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CHAPTER 1.4: SCALAR AND VECTORCHAPTER 1.4: SCALAR AND VECTOR
SCALAR AND VECTOR
Scalars are quantities which have magnitude without direction
Examples of scalars
• temperature• mass• kinetic energy
• time• amount• density• charge
Scalars
Vector
A vector is a quantity that has both magnitude (size) and direction
it is represented by an arrow whereby– the length of the arrow is the magnitude, and– the arrow itself indicates the direction
The symbol for a vector is a letter with an arrow over it
A
Example
Two ways to specify a vector
It is either given by• a magnitude A, and• a direction
Or it is given in the x and y components as
• Ax
• Ay
y
x
A
A
Ay
x
Ax
Ay
y
x
AAx
AyA
Ax = A cos
Ay = A sin
│A │ =√ ( Ax2
+ Ay2
)
The magnitude (length) of A is found by using the Pythagorean Theorem
The length of a vector clearly does not depend on its direction.
y
x
AAx
AyA
The direction of A can be stated as
tan = Ay / Ax
=tan-1(Ay / Ax)
Some Properties of Vectors
Equality of Two Vectors
Two vectors A and B may be defined to be equal if they have the same magnitude and point in the same directions. i.e. A = B
A BA
A
B
B
Negative of a Vector
The negative of vector A is defined as giving the vector sum of zero value when added to A . That is, A + (- A) = 0. The vector A and –A have the same magnitude but are in opposite directions.
A
-A
Scalar Multiplication
The multiplication of a vector Aby a scalar
- will result in a vector B
B = A- whereby the magnitude is changed but not the direction
• Do flip the direction if is negative
B = A
If = 0, therefore B = A = 0, which is also known as a zero vector
(A) = A = (A)
(+)A = A + A
Example
The addition of two vectors A and B
- will result in a third vector C called the resultant
C = A + B
A
BC
Geometrically (triangle method of addition)
• put the tail-end of B at the top-end of A• C connects the tail-end of A to the top-end of B
We can arrange the vectors as we like, as long as we maintain their length and direction
Vector Addition
Example
More than two vectors?
x1
x5
x4
x3
x2
xi
xi = x1 + x2 + x3 + x4 + x5
Example
Vector Subtraction
Equivalent to adding the negative vector
A
-BA - B
B
A BC =
A + (-B)C =
Example
Rules of Vector Addition
commutative
A + B = B + A
A
B
A + BB
A A + B
associative
(A + B) + C = A + (B + C)
B
CA
B
CA A + B
(A + B) + CA + (B + C)
B + C
distributive
m(A + B) = mA + mB
A
B
A + B mA
mB
m(A + B)
Parallelogram method of addition (tailtotail)
A
B
A + B
The magnitude of the resultant depends on the relative directions of the vectors
a vector whose magnitude is 1 and dimensionless
the magnitude of each unit vector equals a unity; that is, │ │= │ │= │ │= 1
i a unit vector pointing in the x direction
j a unit vector pointing in the y direction
k a unit vector pointing in the z direction
and defined as
Unit Vectors
k
j
i
Useful examples for the Cartesian unit vectors [ i, j, ki, j, k ] - they point in the direction of the x, y and z axes respectively
x
y
z
ii
jj
kk
Component of a Vector in 2-D
vector A can be resolved into two components
Ax and Ay
x- axis
y- axis
Ay
Ax
A
θ
A = Ax + Ay
The component of A are
│Ax│ = Ax = A cos θ
│Ay│ = Ay = A sin θ
The magnitude of A
A = √Ax2 + Ay
2
tan = Ay / Ax
=tan-1(Ay / Ax)
The direction of A
x- axis
y- axis
Ay
Ax
A
θ
ExampleExample
The unit vector notation for the vector A is written
A = Axi + Ayj
x- axis
y- axis
Ax
Ay
θ
A
i
j
ExampleExample
Component of a Vector in 3-D
vector A can be resolved into three components
Ax , Ay and Az
A
Ax
Ay
Az
z- axis
y- axis
x- axis
ij
k
A = Axi + Ayj + Azk
if
A = Axi + Ayj + Azk
B = Bxi + Byj + Bzk
A + B = C sum of the vectors A and B can then be obtained as vector C
C = (Axi + Ayj + Azk) + (Bxi + Byj + Bzk)
C = (Ax + Bx)i+ (Ay + By)j + (Az + Bz)kC = Cxi + Cyj + Czk Example
Dot product (scalar) of two vectors
The definition:
θ
B
AA · B = │A││B │cos θ
if θ = 900 (normal vectors) then the dot product is zero
Dot product (scalar product) properties:
if θ = 00 (parallel vectors) it gets its maximum
value of 1
and i · j = j · k = i · k = 0|A · B| = AB cos 90 = 0
|A · B| = AB cos 0 = 1 and i · j = j · k = i · k = 1
A + B = B + A
the dot product is commutative
Use the distributive law to evaluate the dot product if the components are known
A · B = (Axi + Ayj + Azk) · (Bxi + Byj + Bzk)
A. B = (AxBx) i.i + (AyBy) j.j + (AzBz) k.k
A . B = AxBx + AyBy + AzBz
Example
Cross product (vector) of two vectorsThe magnitude of the cross product given by
the vector product creates a new vector
this vector is normal to the plane defined by the
original vectors and its direction is found by using the
right hand rule
│C │= │A x B│ = │A││B │sin θ
θ
A
BC
if θ = 00 (parallel vectors) then the cross
product is zero
Cross product (vector product) properties:
if θ = 900 (normal vectors) it gets its maximum
value
and i x i = j x j = k x k = 0|A x B| = AB sin 0 = 0
|A x B| = AB sin 90 = 1 and i x i = j x j = k x k = 1
the relationship between vectors i , j and k can
be described as
i x j = - j x i = k
j x k = - k x j = i
k x i = - i x k = j
Example
Measurement and Error
THE END
Vectors are represented by an arrow
A- B
BA
A
θ
Conceptual Example
If B is added to A, under what condition does the resultant vector A + B have the magnitude equal to A + B ?
Under what conditions is the resultant vector equal to zero?
*
Example (1Dimension)
x1 = 5
x1 x2
x2 = 3
x = x2 - x1 = 2
x1 + x2
x1
x2
x = x2 - x1
x1 + x2 = 8
MORE EXAMPLEMORE EXAMPLE
Example 1 (2 Dimension)
If the magnitude of vector A and B are equal to 2 cm and 3 cm respectively , determine the magnitude and direction of the resultant vector, C for
B
Aa) A + B
b) 2A + B
SOLUTIONSOLUTION
Solution
a) |A + B| = √A2 + B2
= √22 + 32
= 3.6 cm
The vector direction
tan θ = B / A
θ = 56.3
b) |2A + B| = √(2A)2 + B2
= √42 + 32
= 5.0 cm
The vector direction
tan θ = B / 2A
θ = 36.9
MORE EXAMPLEMORE EXAMPLE
Example 2 (A Vacation Trip)
A car travels 20.0 km due north and then 35.0 km in a direction 600 west of north. Find the magnitude and direction of the car’s resultant displacement.
SOLUTIONSOLUTION
Solution
The magnitude of R can be obtained using the law of cosines as in figure
Since θ =1800 – 600 = 1200 and
C2 = A2 + B2 – 2AB cos θ, we find that
C = √A2 + B2 – 2AB cos θ
C = √202 + 352 – 2(20)(35) cos 1200
C = 48.2 km
C
A
B60
θβ
ContinueContinue
The direction of C measured from the northerly direction can be obtained from the sines law
CB
sinsin
629.0120sin2.48
0.35sinsin 0
C
B
β = 38.90
Therefore, the resultant displacement of the car is 48.2 km in direction 38.90 west of north
If one component of a vector is not zero, can its magnitude be zero? Explain.
*Conceptual Example
MORE EXAMPLEMORE EXAMPLE
Conceptual Example
If A + B = 0, what can you say about the components of the two vectors?
*
Example 1
Find the sum of two vectors A and B lying in the xy plane and given by
A = 2.0i + 2.0j and B = 2.0i – 4.0j
SOLUTIONSOLUTION
Solution
Comparing the above expression for A with the general relation A = Axi + Ayj , we see that Ax= 2.0 and Ay= 2.0. Likewise, Bx= 2.0, and By= -4.0 Therefore, the resultant vector C is obtained by using EquationC = A + B + (2.0 + 2.0)i + (2.0 - 4.0)j = 4.0i -2.0j
or Cx = 4.0 Cy = -2.0
The magnitude of C given by equation
C = √Cx2 + Cy
2 = √20 = 4.5
*
Find the angle θ that C makes with the positive x axis
Exercise
Example
A particle undergoes three consecutive displacements d1 = (1.5i + 3.0j – 1.2k) cm,
d2 = (2.3i – 1.4j – 3.6k) cm d3 = (-1.3i + 1.5j) cm. Find the component and its magnitude.
Solution
R = d1 + d2 + d3
= (1.5 + 2.3 – 1.3)i + (3.0 – 1.4 + 1.5)j + (-1.2 – 3.6 + 0)k
= (2.5i + 3.1j – 4.8k) cm
That is, the resultant displacement has component
Rx = 2.5 cm Ry = 3.1 cm and Rz = -4.8 cm
Its magnitude is
R = √ Rx2 + Ry
2 + Rz2
= 6. 2 cm
Example - 2D [headtotail]
x1
x2
x1 + x2
(1, 0)
(2, 2)
x1 + x2 = (1, 0) + (2, 2)= (3, 2)
Example - 2D [tailtotail]
x1 - x2?
(x2)
x1
x1 + x2x2
(1, 0)
(2, 2)
x1 + x2 = (1, 0) + (2, 2)= (3, 2)
Example of 2D (subtraction)
(1, 0)
(2, 2)
x1
x2
x1 + x2
Example -2D for subtraction
x1
-x2x1 - x2
(1, 0)
(2, 2)
x1 - x2 = (1, 0) - (2, 2)= (-1, -2)
x1 - x2 = x1 + (-x2)
Not given the components?
1 m
22 m45o
X1 = (1, 0)X2 = (x2E, x2N) = (22cos(45o), 22sin(45o)) = (2, 2)
x1
-x2x1 - x2
22 m
1 m
45o
Cosine rule:a2=b2 + c2 - 2bccosA = 1 + 8 - 22(1/ 2)a = 5 m