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CHAPTER 1
1.1. Given the vectors M = − 10 ax + 4ay − 8 az and N = 8ax + 7ay − 2 az , find: a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4 ay + 8az + 16ax + 14ay − 4 az = (26, 10 , 4) Thus a =
b) the magnitude of 5ax + N − 3 M:
(5, 0 , 0) + (8, 7 , −2) − (− 30 , 12 , −24) = (43, − 5 , 22), and |(43, − 5 , 22)| = 48. 6.
c) |M|| 2 N|(M + N): |(− 10 , 4 , −8)||(16, 14 , −4)|(− 2 , 11 , −10) = (13.4)(21.6)(− 2 , 11 , −10) = (− 580. 5 , 3193 , −2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the unit vector in the direction of (A − B): First
A − B = (ax + 2ay + 3az ) − (2ax + 3ay − 2 az ) = (−ax − ay + 5az )
whose magnitude is |A − B| = [(−ax − ay + 5az ) · (−ax − ay + 5az )]^1 /^2 =
3 = 5.20. The unit vector is therefore
aAB = (−ax − ay + 5az )/ 5. 20
b) find the unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of A and B: The midpoint is located at
Pmp = [1 + (2 − 1)/ 2 , 2 + (3 − 2)/ 2 , 3 + (− 2 − 3)/2)] = (1. 5 , 2. 5 , 0 .5)
The unit vector is then
amp =
(1. 5 ax + 2. 5 ay + 0. 5 az ) p (1.5)^2 + (2.5)^2 + (0.5)^2
= (1. 5 ax + 2. 5 ay + 0. 5 az )/ 2. 96
1.3. The vector from the origin to the point A is given as (6, − 2 , −4), and the unit vector directed from the origin toward point B is (2, − 2 , 1)/3. If points A and B are ten units apart, find the coordinates of point B. With A = (6, − 2 , −4) and B = 13 B(2, − 2 , 1), we use the fact that |B − A| = 10, or |(6 − 23 B)ax − (2 − 23 B)ay − (4 + 13 B)az | = 10 Expanding, obtain 36 − 8 B + 49 B^2 + 4 − 83 B + 49 B^2 + 16 + 83 B + 19 B^2 = 100 or B^2 − 8 B − 44 = 0. Thus B = 8 ±
√ 64 − 176 2 = 11.75 (taking positive option) and so
B =
(11.75)ax −
(11.75)ay +
(11.75)az = 7. 83 ax − 7. 83 ay + 3. 92 az
1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit vector in rectangular components that lies in the xy plane, is tangent to the circle at (−
and is in the general direction of increasing values of y: A unit vector tangent to this circle in the general increasing y direction is t = −aφ. Its x and y components are tx = −aφ · ax = sin φ, and ty = −aφ · ay = − cos φ. At the point (−
φ = 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax +
3 ay ).
1.5. A vector field is specified as G = 24xyax + 12(x^2 + 2)ay + 18z^2 az. Given two points, P (1, 2 , −1) and Q(− 2 , 1 , 3), find: a) G at P : G(1, 2 , −1) = (48, 36 , 18) b) a unit vector in the direction of G at Q: G(− 2 , 1 , 3) = (− 48 , 72 , 162), so
aG =
c) a unit vector directed from Q toward P :
aQP =
P − Q
|P − Q|
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x^2 + 2), 18 z^2 )|, or 10 = |(4xy, 2 x^2 + 4, 3 z^2 )|, so the equation is
100 = 16x^2 y^2 + 4x^4 + 16x^2 + 16 + 9z^4
1.6. Find the acute angle between the two vectors A = 2ax + ay + 3az and B = ax − 3 ay + 2az by using the definition of: a) the dot product: First, A · B = 2 − 3 + 6 = 5 = AB cos θ, where A =
22 + 1^2 + 3^2 =
and where B =
12 + 3^2 + 2^2 =
- Therefore cos θ = 5/14, so that θ = 69. 1 ◦. b) the cross product: Begin with
A × B =
ØØ
Ø
ØØ
Ø
ax ay az 2 1 3 1 − 3 2
ØØ
Ø
ØØ
Ø
= 11ax − ay − 7 az
and then |A × B| =
112 + 1^2 + 7^2 =
- So now, with |A × B| = AB sin θ =
find θ = sin−^1
1.7. Given the vector field E = 4zy^2 cos 2xax + 2zy sin 2xay + y^2 sin 2xaz for the region |x|, |y|, and |z| less than 2, find: a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0, with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2, |z| < 2; 4) the plane x = π/ 2 , with |y| < 2, |z| < 2. b) the region in which Ey = Ez : This occurs when 2zy sin 2x = y^2 sin 2x, or on the plane 2z = y, with |x| < 2, |y| < 2, |z| < 1. c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy^2 cos 2x = zy sin 2x = y^2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.
1.11. Given the points M (0. 1 , − 0. 2 , − 0 .1), N (− 0. 2 , 0. 1 , 0 .3), and P (0. 4 , 0 , 0 .1), find:
a) the vector RM N : RM N = (− 0. 2 , 0. 1 , 0 .3) − (0. 1 , − 0. 2 , − 0 .1) = (− 0. 3 , 0. 3 , 0 .4). b) the dot product RM N · RM P : RM P = (0. 4 , 0 , 0 .1) − (0. 1 , − 0. 2 , − 0 .1) = (0. 3 , 0. 2 , 0 .2). RM N · RM P = (− 0. 3 , 0. 3 , 0 .4) · (0. 3 , 0. 2 , 0 .2) = − 0 .09 + 0.06 + 0.08 = 0. 05. c) the scalar projection of RM N on RM P :
RM N · aRM P = (− 0. 3 , 0. 3 , 0 .4) ·
d) the angle between RM N and RM P :
θM = cos−^1
μ RM N · RM P |RM N ||RM P |
= cos−^1
μ
- 05 √
- 34
1.12. Write an expression in rectangular components for the vector that extends from (x 1 , y 1 , z 1 ) to (x 2 , y 2 , z 2 ) and determine the magnitude of this vector.
The two points can be written as vectors from the origin:
A 1 = x 1 ax + y 1 ay + z 1 az and A 2 = x 2 ax + y 2 ay + z 2 az
The desired vector will now be the difference:
A 12 = A 2 − A 1 = (x 2 − x 1 )ax + (y 2 − y 1 )ay + (z 2 − z 1 )az
whose magnitude is
|A 12 | =
p A 12 · A 12 =
(x 2 − x 1 )^2 + (y 2 − y 1 )^2 + (z 2 − z 1 )^2
1.13. a) Find the vector component of F = (10, − 6 , 5) that is parallel to G = (0. 1 , 0. 2 , 0 .3):
F||G =
F · G
|G|^2
G =
b) Find the vector component of F that is perpendicular to G:
FpG = F − F||G = (10, − 6 , 5) − (0. 93 , 1. 86 , 2 .79) = (9. 07 , − 7. 86 , 2 .21)
c) Find the vector component of G that is perpendicular to F:
GpF = G − G||F = G −
G · F
|F|^2
F = (0. 1 , 0. 2 , 0 .3) −
1.14. Given that A + B + C = 0, where the three vectors represent line segments and extend from a common origin, a) must the three vectors be coplanar?
In terms of the components, the vector sum will be
A + B + C = (Ax + Bx + Cx)ax + (Ay + By + Cy )ay + (Az + Bz + Cz )az
which we require to be zero. Suppose the coordinate system is configured so that vectors A and B lie in the x-y plane; in this case Az = Bz = 0. Then Cz has to be zero in order for the three vectors to sum to zero. Therefore, the three vectors must be coplanar.
b) If A + B + C + D = 0, are the four vectors coplanar?
The vector sum is now
A + B + C + D = (Ax + Bx + Cx + Dx)ax + (Ay + By + Cy + Dy )ay + (Az + Bz + Cz + Dz )az
Now, for example, if A and B lie in the x-y plane, C and D need not, as long as Cz + Dz = 0. So the four vectors need not be coplanar to have a zero sum.
1.15. Three vectors extending from the origin are given as r 1 = (7, 3 , −2), r 2 = (− 2 , 7 , −3), and r 3 = (0, 2 , 3). Find: a) a unit vector perpendicular to both r 1 and r 2 :
ap 12 =
r 1 × r 2 |r 1 × r 2 |
b) a unit vector perpendicular to the vectors r 1 − r 2 and r 2 − r 3 : r 1 − r 2 = (9, − 4 , 1) and r 2 − r 3 = (− 2 , 5 , −6). So r 1 − r 2 × r 2 − r 3 = (19, 52 , 32). Then
ap =
c) the area of the triangle defined by r 1 and r 2 :
Area =
|r 1 × r 2 | = 30. 3
d) the area of the triangle defined by the heads of r 1 , r 2 , and r 3 :
Area =
|(r 2 − r 1 ) × (r 2 − r 3 )| =
|(− 9 , 4 , −1) × (− 2 , 5 , −6)| = 32. 0
1.18. A certain vector field is given as G = (y + 1)ax + xay. a) Determine G at the point (3,-2,4):
G(3, − 2 , 4) = −ax + 3ay.
b) obtain a unit vector defining the direction of G at (3,-2,4). |G(3, − 2 , 4)| = [1 + 3^2 ]^1 /^2 =
- So the unit vector is
aG(3, − 2 , 4) =
−ax + 3ay √ 10
1.19. a) Express the field D = (x^2 + y^2 )−^1 (xax + yay ) in cylindrical components and cylindrical variables: Have x = ρ cos φ, y = ρ sin φ, and x^2 + y^2 = ρ^2. Therefore
D =
ρ
(cos φax + sin φay )
Then Dρ = D · aρ =
ρ
[cos φ(ax · aρ) + sin φ(ay · aρ)] =
ρ
cos^2 φ + sin^2 φ
ρ and
Dφ = D · aφ =
ρ
[cos φ(ax · aφ) + sin φ(ay · aφ)] =
ρ
[cos φ(− sin φ) + sin φ cos φ] = 0
Therefore D =
ρ
aρ
b) Evaluate D at the point where ρ = 2, φ = 0. 2 π, and z = 5, expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0. 5 aρ. To express this in cartesian, we use
D = 0.5(aρ · ax)ax + 0.5(aρ · ay )ay = 0.5 cos 36◦ax + 0.5 sin 36◦ay = 0. 41 ax + 0. 29 ay
1.20. If the three sides of a triangle are represented by the vectors A, B, and C, all directed counter- clockwise, show that |C|^2 = (A + B) · (A + B) and expand the product to obtain the law of cosines.
With the vectors drawn as described above, we find that C = −(A + B) and so |C|^2 = C^2 = C · C = (A + B) · (A + B) So far so good. Now if we expand the product, obtain
(A + B) · (A + B) = A^2 + B^2 + 2A · B
where A · B = AB cos(180◦^ − α) = −AB cos α where α is the interior angle at the junction of A and B. Using this, we have C^2 = A^2 + B^2 − 2 AB cos α, which is the law of cosines.
1.21. Express in cylindrical components:
a) the vector from C(3, 2 , −7) to D(− 1 , − 4 , 2): C(3, 2 , −7) → C(ρ = 3. 61 , φ = 33. 7 ◦, z = −7) and D(− 1 , − 4 , 2) → D(ρ = 4. 12 , φ = − 104. 0 ◦, z = 2). Now RCD = (− 4 , − 6 , 9) and Rρ = RCD · aρ = −4 cos(33.7) − 6 sin(33.7) = − 6 .66. Then Rφ = RCD · aφ = 4 sin(33.7) − 6 cos(33.7) = − 2 .77. So RCD = − 6. 66 aρ − 2. 77 aφ + 9az b) a unit vector at D directed toward C: RCD = (4, 6 , −9) and Rρ = RDC · aρ = 4 cos(− 104 .0) + 6 sin(− 104 .0) = − 6 .79. Then Rφ = RDC · aφ = 4[− sin(− 104 .0)] + 6 cos(− 104 .0) = 2.43. So RDC = − 6. 79 aρ + 2. 43 aφ − 9 az Thus aDC = − 0. 59 aρ + 0. 21 aφ − 0. 78 az c) a unit vector at D directed toward the origin: Start with rD = (− 1 , − 4 , 2), and so the vector toward the origin will be −rD = (1, 4 , −2). Thus in cartesian the unit vector is a = (0. 22 , 0. 87 , − 0 .44). Convert to cylindrical: aρ = (0. 22 , 0. 87 , − 0 .44) · aρ = 0.22 cos(− 104 .0) + 0.87 sin(− 104 .0) = − 0 .90, and aφ = (0. 22 , 0. 87 , − 0 .44) · aφ = 0.22[− sin(− 104 .0)] + 0.87 cos(− 104 .0) = 0, so that finally, a = − 0. 90 aρ − 0. 44 az.
1.22. A sphere of radius a, centered at the origin, rotates about the z axis at angular velocity Ω rad/s. The rotation direction is clockwise when one is looking in the positive z direction. a) Using spherical components, write an expression for the velocity field, v, which gives the tan- gential velocity at any point within the sphere: As in problem 1.20, we find the tangential velocity as the product of the angular velocity and the perperdicular distance from the rotation axis. With clockwise rotation, we obtain
v(r, θ) = Ωr sin θ aφ (r < a)
b) Convert to rectangular components: From here, the problem is the same as part c in Problem 1.20, except the rotation direction is reversed. The answer is v(x, y) = Ω [−y ax + x ay ], where (x^2 + y^2 + z^2 )^1 /^2 < a.
1.23. The surfaces ρ = 3, ρ = 5, φ = 100◦, φ = 130◦, z = 3, and z = 4.5 define a closed surface. a) Find the enclosed volume:
Vol =
Z 4. 5
3
Z 130 ◦
100 ◦
Z 5
3
ρ dρ dφ dz = 6. 28
NOTE: The limits on the φ integration must be converted to radians (as was done here, but not shown).
b) Find the total area of the enclosing surface:
Area = 2
Z 130 ◦
100 ◦
Z 5
3
ρ dρ dφ +
Z 4. 5
3
Z 130 ◦
100 ◦
3 dφ dz
Z 4. 5
3
Z 130 ◦
100 ◦
5 dφ dz + 2
Z 4. 5
3
Z 5
3
dρ dz = 20. 7
1.26. Express the uniform vector field, F = 5 ax in a) cylindrical components: Fρ = 5 ax · aρ = 5 cos φ, and Fφ = 5 ax · aφ = −5 sin φ. Combining, we obtain F(ρ, φ) = 5(cos φ aρ − sin φ aφ). b) spherical components: Fr = 5 ax ·ar = 5 sin θ cos φ; Fθ = 5 ax ·aθ = 5 cos θ cos φ; Fφ = 5 ax ·aφ = −5 sin φ. Combining, we obtain F(r, θ, φ) = 5 [sin θ cos φ ar + cos θ cos φ aθ − sin φ aφ].
1.27. The surfaces r = 2 and 4, θ = 30◦^ and 50◦, and φ = 20◦^ and 60◦^ identify a closed surface. a) Find the enclosed volume: This will be
Vol =
Z 60 ◦
20 ◦
Z 50 ◦
30 ◦
Z 4
2
r^2 sin θdrdθdφ = 2. 91
where degrees have been converted to radians. b) Find the total area of the enclosing surface:
Area =
Z 60 ◦
20 ◦
Z 50 ◦
30 ◦
(4^2 + 2^2 ) sin θdθdφ +
Z 4
2
Z 60 ◦
20 ◦
r(sin 30◦^ + sin 50◦)drdφ
Z 50 ◦
30 ◦
Z 4
2
rdrdθ = 12. 61
c) Find the total length of the twelve edges of the surface:
Length = 4
Z 4
2
dr + 2
Z 50 ◦
30 ◦
(4 + 2)dθ +
Z 60 ◦
20 ◦
(4 sin 50◦^ + 4 sin 30◦^ + 2 sin 50◦^ + 2 sin 30◦)dφ
= 17. 49
d) Find the length of the longest straight line that lies entirely within the surface: This will be from A(r = 2, θ = 50◦, φ = 20◦) to B(r = 4, θ = 30◦, φ = 60◦) or
A(x = 2 sin 50◦^ cos 20◦, y = 2 sin 50◦^ sin 20◦, z = 2 cos 50◦)
to B(x = 4 sin 30◦^ cos 60◦, y = 4 sin 30◦^ sin 60◦, z = 4 cos 30◦)
or finally A(1. 44 , 0. 52 , 1 .29) to B(1. 00 , 1. 73 , 3 .46). Thus B − A = (− 0. 44 , 1. 21 , 2 .18) and
Length = |B − A| = 2. 53
1.28. State whether or not A = B and, if not, what conditions are imposed on A and B when
a) A · ax = B · ax: For this to be true, both A and B must be oriented at the same angle, θ, from the x axis. But this would allow either vector to lie anywhere along a conical surface of angle θ about the x axis. Therefore, A can be equal to B, but not necessarily.
b) A × ax = B × ax: This is a more restrictive condition because the cross product gives a vector. For both cross products to lie in the same direction, A, B, and ax must be coplanar. But if A lies at angle θ to the x axis, B could lie at θ or at 180◦^ − θ to give the same cross product. So again, A can be equal to B, but not necessarily.
1.28c) A · ax = B · ax and A × ax = B × ax: In this case, we need to satisfy both requirements in parts a and b – that is, A, B, and ax must be coplanar, and A and B must lie at the same angle, θ, to ax. With coplanar vectors, this latter condition might imply that both +θ and −θ would therefore work. But the negative angle reverses the direction of the cross product direction. Therefore both vectors must lie in the same plane and lie at the same angle to x; i.e., A must be equal to B.
d) A · C = B · C and A × C = B × C where C is any vector except C = 0: This is just the general case of part c. Since we can orient our coordinate system in any manner we choose, we can arrange it so that the x axis coincides with the direction of vector C. Thus all the arguments of part c apply, and again we conclude that A must be equal to B.
1.29. Express the unit vector ax in spherical components at the point: a) r = 2, θ = 1 rad, φ = 0.8 rad: Use
ax = (ax · ar)ar + (ax · aθ)aθ + (ax · aφ)aφ = sin(1) cos(0.8)ar + cos(1) cos(0.8)aθ + (− sin(0.8))aφ = 0. 59 ar + 0. 38 aθ − 0. 72 aφ
b) x = 3, y = 2, z = −1: First, transform the point to spherical coordinates. Have r =
θ = cos−^1 (− 1 /
- = 105. 5 ◦, and φ = tan−^1 (2/3) = 33. 7 ◦. Then
ax = sin(105. 5 ◦) cos(33. 7 ◦)ar + cos(105. 5 ◦) cos(33. 7 ◦)aθ + (− sin(33. 7 ◦))aφ = 0. 80 ar − 0. 22 aθ − 0. 55 aφ
c) ρ = 2.5, φ = 0.7 rad, z = 1.5: Again, convert the point to spherical coordinates. r =
p √ ρ^2 +^ z^2 = 8 .5, θ = cos−^1 (z/r) = cos−^1 (1. 5 /
8 .5) = 59. 0 ◦, and φ = 0.7 rad = 40. 1 ◦. Now
ax = sin(59◦) cos(40. 1 ◦)ar + cos(59◦) cos(40. 1 ◦)aθ + (− sin(40. 1 ◦))aφ = 0. 66 ar + 0. 39 aθ − 0. 64 aφ
1.30. Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft. We assume a constant altitude, a plane earth, a flight along the x axis from 0 to 10 units, no vertical velocity component, and no change in wind velocity with time. Assume ax to be directed to the east and ay to the north. The wind velocity at the operating altitude is assumed to be:
v(x, y) =
(0. 01 x^2 − 0. 08 x + 0.66)ax − (0. 05 x − 0 .4)ay 1 + 0. 5 y^2 a) Determine the location and magnitude of the maximum tailwind encountered: Tailwind would be x-directed, and so we look at the x component only. Over the flight range, this function maximizes at a value of 0. 86 /(1 + 0. 5 y^2 ) at x = 10 (at the end of the trip). It reaches a local minimum of 0. 50 /(1 + 0. 5 y^2 ) at x = 4, and has another local maximum of 0. 66 /(1 + 0. 5 y^2 ) at the trip start, x = 0.
b) Repeat for headwind: The x component is always positive, and so therefore no headwind exists over the travel range.
c) Repeat for crosswind: Crosswind will be found from the y component, which is seen to maximize over the flight range at a value of 0. 4 /(1 + 0. 5 y^2 ) at the trip start (x = 0).
d) Would more favorable tailwinds be available at some other latitude? If so, where? Minimizing the denominator accomplishes this; in particular, the lattitude associated with y = 0 gives the strongest tailwind.
