Examples in Chapter 1
Problem 1.42
Vector A has components Ax=1.30 cm and
Ay= 2.25 cm; vector B has components
Bx=4.10 cm, By=-3.75 cm. Find
a) The components of the vector sum
A+B
b) the magnitude and direction of A+B
c) The components of the vector B-A
d) The magnitude and direction of B-A
The components of the vector sum
A+B

The sum of the xcomponents of A
and B are equal to
the x component of
A+B
Ax  1.30
 Bx  4.10
 A  B
x
 5.40cm
Ay  2.25
 By  3.75
 A  B
y
 1.50
Find the magnitude of A+B
(A+B)x
(A+B)y
(A+B)
 A  B     A  B  
2
x
y
5.4   1.5   31.41
2
2
A  B  31.41  5.60
2
 A B
2
Find the direction of A+B
(A+B)x

(A+B)
(A+B)y


A B
y
tan   
x
A B


y
x
1.5
  tan
 0.27 radians
5.4
180
0.27 
 15.5o
1

Components, magnitude and
direction of B-A
 A  1.30 2.25 ; B  4.10 3.75
 B  A
 B  A
x
y
 (4.10)  (1.30)  2.8
 (3.75)  (2.25)  6
 2.8   6 
2
2
 43.84  B  A  6.62
6
o
  tan
 1.134    64.98
2.8
1
Problem 1.55
Find the vector product A  B where
A  4.00 xˆ  3.00 yˆ
B  5.00 xˆ  2.00 yˆ
What is the magnitude of the vector product?
Your book’s way

The book gives 3 formulae for C=AxB
Cx=AyBz-AzBy
 Cy=AzBx-AxBz
 Cz=AxBy-AyBx


Since A and B have only x- and ycomponents, we find Cz by


Cz=4*(-2)-5*3=-8-15=-23
||AxB||=23
My way: First, a review?

A determinant represents a single number
and is used in linear algebra
For a 2x2 matrix, its determinant is written:
det( A) 
a b
c d
 ad  cb
A 3 x 3 determinant
a
b
c



d
e
f
 


g
h
i



a
b
c
d
e
f
g
h
i
a
b
c
d
e
f
g
h
i
a
b
c
d
e
f
g
h
i
 a   ei  hf

 b   di  gf
 c   dh  eg 

For a cross-product
ˆ
x
ˆ
y
ˆ
z
d
e
f
g
h
i
ˆ
x
ˆ
y
ˆ
z
d
e
f
g
h
i
ˆ
x
ˆ
y
ˆ
z
d
e
f
g
h
i
ˆ
x
ˆ
y
ˆ
z
d
e
f
g
h
i
ˆ   ei  hf
 x

ˆ   di  gf
 y
ˆ   dh  eg 
 z

For our problem
xˆ
yˆ
zˆ
4
3
0
5
2
0
xˆ
yˆ
zˆ
4
3
0  xˆ   3* 0  ( 2) * 0   0 xˆ
5
2
0
xˆ
yˆ
zˆ
4
3
0   yˆ   4 * 0  5* 0   0 yˆ
5
2
0
xˆ
yˆ
zˆ
4
3
0  zˆ   dh  eg   (4 * ( 2)  3*5) zˆ  23 zˆ
5
2
0
Problem 1.61
Biological tissues are typically made up of
98% water. Given that the density of
water is 1 x 103 kg/m3, estimate the
mass of
a) the heart of an adult human
b) a cell with a diameter of 0.5 mm.
c) a honey bee.
Best guesses
Human heart: size of fist (cylinder 4” long
with diameter of 3”)
 Honey bee: 1” or 2.5 cm long cylinder
with 0.25” diameter

Heart Problem



4” = 4*2.54 cm = 10.16 cm or 0.1016 m
3” =3 * 2.54 cm = 7.62 cm
Volume of cylinder= *(d/2)2 *L

 * (7.62/2)2 *10.16=463 cm3 or cc
1m3
4 3
463cc

4.63
x
10
m
3
3
100 cm
m
   m  V  1000* 4.63 x104  0.463kg
V

Note: 0.454 kg =1 pound
A cell

Assume spherical!
 Volume=4*/3*(d/2)3
d  0.5m m  0.5 x106 m
3
4  0.5 x10 
19
V  

5.23
x
10

3 
2

19
16
m  V  1000*5.23 x10  5.23 x10 kg
6
Book answer differs by order of magnitude
Honey bee

V=*(d/2)2 *L=*(.25/2)2 *1=0.049 in3
3
3
3
3
2.54 cm
1m
7
0.049in

8.04
x
10
13 in3
1003 cm3
3
m  V  1000*8.04 x10  8.04 x10 kg or 0.80 g
7
Book assumes ½ in long
4
Problem 1.70
A sailor in a small sailboat encounters
shifting winds. She sails 2.0 km east
then 3.5 km southeast (-450 w.r.t. east)
and then an unknown distance. Her final
position is 5.8 km directly east of starting
point. Find magnitude and direction of
the third leg of the journey.
Step 1: Draw it!
5.80 km
2.0 km
3.5 km
Step 2: Sketch in the details
5.80 km
2.0 km
3.5 km
Ay
Ay
Ay
?
Step 3: Simplify
5.80 km-2.0 km=3.8km
2.0 km
3.5 km
Ay
Ay2  Ay2  2 Ay2  3.52
Ay 
3.5
 2.47km
2
Ay
Ay
?
Step 4: Solve
5.80 km-2.0 km=3.8km
2.0 km
Ay
3.5 km
Ay
Ay
Ay2  Ay2  2 Ay2  3.52
Ay 
3.5
 2.47km
2
?=3.8-2.47=1.33 km
Step 5:
5.80 km-2.0 km=3.8km
2.0 km
Ay
3.5 km
Ay
Ay=2.47 km
Ay2  Ay2  2 Ay2  3.52
Ay 
3.5
 2.47km
2
?=3.8-2.47=1.33 km=Ax
Step 6:
2.47 km

1.33 km
1.332  2.47 2  7.98
Distance= 7.98  2.81km
2.47
  tan
 1.07 radian
1.33
180
1.07 radian
 61.69o
1
