Chap
.
9
Vector Differential Calculus
.
Grad
,
Div
,
Curl
Sec
.
9
.
1
Vectors in 2

Space and 3

Space
Problem Set 9
.
1
.
Page 370
1
.
Components
.
Length
.
Unit vector
. According to the definition of components you have to calculate
the differences of the coordinates of the terminal point
Q
minus the corresponding coordinates of the
initial point
P
of the vector. Thus,
v
1
u
5
u
3
u
2, etc. Since the
z
coordinates of
P
and
Q
are zero, the
vector
v
is a vector in the
xy
plane; it has no component in the
z
direction. Sketch the vector, so that you
see what it looks like as an arrow in the
xyz
coordinate system in space.
Calculate the length of the vector by (2) (with
a
3
u
0)

v

u
v
1
2
U
v
2
2
u
u
5
u
3
U
2
U
u
u
2
u
2
U
2
u
4
U
16
u
20
.
To obtain the unit vector in the direction of
v
, multiply
v
by 1/
v
; this is a scalar multiplication. It gives
u
u
u
1/
v

U
v
u
u
1/ 20
U±
2,
u
4
²
u
±
1/ 5
,
u
4/5
²
.
In the answer on p. A24 we have carried along a component 0, to emphasize that the
xy
plane is
part of the
xyz
space.
15
.
Addition and scalar multiplication
. It makes no difference whether you first multiply and then add,
or whether you first add the given vectors and then multiply their sum by the scalar 5. This problem and
Example 2 in the text illustrate formula (6a).
29
.
Forces
were foremost among the applications that have suggested the concept of a vector, and forming
the resultant of forces has motivated vector addition to a large extent. Thus, each of Problems 2428
amounts to the addition of three vectors. “Equilibrium” means that the resultant of the given forces is the
zero vector. Hence in Prob. 29 you must determine
v
such that
v
U
p
U
q
U
u
u
0
.
Hence
v
u
u
u
p
U
q
U
u
U
u
u
±
2
u
4
U
2,
2
u
4
U
2,
2
U
0
U
7
²
u
±
0,
0,
u
9
²
.
37
.
Force in a rope
. This is typical of many problems in mechanics. Choose an
xy
coordinate system with
the
x
axis pointing horizontally to the right and the
y
axis pointing vertically downward. Then the given
weight is a force
w
u
±
w
1
,
w
2
²
u
±
0,
w
²
pointing vertically downward. You have to determine the
unknown force in the left rope, call it
u
u
±
u
1
,
u
2
²
, and the unknown force
v
u
±
v
1
,
v
2
²
in the right rope.
The three forces are in equilibrium (they have the resultant
0
U
because the system does not move. Thus,
u
U
v
U
w
u
0
.
(A)
You have two choices of giving
u
a direction, and similarly for
v
. The choice is up to you. Suppose you
choose the vectors to point from the point where the weight is attached upward to the points where the
ropes are fixed. Then (see the figure on p. 370)
u
u
±
u
1
,
u
2
²
u
±
u

u
 cos
u
,
u

u
 sin
u
²
v
u
±
v
1
,
v
2
²
u
±

v
 cos
u
,
u

v
 sin
u
²
w
u
±
w
1
,
w
2
²
u
±
0,
w
²
,
(B)
where the components preceded by a minus sign are those that point in the negative direction of the
corresponding coordinate axis (to the left or upward). From (A) and (B) you obtain for the horizontal