Engineering Electromagnetics Solutions Manual

136 Pages·03.283 MB·English

by  Ferdinand Hayt

CHAPTER 1
1.1. Given the vectors M= 10a +4a 8a and N = 8a +7a 2a , find:
x y z x y z
− − −
a) a unit vector in the direction of M+2N.
−
M+2N = 10a 4a +8a +16a +14a 4a = (26,10,4)
x y z x y z
− − −
Thus
(26,10,4)
a = = (0.92,0.36,0.14)
(26,10,4)
| |
b) the magnitude of 5a +N 3M:
x
−
(5,0,0)+(8,7, 2) ( 30,12, 24)= (43, 5,22), and (43, 5,22) = 48.6.
− − − − − | − |
c) M 2N (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 = (a +2a +3a ) (2a +3a 2a ) = ( a a +5a )
x y z x y z x y z
− − − − −
whose magnitude is A B = [( a a +5a ) ( a a +5a )]1/2 = √1+1+25 =
x y z x y z
| − | − − · − −
3√3 = 5.20. The unit vector is therefore
a = ( a a +5a )/5.20
AB x y z
− −
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
P = [1+(2 1)/2, 2+(3 2)/2, 3+( 2 3)/2)]= (1.5,2.5,0.5)
mp
− − − −
The unit vector is then
(1.5a +2.5a +0.5a )
x y z
a = = (1.5a +2.5a +0.5a )/2.96
mp x y z
(1.5)2+(2.5)2+(0.5)2
p
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 = 1B(2, 2,1), we use the fact that B A = 10, or
− − 3 − | − |
(6 2B)a (2 2B)a (4+ 1B)a = 10
| − 3 x− − 3 y − 3 z|
Expanding, obtain
36 8B+ 4B2+4 8B+ 4B2+16+ 8B+ 1B2 = 100
− 9 − 3 9 3 9
or B2 8B 44 = 0. Thus B = 8 √64 176 = 11.75 (taking positive option) and so
± −
− − 2
2 2 1
B = (11.75)a (11.75)a + (11.75)a = 7.83a 7.83a +3.92a
x y z x y z
3 − 3 3 −
1
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 ( √3,1,0),
−
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 = a . Its x and
φ
−
y components are t = a a = sinφ, and t = a a = cosφ. At the point ( √3,1),
x φ x y φ y
− · − · − −
φ = 150 , and so t = sin150 a cos150 a = 0.5(a +√3a ).
◦ ◦ x ◦ y x y
−
1.5. A vector field is specified as G = 24xya +12(x2 +2)a +18z2a . Given two points, P(1,2, 1)
x y z
−
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
− −
( 48,72,162)
a = − = ( 0.26,0.39,0.88)
G
( 48,72,162) −
| − |
c) a unit vector directed from Q toward P:
P Q (3, 1,4)
a = − = − = (0.59,0.20, 0.78)
QP
P Q √26 −
| − |
d) the equation of the surface on which G = 60: We write 60 = (24xy,12(x2 +2),18z2) , or
| | | |
10 = (4xy,2x2+4,3z2) , so the equation is
| |
100= 16x2y2+4x4+16x2+16+9z4
1.6. Find the acute angle between the two vectors A = 2a +a +3a and B = a 3a +2a by using
x y z x y z
−
the definition of:
a) the dot product: First, A B = 2 3+6 = 5 = ABcosθ, where A = √22+12+32 = √14,
· −
and where B = √12+32+22 = √14. Therefore cosθ = 5/14, so that θ = 69.1 .
◦
b) the cross product: Begin with
a a a
x y z
A B = 2 1 3 = 11a a 7a
x y z
× Ø Ø − −
Ø 1 3 2 Ø
Ø − Ø
Ø Ø
and then A B = √112+12+72ØØ= √171. So nØØow, with A B = ABsinθ = √171,
| × | | × |
find θ = sin 1 √171/14 = 69.1
− ◦
° ¢
1.7. Given the vector field E = 4zy2cos2xa +2zysin2xa +y2sin2xa for the region x , y , and z
x y z
| | | | | |
less than 2, find:
a) the surfaces on which E = 0. With E = 2zysin2x = 0, the surfaces are 1) the plane z = 0,
y y
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 E = E : This occurs when 2zysin2x = y2sin2x, or on the plane 2z = y,
y z
with x < 2, y < 2, z < 1.
| | | | | |
c) the region in which E = 0: We would have E = E = E = 0, or zy2cos2x = zysin2x =
x y z
y2sin2x = 0. This condition is met on the plane y = 0, with x < 2, z < 2.
| | | |
2
1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between
two vectors by finding the angle between A = 3a 2a +4a and B = 2a +a 2a . Does this
x y z x y z
− −
ambiguity exist when the dot product is used?
We use the relation A B = A B sinθn. With the given vectors we find
× | || |
2a +a
A B = 14a +7a = 7√5 y z = √9+4+16√4+1+4 sinθn
y z
× √5
∑ ∏
n
±
| {z }
where n is identified as shown; we see that n can be positive or negative, as sinθ can be
positive or negative. This apparent sign ambiguity is not the real problem, however, as we
really want the magnitude of the angle anyway. Choosing the positive sign, we are left with
sinθ = 7√5/(√29√9) = 0.969. Two values of θ (75.7 and 104.3 ) satisfy this equation, and
◦ ◦
hence the real ambiguity.
In using the dot product, we find A B = 6 2 8 = 4 = A B cosθ = 3√29cosθ, or
· − − − | || |
cosθ = 4/(3√29) = 0.248 θ = 75.7 . Again, the minus sign is not important, as we
◦
− − ⇒ −
care only about the angle magnitude. The main point is that only one θ value results when
using the dot product, so no ambiguity.
1.9. A field is given as
25
G = (xa +ya )
(x2+y2) x y
Find:
a) a unitvectorin thedirectionof G at P(3,4, 2): Have G = 25/(9+16) (3,4,0) = 3a +4a ,
p x y
− ×
and G = 5. Thus a = (0.6,0.8,0).
p G
| |
b) the angle between G and a at P: The angle is found through a a = cosθ. So cosθ =
x G x
·
(0.6,0.8,0) (1,0,0) = 0.6. Thus θ = 53 .
◦
·
c) the value of the following double integral on the plane y = 7:
4 2
G a dzdx
y
·
Z0 Z0
4 2 25 4 2 25 4 350
(xa +ya ) a dzdx= 7dzdx= dx
x2+y2 x y · y x2+49 × x2+49
Z0 Z0 Z0 Z0 Z0
1 4
= 350 tan 1 0 = 26
−
× 7 7 −
∑ µ ∂ ∏
1.10. Byexpressingdiagonalsasvectorsandusingthedefinitionofthedotproduct, findthesmallerangle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A = b(a + a + a ) and B =
x y z
b(a a +a ). Now use A B = A B cosθ, or b2(1 1+1) = (√3b)(√3b)cosθ cosθ =
x y z
− · | || | − ⇒
1/3 θ = 70.53 . This result (in magnitude) is the same for any two diagonal vectors.
◦
⇒
3
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 R : R = ( 0.2,0.1,0.3) (0.1, 0.2, 0.1) = ( 0.3,0.3,0.4).
MN MN
− − − − −
b) the dot product R R : R = (0.4,0,0.1) (0.1, 0.2, 0.1) = (0.3,0.2,0.2). R
MN MP MP MN
· − − − ·
R = ( 0.3,0.3,0.4) (0.3,0.2,0.2) = 0.09+0.06+0.08 = 0.05.
MP
− · −
c) the scalar projection of R on R :
MN MP
(0.3,0.2,0.2) 0.05
R a = ( 0.3,0.3,0.4) = = 0.12
MN RMP
· − · √0.09+0.04+0.04 √0.17
d) the angle between R and R :
MN MP
R R 0.05
θM = cos−1 MN · MP = cos−1 = 78◦
µ|RMN||RMP|∂ µ√0.34√0.17∂
1.12. Write an expression in rectangular components for the vector that extends from (x ,y ,z ) to
1 1 1
(x ,y ,z ) and determine the magnitude of this vector.
2 2 2
The two points can be written as vectors from the origin:
A = x a +y a +z a and A = x a +y a +z a
1 1 x 1 y 1 z 2 2 x 2 y 2 z
The desired vector will now be the difference:
A = A A = (x x )a +(y y )a +(z z )a
12 2 1 2 1 x 2 1 y 2 1 z
− − − −
whose magnitude is
A = A A = (x x )2+(y y )2+(z z )2 1/2
12 12 12 2 1 2 1 2 1
| | · − − −
p £ §
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 (10, 6,5) (0.1,0.2,0.3)
F = · G = − · (0.1,0.2,0.3) = (0.93,1.86,2.79)
||G G 2 0.01+0.04+0.09
| |
b) Find the vector component of F that is perpendicular to G:
F = F F = (10, 6,5) (0.93,1.86,2.79)= (9.07, 7.86,2.21)
pG G
− || − − −
c) Find the vector component of G that is perpendicular to F:
G F 1.3
G = G G = G · F = (0.1,0.2,0.3) (10, 6,5) = (0.02,0.25,0.26)
pF − ||F − F 2 − 100+36+25 −
| |
4
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 = (A +B +C )a +(A +B +C )a +(A +B +C )a
x x x x y y y y z z z z
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 A = B = 0. Then C has to be zero in order for the
z z z
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 = (A +B +C +D )a +(A +B +C +D )a +(A +B +C +D )a
x x x x x y y y y y z z z z z
Now, for example, if A and B lie in the x-y plane, C and D need not, as long as C +D = 0.
z z
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 = (7,3, 2), r = ( 2,7, 3), and r =
1 2 3
− − −
(0,2,3). Find:
a) a unit vector perpendicular to both r and r :
1 2
r r (5,25,55)
1 2
a = × = = (0.08,0.41,0.91)
p12
r r 60.6
1 2
| × |
b) a unit vector perpendicular to the vectors r r and r r : r r = (9, 4,1) and r r =
1 2 2 3 1 2 2 3
− − − − −
( 2,5, 6). So r r r r = (19,52,32). Then
1 2 2 3
− − − × −
(19,52,32) (19,52,32)
a = = = (0.30,0.81,0.50)
p
(19,52,32) 63.95
| |
c) the area of the triangle defined by r and r :
1 2
1
Area= r r = 30.3
1 2
2| × |
d) the area of the triangle defined by the heads of r , r , and r :
1 2 3
1 1
Area= (r r ) (r r ) = ( 9,4, 1) ( 2,5, 6) = 32.0
2 1 2 3
2| − × − | 2| − − × − − |
5
1.16. If A represents a vector one unit in length directed due east, B represents a vector three units in
length directed due north, and A+B = 2C D and 2A B = C+2D, determine the length and
− −
direction of C. (difficulty 1)
Take north as the positive y direction, and then east as the positive x direction. Then we may write
A+B = a +3a = 2C D
x y
−
and
2A B = 2a 3a = C+2D
x y
− −
Multiplying the first equation by 2, and then adding the result to the second equation eliminates
D, and we get
4 3
4a +3a = 5C C = a + a
x y x y
⇒ 5 5
The length of C is C = (4/5)2+(3/5)2 1/2 = 1
| |
£ §
C lies in the x-y plane at angle from due north (the y axis) given by α = tan 1(4/3) = 53.1 (or
− ◦
36.9 from the x axis). For those having nautical leanings, this is very close to the compass point
◦
NE3E (not required).
4
1.17. Point A( 4,2,5) and the two vectors, R = (20,18, 10) and R = ( 10,8,15), define a
AM AN
− − −
triangle.
a) Find a unit vector perpendicular to the triangle: Use
R R (350, 200,340)
AM AN
a = × = − = (0.664, 0.379,0.645)
p
R R 527.35 −
AM AN
| × |
The vector in the opposite direction to this one is also a valid answer.
b) Find a unit vector in the plane of the triangle and perpendicular to R :
AN
( 10,8,15)
a = − = ( 0.507,0.406,0.761)
AN
√389 −
Then
a = a a = (0.664, 0.379,0.645) ( 0.507,0.406,0.761)= ( 0.550, 0.832,0.077)
pAN p AN
× − × − − −
The vector in the opposite direction to this one is also a valid answer.
c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit
vector in the required direction is (1/2)(a +a ), where
AM AN
(20,18, 10)
a = − = (0.697,0.627, 0.348)
AM
(20,18, 10) −
| − |
Now
1 1
(a +a ) = [(0.697,0.627, 0.348)+( 0.507,0.406,0.761)]= (0.095,0.516,0.207)
AM AN
2 2 − −
Finally,
(0.095,0.516,0.207)
a = = (0.168,0.915,0.367)
bis
(0.095,0.516,0.207)
| |
6
1.18. A certain vector field is given as G = (y+1)a +xa . a) Determine G at the point (3,-2,4):
x y
G(3, 2,4) = a +3a .
x y
− −
b) obtain a unit vector defining the direction of G at (3,-2,4).
G(3, 2,4) = [1+32]1/2 = √10. So the unit vector is
| − |
a +3a
x y
a (3, 2,4) = −
G
− √10
1.19. a) Express the field D = (x2+y2) 1(xa +ya ) in cylindricalcomponents and cylindricalvariables:
− x y
Have x = ρcosφ, y = ρsinφ, and x2+y2 = ρ2. Therefore
1
D = (cosφa +sinφa )
x y
ρ
Then
1 1 1
D = D a = [cosφ(a a )+sinφ(a a )] = cos2φ+sin2φ =
ρ ρ x ρ y ρ
· ρ · · ρ ρ
£ §
and
1 1
D = D a = [cosφ(a a )+sinφ(a a )] = [cosφ( sinφ)+sinφcosφ] = 0
φ φ x φ y φ
· ρ · · ρ −
Therefore
1
D = a
ρ
ρ
b) Evaluate D at the point where ρ = 2, φ = 0.2π, and z = 5, expressing the result in cylindrical
and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0.5a . To
ρ
express this in cartesian, we use
D = 0.5(a a )a +0.5(a a )a = 0.5cos36 a +0.5sin36 a = 0.41a +0.29a
ρ x x ρ y y ◦ x ◦ y x y
· ·
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.
| | ·
Withthe vectorsdrawn as described above, we findthatC = (A+B) andso C 2 = C2 = C C =
− | | ·
(A+B) (A+B) So far so good. Now if we expand the product, obtain
·
(A+B) (A+B) = A2+B2+2A B
· ·
where A B = ABcos(180 α) = ABcosα where α is the interior angle at the junction of A
◦
· − −
and B. Using this, we have C2 = A2+B2 2ABcosα, which is the law of cosines.
−
7
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 R = ( 4, 6,9) and R = R a = 4cos(33.7) 6sin(33.7) = 6.66. Then
CD ρ CD ρ
− − · − − −
R = R a = 4sin(33.7) 6cos(33.7) = 2.77. So R = 6.66a 2.77a +9a
φ CD φ CD ρ φ z
· − − − −
b) a unit vector at D directed toward C:
R = (4,6, 9) and R = R a = 4cos( 104.0)+6sin( 104.0) = 6.79. Then R =
CD ρ DC ρ φ
− · − − −
R a = 4[ sin( 104.0)]+6cos( 104.0) = 2.43. So R = 6.79a +2.43a 9a
DC φ DC ρ φ z
· − − − − −
Thus a = 0.59a +0.21a 0.78a
DC ρ φ z
− −
c) a unit vector at D directed toward the origin: Start with r = ( 1, 4,2), and so the
D
− −
vector toward the origin will be r = (1,4, 2). Thus in cartesian the unit vector is a =
D
− −
(0.22,0.87, 0.44). Convert to cylindrical:
−
a = (0.22,0.87, 0.44) a = 0.22cos( 104.0)+0.87sin( 104.0) = 0.90, and
ρ ρ
− · − − −
a = (0.22,0.87, 0.44) a = 0.22[ sin( 104.0)] + 0.87cos( 104.0) = 0, so that finally,
φ φ
− · − − −
a = 0.90a 0.44a .
ρ z
− −
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) = Ωrsinθa (r < a)
φ
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[ ya +xa ], where (x2+y2+z2)1/2 < a.
x y
−
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:
4.5 130◦ 5
Vol = ρdρdφdz = 6.28
Z3 Z100◦ Z3
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:
130◦ 5 4.5 130◦
Area= 2 ρdρdφ + 3dφdz
Z100◦ Z3 Z3 Z100◦
4.5 130◦ 4.5 5
+ 5dφdz + 2 dρdz = 20.7
Z3 Z100◦ Z3 Z3
8
1.23c) Find the total length of the twelve edges of the surfaces:
30 30
◦ ◦
Length= 4 1.5 + 4 2 + 2 2π 3 + 2π 5 = 22.4
× × × 360 × × 360 × ×
∑ ◦ ◦ ∏
d) Find the length of the longest straight line that lies entirely within the volume: This will be
between the points A(ρ = 3, φ = 100 , z = 3) and B(ρ = 5, φ = 130 , z = 4.5). Performing
◦ ◦
point transformations to cartesian coordinates, these become A(x = 0.52, y = 2.95, z = 3)
−
and B(x = 3.21, y = 3.83, z = 4.5). Taking A and B as vectors directed from the origin, the
−
requested length is
Length= B A = ( 2.69,0.88,1.5) = 3.21
| − | | − |
1.24. Two unit vectors, a and a lie in the xy plane and pass through the origin. They make angles φ
1 2 1
and φ with the x axis respectively.
2
a) Express each vector in rectangular components; Have a = A a + A a , so that A =
1 x1 x y1 y x1
a a = cosφ . Then, A = a a = cos(90 φ ) = sinφ . Therefore,
1 x 1 y1 1 y 1 1
· · −
a = cosφ a +sinφ a and similarly, a = cosφ a +sinφ a
1 1 x 1 y 2 2 x 2 y
b) take the dot product and verify the trigonometric identity, cos(φ φ ) = cosφ cosφ +
1 2 1 2
−
sinφ sinφ : From the definition of the dot product,
1 2
a a = (1)(1)cos(φ φ )
1 2 1 2
· −
= (cosφ a +sinφ a ) (cosφ a +sinφ a ) = cosφ cosφ +sinφ sinφ
1 x 1 y 2 x 2 y 1 2 1 2
·
c) take the cross product and verify the trigonometric identity sin(φ φ ) = sinφ cosφ
2 1 2 1
− −
cosφ sinφ : From the definition of the cross product, and since a and a both lie in the x-y
2 1 1 2
plane,
a a a
x y z
a a = (1)(1)sin(φ φ )a = cosφ sinφ 0
1 2 1 2 z 1 1
× − Ø Ø
Øcosφ sinφ 0 Ø
2 2
Ø Ø
Ø Ø
= [sinφ2cosφ1 cosφ2sinØφ1] az Ø
− Ø Ø
thus verified.
1.25. Given point P(r = 0.8, θ = 30 , φ = 45 ), and
◦ ◦
1 sinφ
E = cosφa + a
r2 r sinθ φ
µ ∂
a) Find E at P: E = 1.10a +2.21a .
ρ φ
b) Find E at P: E = √1.102+2.212 = 2.47.
| | | |
c) Find a unit vector in the direction of E at P:
E
a = = 0.45a +0.89a
E r φ
E
| |
9
1.26. Express the uniform vector field, F = 5a in
x
a) cylindrical components: F = 5a a = 5cosφ, and F = 5a a = 5sinφ. Combining, we
ρ x ρ φ x φ
· · −
obtain F(ρ,φ) = 5(cosφa sinφa ).
ρ φ
−
b) sphericalcomponents: F = 5a a = 5sinθcosφ; F = 5a a = 5cosθcosφ; F = 5a a =
r x r θ x θ φ x φ
· · ·
5sinφ. Combining, we obtain F(r,θ,φ) = 5[sinθcosφa +cosθcosφa sinφa ].
r θ φ
− −
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
60◦ 50◦ 4
Vol = r2sinθdrdθdφ = 2.91
Z20◦ Z30◦ Z2
where degrees have been converted to radians.
b) Find the total area of the enclosing surface:
60◦ 50◦ 4 60◦
Area= (42+22)sinθdθdφ+ r(sin30 +sin50 )drdφ
◦ ◦
Z20◦ Z30◦ Z2 Z20◦
50◦ 4
+2 rdrdθ = 12.61
Z30◦ Z2
c) Find the total length of the twelve edges of the surface:
4 50◦ 60◦
Length= 4 dr + 2 (4+2)dθ+ (4sin50 +4sin30 +2sin50 +2sin30 )dφ
◦ ◦ ◦ ◦
Z2 Z30◦ Z20◦
= 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 = 2sin50 cos20 ,y = 2sin50 sin20 ,z = 2cos50 )
◦ ◦ ◦ ◦ ◦
to
B(x = 4sin30 cos60 ,y = 4sin30 sin60 ,z = 4cos30 )
◦ ◦ ◦ ◦ ◦
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 a = B a : For this to be true, both A and B must be oriented at the same angle, θ, from
x x
· ·
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 a = B a : This is a more restrictive condition because the cross product gives a vector.
x x
× ×
For both cross products to lie in the same direction, A, B, and a must be coplanar. But if A
x
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.
10