1.1
Basic Vector Notation:
Here, the resultant is A
and the components are: A x and A y such that:
A
= A x + A y where:
A x = A x i and
A y =
A y j
Where i and j are unit vectors, and:
A
x
= A cos q , A y
= A sin q , q
= tan -1 (A y
/ A x )
A
x
and A y are scalar quantities called the components
of the vector A. Again, the
magnitude of A can be written:
‖ A ‖ = Ö (A x 2
+ A y 2 )
It is also possible to have more complex vector configurations
which require more analysis, such as shown in the diagram below:
In the figure above the task is to identify the components for each of
the resultants, A, B and C.
Given the units are provided to scale, e.g. the resultant A has two
equal components of 2 units each, this is possible.
Then for resultant A:
A = A x i + A y
j
A
x
= - 2, A y
= 2 ,
‖ A ‖ =
Ö (A
x
2 + A y 2 ) = Ö (-2) 2 + (2) 2 )
‖ A ‖ =
Ö (8) = 2 Ö 2
q
= tan -1 (A y
/ A x ) = tan -1 (2
/ -2 ) = tan -1 (-1)
q = - 45
degrees
Similarly:
B = B x i + B y
j
B
x
= - 2, B y = -3 ,
‖ B ‖ =
Ö (B
x
2 + B y 2 ) = Ö (-2) 2 + (-3) 2 )
‖ B ‖ =
Ö (13)
q = tan -1 (B y
/ B x ) = tan -1 (-3 / -2 )
= tan -1 (3/2)
q = 56.
3 degrees
Finally,
we write for resultant C:
‖ C ‖ = Ö (C x 2
+ C y 2 )
= Ö (4) 2 + (3) 2 )
‖ C ‖ = Ö (25) = 5
units
q = tan -1 (C y
/ C x ) = tan -1 (3 /4 )
= tan -1 (3/4)
q = 36.8
degrees
More generally, we can have a situation with several vectors to
be added which need not occur in the same plane, e.g.:
So in this case: T = A + B + C
Whence:
Tx = A x + B x + C x
Ty = A y + B y + C y
Tz = A z + B z + C z
Then once the components
of T have been found the
magnitude of the vector may be obtained:
‖ T ‖ = Ö (T x 2
+ T y 2 + T z 2 )
By definition the components of such a vector are numbers which
multiply the unit vectors . So if the
components of T are T x,
T y, and T z then:
T = T x i
+ T y j
+ T z k
We can also have the situation for which T = A + B, but for which the two vectors, A and B, are
not necessarily in the same plane. Hence each vector can be described in 3
dimensions using unit vectors. Then one can write:
T = (A x i
+ A y j
+ A z k ) + (B x i
+ B y j
+ B z k )
And since vector addition is commutative, the preceding can also
be written:
T = (A x + B x ) i
+ (A y + B y ) j + (A
z + B z ) k
We note the length of any vector in 3 dimensions, say referenced
along the diagonal of a three- dimensional box, say:
T = a i
+ b j + c k
Is easily determined by applying the Pythagorean theorem twice,
once to the diagonal of a face the rectangular box, then to the overall
diagonal, e.g.
In summary:
‖ a i
+ b j + c k ‖ =
Ö (a
2
+ b 2 + c
2
)
Direction cosines:
If the diagonal
vector T
= a i
+ b j + c k makes angles a, b and g respectively, i.e. with the x-, y- and z- axes, then: cos a , cos b and cos g are
called the direction cosines, where:
cos
a = a / Ö (a 2 + b 2 + c 2 )
cos
b = b / Ö (a 2 + b 2 + c 2 )
cos
g = b / Ö (a 2 + b 2 + c 2 )
It can also be shown
that:
cos
2 a +
cos 2 b + cos 2
g = 1
Problems:
1) For two vectors A = 3i – 4j
and B = -i – j, find the magnitude and direction of: A + B, B – A.
2) Using the method of components, find the vector sum
of the two vectors A and B if A makes an angle of 45
degrees with the x-axis and has a length of 6 units, and vector B makes
an angle f 135 degrees with the x-axis and has a length of 8 units.
3) For the 3D rectangular box shown earlier, for which: T = a i + b j + c k, assuming sides a = 6, b = 3, and c = 2 :
a)Find the direction
cosines.
b) Show that: cos 2
a +
cos 2 b + cos 2
g = 1
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