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).
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
2. Vectors can be described in Cartesian coordinates as:

 

v  ai  bj  ck
3. The coefficients a , b , and c are the x , y , and z components of the vector.


 
4. The entities i , j , and k are themselves vectors. The vector i points in the x


direction, j points in the y direction, and k points in the z direction. See Figure 2.
y
 

Figure 2: The i , j and k unit vectors. These
are true vectors in the sense that they have both
magnitude and direction.

j

k

v
x

i
z

 
5. The vectors i , j , and k are called “basis” vectors.
6. The magnitude of a vector is its length, which is given by:

v  v  a 2  b2  c 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 that
the magnitude of j is 1.
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Problem 2: Refer to Figure 3 and 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. Use the Pythagorean theorem to obtain d 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
1
2
2
2
2
7. The dot product between two vectors,
and
is
 
v1  v2  a1a2  b1b2  c1c2
8. The dot product is a scalar value.
 
9. The value of the dot product is v1 v2 cos  , where  is the angle between the two
vectors.
10. 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.
11. A unit vector is a vector with magnitude of one.


 



12. The basis vectors i , j , and k are unit vectors because, for example, i  1i  0 j  0k

so that i  12  0 2  0 2  1 .
13. 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 .
2
14. 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. (e.g., 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, as in Figure 4).
Light Source

v
Figure 4: Projection of a vector.

The vector v casts a shadow
 
(white vector) v  u from the light
source in the direction of the unit

vector u .
 
v u

u
15. 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
 
 
16. The magnitude of the cross-product has the value of v1  v2  v1 v2 sin  , where  is the
angle between the two vectors.
17. 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.
18. 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 seen the gradient operator in their math class, they may not yet
understand what it means physically.

f  f  f 
19. The gradient of a function f r  is defined as f  i 
j k.
x
y
z
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20. The gradient is a vector.
21. However, the function itself is not a vector in this case.
22. The gradient is a function of the spatial coordinates.
23. 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
Problem 3: Assume in the paper mill example that the odor drops off with distance from
2
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
2
2
center of the mill. Hint: You should write r  x  y and then take the derivatives


x i  yj
O  2
r4
with respect to x and y . Answer:
.




Problem 4: For the same paper mill, what are the i 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?
4

24. A surface in space is defined by the equation f x, y, z   C , or equivalently f r   C ,
where C is a constant.
Problem 5: Describe the surface that is defined by z  4 .
Solution: If z  4 , we have f x, y, z   z . This surface is a plane that passes through
the z-axis at 4 and is parallel to the x-y plane.
y
Figure 5: The surface defined by z=4.
x
z
Problem 6: Describe the surface that is defined by y  e  x
Solution: If y  e  x
2
 z2
2
 z2
 0.
 0,
then f x, y, z   y  e  x  z  , and
the constant C is zero. We
can solve for y in terms of
2
2
Figure 6: Bi-Gaussian surface.
x and z to get y  e  x  z  .
This shape corresponds to a
bi-Gaussian surface and can
be plotted with standard
graphing software, as shown in
Figure 6.
2
2
25. The gradient of a function f x, y, z  , evaluated at a specific point, x0 , y0 , z0  is a
vector that is perpendicular to the surface f x, y, z   C that passes through the
point x0 , y0 , z0 .
Problem 7: What vector is perpendicular to a surface y  e  x
(1,1,1).
2
 z2
  C that at the point
26. 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.
5

27. 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.

28. 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
29. The divergence is a scalar quantity because it is a dot product.
30. 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
31. 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
32. Thus,
 2
2
2 
 2 f   2  2  2  f
y
z 
 x
33. 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
34. The curl of a vector is another vector.
35. The curl of a vector can be written as:

i



curl v     v 
x
v1

j

y
v2

k

z
v3
6
and this form provides one with an easy way to remember where the “–“ signs must go.
Problem 7: Which of these are not legitimate vector operations. Why? Assume f is a
scalar function.

a.   v
b.   f
c.   f

d.     v 

e.     v
f. f   f 
2
g.  f 
2
h.    f 
Problem 8: For each legitimate operation in Problem 7, state what kind of entity (vector
or scalar) is the result.
Problem 9: Of the expressions in Problem 7, which ones will always be equal to zero

(either the zero vector 0 or the zero scalar 0 ).
7