = v cos θ = (14.4 m/s) cos 34◦ = 11.9 m/s

Problem 163. An bird is flying at 20 m/s in a direction 30◦ south of west. If we take thex-axis pointing east and the y-axis pointing north, then what is the southward componentof its velocity ~v = 〈vx, vy〉? Round your answer to the nearest m/s.

(a) 10√

3 m/s (b) 10 m/s

*(c) −10√

3 m/s (d) −10 m/s(e) None of these

Solution: The tip of the vector lies in the 3rd quadrant, so the y-component (thesouthward component) is negative. vy = vS = v sin θ = −20 sin(30◦) m/s = −10 m/s

Problem 164. Your Civil War cannon has a muzzle velocity of v0 = 1220 ft/s. If it isfired at an elevation of 6◦ above the horizontal, what is the vertical component v0y of thevelocity? Round your answer to the nearest ft/s.

Solution: v0y = v0 sin θ = (1220 ft/s) sin 6◦ = 128 ft/s

Problem 165. An airplane is flying at 330 km/hr, in a direction 0.19 rad above thehorizontal. At what speed is the airplane gaining altitude?

Solution: We want the vertical component of the airplane’s velocity vector:

vy = v sin θ = (330 km/hr)(sin 0.19 rad) = 62 km/hr

Problem 166. A train is moving at 68 mph in a direction 22◦ north of west. What isthe train’s speed westward?

Solution: We want the westward component of the train’s velocity vector:

vW = v cos θ = (68 mph)(cos 22◦) = 63 mph


Problem 167. (Position) Pirates have buried their treasure on campus, and you haveacquired a copy of their map. (Like everyone else, pirates need to be careful about that“Reply All” button.) It tells you to start at the main entrance to the physics buildingand take 120 steps in a direction 0.38 rad north of east; from there, you should go another190 steps in a direction 0.33 rad south of east. How far from the entrance is the treasure,and in what direction?

Solution: We will call the first displacement vector ~A and the second ~B. We needto find the magnitude and direction of ~C = ~A + ~B. We begin by writing the vectors incomponent form:

~A =⟨‖ ~A‖ cos θA, ‖ ~A‖ sin θA

⟩= 〈120 · cos 0.38 rad, 120 · sin 0.38 rad〉

~B =⟨‖ ~B‖ cos θB, ‖ ~B‖ sin θB

⟩= 〈190 · cos(−0.33 rad), 190 · sin(−0.33 rad)〉

Remember that cos(−θ) = cos θ and that sin(−θ) = − sin θ. Then the components of~C = ~A+ ~B are:

CE = 120 · cos 0.38 rad + 190 · cos 0.33 rad = 291 steps

CN = 120 · sin 0.38 rad− 190 · sin 0.33 rad = −17 steps

Since CE is positive and CN is negative, the direction is in the fourth quadrant: that is,south of east. We calculate

‖~C‖ =√CE

2 + CN2 = 292 steps

θ = tan−1

∣∣∣∣CNCE∣∣∣∣ south of east = 0.06 rad south of east

Problem 168. (Position) You have discovered the Lost Belgian Mine, which runsdirectly northward into a mountainside. From the opening, you go 47 m in a direction0.28 rad below the horizontal. From there, you go another 66 m in a direction 0.19 radbelow the horizontal. At this point, what is your horizontal distance from the opening,and how far are you above or below it?

Solution: We begin by finding the components of the two vectors:


⟩= 〈47 m · cos(−0.28 rad), 47 m · sin(−0.28 rad)〉


⟩= 〈66 m · cos(−0.19 rad), 66 m · sin(−0.19 rad)〉

Remember that cos(−θ) = cos θ and that sin(−θ) = − sin θ. Then the components of~C = ~A+ ~B are:

Cx = 47 m · cos 0.28 rad + 66 m · cos 0.19 rad = 110 m

Cy = −47 m · sin 0.28 rad− 66 m · sin 0.19 rad = −25 m

Your horizontal distance from the entrance is 110 m, and you are 25 m below it.


Problem 169. A goose is directly east of an airplane. The airplane is flying at a speedof 160 mph, in a direction 15◦ north of east. The goose is flying at a speed of 45 mph,in a direction θ north of west. What must θ be in order for the goose to collide with theairplane? Give your answer in degrees.

Solution: Since the goose is directly east of the airplane, its northbound componentof velocity must match the airplane’s if they are to collide. That means

(45 mph)(sin θ) = (160 mph)(sin 15◦)

Solving for θ gives us

θ = sin−1

((160 mph)(sin 15◦)

45 mph

)= 67◦

Problem 170. You are tracking a submarine by determining its position when it sendsradio messages. When you first detect the sub, it is 57 km west and 24 km north of you.When it sends its next message 37 min later, it is 51 km west and 13 km north of you.What are the magnitude and direction of the sub’s average velocity during that time?(Give the magnitude in km/hr, and the direction in radians, relative to east or west:e.g. θ rad north of east, with 0 < θ < π/2.)

Solution: Don’t forget to convert 37 min to hours. A look at the numbers showsthat the submarine is moving to the east and south. The eastbound component of thevelocity is:

vE =57 km− 51 km

37 min· 60 min

1 hr≈ 9.7 km/hr

The southbound component of the velocity is:

vS =24 km− 13 km

37 min· 60 min

1 hr≈ 17.8 km/hr

(We present these intermediate results because it’s a good idea to look at them and makesure that they seem reasonable. You should not use these rounded numbers in your latercalculations.)

The magnitude of the velocity is√vE2 + vS2 = 20 km/hr

The direction is

tan−1

(vSvE

)= 1.1 rad south of east


Problem 171. A cockroach has a tiny embedded radio transponder, allowing you totrack its position and velocity. Just before your lab partner attempts to stomp on it, itis moving at 8 cm/s to the east and 5 cm/s to the south. 1.3 seconds later, it is scurryingaway at 31 cm/s to the west and 19 cm/s to the south. What are the components of theroach’s average acceleration during that time?

Solution: A look at the numbers shows that the roach is accelerating westward andsouthward. The westward component of acceleration is

aW =31 cm/s− (−8 cm/s)

1.3 s= 30 cm/s2

The southward component is

aS =19 cm/s− 5 cm/s

1.3 s= 11 cm/s2

Problem 172. A car is travelling eastward along a highway at 60 mph. It passes undera hawk that is flying northward at 42 mph. From the point of view of an observer in thecar, what are the hawk’s speed and direction? Give the direction in degrees relative tonorth, e.g. θ◦ east of north.

Solution: For an observer in the car, objects fixed to the ground are moving westwardat 60 mph. To that observer, the hawk is moving westward at vW = 60 mph andnorthward at vN = 42 mph. In that frame of reference, the hawk’s speed is

v =√vW 2 + vN 2 =

√(60 mph)2 + (42 mph)2 = 73 mph

The direction, relative to north, is

tan−1

(vWvN

)= tan−1

(60 mph

42 mph

)= 55◦ west of north

Problem 173. You can paddle a kayak in still water at 2.6 m/s. The Santa Cruz Riveris 55 m wide, and flows northward at 1.9 m/s. If you launch your kayak on the eastbank, and point it directly westward, what will your speed and direction be relative toan observer standing on the bank? Give your direction in radians relative to west, e.g. θrad south of west.

Solution: Relative to a fixed observer on the bank, you are moving westward atvW = 2.6 m/s and northward at vN = 1.9 m/s. In that frame of reference, your speed is

v =√vW 2 + vN 2 =

√(2.6 m/s)2 + (1.9 m/s)2 = 3.2 m/s

Your direction relative to west is

tan−1

(vNvW

)= tan−1

(1.9 m/s

2.6 m/s

)= 0.63 rad north of west

The width of the river is irrelevant to the problem.


Problem 174. An airplane can fly at 410 km/hr in still air. The pilot wishes to fly itfrom Smithpur to Jonesgorod, 530 km to the north. The wind is blowing at 55 km/hrfrom west to east. In what direction should the pilot point the plane to fly directly toJonesgorod? How long will the journey take? Give your direction in radians relative tonorth, e.g. θ rad east of north.

Solution: The pilot wants to fly directly north. He must point the airplane west ofnorth, so that his still-air velocity has a westward component that offsets the 55 km/hreastward wind. If θ is the pilot’s heading west of north, then

vW = (410 km/hr) sin θ = 55 km/hr

Solving this for θ yields

θ = sin−1

(55 km/hr

410 km/hr

)= 0.13 rad west of north

The northward component of the plane’s velocity is (410 km/hr) cos θ. To fly 530 km tothe north requires a time of

t =530 km

(410 km/hr) cos θ= 1.3 hr

Problem 175. A river has a current with a speed of 5 ft/s. You can swim at 4 ft/s instill water. At what angle to the cross-river direction do you need to swim to reach apoint directly across the river from you?

Solution: No solution. Whatever the angle is, your upstream component of velocitycan be no greater than the total magnitude of your velocity: 4 ft/s. Thus you will findyourself going downstream at 1 ft/s or more.

Problem 176. A train is going directly north at 70 mph. An action-movie star is atopthe train, running southward at 12 mph. Relative to someone standing on the ground,how fast is the action-movie star moving, and in what direction?

Solution: Relative to someone on the ground, the AMS is moving northward at70− 12 = 58 mph.


1.7.2 Introduction to force vectors

Working definition: You can think of a force as a push or a pull. Since a push orpull has both magnitude and direction, it follows that a force is a vector. 1

We’ll learn more about forces later on in the course when you learn about Newton’s Laws.In particular, Newton’s 2nd Law Fnet = ma requires computing the net force (sometimescalled the resultant force) acting on an object of mass m. Since a force is a vector,the net force is just the resultant vector. In more than one spatial dimension we findthe resultant vector by breaking the individual force vectors acting on the object intocomponent vectors and then finding the net force in each of the x and y directions.

The purposes of these exercises is to give you an early exposure to the concept of break-ing forces into their respective components and summing them (keeping proper ± signconventions) to find the net force in the the x and y directions, denoted Fnet,x and Fnet,y.

Problem 177. A rope is used to pull a blockup a ramp. The ramp makes an angle of 25◦

to the horizontal; the rope makes an angle of33◦ with the ramp. The rope is pulled with aforce ~F whose magnitude is 317 N. What isthe horizontal component of ~F? Round youranswer to the nearest newton.*(a) 168 N (b) 185 N(c) 203 N (d) 224 N(e) None of these

-x

6y

!!!!

!!!!

!!!!

!!!

25◦LLL!!

LLL��7

33◦

F = 317 N

Solution: Draw a free-body diagram andadd the angles to find the net angle θ between~F and the horizontal: θ = 25◦+33◦ = 58◦. Thehorizontal component is

Fx = 317 cos 58◦ = 168 N

The vertical component is

Fy = 317 sin 58◦ = 269 N -x

6y

!!!!

!

25◦��7

~F

33◦

θ = 58◦

1Actually, this is not obvious and must be proven mathematically.



to the horizontal; the rope makes an angle of33◦ with the ramp. The rope is pulled witha force ~F whose magnitude is 317 N. Whatis the vertical component of ~F? Round youranswer to the nearest newton.(a) 218 N (b) 242 N*(c) 269 N (d) 296 N(e) None of these -

x

6y

!!!!

!!!!

!!!!

!!!

25◦LLL!!

LLL��7

33◦

F = 317 N


Fx = 317 cos 58◦ = 168 N


Fy = 317 sin 58◦ = 269 N -x

6y

!!!!

!

25◦��7

~F

33◦

θ = 58◦


to the horizontal; the rope makes an angle of31◦ with the ramp. The rope is pulled with aforce ~F whose magnitude is 428 N. What isthe horizontal component of ~F? Round youranswer to the nearest newton.(a) 209 N (b) 232 N*(c) 258 N (d) 283 N(e) None of these

-x

6y

!!!!

!!!!

!!!!

!!!

22◦LLL!!

LLL��7

31◦

F = 428 N


Fx = 428 cos 53◦ = 258 N


Fy = 428 sin 53◦ = 342 N -x

6y

!!!!

!

22◦��7

~F

31◦

θ = 53◦


Problem 180. Two forces ~F1 and ~F2 are acting on aparticle, as shown at right. ~F1 makes an angle of 35◦

with the x-axis and has magnitude 87 N. ~F2 makesan angle of 18◦ with the y-axis and has magnitude181 N. Use the component method to find the netforce in the x-direction. Round your answer to thenearest newton. (The picture is not drawn to scale.)

(a) 12 N (b) 13 N*(c) 15 N (d) 16 N

y

xu��

�3F1 = 87 N

35◦

F1x

F1y

AAAAAAAAK

F2 = 181 N

18◦F2y

F2x

Solution: Let ~F = ~F1 + ~F2. Then

Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦ = 15 N


with the x-axis and has magnitude 87 N. ~F2 makesan angle of 18◦ with the y-axis and has magnitude181 N. Use the component method to find the netforce in the y-direction. Round your answer to thenearest newton. (The picture is not drawn to scale.)

(a) 200 N *(b) 222 N(c) 244 N (d) 269 N

y

xu��

�3F1 = 87 N

35◦

F1x

F1y

AAAAAAAAK

F2 = 181 N

18◦F2y

F2x


Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦ = 222 N


with the x-axis and has magnitude 87 N. ~F2 makesan angle of 18◦ with the y-axis and has magnitude181 N. Find the magnitude of the net force on theparticle. Round your answer to the nearest newton.(The picture is not drawn to scale.)

(a) 200 N *(b) 223 N(c) 245 N (d) 269 N

y

xu��

�3F1 = 87 N

35◦

F1x

F1y

AAAAAAAAK

F2 = 181 N

18◦F2y

F2x


Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦

Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦

F =√F 2x + F 2

y

=[((87 N) cos 35◦ − (181 N) sin 18◦)2 + ((87 N) sin 35◦ + (181 N) cos 18◦)2

]1/2= 223 N



with the x-axis and has magnitude 87 N. ~F2 makesan angle of 18◦ with the y-axis and has magnitude181 N. If ~Fnet is the net force on the particle, findthe angle that ~Fnet makes with the x-axis. Roundyour answer to the nearest degree. (The picture isnot drawn to scale.)

(a) 83◦ *(b) 86◦

(c) 89◦ (d) 92◦

y

xu��

�3F1 = 87 N

35◦

F1x

F1y

AAAAAAAAK

F2 = 181 N

18◦F2y

F2x


Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦ ≈ 15 N

Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦ ≈ 222 N

Since Fx and Fy are both positive, ~F is in the first quadrant. Hence

θ = tan−1

(FyFx

)= tan−1

((87 N) sin 35◦ + (181 N) cos 18◦

(87 N) cos 35◦ − (181 N) sin 18◦

)= 86◦

Problem 184. You are pulling a crate of physics books down the hallway, using a ropethat makes an angle of 34◦ to the horizontal. If you are pulling on the rope with a force of440 N, what is the horizontal component of the force? Round your answer to the nearestnewton.

(a) 373 N (b) 233 N*(c) 365 N (d) 246 N(e) None of these

Solution: The horizontal component of the force is

Fh = F cos θ = (440 N) cos 34◦ = 365 N

Problem 185. You are pulling a crate of physics books down the hallway, using a ropethat makes an angle of 26◦ to the horizontal. If you are pulling on the rope with a force of290 N, what is the horizontal component of the force? Round your answer to the nearestnewton.

(a) 127 N (b) 188 N(c) 221 N *(d) 261 N(e) None of these

Solution: The horizontal component of the force is

Fh = F cos θ = (290 N) cos 26◦ = 261 N


Problem 186. Your car is stuck in a muddy road, and you and a friend are trying topull it out with ropes. You are pulling with a force of 820 N in a direction 0.45 rad eastof north; your friend is pulling with a force of 1040 N in a direction 0.32 rad west ofnorth. What is the northward force on the car?

Solution: We will use a coordinate system in which north is the positive x-directionand west is the positive y-direction. In that system, the angle at which you are pullingis −0.45 rad; the angle at which your friend is pulling is 0.32 rad.

We will need to write the two force vectors in component form, then find their sum, thenfind the magnitude and direction of that sum. The two force vectors are:


⟩= 〈820 N · cos(−0.45) rad, 820 N · sin(−0.45) rad〉


⟩= 〈1040 N · cos 0.32 rad, 1040 N · sin 0.32 rad〉

Remember that cos(−θ) = cos θ. Then the northerly component of ~C = ~A+ ~B is:

CN = 820 N · cos 0.45 rad + 1040 N · cos 0.32 rad = 1730 N

We have rounded this number to 10 N, since that appears to be the accuracy of theoriginal data.

Problem 187. A boat is held in the middle of a river by two ropes, one going to eitherbank. Each rope makes an angle of 79◦ to the upstream direction of the river. Thecurrent exerts a downstream force of 550 lbs on the boat. How much force does eachrope have to exert to hold the boat in place? (It is the same for both ropes.)

Solution: There are two ropes, each exerting a force with a magnitude of F . Theupstream component of each force vector is F cos 79◦. The two ropes working togethermust offset the downstream force of the current:

2F cos 79◦ = 550 lbs

Solving this for F give us

F =550 lbs

2 cos 79◦= 1440 lbs

Problem 188. You are riding your bicycle straight north. The wind is exerting a forceof 48 lbs on you, in a direction 39◦ east of south. What is the southward component ofthe wind’s force?

Solution: The southward component is

FS = F cos θ = (48 lbs)(cos 39◦) = 37 lbs

Problem 189. A dietitian is pulling a crate of romaine lettuce, using a rope that makesan angle of 0.41 rad to the horizontal. To move the crate requires a horizontal force of240 N. What is the force with which the dietitian needs to pull on the rope?

Solution: The number we want is F , the magnitude of the force vector. The horizontalforce is

Fx = F cos θ = F (cos 0.41 rad) = 240 N

Solving this for F give us

F =Fx

cos θ=

240 N

cos 0.41 rad= 262 N


1.8 Dot Product

1.8.1 Using the algebraic definition of dot product

Using the definition of dot product ~v · ~w =∑i

viwi compute the dot product of ~v and ~w

for the following problems.

Problem 190. ~v = 〈1, 1〉 and ~w = 〈2, 3〉

Solution: ~v · ~w = vxwx + vywy = 1 · 2 + 1 · 3 = 5

Problem 191. ~v = 〈1, 1〉 and ~w = 〈1,−1〉

Solution: ~v · ~w = vxwx + vywy = 1 · 1 + 1 · (−1) = 0

Problem 192. ~v = 〈−7, 3〉 and ~w = 〈2,−5〉

Solution: ~v · ~w = vxwx + vywy = (−7) · 2 + 3 · (−5) = −29

Problem 193. ~v = 〈1, 1, 1〉 and ~w = 〈1,−1, 1〉

Solution: ~v · ~w = vxwx + vywy + vzwz = 1 · 1 + 1 · (−1) + 1 · 1 = 1

Problem 194. ~v = 〈1, 2, 3〉 and ~w = 〈−7, 5, 2〉

Solution: ~v · ~w = vxwx + vywy + vzwz = 1 · (−7) + 2 · 5 + 3 · 2 = 9

Problem 195. ~v = 〈−1, 0, 2〉 and ~w = 〈4,−3, 7〉

Solution: ~v · ~w = vxwx + vywy + vzwz = (−1) · 4 + 0 · (−3) + 2 · 7 = 10

Problem 196. ~v = 〈1,−5, 6〉 and ~w = 〈13, 2,−8〉

Solution: ~v · ~w = vxwx+vywy+vzwz = 1 ·13+(−5) ·2+6 · (−8) = 13−10−48 = −45

Problem 197. ~v = 〈13, 2,−7〉 and ~w = 〈7, 6, 1〉

Solution: ~v · ~w = vxwx+vywy +vzwz = 13 ·7+2 ·6+(−7) ·1 = 12 ·7+12 = 12 ·8 = 96

Problem 198. ~v = 〈1, 2, 3〉 and ~w = 〈3, 2, 1〉

Solution: ~v · ~w = vxwx + vywy + vzwz = 1 · 3 + 2 · 2 + 3 · 1 = 10

Problem 199. ~v = 〈1, 2, 3〉 and ~w = 〈1, 2, 3〉

Solution: ~v · ~w = vxwx + vywy + vzwz = 1 · 1 + 2 · 2 + 3 · 3 = 14


Using the definition of dot product for the unit coordinate vectors (the ijk-basis) computethe following dot products. Recall: The ijk-basis satisfies the following relationships:

ı · = ı · k = · k = 0 ; ı · ı = · = k · k = 1 .

Note: You may need to use the distributivity property and the foil method to computethe dot products.

Problem 200. ı ·

Solution: Since the basis vectors are orthogonal to one another ı · = 0.

Problem 201. · ı

Solution: Since the basis vectors are orthogonal to one another · ı = 0. Moreover,the dot product is commutative, so · ı = ı · = 0.

Problem 202. ·

Solution: Since the basis vectors are orthonormal to one another · = 1.

Problem 203. ı · k

Solution: Since the basis vectors are orthogonal to one another ı · k = 0.

Problem 204. ı · (2+ 3k)

Solution: Since the dot product for vectors satisfies the distributive property wedistribute the first vector over the sum inside the parenthesis

ı · (2+ 3k) = 2(ı · ) + 3(ı · k) = 2(0) + 3(0) = 0

Problem 205. ı · (ı+ 2− 7k)

Solution: Since the dot product for vectors satisfies the distributive property wedistribute the first vector over the sum inside the parenthesis

ı · (ı+ 2− 7k) = ı · ı+ 2(ı · )− 7(ı · k) = 1 + 2(0)− 7(0) = 1


Problem 206. (aı+ b) · (cı+ d)

Solution: Use the foil method:

(aı+ b) · (cı+ d) = ac(ı · ı) + ad(ı · ) + bc( · ı) + bd( · )= ac(1) + ad(0) + bc(0) + bd(1)

= ac+ bd

Notice that if we use angle-bracket notation the calculation is much simpler:

〈a, b〉 · 〈c, d〉 ≡ ac+ bd .

You may be wondering at this point why we bother to use ijk-notation at all. Butangle-bracket notation is only useful for working in a rectangular coordinate system.In say a polar, spherical, or a more exotic coordinate system such as oblate-spherical,or parabolic-elliptical coordinate system the angle-bracket notation would have to bemodified using an appropriate orthonormal bases {e1, e2, e3} similar to the ijk-bases towork out the rules for the dot product. Such an approach would be an exercise in futility.

Problem 207. (aı+ b) · (aı− b)

Solution: Use the foil method:

(aı+ b) · (aı− b) = a2(ı · ı)− ab(ı · ) + ba( · ı)− b2( · ) = a2 − b2

Problem 208. (ı− 2) · (7ı+ 3)

Solution: Use the foil method: (ı− 2) · (7ı+ 3) = 7 + 0 + 0− 6 = 1

Problem 209. (ı+ ) · (ı+ + k)

Solution: Use the distributive property:

(ı+ ) · (ı+ + k) = ı · (ı+ + k) + · (ı+ + k) = ı · ı+ · = 1 + 1 = 2


1.8.2 Using the geometric form of dot product

For the following problems use the geometric form of the dot product ~v· ~w = ||~v|| ||~w|| cos θto find the angle between ~v and ~w.

Problem 210. ~v = 〈1, 1〉 and ~w = 〈1,−1〉

Problem 211. ~v = 〈1, 1〉 and ~w = 〈0, 1〉

Problem 212. ~v = 〈3, 4〉 and ~w = 〈5,−12〉

Problem 213. ~v = 〈1, 1, 1〉 and ~w = 〈1,−1, 1〉

Problem 214. ~v = 〈−1, 0, 3〉 and ~w = 〈4,−3, 2〉

Problem 215. ı and (ı+ 2− 7k)

Problem 216. (ı− 2) and (7ı+ 3)

Problem 217. (aı+ b) and (aı− b)

Problem 218. (aı+ b) and (cı+ d)

Problem 219. (ı+ ) and (ı+ + k)


1.8.3 Orthogonal vectors

For the following problems determine if the vectors ~v and ~w are orthogonal.Recall: Two nonzero vectors are orthogonal ⇐⇒ ~v · ~w = 0.

Problem 220. ~v = 2ı and ~w = ı−

Problem 221. ~v = 〈4, 3〉 and ~w =⟨

12,−2

3

⟩

Problem 222. ~v = 〈cos θ, sin θ,−1〉 and ~w = 〈cos θ, sin θ, 1〉

Problem 223. ~v = 〈cos θ, sin θ,−1〉 and ~w = 〈sin θ,− cos θ, 1〉

Problem 224. ~v = −2ı+ 3− k and ~w = 2ı+ − k

For the following problems determine the value/values of the parameter α that makesthe vectors ~v and ~w perpendicular. If no such value exists, then write ”none”.

Problem 225. ~v = 〈α, 3〉 and ~w = 〈−3, 2〉

Problem 226. ~v =⟨α2, 3⟩

and ~w =⟨

12,−2

3

⟩

Problem 227. ~v = 〈cosα, sinα,−1〉 and ~w = 〈cosα, sinα, 1〉

Problem 228. ~v = 〈cosα, sinα,−1〉 and ~w = 〈sinα,− cosα, 1〉

Problem 229. ~v =⟨

cosα, sinα,−12

⟩and ~w =

⟨sinα, cosα, 1

2

⟩


1.8.4 Direction cosines

For the following problems find the direction angles of the vector ~v. Recall: Any vector~v can be written in terms of direction cosines by first finding the magnitude of ~v, andthen multiplying it by the unit vector composed of the direction cosines:

~v = ||~v||〈cosα, cos β, cos γ〉 ⇒ ~uv =~v

||~v||= 〈cosα, cos β, cos γ〉 ,

where α is the angle between the x-axis and the vector ~v, β is the angle betweenthe y-axis and the vector ~v, and γ is the angle between the z-axis and the vector ~v.NOTE: 〈cosα, cos β, cos γ〉 is a unit vector, in fact it is the 3-D analogy of 〈cos θ, sin θ〉 =⟨

cos θ, cos(π

2− θ)⟩

.

Problem 230. ~v = 〈1, 2, 2〉

Problem 231. ~v = 〈−1, 5, 2〉

Problem 232. ~v = 〈3,−7, 9〉

Problem 233. ~v = 〈−12,−1, 3〉

Problem 234. ~v = 〈−7, 2, 5〉


1.8.5 Projections

For the following problems find the projection of the vector ~v onto the ~w-direction. Don’tforget to normalize ~w to be a unit vector. The formula for the projection is

proj~w(~v) =

(~v · ~w

||~w||

)~w

||~w||=

(~v · ~w||~w||2

)~w .

Problem 235. ~v = 〈2,−1〉 and ~w = 〈4, 3〉

Problem 236. ~v = 〈7, 3〉 and ~w = 〈5, 12〉

Problem 237. ~v = 〈−5, 2, 7〉 and ~w = 〈1, 1, 1〉

Problem 238. ~v =⟨1

2,−5

3, 11⟩

and ~w = 〈−1, 0, 1〉

Problem 239. ~v = 〈v1, v2, v3〉 and ~w = 〈w1, w2, w3〉, where ||~w|| 6= 0.


For the following problems starting with the given nonzero vectors ~v and ~w, convert theseinto a pair of orthonormal unit vectors: u~w and u~w⊥ , where

u~w =~w

||~w||, u~w⊥ =

~w⊥||~w⊥||

, and ~w⊥ = ~v −(~v · ~w||~w||2

)~w .

A special case of this is if ~w = ı, ~w⊥ = , and ~v = a ı+ b . Notice that we can express ~v

as ~v = (~v · ı) ı+ (~v · ) =

(~v · ~w

||~w||

)~w

||~w||+

(~v · ~w⊥||~w⊥||

)~w⊥||~w⊥||

.

Problem 240. ~v = 〈2,−1〉 and ~w = 〈4, 3〉

Problem 241. ~v = 〈7, 3〉 and ~w = 〈5, 12〉

Problem 242. ~v = 〈−5, 2, 7〉 and ~w = 〈1, 1, 1〉

Problem 243. ~v =⟨1

2,−5

3, 11⟩

and ~w = 〈−1, 0, 1〉

Problem 244. ~v = 〈v1, v2, v3〉 and ~w = 〈w1, w2, w3〉, where ||~w|| 6= 0.


1.8.6 Mixing it up

Problem 245. Use the component definition of the dot product to find the componentof ~F = 〈3,−5〉 in the direction of the vector ~s = 〈1,−6〉.

(a) 33 *(b) 33/√

37

(c) −15/√

17 (d) 0(e) None of these

Solution: The first step is to convert ~s to a unit vector: ~us =~s

‖~s ‖. Next, project the

force onto the s-direction using the dot product: projs(F ) = ~F · ~us. See my solutions onthe web for more detail.

Problem 246. Use the component definition of the dot product to find the componentof ~F = 〈1,−1〉 in the direction of the vector ~s = 〈1, 1〉.

*(a) 0 (b)√

2

(c) −1/√

17 (d) -2(e) None of these

Solution: This one has a short cut. Notice that the vectors are orthogonal: ~F ·~s = 0.The projection of a vector onto a direction that is orthogonal to the vector’s direction isalways zero.

Problem 247. Use the geometric definition of the dot product to find the angle betweenthe vectors ~A = 〈3, 1, 5〉 and ~B = 〈1,−2, 3〉.*(a) 43.7◦ (b) 256.8◦

(c) 59.3◦ (d) 3.4◦

(e) None of these

Solution: Use the geometric definition of dot product and solve for θ. See mysolutions on the web for the details.

Problem 248. Use the component definition of the dot product to find the componentof ~F = 〈−4, 2〉 in the direction of the vector ~s = 〈−1, 3〉.

(a) 10 (b) 1/√

2

*(c)√

10 (d)√

5(e) None of these

Solution: The first step is to convert ~s to a unit vector:

~us =~s

‖~s ‖

Next, project the force onto the s-direction using the dot product:

projs(F ) = ~F · ~us =~F · ~s‖~s ‖

=(−4)(−1) + (2)(3)√

10=

10√10

=√

10


Problem 249. Use the geometric definition of the dot product to find the angle betweenthe vectors ~A = 〈3, 1, 4〉 and ~B = 〈2, 8, 5〉. Round your answer to the nearest 0.1◦.

(a) 43.7◦ (b) 145.3◦

(c) 34.7◦ *(d) 46.3◦

(e) None of these

Solution: Use the geometric definition of the dot product and solve for θ:

~A · ~B = ‖ ~A‖‖ ~B‖ cos θ

θ = cos−1

(~A · ~B‖ ~A‖‖ ~B‖

)= cos−1

(3 · 2 + 1 · 8 + 4 · 5√

26 · 93

)= cos−1

(34√2418

)= 46.3◦

1.8.7 Application Problems

Problem 250. (work problem) A block is pulled by a taunt rope d meters across asmooth floor by a constant force of F Newtons in the direction of the rope. Let θ be theangle between the rope and the horizontal. Find the work done as a function of the ofthe distance, the applied force, and the angle between the rope and the horizontal.

Problem 251. (projection problem) An small airport has a runway in the direction~v = 〈2, 5〉. The wind velocity is measured to be ~w = 〈4, 1〉 m/s. The FAA’s law for thisairport states that a plane cannot takeoff with a tail wind in excess of 3 m/s measuredin the direction of the runway. Will the plane be allowed to takeoff? You must justifyyour answer with a calculation.

Problem 252. (work problem) A horse pulls a cart up a ramp that is d meters longand inclined at an angle of θ1 degrees above the horizontal. The cart is attached to thehorse by a rope that make an angle of θ2 degrees between the rope and the horizontal(not the incline plane). The force exerted on the rope by the horse is ~F . Find the workdone on the cart by the horse.


1.9 Cross Product

For the following problems evaluate the following expressions using the ijk-permutationcircle.

Problem 253. ı×

Problem 254. × k

Problem 255. k × ı

Problem 256. × ı

Problem 257. k ×

Problem 258. ı× k

Problem 259. k × k

Problem 260. ı× (−)

Problem 261. ı× (× k)

Problem 262. (ı× )× k

Problem 263. (ı× ı)× k

Problem 264. ı× (ı× k)

Problem 265. (× )× k

Problem 266. × (× k)


Problem 267. (Use the Foil Method for cross products) (aı+ b)× (cı+ d)

Problem 268. Based on the results from the previous problems does the cross-productcommute?Recall: Two vectors commute under cross multiplication if ~v × ~w = ~w × ~v.

Problem 269. Based on the results of previous problems does the cross-product satisfythe associative property? Recall: Two vectors satisfy the associative property undercross multiplication if (~v1 × ~v2)× ~v3 = ~v1 × (~v2 × ~v3).

Problem 270. × (3ı+ 4k) (Use the distributive property)

Problem 271. ı× (2ı− 7k) (Use the distributive property)

Problem 272. × (2ı− 7k) (Use the distributive property)

For the following problems: Given ~v and ~w evaluate the following expressions using thedefinition of cross products:

~v × ~w =

∣∣∣∣∣∣ı kv1 v2 v3

w1 w2 w3

∣∣∣∣∣∣ = ı

∣∣∣∣v2 v3

w2 w3

∣∣∣∣− ∣∣∣∣v1 v3

w1 w3

∣∣∣∣+ k

∣∣∣∣v1 v2

w1 w2

∣∣∣∣

Problem 273. Compute ~v × ~w for ~v = 〈−1, 1, 2〉 and ~w = 〈0, 1, 1〉.

Problem 274. Compute ~v × ~w for ~v = 〈1,−1, 0〉 and ~w = 〈1, 1, 0〉.

Problem 275. Compute ~v × ~w for ~v = 〈1, 2, 3〉 and ~w = 〈3, 2,−1〉.

Problem 276. Compute ~v × ~w for ~v = 〈5,−3, 2〉 and ~w = 〈7,−1, 4〉.

Problem 277. Compute ~v × (~v × ~w) for ~v = 〈5,−3, 2〉 and ~w = 〈7,−1, 4〉.


For the following problems find the area of the parallelogram having ~v and ~w as adjacentsides.Recall: The area of the parallelogram is ||~v × ~w||.

Problem 278. Compute ||~v × ~w|| for ~v = 〈−1, 1, 2〉 and ~w = 〈0, 1, 1〉.

Problem 279. Compute ||~v × ~w|| for ~v = 〈1,−1, 0〉 and ~w = 〈1, 1, 0〉.

Problem 280. Compute ||~v × ~w|| for ~v = 〈1, 2, 3〉 and ~w = 〈3, 2,−1〉.

Problem 281. Compute ||~v × ~w|| for ~v = 〈5,−3, 2〉 and ~w = 〈7,−1, 4〉.

Problem 282. Compute ||~v × (~v × ~w)|| for ~v = 〈5,−3, 2〉 and ~w = 〈7,−1, 4〉.

For the following problems find the area of the triangle with the given vertices.

(Hint:1

2||~v × ~w|| is the area of the triangle having ~v and ~w as adjacent sides).

Problem 283. ~v = 〈−1, 0, 1〉 and ~w = 〈0, 1, 1〉.

Problem 284. ~v = 〈2,−1, 0〉 and ~w = 〈1, 2, 0〉.

Problem 285. ~v = 〈0, 2, 3〉 and ~w = 〈0, 2,−1〉.

Problem 286. ~v = 〈−1,−3, 2〉 and ~w = 〈7,−1, 4〉.

Problem 287. ~v = 〈5, 0,−3〉 and ~w = 〈1, 0, 1〉.


For the following problems use a certain property of the vector cross product to find aunit vector perpendicular to the plane containing the vectors ~v and ~w.

Problem 288. ~v = 〈−1, 0, 1〉 and ~w = 〈0, 1, 1〉.

Problem 289. ~v = 〈2,−1, 0〉 and ~w = 〈1, 2, 0〉.

Problem 290. ~v = 〈0, 2, 3〉 and ~w = 〈0, 2,−1〉.

Problem 291. ~v = 〈−1,−3, 2〉 and ~w = 〈7,−1, 4〉.

Problem 292. ~v = 〈5, 0,−3〉 and ~w = 〈1, 0, 1〉.

Problem 293. Find the volume with the parallelepiped with vertices:

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

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


1.10 Advanced-Level Problems (Dot and Cross Product)

1.10.1 Dot product

Problem 294. (Derive Problem) Show that the vectors (~v · ~w)~u− (~u · ~w)~v and ~w areperpendicular.

Solution: Work it out!

Problem 295. (Derive Problem) Prove the identity ||~v± ~w||2 = ||~v||2±2~v · ~w+ ||~w||2.Hint: For any vector ~v, ||~v||2 = ~v · ~v. Use this together with the foil method.


Problem 296. (Derive Problem) Prove the Cauchy-Schwartz inequality |~v · ~w| ≤||~v|| ||~w||.Hint: What are the bounds on cos θ ?


Problem 297. (Derive Problem) Prove the triangle inequality ||~v+ ~w|| ≤ ||~v||+ ||~w||.


Problem 298. (Derive Problem) Prove the parallelogram law ||~v+ ~w||2 + ||~v− ~w||2 =2(||~v||2 + ||~w||2). Give a geometric interpretation of the parallelogram law.


Problem 299. (Derive Problem) Find the angle between the diagonal of a cube andone of its edges.



Problem 300. (Derive Problem [point masses and gravitation])In physics, the law of gravitation says that if P and Q are (point) masses with mass Mand m, respectively, then P is attracted to Q by the force

~F =

(GMm

||~r||2

)~r

||~r||,

where ~r is the directed line segment from P to Q and G is the universal gravitationalconstant. Moreover, if Q1, Q2, . . . , Qk are point masses with mass m1,m2, . . . ,mk, re-spectively, then the force on P due to all of the Qi’s is

~F =k∑i=1

(GMmi

||~ri||2

)~ri||~ri||

=k∑i=1

(GMmi ~ri||~ri||3

),

where ~ri is the directed line segment from P to Qi.

a) Let point P of mass M be located at P = (0, d), d > 0 in the coordinate plane. Fori ∈ {−n,−(n−1), . . . ,−1, 0, 1, . . . , (n−1), n}, let Qi be located at point (id, 0) and havemass mi = mi. Find the gravitational force on P due to all of the Qi’s.

b) Is the limit as n→∞ of the magnitude of the force on P finite? Verify your answerwith an appropriate calculation.


1.10.2 Cross product

Problem 301. (Derive Problem) Prove the identity (~a+~b)× (~a−~b) = −2(~a×~b)


Problem 302. (Derive Problem) Prove the identity ~u× (~v× ~w) = (~u · ~w)~v− (~u ·~v) ~w .Hint: Compute each side of the equality separately and show that they are equal.


Problem 303. (Derive Problem)(a) Prove the identity

~u× (~v × ~w) + ~w × (~u× ~v) + ~v × (~w × ~u) = 0 .

(b) Referring to the identity in part (a), under what conditions does the cross productsatisfy the associative property:

~u× (~v × ~w) = (~u× ~v)× ~w ?



Problem 304. (Derive Problem [Relativistic sums])Einstein’s special theory of relativity roughly says that with respect to a reference frame(coordinate system) no material object can travel as fast as c, the speed of light. So, if ~xand ~y are two velocities such that ‖~x‖ < c and ‖~y‖ < c, then the relativistic sum ~x⊕~y of~x and ~y must have its length less than c. Einstein’s special theory of relativity says that

~x⊕ ~y =~x+ ~y

1 +~x · ~yc2

+1

c2γx

γx + 1

~x× (~x× ~y)

1 +~x · ~yc2

,

where

γx =1√

1− ~x · ~xc2

It can be shown that if ‖~x‖ < c and ‖~y‖ < c, then ‖~x ⊕ ~y‖ < c. We now look at twospecial cases:(a) Prove that if ~x and ~y are orthogonal, ‖~x‖ < c, and ‖~y‖ < c, then ‖~x⊕ ~y‖ < c.(b) Prove that if ~x and ~y are parallel, ‖~x‖ < c, and ‖~y‖ < c, then ‖~x⊕ ~y‖ < c.(c) Compute lim

c→∞~x⊕ ~y.

Difficulty rating: (a) (difficult level) (b) (intermediate level) and (c) (easy)Hint: For one of the parts you’ll find the identity ~u× (~v× ~w) = (~u · ~w)~v− (~u ·~v)~w useful.


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Vector Primer Problem Set Solutions - Pima Community College

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