Proprietary of Prof. Lee, Yeon Ho, 2014
Problems for Chapter 1
1-1
Verify graphically the associative law of vector addition, as given in Eq. (1-5a), for three vectors shown in
Fig. 1-29
ANS
1-2
Two vectors A and B form a parallelogram in three-dimensional space. Verify that the projection of the
parallelogram onto the xy-plane is also a parallelogram.
ANS
Let the four sides of the parallelogram correspond to A, B, A , and B , where A A and B B .
(projections of A) (projections of A ), and (projections of B) (projections of B ) on the xy-plane.
The projection of the parallelogram is also a parallelogram
1-3
With reference to Example 1-2 in which the distributive law of dot product was verified for three vectors
lying in a common plane, verify the distributive law for three arbitrary vectors.
ANS
According to two-step projection,
A
B  C  A B  C
Three vectors A , B, and C are on the same plane. Thus, using the result of Example 1-2, we can write
A
B  C  A B  A C
Using two-step projection
A BA C  A BA C
Thus,
A
1-4
B  C  A B  A C
With reference to Example 1-3 in which the distributive law of cross product was verified for three vectors(A
is normal to the plane formed by B and C), verify the distributive law for three arbitrary vectors.
Problems chapter 1 |1
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ANS
S, a plane normal to A
B , projection of B onto S
C , projection of C onto S
We note
AB  AB,
(1)
(2)
AC  AC
Similarly
A  B  C  A  B  C
(3)
B and C are normal to A. Using the result of Example of 1-3
A   B  C    A  B   A  C 
(4)
Inserting (1), (2) and (4) into (3)
A   B  C    A  B   A  C 
1-5
Two vectors A and B form a parallelogram.
(a) Find expressions for two diagonals in terms of A and B
(b) Under what condition are the diagonals perpendicular to each other?
(c) Two diagonal vectors are displaced such that they meet at the tail. Find the area of the new parallelogram.
ANS
(a)
A  B and A  B
(b)
Condition for perpendicular diagonals
:  A  B  A  B  0
From the distributive law
: A2  B 2  B A  A B  0
Using B A  A B
: A2  B 2
The parallelogram should be an equilateral parallelogram, called a rhombus.
Problems chapter 1 |2
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(c)
Area of the parallelogram formed by two diagonal vectors
area   A  B    A  B    A  A    A  B    B  A    B  B 
Using A  A  0 , B  B  0 and  A  B     B  A 
area  2 A  B
The area is twice as large as the area of the parallelogram formed by A and B .
1-6
Three vectors A, B, and A  B form a triangle. Show the followings:
(a) Parallelograms formed by any two vectors have the same area
(b) Law of sines,
A
B
C


sin A sin B
sin C
(Sides A, B and C subtend angles A , B , and C , respectively)
ANS
(a)
Areas of the three parallelograms
(b)
AB
(1)
A   A  B   A   B   A  B
(2)
B   A  B  B  A  A  B
(3)
Let C  A  B . From (1), (2) and (3)
AB  AC  BC
 AB sin C  AC sin B  BC sin A
(4)
From the first and third terms in (4)
:
A
C

sin A sin C
(5)
From the second and third terms in (4)
:
A
B

sin  A sin B
(6)
Combining (5) and (6)
A
B
C


sin A sin B
sin C
1-7
Two vectors A  2a x  ay  3a z and B   a x  2ay  a z are at the same point in Cartesian coordinates.
Find (a) a A (b) B  A (c) A B (d) A  B (e) AB (f) vector component of B in the direction of A (g)
vector component of B in the direction perpendicular to A
ANS
(a)
A A 
aA 
(b)
 2  1  3  14
2
2
2
A
2
1
3

ax 
ay 
az
A
14
14
14
B  A   1  2  a x   2  1 ay  1  3  az
 3 a x  ay  2 a z
Problems chapter 1 |3
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(c)
(d)
(e)
A B   2   1  1  2    3  1  3
ax
ay
AB  2
1
3  5 a x  5 ay  5 a z
1
2
1
B 
az
 1   2  1 
2
2
2
6
3

 A B
1 
o
AB  cos 1 
  cos 
  70.9
 AB 
 14 6 
(f)
B aA  B
A 1
3
 A B
A A
14
The answer is
(g)
3
14
3
2ax  ay  3az 
14
a A or
From B  B  B  , where B  B a A as in part (f)
6 
3 
9 



B   B  B   1 
 ax   2 
 ay  1 
 az
14 
14 
14 



20
25
5

ax 
ay 
az
14
14
14
1-8
Given a vector A  a x  2 ay  2 a z at a point with position vector r  2 ax  3 ay  3 az . Find the angles
that (a) A (b) r makes with three Cartesian coordinate axes.
ANS
(a)
(b)
A  1  22  22  3
A ax 1
cos x 

A
3
A ay
2
cos y 

A
3
A az
2
cos z 

A
3
r  22  32 
y  48.2o
 z  48.2o
2
r ax
2

r
4
r ay
3
cos y 

r
4
1-9

 3  4
cos x 
cos z 
 x  70.5o
r az
3

r
4
 x  60o

y  41.4o

z  64.3o
A straight line passes through two points p1 : 1,1, 0  and p2 :  3, 2, 2  in Cartesian coordinates. Find
(a) unit vector an along the direction from p1 to p 2
(b) position vector of a point on the line, which is at a distance t from p1 .
Problems chapter 1 |4
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ANS
(a)
Position vector of p1
(b)
: r1  a x  ay
Position vector of p 2
: r2  3 a x  2 ay  2 a z
Vector from p1 to p 2
: r2  r1  2 a x  ay  2 az
The unit vector
: an 
Position vector of a point on the line
: r  r1  t an  1 
2
1
2
a x  ay  a z
3
3
3


2 
1 

2 
t  ax  1  t  ay   t  az
3 
3 

3 
1-10 Two straight lines intersect at right angles at point p1 in Cartesian coordinates. The first line passes through
point p 2 :  2,1,1 , while the second one passes through points p3 : 1, 1, 3  and p 4 :  1,1, 2  . Find the
distance between the two points
(a) p1 and p 2
(b) p1 and p 3
ANS
(a)
Distance vector from p 3 to p 2
: R 2  3  r2  r3  a x  2 ay  2 a z
Distance vector from p 3 to p 4
: R 4  3  r4  r3  2 a x  2 ay  a z
Unit vector of R 43
: a4  3  
2
2
1
a x  ay  a z
3
3
3
The distance between p 2 and p1
2
1 
 2
p2 p1  R 2  3  a4 3   a x  2 ay  2 az     a x  ay  az 
3
3 
 3

(b)
1
 2 a x  5 ay  6a z  
3
65
3
The distance between p 3 and p1
2
1
 2

p 3 p1  R 2  3 a 4  3   a x  2 ay  2 a z    a x  ay  a z 
3
3 
 3

1
4
 2  4  2  
3
3
1-11 For a plane defined by three points p1 :  2, 0, 0  , p2 :  0,1, 0  , and p3 :  0, 0,1 in Cartesian coordinates,
find
(a) unit vector normal to the plane, directed toward the origin
(b) perpendicular distance from the origin to the plane
ANS
(a)
Position vectors of three points
Distance vectors
: r1  2ax , r2  ay , and r3  az
: R 2 1  r2  r1  ay  2a x
: R 31  r3  r1  az  2ax
Problems chapter 1 |5
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The cross product of R 21 and R 31 is perpendicular to the plane
ax
ay
az
R 2 1  R 3 1  2
1
0  a x  2 ay  2 a z
2
0
1
1
2
2
a x  ay  a z
3
3
3
1
2
2
A unit normal directed toward the origin : an   ax  ay  az
3
3
3
: an 
A unit normal vector
(b)
r1 an  r2 an  r3 an 
2
3
1-12 A plane intersects three Cartesian coordinate axes at x  a , y  b and z  c , respectively. Find
(a) unit normal to the plane
(b) expression for the plane
ANS
(a)
Position vectors of three intersection points
: r1  a ax , r2  b ay , and r3  c az
:  r2  r1    r3  r1   b ay  a ax   c az  a ax 
A normal to the plane
 bc a x  ac ay  ab a z
(b)
bc a x  ac ay  ab az
A unit surface normal
: an 
Position vector of a point on the plane
: r  x a x  y ay  z a z
bc   ac   ab 
2
2
2
(1)
Distance vector from a point a , 0, 0  to an arbitrary point on the plane
R  r  r1
  x ax  y ay  z az   a ax    x  a  ax  y ay  z az
(2)
Two vectors in (1) and (2) are always perpendicular to each other
R an    x  a  a x  y ay  z az 
bc a x  ac ay  ab a z
bc   ac   ab 
2
2
2
0
or
 x  a  bc  yac  zab  0
or
x y z
  1
a b c
1-13 A plane S passes through the point p1 :  3,1, 2  and is perpendicular to the vector k  a x  2ay  4a z in
Cartesian coordinates. Find an expression for the plane.
ANS
Position vector of p1
Problems chapter 1 |6
: r1  3a x  ay  2a z
Proprietary of Prof. Lee, Yeon Ho, 2014
Position vector of a point on the plane
: r  x a x  y ay  z a z .
A distance vector on S
: R  r  r1   x  3  a x  y  1 ay   z  2  az
R is always perpendicular to k
R k   x  3 ax  y  1 ay   z  2 az   ax  2ay  4az   0
Rewriting it
 x  3  2 y  1  4  z  2   0 , or x  2y  4z  13
1-14 Two vectors A and B define a parallelogram P in three-dimensional space. Projections of A and B onto the
xy-plane define another parallelogram P. Show that the area of P is given by az  A  B  .
ANS
Let
A  Ax a x  Ay ay  Az a z
B  B x a x  B y ay  B z a z
The cross product, corresponding to the area of P
A  B   Ay B z  Az By  ax   Az B x  Ax B z  ay   Ax By  Ay B x  az
(1)
Projections of A and B onto the z  0 plane
A  Ax ax  Ay ay
B  B x a x  B y ay
The area of the parallelogram P '
A  B  Ax By  Ay B x
(2)
(2) is equivalent to the z-component of A  B in (1).
The area of P ' is therefore az  A  B 
1-15 A parallelogram P is formed by two vectors A  3 a x  ay  2 az and B  a x  2 ay  a z . Projections of P
onto the x  0 , y  0 and z  0 planes result in three parallelograms P1 , P2 , and P3 on their respective
planes. Find
(a) unit vector an normal to P
(b) areas of P, P1 , P2 , and P3
(c) relation between an and the three parallelograms P1 , P2 , and P3 .
ANS
(a)
A vector normal to P : A  B   3 ax  ay  2 az    ax  2 ay  az   5 ax  ay  7 az
(b)
5
1
: an  
az
(2)
Area of P
: A  B  5 ax  ay  7 az  5 3
(3)
Problems chapter 1 |7
5 3
ay 
7
Unit normal to P
5 3
ax 
(1)
5 3
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Projections of A and B onto the x  0 plane
: A1  B1    ay  2 az    2 ay  az   5 ax  5
Area of P1
Projections of A and B onto the y  0 plane
Projections of A and B onto the z  0 plane
(5)
: A 3  3 a x  ay , B 3  a x  2 a y
: A 3  B 3   3 a x  ay    a x  2 ay   7 a z  7
Area of P3
(4)
: A 2  3 ax  2 az , B2  ax  az
: A 2  B2   3 ax  2 az    ax  az   ay  1
Area of P2
(c)
: A1   ay  2 a z , B1  2 ay  a z
(6)
The absolute values of the scalar components of an correspond to the relative areas of P1 , P2 , and P3
1-16 Given two vector fields A  3 a  (4 sin ) a  and B  2 a     z  2  az in cylindrical coordinates, find
at point p1 :  2,30o ,1
(a) A  B
(b) A B
(c) A  B
ANS
At point p1 , A  3 a  2 a  and B  4 a  5 a z
(a)
A  B  7 a  2 a   5 a z
(b)
A B   3 a  2 a   4 a  5 az   12
a
a
az
(c)
AB  3
2
0  10 a  15 a   8 a z
4
0
5
1-17 A triangle is defined by three points p1 : (2,0,1) , p2 : (1,1,3) , and p3 : (4,2, 1) in Cartesian coordinates.
Find the area of the triangle.
ANS
Distance vector from p1 to p 2
R 2 1  ( 1  2) a x  (1  0) ay  (3  1) a z  3 a x  ay  2 a z
Distance vector from p1 to p 3
R 3 1  (4  2) a x  (2  0) ay  ( 1  1) a z  2 a x  2 ay  2 a z
For the area of the triangle, compute
ax
ay
az
R 2 1  R 3 1  3
1
2  6 a x  2 ay  8 a z
2
2
2
The area of the triangle is
1
1
R 21  R 31 
36  4  64  5.1
2
2
Problems chapter 1 |8
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1-18 A vector field is given as H  r   (16 / 2 ) a  in cylindrical coordinates. It specifies two vectors H  r1   A
and H  r2   B at two points p1 : (2,30o ,5) and p2 : (2,90o ,5) , respectively. Find
(a) expressions for A and B in cylindrical coordinates
(c) A  aB at point p1 , where a B is the unit vector of B
ANS
A  4 a and B  4 a at their respective points
(a)
(b)
: a B  a  ay
At point p 2
a B is displaced from point p 2 to p1 , and is expanded by the base vectors at p1 such that
aB  ay  sin 30o a  cos 30o a  
1
3
a 
a
2
2
Thus
a
1
A  aB 
4
2
1
a
az
0
0  2 3 az
3
0
1-19 Given two points p1 : (2,90o ,5) and p2 : (2,60o ,5) in cylindrical coordinates, find the expression, in
cylindrical coordinates, for
(a) position vectors of p1 and p 2 (or r1 and r2 )
(b) distance vector R 12 = r1  r2 expanded by the base vectors at p1
ANS
r1  2 a  5 a z
(a)
(1)
r2  2 a  5 a z
(2)
(b)
Projections of r2 onto ax , a y and az are
r2 a x  2 cos(60o )  1
o
(3a)
r2 ay  2 sin(60 )  3
(3b)
r2 az  5
(3c)
Thus, using (3), r2 in Cartesian coordinates is
r2  ax  3 ay  5 az
(4)
At point p1
a  ay
(5a)
a   ax
(5b)
With the help of (5), projections of r2 in (4) onto a  , a  and az at p1 are
r2 a  r2
a   3
(6a)
r2 a   r2
  a x   1
(6b)
y
Problems chapter 1 |9
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r2 az  5
(6c)
Using (6), r2 is expanded by the base vectors at p1 as
r2  3 a  a  5 az
(7)
Using (1) and (7), the distance vector is expanded by the base vectors at p1 as
R 12 = r1  r2   2 a  5 az  
 3 a  a  5 a   2  3  a  a


z


1-20 A half cylinder is defined by the surfaces   4 ,   30o ,   210o , z  0 and z  5 as shown in Fig. 133. Find the differential area vector ds at the following points: (a) p1 (b) p 2 with   2 (c) p 3 (d) p 4 .
Fig. 1-33 half cylinder(Problem 1-20)
ANS
ds  4d dz a
(a)
(b)
ds  2d d  az
(c)
ds  d dz a 
(d)
ds  d dz a 
1-21 With reference to the object shown in Fig. 1-34, which is defined by the spherical surface of R  6 , conical
surface of half-angle   30o , half planes of   45o and   225o in spherical coordinates, find the
differential area vectors at the following points:
(a) p1 with   20o
(b) p 2 with R  4
(c) p 3 with R  4
(d) p 4 with R  5 .
P r o b l e m s c h a p t e r 1 | 10
Proprietary of Prof. Lee, Yeon Ho, 2014
Fig. 1-34 A volume in spherical coordinates (Problem 1-21)
ANS
ds  36 sin 20o d d  a R  12.3 d d  a R
(a)
(b)
ds  4 sin 30o dRd  a   2 dRd  a 
(c)
ds  4 dRd  a 
(d)
ds  5 dRd  a 
1-22 A vector field is defined as E  2R aR  R 2 cos  a   100 sin  cos  a  , in spherical coordinates, inside and
outside the object shown in Fig. 1-34. At three points p1 : (6,20o ,60o ) , p2 : (4,30o ,50o ) and
p 3 : (4,20o ,45o ) on the object, find
(a) E
(b) vector components of E parallel and perpendicular to the surface.
ANS
(a)
At p1 :  6,20o , 60o 
E1  12 aR  36 cos 60o a  100 sin 20o cos 60o a
(1)
 12 aR  18 a  17.1 a
At p2 :  4,30o ,50o 
E 2  8 aR  16 cos 50o a  100 sin 30o cos 50o a
(2)
 8 aR  10.3 a  32.1 a
At p3 :  4,20o , 45o 
E  8 aR  16 cos 45o a  100 sin 20o cos 45o a
(3)
 8 aR  11.3 a  24.2 a
(b)
Unit surface normal at p1 is a R
 E1,  12 a R
Vector component perpendicular to the sphere
: E1 aR  12
Vector component parallel to the sphere
: E1,  E1  E1,  18 a  17.1 a
P r o b l e m s c h a p t e r 1 | 11
Proprietary of Prof. Lee, Yeon Ho, 2014
Unit surface normal at p 2 is a
The perpendicular component
: E 2,  10.3 a 
The parallel component
: E 2,  8 a R  32.1 a 
Unit surface normal at p 3 is a 
The perpendicular component
: E 3,  24.2 a 
The parallel component
: E 3,  8 a R  11.3 a 
1-23 Given a point p1;(2, 3,3) in Cartesian coordinates, find an expression for p1 in (a) cylindrical coordinates
(b) spherical coordinates
ANS
(a)
Using Eq. (1-98)

x 2  y 2  22 
  tan 1
 3  7
2
 3
y
 tan 1 
 40.9o
 2 
x


z 3
(b)
Using Eq. (1-103)
R
x 2  y 2  z 2  22 
 3  3  4
2
2

  tan 1  x 2  y 2 / z   tan 1  22 



  tan 1
 3  / 3  41.4
2
o
 3
y
 tan 1 
 40.9o
 2 
x


1-24 Find expressions for the unit vector ax in terms of spherical coordinates at the following points:
(a) (x , y, z )  (1, 2,3)
(b) (, , z )  (2,30o , 2)
(c) (R , , )  (2,45o ,60o )
ANS
(a)
Spherical coordinates of the point are
R  12   2   32  14
2
5
  tan 1  x 2  y 2 / z   tan 1


3
 cos   3 / 14 , sin   5 / 14
y
 tan 1  2 
x
 cos   1 / 5 , sin   2 / 5
  tan 1
From Eq. (1-101)
P r o b l e m s c h a p t e r 1 | 12
Proprietary of Prof. Lee, Yeon Ho, 2014
 AR   sin 1 cos 1
  
 A    cos 1 cos 1
 A    sin 1
 

sin 1 sin 1
cos 1 sin 1
cos 1


 5 / 14 1 / 5

  3 / 14 1 / 5


 2 / 5



cos 1  1 
 sin 1  0 
 
0  0 
 sin  sin 

1
1
cos 1 sin 1

cos 1
cos 1 
 1 
 sin 1  0 
 
 0 
0  

Thus
ax 
(b)
1
14
3
aR 
70
2
a 
5
a
Cartesian coordinates of the point
 3,1, 2 
Spherical coordinates of the point
cos   1 / 3 , sin   2 / 6
cos  
3 / 2 , sin   1/ 2
From Eq. (1-101)
 AR   sin 1 cos 1
  
 A    cos 1 cos 1
 A    sin 1
 


sin 1 sin 1
cos 1 sin 1
cos 1


cos 1  1 
 sin 1  0 
 
0  0 
 sin  sin 
 cos  sin 
 2/ 6
3 /2

  1/ 3
3 /2


 1 / 2 

1
1
1
1
cos 1
cos 1 
 1 
 sin 1  0 
 
0  0 

Thus
ax 
(c)
1
2
aR 
1
1
a  a
2
2
At the given point
  45o ,   60o
From Eq. (1-101)
 AR   sin 1 cos 1
  
 A    cos 1 cos 1
 A    sin 1
 


sin 1 sin 1
cos 1 sin 1
cos 1


 1 / 2 1 / 2  sin 1 sin 1

  1 / 2 1 / 2  cos 1 sin 1


cos 1
  3 / 2


Thus
ax 
1
2 2
aR 
1
2 2
a 
3
a
2
P r o b l e m s c h a p t e r 1 | 13
cos 1  1 
 sin 1  0 
 
0  0 
cos 1 
 1 
 sin 1  0 
 
 0 
0  

Proprietary of Prof. Lee, Yeon Ho, 2014
1-25 Given that A  a x  a y  0.5 a z at point (2,2,1) in Cartesian coordinates, transform A into (a) cylindrical
and (b) spherical coordinates
ANS
(a)
From Eq. (1-98)
x 2  y2 

22  22  2 2
  tan 1  y / x   tan 1
2
 45o
2
z 1
From Eq. (1-96)
 
 cos  45o  sin  45o  0   1 
 A  
 1 
 A     sin 45o
cos  45o  0   



  
 1 
 Az  
0
0
1  2 


 
Thus
A  2 a 
(b)
1
az
2
From Eq. (1-103) and Pythagorean theorem
R
x 2  y 2  z 2  22  22  1  3
2 2
1
and sin  
3
3
1
1
and sin  
cos  
2
2
cos  
From Eq. (1-101)
 AR   sin  cos  sin  sin  cos    Ax 
  
 
 A    cos  cos  cos  sin   sin    Ay 
 A    sin 
cos 
0   Az 
 
2 2 1

2
 3
 1 1

 3 2

1
 
2

2 2 1
3
2
1 1
3 2
1
2

   
 1   3 
2 2  1   2 
    

3  1  0 
 2 0
0    

1
3
Thus
A
1-26
3
aR
2
Given two points p1 : 1,1, 3  and p2 :  1, 2,1 in Cartesian coordinates, find a unit vector
(a) at p1 directed toward p 2 in Cartesian coordinates
(b) at p 2 directed toward p1 in Cartesian coordinates
(c) at p1 directed toward p 2 in cylindrical coordinates
P r o b l e m s c h a p t e r 1 | 14
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(d) at p 2 directed toward p1 in cylindrical coordinates
ANS
Position vectors of p1 and p 2 in Cartesian coordinates
r1  a x  ay  3 a z
r2   a x  2 ay  a z
Cylindrical coordinates of p1 are, from Eq. (1-98)
1  12  12  2
cos 1 
1
2
(1a)
sin 1 
,
1
(1b)
2
z1  3
(1c)
Cylindrical coordinates of p 2 are, from Eq. (1-98)
 1  22 
2
2 
cos 2  
1
5
,
(2a)
5
sin 2 
2
(2b)
5
z2  1
(a)
(2c)
Distance vector from p1 to p 2 in Cartesian coordinates is
A  r2  r1   ax  2 ay  az    ax  ay  3 az   2 ax  ay  2 az
Its unit vector is
aA  
(b)
2
1
2
ax  ay  az
3
3
3
(3)
Distance vector from p 2 to p1 in Cartesian coordinates is
B  r1  r2   ax  ay  3 az    ax  2 ay  az   2 ax  ay  2 az
Is unit vector is
aB 
(c)
2
1
2
a x  ay  a z
3
3
3
(4)
Coordinate transformation of (3) into cylindrical coordinates is obtained from Eq. (1-96), with the help of (1),
as
 1
 2

 A 
 A   1  1
  3 
2

 Az 
 0


1
2
1
2
0

0
  2 

0  1 
 
  2 
1


Thus
aA 
1
3 2
 a  3 a  2 2 a 


z
P r o b l e m s c h a p t e r 1 | 15
(5)
Proprietary of Prof. Lee, Yeon Ho, 2014
(d)
Coordinate transformation of (4) into cylindrical coordinates is obtained from Eq. (1-96), with the help of (2),
as
 1
 5

B 
B   1  2
  3 
5

 B z 
 0


2

5
1
5
0

0
 2 

0   1
 
  2 
1


Thus
aB 
1
3 5
 4 a  3 a  2 5 a 


(6)
z
Although (3) and (4) clearly show that the two unit vectors are directed in the opposite directions, (5) and (6)
are too complicated to show the opposite directions.
1-27 In the presence of a vector field A   2z  10  a x  x 2 ay  y az , two points are defined as p1 : (1, 3,1) in
Cartesian coordinates and p2 : (4,60o ,3) in cylindrical coordinates. Find the vector component of A at p2
directed toward p1 in terms of (a) Cartesian and (b) cylindrical coordinates
ANS
(a)
Cartesian coordinates of p 2 are, from Eq. (1-97),
x   cos   4 cos 60o  2
y   sin   4 sin 60o  2 3
z 3
Thus, point p 2 in Cartesian coordinates is

p2 : 2, 2 3, 3

(1)
Using (1), vector A at p 2 is


A  2  3   10  ax   2  ay  2 3 az  16 a x  4 ay  2 3 az
2
(2)
Position vectors of p1 and p 2 in Cartesian coordinates are
r1   ax  3 ay  az
r2  2ax  2 3 ay  3az
Distance vector from p 2 to p1
R 12  r1  r2  3 ax  3 ay  2az
Unit vector of R 12 is
aR 1 2  
3
3
2
ax 
ay  a z
4
4
4
From (2) and (3), the scalar component of A at p 2 in the direction of aR
P r o b l e m s c h a p t e r 1 | 16
(3)
1 2
is
Proprietary of Prof. Lee, Yeon Ho, 2014

A aR 1 2  16 ax  4 ay  2 3 az


   34 a  43 a  24 a   12

x
y
The vector component of A at p 2 in the direction of aR
z
1 2

is
 3
3
2 
A '  12 aR 1 2  12   ax 
ay  az   9 ax  3 3ay  6 az
 4
4
4 

(b)
(4)
Transformation of A ' in (4) into cylindrical coordinates is obtained from Eq. (1-96) as
 cos  60o  sin  60o  0   9   1 / 2
 A  

 A     sin 60o cos 60o 0  3 3     3 / 2





 
  

 
 Az  
0
0
1   6   0


3 / 2 0  9 


1 / 2 0  3 3 
0
1  6 

 
Thus, in cylindrical coordinates
A '  9 a  3 3 a  6 az
(5)
1-28 Perform coordinate transformations of the following quantities from Cartesian to spherical coordinates
(a) unit vector ax
(b) point p : (x , y, z )
(c) differential length vector dl
(d) position vector r
ANS
(a)
From Eq. (1-101)
 AR   sin  cos  sin  sin  cos   1 
  
 
 A    cos  cos  cos  sin   sin   0 
 A    sin 
cos 
0  0 
 
In spherical coordinates
a x  sin  cos  a R  cos  cos  a   sin  a 
(b)
From Eq. (1-103)
R 
x 2  y2  z 2
  tan 1  x 2  y 2 / z 


1 y
  tan
x
(c)
From Eq. (1-102)
x  R sin  cos 
y  R sin  sin 
(1)
z  R cos 
Differentiating both sides of (1), we obtain
dx  dR  sin  cos   R cos  cos  d    R sin  sin  d  
dy  dR  sin  sin   R cos  sin  d    R sin  cos  d  
dz  dR  cos   R sin  d  
P r o b l e m s c h a p t e r 1 | 17
(2)
Proprietary of Prof. Lee, Yeon Ho, 2014
Using (2) in the expression dl  dx a x  dy ay  dz az , and then inserting dl into Eq. (1-101), we have
 AR   sin  cos  sin  sin  cos   dx 
  
 
 A    cos  cos  cos  sin   sin  dy 
 A    sin 
cos 
0  dz 
 
(3)
Noting that AR , A , and A  are the R-, - and -components of dl , (3) becomes
dl  dR a R  Rd  a   R sin  a 
(d)
Position vector in Cartesian coordinates
r  x a x  y ay  z a z
From Eq. (1-103)
R 
x 2  y2  z 2
At a point  R , ,  in spherical coordinates, r is parallel to a R having a magnitude of R .
Thus, r  R aR
1-29 A top shown in Fig. 1-35 is defined by a conical surface of half-angle 30o and a disk of radius 5[cm]. It
spins about the z-axis at an angular speed rad/s. Find an expression for the linear velocity v at any point
on the surface of the top in
(a) spherical coordinates
(b) Cartesian coordinates
Fig. 1-35 A top spinning at an angular speed 
ANS
(a)
At a point on the conical surface, with   30o , the linear velocity is
v  R sin 300 a 
1
R a
2
( R  10 )
(1)
( 0    30o )
(2)
At a point on the top plate, we have
v   5 3 tan  a
P r o b l e m s c h a p t e r 1 | 18
Proprietary of Prof. Lee, Yeon Ho, 2014
(b)
Coordinate transformation of v in (1) into Cartesian coordinates is, from Eq. (1-100),
v x  sin  cos  cos  cos   sin   0 
v    sin  sin  cos  sin  cos    0 
 y 


v z   cos 
 sin 
0   R / 2 
Rewriting it
v
R
  sin  ax  cos  ay 
2
(3)
From Eq. (1-103)
R 
x 2  y2  z 2
x 2  y2 / z
tan  
tan  
(4)
y
x
(5)
cos   x / x 2  y 2 , sin   y / x 2  y 2

(6)
Inserting (4) and (6) into (3)
v
 x 2  y2  z 2
2 x 2  y2
 y a  x a 
x
(7)
y
Coordinate transformation of v in (2) into Cartesian is

0
v x  sin  cos  cos  cos   sin  

v    sin  sin  cos  sin  cos   
0

 y 

v z   cos 
 sin 
0    5 3 tan  
Rewriting it
v   5 3 tan    sin  ax  cos  ay 
(8)
Inserting (5) and (6) into (8)
v
5 3
 y a x  x a y 
z
( x 2  y2  5 )
(9)
We see from (8) and (9) that v in Cartesian coordinates is unnecessarily complicated.
1-30 Consider Fig. 1-36, in which r is the position vector of a point on the conical surface of   30o and r  is
the position vector of a point at z  4 on the z-axis. In the presence of a vector field
A  r   r  r  / r  r  , find an expression for A, at a point p on the conical surface, in terms of radial
distance R and unit vector a  .
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Proprietary of Prof. Lee, Yeon Ho, 2014
Fig. 1-36 Vector field on a conical surface
ANS
Position vectors are
r  R aR in spherical coordinates
(1)
r   4 az in Cartesian coordinates
(2)
We define R  r  r  and express it in terms of spherical coordinates
First, r is expanded by the base vectors in Cartesian coordinates
r  R sin  cos  a x  R sin  sin  ay  R cos  a z

R
R
R 3
cos  a x  sin  ay 
az
2
2
2
The distance vector in Cartesian coordinates
R  r  r

R 3

R
R
cos  a x  sin  ay  
 4  az


2
2
 2

Using Eq. (1-101), the coordinate transformation of R into spherical coordinates is
 AR   sin  cos  sin  sin  cos     R / 2  cos  
  

 A    cos  cos  cos  sin   sin     R / 2  sin  
 A    sin 
cos 
0   R 3 / 2  4 
 




 (1 / 2) cos 
(1 / 2)sin  ( 3 / 2)   R / 2  cos  



 ( 3 / 2) cos  ( 3 / 2)sin  (1 / 2)    R / 2  sin  



 sin 
cos 
0   R 3 /2  4 

 


Thus


R  R  2 3 aR  2 a
Using (1) and (3)
A r
 r  r 
r  r
 R aR 

(3)
R  2 3  a  2 a
P r o b l e m s c h a p t e r 1 | 20
R
R

(4)
Proprietary of Prof. Lee, Yeon Ho, 2014
Rewriting (4)
A
2R
a 
R
2R a 
R  2 3   4
2
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