Cap_3 VEctores

8/17/2019 Cap_3 VEctores

1/62

1. The vectors should be parallel to achieve a resultant 7 m long (the unprimed case shown below), antipar-allel (in opposite directions) to achieve a resultant 1 m long (primed case shown), and perpendicular to

achieve a resultant√

32 + 42 = 5 m long (the double-primed case shown). In each sketch, the vectorsare shown in a “head-to-tail” sketch but the resultant is not shown. The resultant would be a straightline drawn from beginning to end; the beginning is indicated by A (with or without primes, as the casemay be) and the end is indicated by B.

b

A-

b

-

B

b

A

-

b

B

b

A

- b

6

B


2/62

2. A sketch of the displace-ments is shown. The re-

sultant (not shown) wouldbe a straight line fromstart (Bank) to finish(Walpole). With a carefuldrawing, one should findthat the resultant vectorhas length 29.5 km at 35◦

west of south.

West East

South

Bank

@

@

@

@

@

@

@

@

@

@ R

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H Y

B

B

B

B

B

B

B N

Walpole


3/62

3. The x component of a is given by ax = 7.3cos250◦ = −2.5 and the y component is given by ay =

7.3sin250◦ = −6.9. In considering the variety of ways to compute these, we note that the vector is

70◦ below the −

x axis, so the components could also have been found from ax = −

7.3cos70◦ anday = −7.3sin70

◦. In a similar vein, we note that the vector is 20◦ from the −y axis, so one could use

ax = −7.3sin20◦ and ay = −7.3cos20

◦ to achieve the same results.


4/62

4. The angle described by a full circle is 360◦ = 2π rad, which is the basis of our conversion factor. Thus,

(20.0◦) 2π rad360◦

= 0.349 rad

and (similarly) 50.0◦ = 0.873 rad and 100◦ = 1.75 rad. Also,

(0.330 rad) 360◦

2π rad = 18.9◦

and (similarly) 2.10 rad = 120◦ and 7.70 rad = 441◦.


5/62

5. The textbook’s approach to this sort of problem is through the use of Eq. 3-6, and is illustrated in SampleProblem 3-3. However, most modern graphical calculators can produce the results quite efficiently using

rectangular ↔ polar “shortcuts.”

(a)

(−25)2 + 402 = 47.2 m

(b) Recalling that tan(θ) = tan(θ + 180◦), we note that the two possibilities for tan−1 (40/− 25)are −58◦ and 122◦. Noting that the vector is in the third quadrant (by the signs of its x and ycomponents) we see that 122◦ is the correct answer. The graphical calculator “shortcuts” mentionedabove are designed to correctly choose the right possibility.


6/62

6. The x component of r is given by 15 cos30◦ = 13 m and the y component is given by 15 sin 30◦ = 7.5 m.


7/62

7. The point P is displaced vertically by 2R, where R is the radius of the wheel. It is displaced horizontallyby half the circumference of the wheel, or πR. Since R = 0.450 m, the horizontal component of the

displacement is 1.414 m and the vertical component of the displacement is 0.900 m. If the x axis ishorizontal and the y axis is vertical, the vector displacement (in meters) is r = (1.414 î + 0.900 ̂j). Thedisplacement has a magnitude of

|r| =

(πR)2 + (2R)2 = R π2 + 4 = 1.68 m

and an angle of

tan−1

2R

πR

= tan−1

2

π

= 32.5◦

above the floor. In physics there are no “exact” measurements, yet that angle computation seemed toyield something exact . However, there has to be some uncertainty in the observation that the wheelrolled half of a revolution, which introduces some indefiniteness in our result.


8/62

8. Although we think of this as a three-dimensional movement, it is rendered effectively two-dimensionalby referring measurements to its well-defined plane of the fault.

(a) The magnitude of the net displacement is

| AB| =

|AD|2 + |AC |2 =

172 + 222 = 27.8 m .

(b) The magnitude of the vertical component of AB is |AD| sin52.0◦ = 13.4 m.


9/62


10/62

10. We label the displacement vectors A, B and C (and denote the result

of their vector sum as r). Wechoose east as the î direction (+xdirection) and north as the ̂j di-rection (+y direction). All dis-tances are understood to be inkilometers. We note that the an-gle between C and the x axis is60◦. Thus,

-

6

?

eastwest

north

south

e

A- e

6 B

e

C

A = 50 î B = 30 ̂j C = 25 cos (60◦) î + 25 sin (60◦) ̂j

r = A + B + C = 62.50 î + 51.65 ̂j

which means

that its magnitude is

|r| =

62.502 + 51.652 ≈ 81 km .

and its angle (counterclockwise from +x axis) is tan−1 (51.65/62.50) ≈ 40◦, which is to say that it points40◦ north of east . Although the resultant r is shown in our sketch, it would be a direct line from the“tail” of A to the “head” of C .


11/62

11. The diagram shows the displacement vectors for the two segments of her walk, labeled A and B, andthe total (“final”) displacement vector, labeled r. We take east to be the +x direction and north to be

the +y direction. We observe that the angle between A and the x axis is 60◦. Where the units are notexplicitly shown, the distances are understood

to be in meters. Thus, the componentsof A are Ax = 250cos60◦ = 125 andAy = 250sin30

◦ = 216.5. The compo-

nents of B are Bx = 175 and By = 0.The components of the total displace-ment are rx = Ax + Bx = 125 + 175 =300 and ry = Ay + By = 216.5 + 0 =216.5. ....

..

..

..

.

..

..

..

..

.

..

..

..

..

.

..

..

..

..

.

..

..

..

..

.

..

..

..

..

.

..

..

..

..

.

..

..

..

..

..

.

..

..

..

......................................................................................................................................................................

......................

.....................................................................................................................................................................................

......................

.....................................................................................................................................................................................................................................................................................................................................................................

......................North

East

A

B

r

(a) The magnitude of the resultant displacement is

|r| =

r2x

+ r2y

=

3002 + 216.52 = 370 m .

(b) The angle the resultant displacement makes with the +x axis is

tan−1ry

rx

= tan−1

216.5

300

= 36◦ .

(c) The total distance walked is d = 250 + 175 = 425 m.

(d) The total distance walked is greater than the magnitude of the resultant displacement. The diagram

shows why: A and B are not collinear.


12/62

12. We label the displacement vectors A, B and C (and denote the result

of their vector sum as r). Wechoose east as the î direction (+xdirection) and north as the ̂j di-rection (+y direction). All dis-tances are understood to be inkilometers. Therefore,

-

6

?

eastwest

north

south

e

A

6

e

B

e

C

?

A = 3.1 ̂j B = −2.4 î C = −5.2 ̂j

r = A + B + C = −2.1 î − 2.4 ̂j

which means

that its magnitude is|r| =

(−2.1)2 + (−2.4)2 ≈ 3.2 km .

and the two possibilities for its angle are

tan−1−2.4

−2.1

= 41◦, or 221◦ .

We choose the latter possibility since r is in the third quadrant. It should be noted that many graphicalcalculators have polar ↔ rectangular “shortcuts” that automatically produce the correct answer for angle(measured counterclockwise from the +x axis). We may phrase the angle, then, as 221◦ counterclockwisefrom East (a phrasing that sounds peculiar, at best) or as 41◦ south from west or 49◦ west from south.

The resultant r is not shown in our sketch; it would be an arrow directed from the “tail” of A to the“head” of C .


13/62

13. We write r = a + b. When not explicitly displayed, the units here are assumed to be meters. Thenrx = ax + bx = 4.0− 13 = −9.0 and ry = ay + by = 3.0 + 7.0 = 10. Thus r = (−9.0 m) î + (10 m) ̂j . The

magnitude of the resultant is

r =

r2x

+ r2y

=

(−9.0)2 + (10)2 = 13 m .

The angle between the resultant and the +x axis is given by tan−1 (ry/rx) = tan−1 10/(−9.0) which is

either −48◦ or 132◦. Since the x component of the resultant is negative and the y component is positive,characteristic of the second quadrant, we find the angle is 132◦ (measured counterclockwise from +xaxis).


14/62

14. The x, y and z components (with meters understood) of r are

rx = cx + dx = 7.4 + 4.4 = 12ry = cy + dy = −3.8 − 2.0 = −5.8

rz = cz + dz = −6.1 + 3.3 = −2.8 .


15/62

15. The vectors are shown on the diagram. The x axis runs from west

to east and the y axis runfrom south to north. Thenax = 5.0 m, ay = 0, bx =−(4.0m)sin35◦ = −2.29m,and by = (4.0m)cos35

◦ =3.28m.

..........................................................................................................................................................................................................................................................................................................................................

............................

.

.......................................................................................................................................................................................................................

..

..

.

..

..............................................

.

.

.

..

.

.

..

.

.

.

.

..

.

.

.

.

.

..

......

.....................................................................................................................................................................................................................................

.................................................

......

............................

a

ba + b

E

N

50.5◦

35◦

(a) Let c = a+ b. Then cx = ax +bx = 5.0 m−2.29m = 2.71 m and cy = ay +by = 0+ 3.28m = 3.28m.The magnitude of c is

c =

c2x

+ c2y

=

(2.71m)2 + (3.28m)2 = 4.3 m .

(b) The angle θ that c = a + b makes with the +x axis is

θ = tan−1 cy

cx= tan−1

3.28 m

2.71 m = 50.4◦ .

The second possibility (θ = 50.4◦ + 180◦ = 126◦) is rejected because it would point in a directionopposite to c.

(c) The vector b−a is found by adding −a to b. The result is shown

on the diagram to theright. Let c = b −a. Then cx = bx −

ax = −

2.29 m−

5.0 m =−7.29m and cy = by −ay = 3.28 m. The magni-

tude of c is c =

c2x

+ c2y

= 8.0 m . ...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

....................................................................................................

..

.

.

.

..

.

.

..

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

..................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................

............

−a

b −a + b

W

N

(d) The tangent of the angle θ that c makes with the +x axis (east) is

tan θ = cy

cx=

3.28 m

−7.29 m = −4.50, .

There are two solutions: −24.2◦ and 155.8◦. As the diagram shows, the second solution is correct.The vector c = −a + b is 24◦ north of west.


16/62

16. All distances in this solution are understood to be in meters.

(a) a + b = (3.0 î + 4.0 ̂j) + (5.0 î − 2.0 ̂j) = 8.0 î + 2.0 ̂j .(b) The magnitude of a + b is

|a + b| =

8.02 + 2.02 = 8.2 m .

(c) The angle between this vector and the +x axis is tan−1(2.0/8.0) = 14◦.

(d) b − a = (5.0 î − 2.0 ̂j) − (3.0 î + 4.0 ̂j) = 2.0 î − 6.0 ̂j .

(e) The magnitude of the difference vector b − a is

| b − a| =

2.02 + (−6.0)2 = 6.3 m .

(f) The angle between this vector and the +x axis is tan−1(−6.0/2.0) = −72◦. The vector is 72◦

clockwise from the axis defined by î.


17/62

17. All distances in this solution are understood to be in meters.

(a) a + b = (4.0 + (−

1.0)) î + ((−

3.0) + 1.0) ̂j + (1.0 + 4.0) k̂ = 3.0 î−

2.0 ̂j + 5.0 k̂ .

(b) a− b = (4.0− (−1.0)) î + ((−3.0)− 1.0) ̂j + (1.0− 4.0) k̂ = 5.0 î− 4.0 ̂j− 3.0 k̂ .

(c) The requirement a− b + c = 0 leads to c = b− a, which we note is the opposite of what we found

in part (b). Thus, c = −5.0 î + 4.0 ̂j + 3.0 k̂ .


18/62

18. Many of the operations are done efficiently on most modern graphical calculators using their built-invector manipulation and rectangular

↔ polar “shortcuts.” In this solution, we employ the “traditional”

methods (such as Eq. 3-6).

(a) The magnitude of a is

42 + (−3)2 = 5.0 m.(b) The angle between a and the +x axis is tan−1(−3/4) = −37◦. The vector is 37◦ clockwise from the

axis defined by î .

(c) The magnitude of b is√

62 + 82 = 10 m.

(d) The angle between b and the +x axis is tan−1(8/6) = 53◦.

(e) a + b = (4 + 6) î + ((−3) + 8) ̂j = 10 î + 5 ̂j , with the unit meter understood. The magnitude of thisvector is

√ 102 + 52 = 11 m; we rounding to two significant figures in our results.

(f) The angle between the vector described in part (e) and the +x axis is tan−1(5/10) = 27◦.

(g) b−

a = (6−

4) î + (8−

(−

3)) ̂j = 2 î + 11 ̂j , with the unit meter understood. The magnitude of thisvector is √ 22 + 112 = 11 m, which is, interestingly, the same result as in part (e) (exactly, not justto 2 significant figures) (this curious coincidence is made possible by the fact that a ⊥ b).

(h) The angle between the vector described in part (g) and the +x axis is tan−1(11/2) = 80◦.

(i) a − b = (4− 6) î + ((−3) − 8) ̂j = −2 î − 11 ̂j , with the unit meter understood. The magnitude of this vector is

(−2)2 + (−11)2 = 11 m.

(j) The two possibilities presented by a simple calculation for the angle between the vector described inpart (i) and the +x direction are tan−1(11/2) = 80◦, and 180◦ + 80◦ = 260◦. The latter possibilityis the correct answer (see part (k) for a further observation related to this result).

(k) Since a− b = (−1)( b−a), they point in opposite (antiparallel) directions; the angle between themis 180◦.


19/62

19. Many of the operations are done efficiently on most modern graphical calculators using their built-invector manipulation and rectangular

↔ polar “shortcuts.” In this solution, we employ the “traditional”

methods (such as Eq. 3-6). Where the length unit is not displayed, the unit meter should be understood.

(a) Using unit-vector notation,

a = 50 cos (30◦) î + 50 sin (30◦) ̂j

b = 50 cos (195◦) î + 50 sin(195◦) ̂j

c = 50 cos (315◦) î + 50 sin(315◦) ̂j

a + b + c = 30.4 î− 23.3 ̂j .

The magnitude of this result is

30.42 + (−23.3)2 = 38 m.(b) The two possibilities presented by a simple calculation for the angle between the vector described

in part (a) and the +x direction are tan−1(−

23.2/30.4) = −

37.5◦, and 180◦ + (−

37.5◦) = 142.5◦.

The former possibility is the correct answer since the vector is in the fourth quadrant (indicatedby the signs of its components). Thus, the angle is −37.5◦, which is to say that it is roughly 38◦clockwise from the +x axis. This is equivalent to 322.5◦ counterclockwise from +x.

(c) We find a− b+c = (43.3− (−48.3)+35.4) î− (25− (−12.9)+(−35.4)) ̂j = 127 î + 2.6 ̂j in unit-vectornotation. The magnitude of this result is

√ 1272 + 2.62 ≈ 1.3× 102 m.

(d) The angle between the vector described in part (c) and the +x axis is tan−1(2.6/127) ≈ 1◦.(e) Using unit-vector notation, d is given by

d = a + b− c= −40.4 î + 47.4 ̂j

which has a magnitude of (−40.4)2 + 47.42 = 62 m.

(f) The two possibilities presented by a simple calculation for the angle between the vector describedin part (e) and the +x axis are tan−1(47.4/(−40.4)) = −50◦, and 180◦+ (−50◦) = 130◦. We choosethe latter possibility as the correct one since it indicates that d is in the second quadrant (indicatedby the signs of its components).


20/62

20. Angles are given in ‘standard’ fashion, so Eq. 3-5 applies directly. We use this to write the vectors inunit-vector notation before adding them. However, a very different-looking approach using the special

capabilities of most graphical calculators can be imagined. Where the length unit is not displayed in thesolution below, the unit meter should be understood.

(a) Allowing for the different angle units used in the problem statement, we arrive at

E = 3.73 î + 4.70 ̂j F = 1.29 î − 4.83 ̂j G = 1.45 î + 3.73 ̂j H = −5.20 î + 3.00 ̂j

E + F + G + H = 1.28 î + 6.60 ̂j .

(b) The magnitude of the vector sum found in part (a) is√

1.282 + 6.602 = 6.72 m. Its angle measured

counterclockwise from the +x axis is tan−1(6.6/1.28) = 79◦ = 1.38 rad.


21/62

21. It should be mentioned that an efficient way to work this vector addition problem is with the cosinelaw for general triangles (and since a, b and r form an isosceles triangle, the angles are easy to figure).

However, in the interest of reinforcing the usual systematic approach to vector addition, we note thatthe angle b makes with the +x axis is 135◦ and apply Eq. 3-5 and Eq. 3-6 where appropriate.

(a) The x component of r is 10cos30◦ + 10cos 135◦ = 1.59 m.

(b) The y component of r is 10sin30◦ + 10sin 135◦ = 12.1 m.

(c) The magnitude of r is√

1.592 + 12.12 = 12.2 m.

(d) The angle between r and the +x direction is tan−1 (12.1/1.59) = 82.5◦.


22/62

22. If we wish to use Eq. 3-5 in an unmodified fashion, we should note that the angle between C and the+x axis is 180◦ + 20◦ = 200◦.

(a) The x component of B is given by C x − Ax = 15cos200◦ − 12cos40◦ = −23.3 m, and the y

component of B is given by C y − Ay = 15sin200◦− 12sin40◦ = −12.8 m. Consequently, its

magnitude is

(−23.3)2 + (−12.8)2 = 26.6 m.

(b) The two possibilities presented by a simple calculation for the angle between B and the +x axisare tan−1((−12.8)/(−23.3)) = 28.9◦, and 180◦ + 28.9◦ = 209◦. We choose the latter possibility

as the correct one since it indicates that B is in the third quadrant (indicated by the signs of itscomponents). We note, too, that the answer can be equivalently stated as −151◦.


23/62

23. We consider A with (x, y) components given by (A cos α, A sin α). Similarly, B → (B cos β, B sin β ). Theangle (measured from the +x direction) for their vector sum must have a slope given by

tan θ = A sin α + B sin β

A cos α + B cos β .

The problem requires that we now consider the orthogonal direction, where tan θ + 90◦ = − cot θ. If this(the negative reciprocal of the above expression) is to equal the slope for their vector difference , then wemust have

−A cos α + B cos β

A sin α + B sin β =

A sin α−B sin β

A cos α−B cos β .

Multiplying both sides by A sin α + B sin β and doing the same with A cos α−B cos β yields

A2 cos2 α−B2 cos2 β = A2 sin2 α−B2 sin2 β .

Rearranging, using the cos2

φ + sin

2

φ = 1 identity, we obtain

A2 = B2 =⇒ A = B .

In a later section, the scalar (dot) product of vectors is presented and this result can be revisited witha more compact derivation.


24/62

24. If we wish to use Eq. 3-5 directly, we should note that the angles for Q, R and S are 100◦, 250◦ and 310◦,respectively, if they are measured counterclockwise from the +x axis.

(a) Using unit-vector notation, with the unit meter understood, we have

P = 10 cos (25◦) î + 10 sin(25◦) ̂j

Q = 12 cos (100◦) î + 12 sin (100◦) ̂j

R = 8cos(250◦) î + 8 sin(250◦) ̂j

S = 9cos(310◦) î + 9 sin(310◦) ̂j

P + Q + R + S = 10.0 î + 1.6 ̂j

(b) The magnitude of the vector sum is√

102 + 1.62 = 10.2 m and its angle is tan−1 (1.6/10) ≈ 9.2◦measured counterclockwise from the +x axis. The appearance of this solution would be quitedifferent using the vector manipulation capabilities of most modern graphical calculators, althoughthe principle would be basically the same.


25/62

25. Without loss of generality, we assume a points along the +x axis, and that b is at θ measured counter-clockwise from a. We wish to verify that

r2 = a2 + b2 + 2ab cos θ

where a = |a| = ax (we’ll call it a for simplicity) and b = | b| =

b2x

+ b2y

. Since r = a + b then

r = |r| =

(a + bx)2 + b2y. Thus, the above expression becomes

(a + bx)2 + b2y

2= a2 +

b2x

+ b2y

2+ 2ab cos θ

a2 + 2abx + b

2

x + b2

y = a2 + b2

x + b2

y + 2ab cos θ

which makes a valid equality since (the last term) 2ab cos θ is indeed the same as 2abx (on the left-handside). In a later section, the scalar (dot) product of vectors is presented and this result can be revisited

with a somewhat different-looking derivation.


26/62

26. The vector equation is R = A + B + C + D. Expressing B and D in unit-vector notation, we have1.69 î + 3.63 ̂j and −2.87 î + 4.10 ̂j, respectively. Where the length unit is not displayed in the solution

below, the unit meter should be understood.

(a) Adding corresponding components, we obtain R = −3.18 î + 4.72 ̂j.

(b) and (c) Converting this result to polar coordinates (using Eq. 3-6 or functions on a vector-capablecalculator), we obtain

(−3.18, 4.72) −→ (5.69 124◦)

which tells us the magnitude is 5.69 m and the angle (measured counterclockwise from +x axis) is124◦.


27/62

27. (a) There are 4 such lines, one from each of the corners on the lower face to the diametrically oppositecorner on the upper face. One is shown on the diagram. Using an xyz coordinate system as shown

(with the origin at the back lower left corner) The position vector of the “starting point” of thediagonal shown is a î and the position vector of the ending point is a ̂j + a k̂, so the vector alongthe line is the difference a ̂j + a k̂ − a î.

The point diametrically oppositethe origin has position vectora î +a ̂j +a k̂ and this is the vec-tor along the diagonal. Anothercorner of the bottom face is ata î+a ̂j and the diametrically op-posite corner is at a k̂, so anothercube diagonal is a k̂ − a î − a ̂j.The fourth diagonal runs from a ̂j

to a î + a k̂, so the vector alongthe diagonal is a î + a k̂ − a ̂j.

.

..

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.............

.

.

.

.

.

.

.

..

.

...........................................................................................................................................................................................................................................................

................................................................................................

.........................................................................................................................................................................................................................................................................................................................................................

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

..............

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

..

.............

.............

.............

.............

.............

.............

............. ............. ............. ............. ............. ............. ............. ............. ............. ..........

.............

.............

.

.

.

.

.

.

.

.

.

.

.

..

.

.

..

.

.

.

.

.

.

.............

.............

.............

.............

.............

.............

.............

.............

.............

.............

.............

.............

.............

.............

.............

.............

.............

a

a

a

x

y

z

(b) Consider the vector from the back lower left corner to the front upper right corner. It is a î +

a ̂j + a k̂. We may think of it as the sum of the vector a î parallel to the x axis and the vectora ̂j + a k̂ perpendicular to the x axis. The tangent of the angle between the vector and the x axisis the perpendicular component divided by the parallel component. Since the magnitude of theperpendicular component is

√ a2 + a2 = a

√ 2 and the magnitude of the parallel component is a,

tan θ =a√

2/a =

√ 2. Thus θ = 54.7◦. The angle between the vector and each of the other two

adjacent sides (the y and z axes) is the same as is the angle between any of the other diagonalvectors and any of the cube sides adjacent to them.

(c) The length of any of the diagonals is given by√ a2 + a2 + a2 = a

√ 3.


28/62

28. Reference to Figure 3-18 (and the accompanying material in that section) is helpful. If we convert B to the

magnitude-angle notation (as A already is) we have B = (14.4 33.7◦) (appropriate notation especially

if we are using a vector capable calculator in polar mode). Where the length unit is not displayed in thesolution, the unit meter should be understood. In the magnitude-angle notation, rotating the axis by+20◦ amounts to subtracting that angle from the angles previously specified. Thus, A = (12.0 40.0◦)

and B = (14.4 13.7◦), where the ‘prime’ notation indicates that the description is in terms of the newcoordinates. Converting these results to (x, y) representations, we obtain

A = 9.19 î

+ 7.71 ̂j

B = 14.0 î

+ 3.41 ̂j

with the unit meter understood, as already mentioned.


29/62

29. We apply Eq. 3-20 and Eq. 3-27.

(a) The scalar (dot) product of the two vectors is a · b = ab cosφ = (10)(6.0)cos60◦

= 30.

(b) The magnitude of the vector (cross) product of the two vectors is |a× b| = ab sinφ = (10)(6.0)sin60◦ =52.


30/62

30. We consider all possible products and then simplify using relations such as î · k̂ = 0 and î · î = 1. Thus,

a · b =

ax î + ay ̂j + az k̂

·

bx î + by ̂j + bz k̂

= axbx î · î + axby î · ̂j + axbz î · k̂ + aybx ̂j · î + aybj ̂j · ̂j + · · ·

= axbx(1) + axby(0) + axbz(0) + aybx(0) + aybj(1) + · · ·

which is seen to reduce to the desired result (one might wish to show this in two dimensions beforetackling the additional tedium of working with these three-component vectors).


31/62

31. Since ab cos φ = axbx + ayby + azbz,

cos φ = axbx + ayby + azbz

ab .

The magnitudes of the vectors given in the problem are

a = |a| =

(3.0)2 + (3.0)2 + (3.0)2 = 5.2

b = | b| =

(2.0)2 + (1.0)2 + (3.0)2 = 3.7 .

The angle between them is found from

cosφ = (3.0)(2.0) + (3.0)(1.0) + (3.0)(3.0)

(5.2)(3.7) = 0.926 .

The angle is φ = 22◦.


32/62

32. We consider all possible products and then simplify using relations such as î× î = 0 and the importantfundamental products

î × ̂j = − ̂j × î = k̂

̂j × k̂ = − k̂ × ̂j = î

k̂ × î = − î × k̂ = ̂j .

Thus,

a× b =ax î + ay ̂j + az k̂

×

bx î + by ̂j + bz k̂

= axbx î × î + axby î × ̂j + axbz î × k̂ + aybx ̂j × î + aybj ̂j × ̂j + · · ·

= axbx(0) + axby( k̂) + axbz(− ̂j) + aybx(− k̂) + aybj(0) + · · ·

which is seen to simplify to the desired result.


33/62

33. The area of a triangle is half the product of its base and altitude. The base is the side formed by vector

a. Then the altitude is b sinφ and the area is A = 12ab sinφ = 1

2|a × b|.


34/62

34. Applying Eq. 3-23, F = qv × B (where q is a scalar) becomes

F x î + F y ̂j + F z k̂ = q (vyBz − vzBy) î + q (vzBx − vxBz) ̂j + q (vxBy − vyBx) k̂

which – plugging in values – leads to three equalities:

4.0 = 2 (4.0Bz − 6.0By)

−20 = 2 (6.0Bx − 2.0Bz)

12 = 2 (2.0By − 4.0Bx)

Since we are told that Bx = By , the third equation leads to By = −3.0. Inserting this value into thefirst equation, we find Bz = −4.0. Thus, our answer is

B = −3.0 î − 3.0 ̂j − 4.0 k̂ .


35/62

35. Both proofs shown below utilize the fact that the vector (cross) product of a and b is perpendicular to

both a and b. This is mentioned in the book, and is fundamental to its discussion of the right-hand rule.

(a) ( b × a) is a vector that is perpendicular to a, so the scalar product of a with this vector is zero.This can also be verified by using Eq. 3-30, and then (with suitable notation changes) Eq. 3-23.

(b) Let c = b×a. Then the magnitude of c is c = ab sinφ. Since c is perpendicular to a the magnitude

of a ×c is ac. The magnitude of a× ( b×a) is consequently |a× ( b×a)| = ac = a2b sin φ. This toocan be verified by repeated application of Eq. 3-30, although it must be admitted that this is muchless intimidating if one is using a math software package such as MAPLE or Mathematica.


36/62

36. If a vector capable calculator is used, this makes a good exercise for getting familiar with those features.Here we briefly sketch the method. Eq. 3-30 leads to

2 A× B = 2

2 î + 3 ̂j − 4 k̂×

−3 î + 4 ̂j + 2 k̂

= 44 î + 16 ̂j + 34 k̂ .

We now apply Eq. 3-23 to evaluate 3 C ·

2 A× B

:

3

7 î − 8 ̂j·

44 î + 16 ̂j + 34 k̂

= 3 ((7)(44) + (−8)(16) + (0)(34)) = 540 .


37/62

37. From the figure, we note that c ⊥ b, which implies that the angle between c and the +x axis is 120◦.

(a) Direct application of Eq. 3-5 yields the answers for this and the next few parts. ax = a cos0◦

=a = 3.00 m.

(b) ay = a sin0◦ = 0.

(c) bx = b cos30◦ = (4.00m)cos30◦ = 3.46 m.

(d) by = b sin30◦ = (4.00m)sin30◦ = 2.00 m.

(e) cx = c cos120◦ = (10.0m)cos120◦ = −5.00 m.

(f) cy = c sin30◦ = (10.0m)sin120◦ = 8.66 m.

(g) In terms of components (first x and then y), we must have

−5.00 m = p(3.00 m) + q (3.46m)

8.66 m = p(0) + q (2.00m) .

Solving these equations, we find p = −6.67

(h) and q = 4.33 (note that it’s easiest to solve for q first). The numbers p and q have no units.


38/62

38. We apply Eq. 3-20 with Eq. 3-23. Where the length unit is not displayed, the unit meter is understood.

(a) We first note that a = |a| = √ 3.22 + 1.62 = 3.58 m and b = | b| = √ 0.52 + 4.52 = 4.53 m. Now,a · b = axbx + ayby = ab cosφ

(3.2)(0.5) + (1.6)(4.5) = (3.58)(4.53) cosφ

which leads to φ = 57◦ (the inverse cosine is double-valued as is the inverse tangent, but we knowthis is the right solution since both vectors are in the same quadrant).

(b) Since the angle (measured from +x) for a is tan−1 (1.6/3.2) = 26.6◦, we know the angle for cis 26.6◦ − 90◦ = −63.4◦ (the other possibility, 26.6◦ + 90◦ would lead to a cx < 0). Therefore,cx = c cos−63.4◦ = (5.0)(0.45) = 2.2 m.

(c) Also, cy = c sin−63.4◦ = (5.0)(−0.89) = −4.5 m.(d) And we know the angle for d to be 26.6◦ + 90◦ = 116.6◦, which leads to dx = d cos 116.6◦ =

(5.0)(−0.45) = −2.2 m.(e) Finally, dy = d sin116.6◦ = (5.0)(0.89) = 4.5 m.


39/62

39. The solution to problem 27 showed that each diagonal has a length given by a√

3, where a is the lengthof a cube edge. Vectors along two diagonals are b = a î + a ̂j + a k̂ and c =

−a î + a ̂j + a k̂. Using

Eq. 3-20 with Eq. 3-23, we find the angle between them:

cos φ = bxcx + bycy + bzcz

bc =

−a2 + a2 + a2

3a2 =

1

3 .

The angle is φ = cos−1(1/3) = 70.5◦.


40/62

40. (a) The vector equation r = a− b−v is computed as follows: (5.0− (−2.0)+ 4.0)̂i + (4.0− 2.0 + 3.0)̂ j +

((−6.0) − 3.0 + 2.0)k̂. This leads to r = 11 î + 5.0 ̂j − 7.0 k̂.

(b) We find the angle from +z by “dotting” (taking the scalar product) r with k̂. Noting that r =|r| =

112 + 52 + (−7)2 = 14, Eq. 3-20 with Eq. 3-23 leads to

r · k̂ = −7.0 = (14)(1) cosφ =⇒ φ = 120◦ .

(c) To find the component of a vector in a certain direction, it is efficient to “dot” it (take the scalarproduct of it) with a unit-vector in that direction. In this case, we make the desired unit-vector by

b̂ = b

| b|=

−2 î + 2 ̂j + 3 k̂

(−2)2 + 22 + 32.

We therefore obtain

ab = a · b̂ = (5)(−2) + (4)(2) + (−6)(3)

(−2)2 + 22 + 32 = −4.9 .

(d) One approach (if we all we require is the magnitude) is to use the vector cross product, as theproblem suggests; another (which supplies more information) is to subtract the result in part (c)

(multiplied by b̂) from a. We briefly illustrate both methods. We note that if a cos θ (where θ is

the angle between a and b) gives ab (the component along b̂) then we expect a sin θ to yield theorthogonal component:

a sin θ = |a× b|

b = 7.3

(alternatively, one might compute θ form part (c) and proceed more directly). The second methodproceeds as follows:

a− ab b̂ = (5.0 − 2.35)̂i + (4.0 − (−2.35))̂ j + ((−6.0) − (−3.53))k̂= 2.65 î + 6.35 ̂j − 2.47 k̂

This describes the perpendicular part of a completely. To find the magnitude of this part, wecompute

2.652 + 6.352 + (−2.47)2 = 7.3

which agrees with the first method.


41/62

41. The volume of a parallelepiped is equal to the product of its altitude and the area of its base. Take thebase to be the parallelogram formed by the vectors b and c. Its area is bc sinφ, where φ is the angle

between b and c. This is just the magnitude of the vector (cross) product b × c. The altitude of theparallelepiped is a cos θ, where θ is the angle between a and the normal to the plane of b and c. Since b×c is normal to that plane, θ is the angle between a and b×c. Thus, the volume of the parallelepipedis V = a| b × c| cos θ = a · ( b × c).


42/62

42. We apply Eq. 3-30 and Eq. 3-23.

(a) a× b = (axby − aybx) k̂ since all other terms vanish, due to the fact that neither a nor b have anyz components. Consequently, we obtain ((3.0)(4.0) − (5.0)(2.0)) k̂ = 2.0 k̂ .

(b) a · b = axbx + ayby yields (3)(2) + (5)(4) = 26.(c) a + b = (3 + 2) î + (5 + 4) ̂j , so that

a + b

· b = (5)(2) + (9)(4) = 46.

(d) Several approaches are available. In this solution, we will construct a b̂ unit-vector and “dot” it(take the scalar product of it) with a. In this case, we make the desired unit-vector by

b̂ = b

| b|=

2 î + 4 ̂j√ 22 + 42

.

We therefore obtain

ab = a · b̂ = (3)(2) + (5)(4)√ 22 + 42

= 5.8 .


43/62

43. We apply Eq. 3-30 and Eq.3-23. If a vector capable calculator is used, this makes a good exercise forgetting familiar with those features. Here we briefly sketch the method.

(a) We note that b× c = −8 î + 5 ̂j + 6 k̂. Thus, a · b× c

= (3)(−8) + (3)(5) + (−2)(6) = −21.

(b) We note that b + c = 1 î − 2 ̂j + 3 k̂. Thus, a · b + c

= (3)(1) + (3)(−2) + (−2)(3) = −9.

(c) Finally, a× b + c

= ((3)(3)−(−2)(−2)) î+((−2)(1)−(3)(3)) ̂j+((3)(−2)−(3)(1)) k̂ = 5 î−11 ̂j−9 k̂.


44/62

44. The components of a are ax = 0, ay = 3.20cos63◦ = 1.45, and az = 3.20sin63

◦ = 2.85. The components

of b are bx = 1.40cos48◦ = 0.937, by = 0, and bz = 1.40sin48◦ = 1.04.

(a) The scalar (dot) product is therefore

a · b = axbx + ayby + azbz = (0)(0.937) + (1.45)(0) + (2.85)(1.04) = 2.97 .

(b) The vector (cross) product is

a× b = (aybz − azby) î + (azbx − axbz) ̂j + (axby − aybx) k̂

= ((1.45)(1.04)− 0)) î + ((2.85)(0.937) − 0)) ̂j + (0 − (1.45)(0.94)) k̂

= 1.51 î + 2.67 ̂j − 1.36 k̂ .

(c) The angle θ between a and b is given by

θ = cos−1

a · b

ab

= cos−1

2.96

(3.30)(1.40)

= 48◦ .


45/62

45. We observe that̂i× î

=

̂i

̂i sin0◦ vanishes because sin 0◦ = 0. Similarly, ̂j× ̂j = k̂× k̂ = 0. When the

unit vectors are perpendicular, we have to do a little more work to show the cross product results. First,the magnitude of the vector î× ̂j is

̂i× ̂j

=

̂i

̂ j sin90◦

which equals 1 because sin 90◦ = 1 and these are all units vectors (each has magnitude equal to 1). This

is consistent with the claim that î × ̂j = k̂ since the magnitude of k̂ is certainly 1. Now, we use theright-hand rule to show that î× ̂j is in the positive z direction. Thus î× ̂j has the same magnitude anddirection as k̂, so it is equal to k̂. Similarly, k̂× î = ̂j and ̂j× k̂ = î. If, however, the coordinate systemis left-handed, we replace k̂ → −k̂ in the work we have shown above and get

î× î = ̂j× ̂j = k̂× k̂ = 0 .

just as before. But the relations that are different are

î× ̂j = −k̂ k̂× î = − ̂j ̂j× k̂ = −î .


46/62

46. (a) By the right-hand rule, A× B points upward if A points north and B points west. If A and B havemagnitude = 1 then, by Eq. 3-27, the result also has magnitude equal to 1.

(b) Since cos 90◦ = 0, the scalar dot product between perpendicular vectors is zero. Thus, A · B = 0 is A points down and B points south.

(c) By the right-hand rule, A× B points south if A points east and B points up. If A and B have unitmagnitude then, by Eq. 3-27, the result also has unit magnitude.

(d) Since cos 0◦ = 1, then A · B = AB (where A is the magnitude of A and B is the magnitude of B).If, additionally, we have A = B = 1, then the result is 1.

(e) Since sin 0◦ = 0, A× B = 0 if both A and B point south.


47/62

47. Let A denote the magnitude of A; similarly for the other vectors. The vector equation is A + B = C where B = 8.0 m and C = 2A. We are also told that the angle (measured in the ‘standard’ sense) for

A is 0◦ and the angle for C is 90◦, which makes this a right triangle (when drawn in a “head-to-tail”fashion) where B is the size of the hypotenuse. Using the Pythagorean theorem,

B = A2 + C 2 =⇒ 8.0 =

A2 + 4A2

which leads to A = 8/√

5 = 3.6 m.


48/62

48. We choose +x east and +y north and measure all angles in the “standard” way (positive ones are

counterclockwise from +x). Thus, vector d1 has magnitude d1 = 4 (with the unit meter and three

significant figures assumed) and direction θ1 = 225◦. Also, d2 has magnitude d2 = 5 and direction

θ2 = 0◦, and vector d3 has magnitude d3 = 6 and direction θ3 = 60

◦.

(a) The x-component of d1 is d1 cos θ1 = −2.83 m.

(b) The y -component of d1 is d1 sin θ1 = −2.83 m.

(c) The x-component of d2 is d2 cos θ2 = 5.00 m.

(d) The y -component of d2 is d2 sin θ2 = 0.

(e) The x-component of d3 is d3 cos θ3 = 3.00 m.

(f) The y -component of d3 is d3 sin θ3 = 5.20 m.

(g) The sum of x-components is −

2.83 + 5.00 + 3.00 = 5.17 m.

(h) The sum of y -components is −2.83 + 0 + 5.20 = 2.37 m.

(i) The magnitude of the resultant displacement is√

5.172 + 2.372 = 5.69 m.

(j) And its angle is θ = tan−1(2.37/5.17) = 24.6◦ which (recalling our coordinate choices) means itpoints at about 25◦ north of east.

(k) and () This new displacement (the direct line home) when vectorially added to the previous(net) displacement must give zero. Thus, the new displacement is the negative, or opposite, of theprevious (net) displacement. That is, it has the same magnitude (5.69 m) but points in the oppositedirection (25◦ south of west).


49/62

49. Reading carefully, we see that the (x, y) specifications for each “dart” are to be interpreted as (∆x, ∆y)descriptions of the corresponding displacement vectors. We combine the different parts of this problem

into a single exposition. Thus, along the x axis, we have (with the centimeter unit understood)

30.0 + bx − 20.0 − 80.0 = −140.0 ,

and along y axis we have40.0 − 70.0 + cy − 70.0 = −20.0 .

Hence, we find bx = −70.0 cm and cy = 80.0 cm. And we convert the final location (−140,−20) intopolar coordinates and obtain (141 −172◦), an operation quickly done using a vector capable calculatorin polar mode. Thus, the ant is 141 cm from where it started at an angle of −172◦, which means 172◦

clockwise from the +x axis or 188◦ counterclockwise from the +x axis.


50/62

50. We find the components and then add them (as scalars, not vectors). With d = 3.40 km and θ = 35.0◦

we find d cos θ + d sin θ = 4.74 km.


51/62

51. We choose +x east and +y north and measure all angles in the “standard” way (positive ones counter-

clockwise from +x, negative ones clockwise). Thus, vector d1 has magnitude d1 = 3.66 (with the unit

meter and three significant figures assumed) and direction θ1 = 90◦. Also, d2 has magnitude d2 = 1.83

and direction θ2 = −45◦, and vector d3 has magnitude d3 = 0.91 and direction θ3 = −135◦. We add thex and y components, respectively:

x : d1 cos θ1 + d2 cos θ2 + d3 cos θ3 = 0.651 m

y : d1 sin θ1 + d2 sin θ2 + d3 sin θ3 = 1.723 m .

(a) The magnitude of the direct displacement (the vector sum d1+ d2+ d3 ) is√

0.6512 + 1.7232 = 1.84 m.

(b) The angle (understood in the sense described above) is tan−1(1.723/0.651) = 69◦. That is, the firstputt must aim in the direction 69◦ north of east.


52/62

52. (a) We write b = b ̂j where b > 0. We are asked to consider

bd

=

bd

̂j

in the case d > 0. Since the coefficient of ̂j is positive, then the vector points in the +y direction.

(b) If, however, d 0, we find that a × ( b/d) has magnitude ab/d and is pointed in the +z direction.


53/62

53. (a) With a = 17.0 m and θ = 56.0◦ we find ax = a cos θ = 9.51 m.

(b) And ay = a sin θ = 14.1 m.

(c) The angle relative to the new coordinate system is θ = 56−18 = 38◦. Thus, ax

= a cos θ = 13.4 m.

(d) And ay

= a sin θ = 10.5 m.


54/62

54. Since cos0◦ = 1 and sin 0◦ = 0, these follows immediately from Eq. 3-20 and Eq. 3-27.


55/62

55. (a) The magnitude of the vector a = 4.0 d is (4.0)(2.5) = 10 m.

(b) The direction of the vector a

= 4.

0 d

is the same as the direction of d

(north).(c) The magnitude of the vector c = −3.0 d is (3.0)(2.5) = 7.5 m.

(d) The direction of the vector c = −3.0 d is the opposite of the direction of d. Thus, the direction of cis south.


56/62

56. The vector sum of the displacements dstorm and dnew must give the same result as its originally intendeddisplacement do = 120 ̂j where east is î, north is ̂j, and the assumed length unit is km. Thus, we write

dstorm = 100 î and dnew = A î + B ̂j .

(a) The equation dstorm + dnew = do readily yields A = −100 km and B = 120 km. The magnitude of dnew is therefore

√ A2 + B2 = 156 km.

(b) And its direction is tan−1(B/A) = −50.2◦ or 180◦ + (−50.2◦) = 129.8◦. We choose the latter valuesince it indicates a vector pointing in the second quadrant, which is what we expect here. Theanswer can be phrased several equivalent ways: 129.8◦ counterclockwise from east, or 39.8◦ westfrom north, or 50.2◦ north from west.


57/62

57. (a) The height is h = d sin θ, where d = 12.5 m and θ = 20.0◦. Therefore, h = 4.28 m.

(b) The horizontal distance is d cos θ = 11.7 m.


58/62

58. (a) We orient î eastward, ̂j northward, and k̂ upward. The displacement in meters is consequently

1000 î + 2000 ̂j− 500 k̂ .

(b) The net displacement is zero since his final position matches his initial position.


59/62

59. (a) If we add the equations, we obtain 2a = 6c, which leads to a = 3c = 9 î + 12 ̂j .

(b) Plugging this result back in, we find b

= c

= 3ˆi + 4

ˆ j.


60/62

60. (First problem in Cluster 1)The given angle θ = 130◦ is assumed to be measured counterclockwise from the +x axis. Angles (if

positive) in our results follow the same convention (but if negative are clockwise from +x).

(a) With A = 4.00, the x-component of A is A cos θ = −2.57.

(b) The y -component of A is A sin θ = 3.06.

(c) Adding A and B produces a vector we call R with components Rx = −6.43 and Ry = −1.54.Using Eq. 3-6 (or special functions on a calculator) we present this in magnitude-angle notation: R = (6.61 − 167◦).

(d) From the discussion in the previous part, it is clear that R = −6.43 î − 1.54 ̂j .

(e) The vector C is the difference of A and B. In unit-vector notation, this becomes

C = A− B = −2.57 î − 3.06 ̂j− −3.86 î − 4.60 ̂jwhich yields C = 1.29 î + 7.66 ̂j .

(f) Using Eq. 3-6 (or special functions on a calculator) we present this in magnitude-angle notation: C = (7.77 80.5◦).

(g) We note that C is the “constant” in all six pictures. Remembering that the negative of a vector

simply reverses it, then we see that in form or another, all six pictures express the relation C = A− B.


61/62

61. (Second problem in Cluster 1)

(a) The dot (scalar) product of 3 A and B is found using Eq. 3-23:

3 A · B = 3(4.00cos130◦) (−3.86) + 3 (4.00sin130◦) (−4.60) = −12.5 .

(b) We call the result D and combine the scalars ((3)(4) = 12). Thus, D = (4 A)× (3 B) becomes, usingEq. 3-30,

12 A× B = 12 ((4.00cos130◦) (−4.60) − (4.00sin130◦) (−3.86)) k̂

which yields D = 284 k̂ .

(c) Since D has magnitude 284 and points in the +z direction, it has radial coordinate 284 and angle-measured-from-z-axis equal to 0◦. The angle measured in the xy plane does not have a well-definedvalue (since this vector does not have a component in that plane)

(d) Since A is in the xy plane, then it is clear that A ⊥ D. The angle between them is 90◦.

(e) Calling this new result G we have

G = (4.00cos130◦) î + (4.00sin130◦) ̂j + (3.00)k̂

which yields G = −2.57 î + 3.06̂ j + 3.00 k̂ .

(f) It is straightforward using a vector-capable calculator to convert the above into spherical coordi-nates. We, however, proceed “the hard way”, using the notation in Fig. 3-44 (where θ is in the xyplane and φ is measured from the z axis):

| G| = r =

(−2.57)2 + 3.062 + 3.002 = 5.00

φ = tan−1(4.00/3.00) = 53.1◦

θ = 130◦ given in problem 60 .


62/62

62. (Third problem in Cluster 1)

(a) Looking at the xy plane in Fig. 3-44, it is clear that the angle to A (which is the vector lying

in the plane, not the one rising out of it, which we called G in the previous problem) measuredcounterclockwise from the −y axis is 90◦ +130◦ = 220◦. Had we measured this clockwise we wouldobtain (in absolute value) 360◦ − 220◦ = 140◦.

(b) We found in part (b) of the previous problem that A× B points along the z axis, so it is perpendicularto the −y direction.

(c) Let u = − ̂j represent the −y direction, and w = 3 k̂ is the vector being added to B in this problem.

The vector being examined in this problem (we’ll call it Q) is, using Eq. 3-30 (or a vector-capablecalculator),

Q = A×

B + w

= 9.19 1̂ + 7.71 ̂j + 23.7 k̂

and is clearly in the first octant (since all components are positive); using Pythagorean theorem,its magnitude is Q = 26.52. From Eq. 3-23, we immediately find u · Q = −7.71. Since u has unitmagnitude, Eq. 3-20 leads to

cos−1

u · Q

Q

= cos−1

−7.71

26.52

which yields a choice of angles 107◦ or −107◦. Since we have already observed that Q is in the firstoctant, the the angle measured counterclockwise (as observed by someone high up on the +z axis)

from the −y axis to Q is 107◦.

Documents

Cap_3 VEctores