Steven A. Jones
March 8, 2006
BIEN 501, Physiological Modeling
Vector Concepts for Fluid Mechanics
1. A vector has magnitude and direction and can be described as an arrow pointing in
the assigned direction with a length equal to its magnitude (Figure 1).
2. Vectors can be described in Cartesian coordinates as:

 

v  ai  bj  ck
Eq. 1
y

Figure 1: A vector v , represented by the bold
arrow. The tip of the arrow can be reached from
the origin by moving a distance a in the x
direction, followed by a distance b in the y
direction, followed by a distance c in the z
direction.

v
c
b
a
x
z
3. The coefficients a , b , and c are the x , y , and z components of the vector.

 


i
,
j
,
and
k
i
x
4.
are themselves vectors.
is a vector pointing in the direction, j is a

y
k
vector pointing in the
direction, and is a vector pointing in the z direction.
See Figure 2.
y

 
j
i
k
Figure 2: The ,
and
unit vectors. These
are true vectors in the sense that they have both
magnitude and direction. For example, the
magnitude of i is 1 and its direction is along the
x axis.

j

k

v
x

i
z
5. The magnitude of a vector is its length, which is given by:

v  v  a 2  b2  c 2
Eq. 2
,
which is simply a statement of the Pythagorean theorem.





j

0
i

1
j

0
k
j
Problem 1: Given that the vector
can be represented as
, show, from

Eq. 2, that the magnitude of j is 1.
Steven A. Jones
March 8, 2006
BIEN 501, Physiological Modeling
Problem 2: Show that the equation for the magnitude of a vector is simply a statement
of the Pythagorean theorem as follows. First note that b and c are perpendicular to
one another because b is parallel to the y axis and c is parallel to the z axis (refer to
Figure 1). Use the Pythagorean theorem to obtain d (in Figure 3) in terms of b and c .
Now, since b and c are both perpendicular to the x axis, d is also perpendicular to the
x axis (i.e., it lies in a plane that is perpendicular to the x axis. Use this information to

write the length of the vector v in terms of a and d . Then back substitute for d to

obtain the length of v in terms of a , b and c .
y
Figure 3: Geometry to be used to demonstrate
that the equation for the magnitude of a vector is
a simple statement of the Pythagorean theorem.

v
c
a
d
b
x
z








v

a
i

b
j

c
k
v

a
i

b
j

c
k
1
1
1
2
2
2
6. The dot product between two vectors, 1
and 2
is
 
v1  v2  a1a2  b1b2  c1c2
Eq. 3


Problem 3: Show that the dot product of i and j is zero. Similarly show that the dot










product of i and k is zero. (Hint: write i  1i  0 j  0k and j  0i  1 j  0k and then
use Eq. 3 for the dot product.








Problem 4: Show that the dot product of u  1i  1 j  1k and j  1i  0.5 j  0.5k is
zero.
7. The dot product of two vectors is a scalar value. It has only magnitude, not
direction.
8. The value of the dot product is:
 
v1 v2 cos 
,
Eq. 4
where  is the angle between the two vectors.

  
Problem 5: Note that the vectors i and v  1i  1 j are in directions that are 45 apart
from one another. Calculate their dot products first by Eq. 3 and then by Eq. 4 and
show that they are equal.
Steven A. Jones
March 8, 2006
BIEN 501, Physiological Modeling
9. The dot product of one vector with another is physically the projection of that vector
on the other vector, multiplied by the length of the other vector.
10. The dot product of a vector with itself is the square of the vector’s magnitude. Thus,
2
v  v  v  v2  v  v  v  v .
11. A unit vector is a vector with magnitude of one.
12. Any vector can be normalized to become a unit vector by simply dividing by its own
magnitude.

 

 v
ai  bj  ck
u  
v
a 2  b2  c 2 .
13. Recalling 9 above, the dot product of any vector with a unit vector is the projection of
that vector in the direction of the unit vector. (The projection is the shadow cast by
the vector by a light source that is in the plane of the two vectors and that is
perpendicular to the unit vector).
14. The cross product of two vectors is a vector and can be written as:

i
 
v1  v2  a1
a2

j
b1
b2

k



c1  b1c2  c1b2  i  a1c2  c1a 2  j  a1b2  c1a 2  k
c2
14. The magnitude of the cross-product has the value of
the angle between the two vectors.
 
 
v1  v2  v1 v2 sin 
, where  is
15. The direction of the cross-product is perpendicular to both of the two original vectors,
or, equivalently, perpendicular to the plane in which the two original vectors lie.
Problem 6: Find the magnitude and direction of the cross product of the two vectors
given in Problem 5.
Problem 7: Show that the dot product of the result from Problem 6 with the two vectors
given in Problem 5 is zero. Hence, show that the cross product of these to vectors is
perpendicular to the original two vectors.
The above concepts are ones that students should know from their statics course. The
students will need this much review, and they will need to be reminded that they already
know these concepts from statics. Next come some concepts that will be new in the
sense of understanding vector applications. In other words, while students may have
Steven A. Jones
March 8, 2006
BIEN 501, Physiological Modeling
seen the gradient operator in their math class, they may not yet understand what it
means physically.




16. Define a vector r  xi  yj  zk . (The operator  means “is defined as,” as opposed
to “is equal to.” The difference between  and  is subtle, but often important).

Note that this vector may also be represented by r  x, y, z  . A function of the three
spatial coordinates x, y, z  , i.e. f x, y, z  , can then be represented as f r  . The
f  f  f 

gradient of the function f r  is defined as f  i 
j k.
x
y
z


Problem 8: If f r   x sin  yz  , find f r  .



(Answer: f r   sin  yz  i  xz cos yz  j  xy cos yz  k )
17. The gradient is a vector function. In other words, for a given set of values of x , y ,

 

and z , the gradient can be represented by f r   ai  bj  ck .


Problem 9: For the answer to Problem 8, what is the vector f r  at r  2,  , 14  .

1 
1 
(Answer:
i 
j   2 k ).
2
2 2
18. However, the function itself is not a vector.


Problem 10: What is the value of the function f r  at r  2,  , 14  .
19. The gradient is a function of the spatial coordinates.
20. Each term of the gradient is the rate of change of the function as one moves in the
given coordinate direction, assuming that there is no change in position with respect
to the other two coordinate directions. To illustrate the gradient, consider the
situation shown in Figure 4. A paper mill produces a sulfurous odor that diminishes
in pungency with distance from the mill. A person at point A experiences a relatively
large rate of reduction of the odor as he moves in the x direction. The same person
experiences a relatively small rate of reduction of odor as he moves in the y
direction. Thus, the x component of the gradient is large and the y y component of
the gradient is small.
Figure 4: The concept of
gradient illustrated by the rate
at which the odor from a paper
mill decreases as one moves
in a particular direction.
y
A
x
Paper Mill
Steven A. Jones
March 8, 2006
BIEN 501, Physiological Modeling
Problem 11: Assume in the paper mill example that the odor drops off with distance
2
from the mill according to the equation O  A r . Give the equation for the gradient at
any location x, y  from the mill, assuming that the origin of the x and y axes is at the
center of the mill. Hint: You should write r 
x 2  y 2 and then take the derivatives
 
xi  yj
with respect to x and y . Answer: O  2 A
.
r4




i
Problem 12: For the same paper mill, what are the and j components of the
gradient found in Problem 3 at a point x, y   2,0 ? How do they agree with the above
statement that the odor should drop off more quickly along the x direction than along
the y direction at this location?
21. Velocity is a vector quantity. It’s magnitude is the speed of an object. It’s direction is
the direction in which the object is moving.

22. The gradient operator can be described as a vector,    i   j   k . Thus, the
x
y
z
gradient of a function is the gradient operator applied to the function.

23. The divergence of a vector v (such as velocity) is the dot product of the gradient
operator with the vector.

 v v
v
div(v )    v  1  2  3
x
y
z
24. The divergence is a scalar quantity because it is a dot product.
25. One may take the divergence of the gradient of a function. This scalar quantity is
known as the Laplacian operator.

          f  f  f
  f r    2 f ( r )   i 
j  k    i 
j
y
z   x
y
z
 x
  2 f 2 f 2 f
k   2  2  2
y
z
 x
2
26. One may also consider the divergence to be an operator  defined as:
                2
2
2
 2       i 
j  k    i 
j  k   2  2  2
y
z   x
y
z  x
y
z
 x
Steven A. Jones
March 8, 2006
BIEN 501, Physiological Modeling
 2
2
2 
27. Thus,  2 f   2  2  2  f
y
z 
 x
28. The curl of a vector is the cross-product of the gradient operator with the vector.

  v
v    v v    v v  
curl (v )    v   2  3 i   1  3  j   1  2 k
y   z
x 
x 
 z
 y
29. The curl of a vector is another vector.
30. The curl of a vector can be written as:

i



curl v     v 
x
v1

j

y
v2

k

z
v3
and this form provides one with an easy way to remember where the “–“ signs must go.
Problem 13: Which of these are not legitimate vector operations. Why? Assume f is
a scalar function.


2
a.   v
d.     v 
f. f   f 
h.    f 

2
b.   f
e.     v
g.  f 
c.   f
Problem 14: For each legitimate operation in Problem 5, state what kind of entity
(vector or scalar) is the result.
31. The equation f r   c0 , where c0 is a constant, represents a surface in threedimensional space. A vector normal to this surface is f r  .
32. f r  is also the direction of maximum rate of change in f r  .
33. A curve in space is the intersection of two surfaces.
Problem 15: What is the curve represented by the intersection of the following two
surfaces:
x 2  y 2  z 2  25 ; x 2  y 2  z  0 ?
Steven A. Jones
March 8, 2006
BIEN 501, Physiological Modeling
Solution: First notice that the first curve is a sphere of radius 5 and the second curve is
a parabola of rotation. If we substitute z  x 2  y 2 from the second curve into the first
curve, the result is:
z 2  z  25  0 ,
 1  1  100
, one of which is negative and therefore
2
irrelevant. It follows that the curve is in the plane represented by z  101  1 2 and on
which has two solutions, z 
the curve in that plane x 2  y 2 
 101  1 2 .


31. There are several vector identities that should be recognized by the student
(though not necessarily memorized).
a. a  b  c  b  c  a  c  a  b  a  c  b  b  a  c  c  b  a
Proof of the first equality:
i
a  b  a1
b1
j
a2
b2
k
a3  a 2 b3  a3b2 i  a1b3  a3b1 j  a1b2  a 2 b1 k
b3
a  b   c  a2 b3  a3b2 c1  a1b3  a3b1 c2  a1b2  a2 b1 c3
 b2 c3  b3c2 a1  b1c3  b3c1 a 2  b1c2  c1b2 a3
i
j
 b1 b2
c1 c2
k
b3  a  b  c   a
c3
The other equalities follow similarly.
Steven A. Jones
March 8, 2006
BIEN 501, Physiological Modeling
b. a  b  c  a  cb  b  ca
Proof: a  b is a vector perpendicular to both a and b . Any vector that is
perpendicular to a  b must therefore be in the plane of both a and b . In
particular a  b  c must be in that plane. Any vector in that plane can be
described by a linear combination of a and b . Thus, a  b  c  ma  nb .
Furthermore, since c must be perpendicular to a  b  c , it follows that
a  b  c  c  0 so that ma  nb  c  0 , i.e. ma  c  nb  c . Therefore,
a  b  c  ma  nb