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1. C. M. Jenkins, Dept of Physi s, U. of South Alabama 1
Ve tors
Some quantities require just a number to des ribe them.
Volume: This is a one liter oke...
Temperature: It is 93o
outside...
Relative Humidity.... With 96% humidity!
These quantities are alled s alars.
For other quantities a number is not enough.
Some quantities need an number and a di-
re tion.
We agree on a point.
Then I walk a distan e of 5 meters from
that point.
My nal lo ation is anywhere on a ir le of
5 meters from the starting point.
These quantities are alled Ve tors.
Examples of ve tor quantities:
Displa ement (or lo ation).
Velo ity.
A eleration.
For e.

2. C. M. Jenkins, Dept of Physi s, U. of South Alabama 2
Ve tors are represented as:
~
Aor A
Graphi ally, a ve tor is drawn as a line with
an arrow head on it.
The arrowhead end is alled the head.
The other end is alled the tail.
Remember: the ve tor is des ribed by a number and dire tion.
The number orresponds to the length of the ve tor and is alled
the magnitude.
The magnitude of ~
A (the length of the ve tor) is represented
by: A or j ~
Aj.
The dire tion of the ve tor orresponds to the dire tion that the
ve tor is pointing.
The dire tion of the ve tor is represented by the angle with
respe t to the X axis.
Or the dire tion of ve tor ~
A is represented by the unit ve tor:
^
A =
~
A
j ~
Aj
.
Ve tors may be displa ed (i.e. moved).
The ve tor is not hanged if its length and dire tion are not
hanged.

3. C. M. Jenkins, Dept of Physi s, U. of South Alabama 3
Ve tor Addition
Triangle Method
Two ve tors may be added.
This operation is de ned di erently than the addition of two s alars.
We must add the ve tors in a way that adds the lengths (i.e.) the
magnitudes..
But we must also a ount for their dire tions...
Consider adding two ve tors ~
A and ~
B.
The result is the resultant ve tor ~
R:
~
R = ~
A + ~
B
At rst we de ne ve tor addition
by graphi al methods.
The rst method is the Triangle
method:
Draw ve tor ~
A (i.e. from tail to head).
At the head of ~
A, pla e the tail of ~
B.
Draw ve tor ~
B.
The resultant ve tor (~
R) is determined by drawing a ve tor from
the tail of ~
A to the head of ~
B.

4. C. M. Jenkins, Dept of Physi s, U. of South Alabama 4
Ve tor Addition
Parallelogram Method
The se ond method is the parallelogram method.
This method gives the same result as the triangle method...
Consider adding two ve tors ~
Aand
~
B.
The result is the resultant ve tor
~
R:
~
R = ~
A+ ~
B
Draw ve tor ~
A (i.e. from tail to head).
At the head of ~
A, pla e the tail of ~
B.
Draw ve tor ~
B.
Make opies of ~
A and ~
B.
Displa e the opy of ~
B until its tail is tou hing the tail of ~
A.
Displa e the opy of ~
A until its tail is tou hing the head of the
opy of ~
B.
The resultant ve tor ~
R is drawn along the diagonal from the tails
of ve tors ~
A and ~
B to the heads of ~
A and ~
B.

5. C. M. Jenkins, Dept of Physi s, U. of South Alabama 5
Properties of addition of ve tors
Two ve tors may be added in any order: ~
A + ~
B = ~
B + ~
A, i.e.
ve tor addition ommutes.
Three or more ve tors may be grouped in any order when added:
~
A+
~
B + ~
C
=
~
A + ~
B
+~
C , i.e. ve tor addition is asso iative.
Any number of ve tors may be added together:
~
R = ~
A+ ~
B + ~
C + ~
D

6. C. M. Jenkins, Dept of Physi s, U. of South Alabama 6
Inverse of a Ve tor
S alar Inverse
Numbers have an additive inverse or inverse:
For the number a, there exist an additive inverse: -a.
A number a added to its additive inverse -a gives the identity
(zero) as a result.
a + (-a) = 0.
Note this is how the operation of subtra tion is de ned. (i.e.
just remove the parenthesis in the above example).
Ve tor Inverse
The inverse of a ve tor is onstru ted by taking the ve tor and hang-
ing its dire tion by 180o
without hanging its length.

7. C. M. Jenkins, Dept of Physi s, U. of South Alabama 7
Ve tor Subtra tion
We use the inverse ve tor and ve -
tor addition to de ne ve tor sub-
tra tion.
Suppose we wish to subtra t ~
B
from ~
A to get the resultant ve -
tor ~
R:
~
R = ~
A ~
B
Draw ve tor ~
A (i.e. from tail to head).
Constru t the inverse of ve tor ~
B.
At the head of ~
A, pla e the tail of (-~
B).
Draw ve tor (-~
B).
The resultant ve tor (~
R) is determined by drawing a ve tor from
the tail of ~
A to the head of ~
B.

8. C. M. Jenkins, Dept of Physi s, U. of South Alabama 8
Or onstru t a parallelogram:
~
R = ~
A+ ~
B
Draw ve tor ~
A (i.e. from tail to head).
At the head of ~
A, pla e the tail of ~
B.
Draw ve tor ~
B.
Make opies of ~
A and ~
B. Displa e the opy
of ~
B until its tail is tou hing the tail of ~
A.
Displa e the opy of ~
A until its tail is tou h-
ing the head of the opy of ~
B.
The resultant ve tor ~
R is drawn along the di-
agonal from the head of ve tor ~
B (original)
heads of ~
A (original).
This diagram may be simpli ed by drawing ~
A and
~
B with their tails at the same point.
The resultant (~
R = ~
A ~
B) is drawn with its tail at
the head of ~
B to the head of ~
A.
Note that these methods are as a urate as the
graphi al tools (su h as rulers and protra tors) that
are used...
We need numeri al methods to al ulate ve tor ad-
dition and subtra tion.

9. C. M. Jenkins, Dept of Physi s, U. of South Alabama 9
Coordinate Systems
Consider a world onsisting of a plane.
We want to des ribe the lo ation of points in this plane.
Coordinate Systems for that world needs:
A single point to measure all
other points from: origin.
Two di erent dire tions to mea-
sure along.
In our ase two mutually per-
pendi ular dire tions.
Lets all the horizontal di-
re tion (or axis) the X axis, the
verti al dire tion the Y axis.
A rule to tell use the order that
the dire tions position is ommu-
ni ated.
In our ase (X, Y).
The lo ation of (5,3) is indi-
ated in the Figure....
This is alled a Cartesian oordinate system.

10. C. M. Jenkins, Dept of Physi s, U. of South Alabama 10
Components of Ve tors
Let's onsider a ve tor
~
Athat has
a magnitude of 7.0 with an angle
of 50
o with respe t to the X axis.
Lets onstru t ve tor ~
A with the
ve tor sum of two two ve tors: ~
AX
and ~
AY .
~
AX is parallel to the X axis.
~
AY is parallel to the Y axis.
The ve tors ~
AX and ~
AY are alled
the omponents of the ~
A.

11. C. M. Jenkins, Dept of Physi s, U. of South Alabama 11
A Cartesian oordinate system allows us to use right triangles and
therefore trigonometry.
From the Figure you an identify:
The right angle.
The hypotenuse.
Use the angle with respe t to the X axis (50o).
Then the length of ~
AX is: AX = A os(50o).
And the length of ~
AY is: AY = Asin(50o).
These lengths are alled the omponents of ~
A.
The dire tions of these lengths are alled the unit ve tors
of ~
A.
^
i means one unit along the X dire tion.
^
j means one unit along the Y dire tion.
So ~
AX (whi h is parallel to the X axis) is written as: ~
AX = AX^
i =
A os(50o)^
i.
And ~
AY (whi h is parallel to the Y axis) is written as: ~
AY = AY ^
j =
Asin(50o)^
j .
The ve tor ~
A = A os(50o)^
i+ Asin(50o)^
j.

12. C. M. Jenkins, Dept of Physi s, U. of South Alabama 12
Example: Components of Ve tors
Ba k to our example of a ve tor
~
A that has a magnitude of 7.0
with an angle of 50o
with respe t
to the X axis.
Find the omponents of this ve -
tor and write the the ve tor ~
A in
omponent form.
The X omponent of the ve tor is:
Ax = 7:0 os(50
o
) = 7:0(0:6427) = 4:500
The Y omponent of this ve tor
is:
Ay = 7:0 sin(50
o
) = 7:0(0:7660) = 5:362
Using the unit ve tors:
~
A = 4:500^
i + 5:362^
j

13. C. M. Jenkins, Dept of Physi s, U. of South Alabama 13
Example: Magnitude Dire tion of a Ve tor from
Components
Suppose a ve tor is given by its omponents.
How do we nd the magnitude of the ve tor:
Use the Pythagorean theorem: j ~
Aj =
q
A2
x + A2
y
How do we nd the dire tion (i.e. the angle with respe t to the X axis)
of the ve tor?
Use the tangent of the angle with respe t to the X axis: =
Tan 1
(Ay
Ax
)
Note that the angle with respe t to the X axis be omes tri ky if the
ve tor is not lo ated in Quadrant I.
First Quadrant
~
A = 4^
i + 5^
j
Use the Pythagorean theorem to
determine the magnitude:
j ~
Aj =
p42 + 52 = 6:403
Use the tangent of the angle with
respe t to the X axis:
= tan 1
(
5
4
) = 51:34o

14. C. M. Jenkins, Dept of Physi s, U. of South Alabama 14
Example: Magnitude Dire tion of a Ve tor from
Components
(Continued)
Se ond Quadrant
~
A = 4^
i + 5^
j
Use the Pythagorean theorem to
determine the magnitude:
j~
Aj =
p( 4)2 + 52 = 6:403
The value of tangent returned by
your al ulator is:
= tan
1
(
5
4
) = 51:34
o
Use the tangent of the angle with
respe t to the X axis: = 180o
51:34o
= 128:66o

15. C. M. Jenkins, Dept of Physi s, U. of South Alabama 15
Example: Magnitude Dire tion of a Ve tor from
Components
(Continued)
Third Quadrant
~
A = 4^
i 5^
j
Use the Pythagorean theorem to
determine the magnitude:
j~
Aj =
p( 4)2 + ( 5)2 = 6:403
The value of tangent returned by
your al ulator is:
= tan
1
(
5
4
) = 51:34
o
Use the tangent of the angle with
respe t to the X axis: = 180o
+ 51:34o
= 231:34o

16. C. M. Jenkins, Dept of Physi s, U. of South Alabama 16
Example: Magnitude Dire tion of a Ve tor from
Components
(Continued)
Fourth Quadrant
~
A = +4^
i 5^
j
Use the Pythagorean theorem to
determine the magnitude:
j~
Aj =
p42 + ( 5)2 = 6:403
The value of tangent returned by
your al ulator is:
= tan
1
(
5
4
) = 51:34
o
Use the tangent of the angle with
respe t to the X axis: = 360o
51:34o
= 308:66o

17. C. M. Jenkins, Dept of Physi s, U. of South Alabama 17
Addition of Ve tors
We want a numeri al method to add two or more ve tors.
The rule for addition of ve tors is simple:
To add ve tors: algebrai ly add the omponents (i.e.
add the X omponents together, then add the Y om-
ponents together).
Remember the result is a ve tor, whi h requires at least two
numbers to des ribe it.
These two numbers are:
A magnitude and dire tion.
Or an X omponent and Y omponent.
Suppose we want to add two ve tors ~
C =
~
A+ ~
B:
~
A = Ax^
i + Ay^
j
~
B = Bx^
i + By^
j
Using unit ve tors:
~
A = Ax^
i + Ay^
j
~
B = Bx^
i + By^
j
~
C = (Ax + Bx)^
i + (Ay + By)^
j
Ve tor addition using the omponents depi ted graphi ally.

18. C. M. Jenkins, Dept of Physi s, U. of South Alabama 18
Addition of Ve tors
Example: Ve tor Addition
Find the resultant from the sum of ~
A and ~
B, where:
~
A = (7^
i + 5^
j) m
~
B = ( 17^
i + 9^
j) m
Just add the omponents....
~
A = (7^
i + 5^
j) m
+ ~
B = ( 17^
i + 9^
j) m
~
C = ( 10^
i + 14^
j) m
Find the magnitude and dire tion of the resultant ve tor ( ~
C).
Magnitude:
C =
p( 10)2 + 142 = 17:20m
The angle with respe t to the X axis:
= tan
1
(
14
10
) = 54:46
o
Sin e the resultant ve tor is in quadrant II, the angle is: = 180o
54:46o
= 125:54o
.

19. C. M. Jenkins, Dept of Physi s, U. of South Alabama 19
Ve tor Subtra tion using Components
The numeri al method to subtra t ve tors is very similar to adding two
ve tors.
The rule for subtra tion of two ve tors is:
To subtra t ve tors: subtra t the omponents
(i.e. subtra t the X omponents, then subtra t
the Y omponents).
Remember the result is a ve tor, whi h re-
quires at least two numbers to des ribe it.
These two numbers are:
A magnitude and dire tion.
Or an X omponent and Y omponent.
Suppose we want to subtra t two ve tors ~
C = ~
A ~
B:
~
A = Ax^
i + Ay^
j
~
B = Bx^
i + By^
j
Using unit ve tors:
~
A = Ax^
i + Ay^
j
( ~
B = Bx^
i + By^
j)
~
C = (Ax Bx)^
i + (Ay By)^
j

20. C. M. Jenkins, Dept of Physi s, U. of South Alabama 20
Example: Ve tor Addition
Find the resultant ~
C = ~
A ~
B, where:
~
A = (7^
i + 5^
j) m
( ~
B = ( 17^
i + 9^
j) m )
Just subtra t the omponents....
Using unit ve tors:
~
A = (7^
i + 5^
j) m
~
B = ( 17^
i + 9^
j) m
~
C = (24^
i 4^
j) m
Find the magnitude and dire tion of the resultant ve tor ( ~
C).
Magnitude:
C =
p(24)2 + ( 4)2 = 24:33m
The angle with respe t to the X axis:
= tan
1
(
4
24
) = 9:46
o
Sin e the resultant ve tor is in quadrant IV, the angle is: = 360o
9:46o
= 350:53o
.

21. C. M. Jenkins, Dept of Physi s, U. of South Alabama 21
Example: A Ve tor Equation
A hiker walks along a trail in four
legs. The rst leg is 60 m north,
the se ond leg is 130 m east, head-
ing and distan e of the third leg
is 80 m at an angle of 30o
west
of north. The four leg has an un-
known length and heading. The
hiker ends up at a distan e of 132.72
m 13.06o
west of north. Find the
length and heading of the fourth
leg.
Let the hike be represented by the
ve tor equation:
~
R = ~
A + ~
B + ~
C + ~
D
where the unknown ve tor is ~
D. Solving this equation for ~
D:
~
D = ~
R ~
A ~
B ~
C
To al ulate ~
D, resolve all known ve tors into omponents:
~
A = (0^
i + 60^
j) m
~
B = (130^
i + 0^
j) m
~
C = ( 80 sin(30)^
i + 80 os(30)^
j) m
~
D = (Dx
^
i + Dy
^
j) m
~
R = ( 132:72 sin(13:06)^
i + 132:72 os(13:06)^
j) m
Or:
~
A = (0^
i + 60^
j) m
~
B = (130^
i + 0^
j) m
~
C = ( 40^
i + 69:28^
j) m
~
D = (Dx
^
i + Dy
^
j) m
~
R = ( 29:99^
i + 129:29^
j) m

22. C. M. Jenkins, Dept of Physi s, U. of South Alabama 22
Solving for the omponents of ~
D:
~
R = ( 29:99^
i + 129:66^
j) m
( ~
A = (0^
i + 60^
j) m )
( ~
B = (130^
i + 0^
j) m )
( ~
C = ( 40^
i + 69:28^
j) m )
~
D = ( 119:99^
x 0:38^
j) m
So the ve tor des ribing the fourth leg is:
~
D = ( 119:99 ^
x 0:38^
j) m
The length of ~
D (i.e. the distan e) is:
j~
Dj =
p( 119:99)2 + ( 0:38)2 = 119:99 m
The heading is:
= tan 1
(
0:38
119:99
) = 0:181o
Both the X and Y ompo-
nents are negative.
So this ve tor is in the third quad-
rant.
Or 0.181o
South of West.

23. C. M. Jenkins, Dept of Physi s, U. of South Alabama 23
S alar or Dot Produ t
The s alar or dot produ t is a ve tor operation where two ve tors are
multiplied and a s alar results.
The dot produ t is de ned as:
~
A ~
B = j ~
Ajj ~
Bj os()
Where is the angle between the two ve tors.
the s alar produ t ommutes: ~
A ~
B = ~
B ~
A.
The s alar produ t obeys the distributive prop-
erty: ~
A
~
B + ~
C
= ~
A ~
B + ~
A ~
C
Note the dot produ t between two identi al
unit ve tors is:
^
i ^
i = j^
ijj^
ij os(0)
^
i ^
i = 1
Note the dot produ t between two di erent unit ve tors is:
^
i ^
j = j^
ijj^
jj os(90)
^
i ^
j = 0
This gives us the general rule for a dot produ t of:

24. C. M. Jenkins, Dept of Physi s, U. of South Alabama 24
^
i ^
i = 1 ^
j ^
j = 1 ^
k ^
k = 1
^
i ^
j = 0 ^
i ^
k = 0 ^
j ^
k = 0
Geometri ally, the dot produ t is the proje tion of on ve tor onto the
other.
Or, howmu hofoneve torispar-
allel to the se ond.
Thedotprodu tbetweentwove -
torsiseasily al ulatedusing om-
ponents.
~
A ~
B =
Ax^
i+Ay^
j +Az
^
k
Bx^
i+By^
j +Bz
^
k
Just multiply out as a polynomial:
~
A ~
B = AxBx^
i ^
i +AxBy^
i ^
j +AxBz^
i ^
k +
AyBx^
j ^
i +AyBy^
j ^
j +AyBz^
j ^
k +
AzBx
^
k ^
i+AzBy
^
k ^
j +AzBz
^
k ^
k
Now apply the rules for the dot produ t between two unit ve tors:
~
A ~
B = AxBx^
i ^
i
%
1
+AxBy^
i ^
j
%
0
+AxBz^
i ^
k
%
0
+
AyBx^
j ^
i
%
0
+AyBy^
j ^
j
%
1
+AyBz^
j ^
k
%
0
+
AzBx
^
k ^
i
%
0
+AzBy
^
k ^
j
%
0
+AzBz
^
k ^
k
%
1

25. C. M. Jenkins, Dept of Physi s, U. of South Alabama 25
This gives us the result
~
A ~
B = AxBx + AyBy + AzBz
Note any ve tor dotted into itself gives the square of the magnitude
of that ve tor:
~
A ~
A = AxAx + AyAy + AzAz
~
A ~
A = A2
x + A2
y + A2
z
~
A ~
A = j ~
Aj
2

26. C. M. Jenkins, Dept of Physi s, U. of South Alabama 26
Example: The Dot Produ t
Ve tor A: ~
A = 3:00^
i + 4:00 ^
j.
Ve tor B: ~
B = 6:00^
i + 2:00 ^
j.
First ne the dot produ t: ~
A ~
B:
~
A ~
B =
3:00^
i + 4:00 ^
j
6:00^
i + 2:00 ^
j
~
A ~
B = (3:00)(6:00) + (4:00)(2:00)
~
A ~
B = 26:00
Find the angle between ve tors ~
A and ~
B.
~
A ~
B = j ~
Ajj ~
Bj os()
So rst nd the magnitudes of ve tors ~
A and ~
B.
First, ve tor A:
j ~
Aj =
q
A2
x + A2
y
j ~
Aj =
p
(3:00)2 + (4:00)2
j ~
Aj = 5:00
Next, ve tor B:
j ~
Bj =
q
B2
x + B2
y
j ~
Bj =
p
(6:00)2 + (2:00)2
j ~
Bj =
p
40:00

27. C. M. Jenkins, Dept of Physi s, U. of South Alabama 27
Using the de nition of the dot produ t:
~
A ~
B = j ~
Ajj ~
Bj os()
26:00 = (5:00)
p
40:00 os()
26:00
5:00
p
40:00
= os()
= os 1
(0:8222)
= 34:47o

28. C. M. Jenkins, Dept of Physi s, U. of South Alabama 28
Ve tor or Cross Produ t
The ve tor or ross produ t is a ve tor oper-
ation where two ve tors are multiplied and a
ve tor results.
~
C = ~
A ~
B
This ve tor is perpendi ular to the plane de ned by the two ve -
tors that are multiplied.
The magnitude of the ross produ t is de ned as:
j~
A ~
Bj = j~
Ajj~
Bjsin()
Where is the angle between the two ve tors.
The dire tion of the resultant ve tor ~
C is determined by the right hand
rule.
Use your right hand.... Take the ngers and
point them in the dire tion of the rst ve tor
in the ross produ t. Orient your hand so that
the se ond ve tor points out of you palm. The
thumb points in the dire tion of the resultant
ve tor ~
C.
The ross produ t does not ommute: ~
A ~
B = ~
B ~
A.
This may be proven by use of of the right hand rule.

29. C. M. Jenkins, Dept of Physi s, U. of South Alabama 29
The magnitude of the ross produ t between the same unit ve tor is:
j^
i ^
ij = j^
ijj^
ijsin(0)
j^
i^
ij = (1)(1)(0)
j^
i ^
ij = 0
so:
^
i^
i = 0 ^
j ^
j = 0 ^
k ^
k = 0
Note, that this means: ~
A ~
A = 0 .
The ross produ t between two di erent unit
ve tor is:
j^
i^
jj = j^
ijj^
jjsin(90)
j^
i ^
jj = (1)(1)(1)
j^
i ^
jj = 1
The dire tion is determined by the right hand rule.
For this example, the dire tion is ^
k.
From this example and the right hand rule, we an dedu e the following
relationships:
^
i^
j = ^
k ^
k ^
i = ^
j ^
j ^
k = ^
i
Suppose we have two ve tors: ~
A = Ax^
i + Ay^
j + Az
^
k and ~
B = Bx^
i +
By^
j + Bz
^
k
And we want to take the ross produ t of these two ve tors:

30. C. M. Jenkins, Dept of Physi s, U. of South Alabama 30
~
C = ~
A ~
B
~
A ~
B =
Ax^
i+ Ay^
j + Az
^
k
Bx^
i + By^
j + Bz
^
k
Just multiply out as a polynomial (be areful of the order of multipli-
ation) :
~
A ~
B = AxBx^
i^
i + AxBy^
i ^
j + AxBz^
i ^
k +
AyBx^
j ^
i + AyBy^
j ^
j + AyBz^
j ^
k +
AzBx
^
k ^
i+ AzBy
^
k ^
j + AzBz
^
k ^
k
Apply the rules for taking the ross produ t between unit ve tors
~
A ~
B = AxBx^
i ^
i
%0
+ AxBy^
i^
j
%^
k
+ AxBz^
i ^
k
% ^
j
+
AyBx^
j ^
i
% ^
k
+ AyBy^
j ^
j
%0
+ AyBz^
j ^
k
%^
i
+
AzBx
^
k ^
i
%^
j
+ AzBy
^
k ^
j
% ^
i
+ AzBz
^
k ^
k
%0
so the result is:
~
A ~
B = [AyBz AzBy℄^
i+ [AzBx AxBz℄ ^
j + [AxBy AyBx℄ ^
k
The ross produ t between these two ve tors are more easily al ulated
by using the determinant of a 3 3 matrix:
~
A ~
B = Det
^
i ^
j ^
k
Ax Ay Az
Bx By Bz
=
[AyBz AzBy℄^
i+
[AxBz AzBx℄ ( ^
j)+
[AxBy AyBx℄ ^
k

31. C. M. Jenkins, Dept of Physi s, U. of South Alabama 31
Example: The Cross Produ t
Ve tor A: ~
A = 3:00^
i + 4:00^
j + 0:00^
k.
Ve tor B: ~
B = 6:00^
i + 2:00^
j + 0:00^
k.
A) First nd the ross produ t: ~
A ~
B:
~
A ~
B = Det
^
i ^
j ^
k
3:00 4:00 0:00
6:00 2:00 0:00
=
[(4:00)(0:00) (0:00)(2:00)℄^
i+
[(3:00)(0:00) (0:00)(3:00)℄( ^
j)+
[(3:00)(2:00) (4:00)(6:00)℄^
k
So:
~
A ~
B = 0:00^
i + 0:00^
j 18:00^
k
B) Next nd the angles between the two ve tors:
j~
A ~
Bj = j~
Ajj~
Bjsin()
So rst nd the magnitudes of ve tors ~
A and ~
B.
First, ve tor A:
j~
Aj =
q
A2
x + A2
y
j~
Aj =
p(3:00)2 + (4:00)2
j~
Aj = 5:00
Next, ve tor B:
j~
Bj =
q
B2
x + B2
y
j~
Bj =
p(6:00)2 + (2:00)2
j~
Bj =
p
40:00
Using the de nition of the dot produ t:

32. C. M. Jenkins, Dept of Physi s, U. of South Alabama 32
B) Next nd the angles between the two ve tors:
j~
A ~
Bj = j~
Ajj~
Bjsin()
18:00 = 5:00
p
40:00sin()
18:00
5:00
p
40:00
= sin()
= sin 1
(0:5692)
= 34:47o