vectors

1. Vectors

2. Vectors
A vector is a numerical measurement in a specified
direction (usually in 2 or higher dimensional spaces).

3. Vectors
A vector is a numerical measurement in a specified
direction (usually in 2 or higher dimensional spaces).
Geometrically we use arrows to represent vectors.

4. Vectors
A vector is a numerical measurement in a specified
direction (usually in 2 or higher dimensional spaces).
Geometrically we use arrows to represent vectors.
Many real world problems can be modeled using
vector(arrow)-diagrams, for example,
the wind–map in a weather report.
A wind–map

5. Vectors
A vector is a numerical measurement in a specified
direction (usually in 2 or higher dimensional spaces).
Geometrically we use arrows to represent vectors.
Many real world problems can be modeled using
vector(arrow)-diagrams, for example,
the wind–map in a weather report.
The arrows in the map point
to the directions of the wind and
the lengths of the arrows indicate
the wind speeds.
A wind–map

6. Vectors
A vector is a numerical measurement in a specified
direction (usually in 2 or higher dimensional spaces).
Geometrically we use arrows to represent vectors.
Many real world problems can be modeled using
vector(arrow)-diagrams, for example,
the wind–map in a weather report.
The arrows in the map point
to the directions of the wind and
the lengths of the arrows indicate
the wind speeds.
These arrows or vectors may be
defined and manipulated,
vector-equations may be set up and solved with the
relevant vector solutions.
A wind–map

7. We use arrows to represent vectors, the arrows give
the directions and the lengths of the arrows are the
numerical measurements.
Vectors

8. We use arrows to represent vectors, the arrows give
the directions and the lengths of the arrows are the
numerical measurements.
Vectors
We use the symbols u, v, and w (or u, v, and w)
to represent vectors in mathematics.

9. We use arrows to represent vectors, the arrows give
the directions and the lengths of the arrows are the
numerical measurements.
The length of a vector is
called the magnitude of u
or the absolute value of u
and is denoted as |u|.
Vectors
We use the symbols u, v, and w (or u, v, and w)
to represent vectors in mathematics.

10. We use arrows to represent vectors, the arrows give
the directions and the lengths of the arrows are the
numerical measurements.
The length of a vector is
called the magnitude of u
or the absolute value of u
and is denoted as |u|.
Vectors
We use the symbols u, v, and w (or u, v, and w)
to represent vectors in mathematics.
We also write a vector as AB
if A is the base and B is the tip
of the vector.

11. We use arrows to represent vectors, the arrows give
the directions and the lengths of the arrows are the
numerical measurements.
The length of a vector is
called the magnitude of u
or the absolute value of u
and is denoted as |u|.
Vectors
We use the symbols u, v, and w (or u, v, and w)
to represent vectors in mathematics.
We also write a vector as AB
if A is the base and B is the tip
of the vector.
In physics, a force applied in a specified direction of a
given strength is viewed as a vector.

13. Vector Arithmetic

14. Vector Arithmetic
Equality of Vectors
Two vectors u and v with
the same length and same
direction are equal, i.e. u = v.

15. Vector Arithmetic
Equality of Vectors
Two vectors u and v with
the same length and same
direction are equal, i.e. u = v.
Scalar Multiplication
Given a number λ and a vector v,
λv is the extension/compression
of the vector v by a factor λ. 0.

16. v
Vector Arithmetic
Equality of Vectors
Two vectors u and v with
the same length and same
direction are equal, i.e. u = v.
Scalar Multiplication
Given a number λ and a vector v,
λv is the extension/compression
of the vector v by a factor λ.

17. v
2v
Vector Arithmetic
Equality of Vectors
Two vectors u and v with
the same length and same
direction are equal, i.e. u = v.
Scalar Multiplication
Given a number λ and a vector v,
λv is the extension/compression
of the vector v by a factor λ.

18. v
2v
v
2
1
Vector Arithmetic
Equality of Vectors
Two vectors u and v with
the same length and same
direction are equal, i.e. u = v.
Scalar Multiplication
Given a number λ and a vector v,
λv is the extension/compression
of the vector v by a factor λ.

19. v
2v
v
2
1
Vector Arithmetic
Equality of Vectors
Two vectors u and v with
the same length and same
direction are equal, i.e. u = v.
Scalar Multiplication
Given a number λ and a vector v,
λv is the extension/compression
of the vector v by a factor λ.
If IλI > 1, the vector is elongated.
If IλI < 1, the vector is shortened.

20. v
2v
v
2
1
-2v
Vector Arithmetic
Equality of Vectors
Two vectors u and v with
the same length and same
direction are equal, i.e. u = v.
Scalar Multiplication
Given a number λ and a vector v,
λv is the extension/compression
of the vector v by a factor λ.
If IλI > 1, the vector is elongated.
If IλI < 1, the vector is shortened.
If λ < 0, λv points in the opposite
direction of v.

21. Vector Arithmetic
Equality of Vectors
Two vectors u and v with
the same length and same
direction are equal, i.e. u = v.
v
2v
v
2
1
Scalar Multiplication
Given a number λ and a vector v,
λv is the extension/compression
of the vector v by a factor λ.
If IλI > 1, the vector is elongated.
If IλI < 1, the vector is shortened.
If λ < 0, λv points in the opposite
direction of v.
0v = 0, the zero vector of length 0.
-2v

22. The Vector Parallelogram Rules
Given two vectors u and v at the same base–point,
they form a parallelogram.
u
v
Vector Arithmetic
The
Parallelogram
Rules

23. The Vector Parallelogram Rules
Given two vectors u and v at the same base–point,
they form a parallelogram.
u
v
Vector Arithmetic
The
Parallelogram
Rules
Algebraically, the three vector-
arithmetic operations
u + v, u – v, v – u,
correspond to
the diagonals of the
parallelogram:

24. The Vector Parallelogram Rules
Given two vectors u and v at the same base–point,
they form a parallelogram.
u
v
u + v
Algebraically, the three vector-
arithmetic operations
u + v, u – v, v – u,
correspond to
the diagonals of the
parallelogram:
u + v is the diagonal from
the base to the opposite corner.
Vector Arithmetic
The
Parallelogram
Rules

25. The Vector Parallelogram Rules
Given two vectors u and v at the same base–point,
they form a parallelogram.
u
v
u + v
Algebraically, the three vector-
arithmetic operations
u + v, u – v, v – u,
correspond to
the diagonals of the
parallelogram:
u + v is the diagonal from
the base to the opposite corner.
u – v, v – u are the two diagonals
from the tip of one vector to
the other.
u – v
v – u
Vector Arithmetic
The
Parallelogram
Rules

26. The Base to Tip Rule for u + v
Vector Arithmetic

27. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
Vector Arithmetic
u
v
A

28. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v.
Vector Arithmetic
u
v
A

29. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v.
Vector Arithmetic
u
v
A

30. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v.
Vector Arithmetic
u
u + v
v
A

31. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v. We use
the Cosine Law and the angle
180o – A to compute u + v.
Vector Arithmetic
u
u + v
v
A
180o – A

32. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v. We use
the Cosine Law and the angle
180o – A to compute u + v.
Example A. Given l v l = 8 and
l v l = 5, and the angle A is 102o,
draw.
a. Find | u + v |
Vector Arithmetic
u
u + v
v
A
180o – A

33. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v. We use
the Cosine Law and the angle
180o – A to compute u + v.
Example A. Given l v l = 8 and
l v l = 5, and the angle A is 102o,
draw.
a. Find | u + v |
Vector Arithmetic
u
u + v
v
A
180o – A
102o
lul=8
lvl=5

34. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v. We use
the Cosine Law and the angle
180o – A to compute u + v.
Vector Arithmetic
u
u + v
v
A
180o – A
Example A. Given l v l = 8 and
l v l = 5, and the angle A is 102o,
draw.
a. Find | u + v |
102o
lul=8
lvl=5
u + v

35. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v. We use
the Cosine Law and the angle
180o – A to compute u + v.
Vector Arithmetic
u
u + v
v
u + v
A
180o – A
102o
lul=8
lvl=5
180o – 102o = 78o
180 – 102 = 78
Example A. Given l v l = 8 and
l v l = 5, and the angle A is 102o,
draw.
a. Find | u + v |

36. The Base to Tip Rule for u + v
Given two vectors u, v and the angle A between them,
the vector u + v may be obtained by placing the base
of one vector at the tip of the other, then the 3rd side
of the triangle is u + v. We use
the Cosine Law and the angle
180o – A to compute u + v.
Vector Arithmetic
u
u + v
v
u + v
A
180o – A
102o
lul=8
lvl=5
180o – 102o = 78o
180 – 102 = 78 so by the cosine law
|u + v|2 = 82 + 52 – [ 2(8)(5)cos(78) ]
or that |u + v| ≈ 8.51.
Example A. Given l v l = 8 and
l v l = 5, and the angle A is 102o,
draw.
a. Find | u + v |

37. b. Find the angle between u + v and v.
Vector Arithmetic
u + v
lvl=5
78o

38. b. Find the angle between u + v and v.
Vector Arithmetic
u + v
lvl=5
78oWe want the angle B as shown.
B

39. b. Find the angle between u + v and v.
Vector Arithmetic
u + v lul=8
lvl=5
78oWe want the angle B as shown.
B 78o

40. b. Find the angle between u + v and v.
Vector Arithmetic
u + v lul=8
lvl=5
78oWe want the angle B as shown.
B 78o
Using the Cosine law, we have
52 + 8.512 – 82
cos(B) =
2(5)(8.51)
so B ≈ 66.9o

41. b. Find the angle between u + v and v.
Vector Arithmetic
u + v lul=8
lvl=5
78oWe want the angle B as shown.
B
Using the Cosine law, we have
52 + 8.512 – 82
cos(B) =
2(5)(8.51)
so B ≈ 66.9o
The Tip to Tip Rule for Vector Subtraction
78o

42. b. Find the angle between u + v and v.
Vector Arithmetic
u + v lul=8
lvl=5
78oWe want the angle B as shown.
B
Using the Cosine law, we have
52 + 8.512 – 82
cos(B) =
2(5)(8.51)
so B ≈ 66.9o
The Tip to Tip Rule for Vector Subtraction
Given vectors u and v, then u – v = u + (–v) is the
vector from the tip of v to tip
of the u.
u
v
78o

43. b. Find the angle between u + v and v.
Vector Arithmetic
u + v lul=8
lvl=5
78oWe want the angle B as shown.
B
Using the Cosine law, we have
52 + 8.512 – 82
cos(B) =
2(5)(8.51)
so B ≈ 66.9o
The Tip to Tip Rule for Vector Subtraction
Given vectors u and v, then u – v = u + (–v) is the
vector from the tip of v to tip
of the u.
u
v
– v
78o

44. b. Find the angle between u + v and v.
Vector Arithmetic
u + v lul=8
lvl=5
78oWe want the angle B as shown.
B
Using the Cosine law, we have
52 + 8.512 – 82
cos(B) =
2(5)(8.51)
so B ≈ 66.9o
The Tip to Tip Rule for Vector Subtraction
Given vectors u and v, then u – v = u + (–v) is the
vector from the tip of v to tip
of the u.
u
v
u – v
– v
u
78o

45. b. Find the angle between u + v and v.
Vector Arithmetic
u + v lul=8
lvl=5
78oWe want the angle B as shown.
B
Using the Cosine law, we have
52 + 8.512 – 82
cos(B) =
2(5)(8.51)
so B ≈ 66.9o
u
v
u – v
Vector Subtraction
The Tip to Tip Rule for Vector Subtraction
Given vectors u and v, then u – v = u + (–v) is the
vector from the tip of v to tip
of the u.
u
v
u – v
– v
u
or
u – v
78o

46. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5

47. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – v

48. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law,

49. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.

50. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.
It is easy visualize the sum of three or
more to using the base-to-tip rule to
vectors.

51. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.
It is easy visualize the sum of three or
more to using the base-to-tip rule to
vectors.

52. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.
It is easy visualize the sum of three or
more to using the base-to-tip rule to
vectors.

53. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.
the sum
It is easy visualize the sum of three or
more to using the base-to-tip rule to
vectors.

54. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.
the sum
It is easy visualize the sum of three or
more to using the base-to-tip rule to
vectors. Note that the order of the
addition does not matter.

55. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.
the sum
It is easy visualize the sum of three or
more to using the base-to-tip rule to
vectors. Note that the order of the
addition does not matter. Hence the
addition is associative and
commutative.

56. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.
the sum
It is easy visualize the sum of three or
more to using the base-to-tip rule to
vectors. Note that the order of the
addition does not matter. Hence the
addition is associative and
commutative. So for vector u, v and w
(u + v) + w = u + (v + w) and
u + v = v + u.

57. Vector Arithmetic
Example A. (con´t) c. Find | u – v |
102o
lul=8
lvl=5
u – vUsing the Cosine law, we have
|u – v|2 = 82 + 52 – [ 2(8)(5)cos(102) ]
or |u – v| ≈ 10.3.
the sum
It is easy visualize the sum of three or
more to using the base-to-tip rule to
vectors. Note that the order of the
addition does not matter. Hence the
addition is associative and
commutative. So for vector u, v and w
(u + v) + w = u + (v + w) and
u + v = v + u. If α and β are scalars,
then the distributive laws hold,
so α(u + v) = αv + αu and (α + β )u = αu + βu

58. Vector Arithmetic
In physics, forces are represent is by vectors.
So if u and v represent two forces,
then u + v is called the resultant of u and V.
u + vu
v
In other words, u + v is the
combined force with strength of
| u + v | in the direction
of the diagonal of the parallelogram.
u + v
102o
lul=8
lvl=5
Hence in the above example A,
if u and v are two forces pulling on
an object at the same time,
then l u + v l is amount of force
applies to the object and u + v
gives the direction the object would
travel.

59. A vector v placed in the
coordinate system with the
base at the origin (0, 0) is said
to be in the standard position.
Vectors in a Coordinate System

60. A vector v placed in the
coordinate system with the
base at the origin (0, 0) is said
to be in the standard position.
If the tip of the vector v in the
standard position is (a, b),
we write v as <a, b>.
Vectors in a Coordinate System

61. A vector v placed in the
coordinate system with the
base at the origin (0, 0) is said
to be in the standard position.
If the tip of the vector v in the
standard position is (a, b),
we write v as <a, b>.
Vectors in a Coordinate System

62. A vector v placed in the
coordinate system with the
base at the origin (0, 0) is said
to be in the standard position.
If the tip of the vector v in the
standard position is (a, b),
we write v as <a, b>.
The zero vector is 0 =<0, 0>
Vectors in a Coordinate System

63. A vector v placed in the
coordinate system with the
base at the origin (0, 0) is said
to be in the standard position.
If the tip of the vector v in the
standard position is (a, b),
we write v as <a, b>.
The zero vector is 0 =<0, 0>
A vector may also specified by
it's starting point B (the base
point) and it's tip T, it is denoted
as BT.
Vectors in a Coordinate System

64. Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
Vectors in a Coordinate System

65. Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
Vectors in a Coordinate System

66. Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
Vectors in a Coordinate System
B=(3, -1)
T=(-2, 4)
BT

67. Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
BT=T – B
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
Vectors in a Coordinate System
B=(3, -1)
T=(-2, 4)
BT

68. Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
BT=T – B = (-2, 4) – (3, -1)
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
Vectors in a Coordinate System
B=(3, -1)
T=(-2, 4)
BT

69. Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
BT=T – B = (-2, 4) – (3, -1)
=< -2 – 3, 4 – (-1) >
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
Vectors in a Coordinate System
B=(3, -1)
T=(-2, 4)
BT

70. Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
BT=T – B = (-2, 4) – (3, -1)
=< -2 – 3, 4 – (-1) >
=< -5, 5 > in the standard form.
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
Vectors in a Coordinate System
B=(3, -1)
T=(-2, 4)
BT

71. Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
BT=T – B = (-2, 4) – (3, -1)
=< -2 – 3, 4 – (-1) >
=< -5, 5 > in the standard form.
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
BT in the standard
form based at (0, 0)
Vectors in a Coordinate System
B=(3, -1)
T=(-2, 4)
< -5, 5 >
BT

72. Vertical format:
( -2, 4 )
– ( 3, -1 )
Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
BT=T – B = (-2, 4) – (3, -1)
=< -2 – 3, 4 – (-1) >
=< -5, 5 > in the standard form.
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
BT in the standard
form based at (0, 0)
Vectors in a Coordinate System
B=(3, -1)
T=(-2, 4)
< -5, 5 >
BT

73. Vertical format:
( -2, 4 )
– ( 3, -1 )
<-5, 5>
Example B. Let B=(3, -1), and
T=(-2, 4). Find the vector BT
in the standard form.
BT=T – B = (-2, 4) – (3, -1)
=< -2 – 3, 4 – (-1) >
=< -5, 5 > in the standard form.
Given a vector v with base at
the point B=(a, b) and the tip
at the point T=(c, d), then BT
in the standard form is v where
v=BT=T – B= <c – a, d – b>
BT in the standard
form based at (0, 0)
Vectors in a Coordinate System
B=(3, -1)
T=(-2, 4)
< -5, 5 >
BT

74. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector

75. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector

76. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector
Example C. Find |BT| if
B = (1, -3) and T = (5, 1).

77. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector
Example C. Find |BT| if
B = (1, -3) and T = (5, 1).
We need v = BT = T – B in
the standard form.

78. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector
Example C. Find |BT| if
B = (1, -3) and T = (5, 1).
We need v = BT = T – B in
the standard form. Vertical format:
( 5, 1 )
– ( 1, -3)

79. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector
Example C. Find |BT| if
B = (1, -3) and T = (5, 1).
We need v = BT = T – B in
the standard form. Vertical format:
( 5, 1 )
– ( 1, -3)
v = ( 4, 4)

80. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector
Example C. Find |BT| if
B = (1, -3) and T = (5, 1).
We need v = BT = T – B in
the standard form. Vertical format:
( 5, 1 )
– ( 1, -3)
v = ( 4, 4)
So |v| = |<4, 4>|

81. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector
Example C. Find |BT| if
B = (1, -3) and T = (5, 1).
We need v = BT = T – B in
the standard form. Vertical format:
( 5, 1 )
– ( 1, -3)
v = ( 4, 4)
So |v| = |<4, 4>|
= 42 + 42
= 32

82. The magnitude or the
length of the vector
v = <a, b> is given by
|v| = a2 + b2
Magnitude of a Vector
Example C. Find |BT| if
B = (1, -3) and T = (5, 1).
We need v = BT = T – B in
the standard form. Vertical format:
( 5, 1 )
– ( 1, -3)
v = ( 4, 4)
So |v| = |<4, 4>|
= 42 + 42
= 32 = 42  5.64

83. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Algebra of Standard Vectors

84. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Scalar-multiplication:
Let u=<a, b> and λ be a
real number, then
λu= λ<a, b> = <λa, λb>.
Algebra of Standard Vectors

85. Example D.
If u = <2, -1> then 2u = 2<2, -1> = <4, -2>
Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Scalar-multiplication:
Let u=<a, b> and λ be a
real number, then
λu= λ<a, b> = <λa, λb>.
Algebra of Standard Vectors

86. Example D.
If u = <2, -1> then 2u = 2<2, -1> = <4, -2>
Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Scalar-multiplication:
Let u=<a, b> and λ be a
real number, then
λu= λ<a, b> = <λa, λb>.
Algebra of Standard Vectors

87. Example D.
If u = <2, -1> then 2u = 2<2, -1> = <4, -2>
(This corresponds to stretching u to twice its length.)
Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Scalar-multiplication:
Let u=<a, b> and λ be a
real number, then
λu= λ<a, b> = <λa, λb>.
Algebra of Standard Vectors

88. Example D.
If u = <2, -1> then 2u = 2<2, -1> = <4, -2>
(This corresponds to stretching u to twice its length.)
Algebra of Standard Vectors
Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Scalar-multiplication:
Let u=<a, b> and λ be a
real number, then
λu= λ<a, b> = <λa, λb>.

89. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Vector Addition:
Let u=<a, b>, v=<c, d>
then u + v = <a+c, b+d>.
Algebra of Standard Vectors

90. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Vector Addition:
Let u=<a, b>, v=<c, d>
then u + v = <a+c, b+d>.
Algebra of Standard Vectors
Example E. Let u = <2, -1>, v = <-3, -2>,
a. u + v =
v=<-3, -2>
u=<2, -1>

91. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Vector Addition:
Let u=<a, b>, v=<c, d>
then u + v = <a+c, b+d>.
Algebra of Standard Vectors
Example E. Let u = <2, -1>, v = <-3, -2>,
a. u + v = <2+(-3), -1+(-2)> = <-1,-3>
v=<-3, -2>
u=<2, -1>

92. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Vector Addition:
Let u=<a, b>, v=<c, d>
then u + v = <a+c, b+d>.
Algebra of Standard Vectors
Example E. Let u = <2, -1>, v = <-3, -2>,
a. u + v = <2+(-3), -1+(-2)> = <-1,-3>
v=<-3, -2>
u + v =<-1,-3>
u=<2, -1>

93. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Vector Addition:
Let u=<a, b>, v=<c, d>
then u + v = <a+c, b+d>.
We define u – v = u + (- v)
Algebra of Standard Vectors
Example E. Let u = <2, -1>, v = <-3, -2>,
a. u + v = <2+(-3), -1+(-2)> = <-1,-3>
v=<-3, -2>
u + v =<-1,-3>
u=<2, -1>

94. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Vector Addition:
Let u=<a, b>, v=<c, d>
then u + v = <a+c, b+d>.
We define u – v = u + (- v)
Algebra of Standard Vectors
Example E. Let u = <2, -1>, v = <-3, -2>,
a. u + v = <2+(-3), -1+(-2)> = <-1,-3>
v=<-3, -2>
b. u – v = <2 – (-3), -1 – (-2)> = <5, 1>
u + v =<-1,-3>
u=<2, -1>

95. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Vector Addition:
Let u=<a, b>, v=<c, d>
then u + v = <a+c, b+d>.
We define u – v = u + (- v)
Algebra of Standard Vectors
Example E. Let u = <2, -1>, v = <-3, -2>,
a. u + v = <2+(-3), -1+(-2)> = <-1,-3>
u=<2, -1>
v=<-3, -2>
u + v =<-1,-3>
b. u – v = <2 – (-3), -1 – (-2)> = <5, 1>
-v=< 3, 2>

96. Scalar-multiplication and
vector addition are done
coordinate-wise, i.e.
operations are done at
each coordinate.
Vector Addition:
Let u=<a, b>, v=<c, d>
then u + v = <a+c, b+d>.
We define u – v = u + (- v)
Algebra of Standard Vectors
Example E. Let u = <2, -1>, v = <-3, -2>,
a. u + v = <2+(-3), -1+(-2)> = <-1,-3>
u=<2, -1>
v=<-3, -2>
u + v =<-1,-3>
b. u – v = <2 – (-3), -1 – (-2)> = <5, 1>
-v=< 3, 2>
u–v=<5,1>

97. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
Algebra of Standard Vectors

98. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
Algebra of Standard Vectors

99. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
= <-20, 5> + <-9, 18>
= <-29, 23>
Algebra of Standard Vectors

100. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
= <-20, 5> + <-9, 18>
= <-29, 23>
Algebra of Standard Vectors
v
u

101. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
= <-20, 5> + <-9, 18>
= <-29, 23>
Algebra of Standard Vectors
v
u
5u

102. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
= <-20, 5> + <-9, 18>
= <-29, 23>
Algebra of Standard Vectors
v
u
5u -3v

103. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
= <-20, 5> + <-9, 18>
= <-29, 23>
Algebra of Standard Vectors
v
u
5u -3v
5u – 3v

104. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
= <-20, 5> + <-9, 18>
= <-29, 23>
b. |5u – 3v|
= (-29)2 + (23)2
Algebra of Standard Vectors
v
u
5u -3v
5u – 3v

105. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
= <-20, 5> + <-9, 18>
= <-29, 23>
b. |5u – 3v|
= (-29)2 + (23)2
= 1370  37.0
Algebra of Standard Vectors
v
u
5u -3v
5u – 3v

106. Example F. Let u= <-4, 1>, v= <3, -6>, find
a. 5u – 3v
= 5<-4, 1> – 3<3, -6>
= <-20, 5> + <-9, 18>
= <-29, 23>
b. |5u – 3v|
= (-29)2 + (23)2
= 1370  37.0
Algebra of Standard Vectors
v
u
5u -3v
5u – 3v
Your Turn
Given the u and v above, find 3u – 5v and |3u – 5v|
Ans: 3u – 5v = <-27, 33> and |3u – 5v|  42.6

107. Let i = <1, 0>, j = <0, 1>,
then any vector <a, b> may
be written as
a<1, 0>+b<0, 1>= ai + bj
i = <1, 0>
j = <0, 1>
Example G. Compute
using the i and j notation.
<3, -4> – <1, -5>
= (3i – 4j) – (i – 5j)
= 3i – 4j – i + 5j
= 2i + j = <2, 1>
Algebra of Standard Vectors

108. Vectors in a Coordinate System
Ex. A. Given the following vectors (forces).
Draw the following vector combinations.
A
F
D
C
B
E
1. 3A
2. ½ D
3. –½ F
4. –2B
5. A + C
6. B – D
7. F – D
8. E + C

109. Ex. B. Given the following two vectors (forces) u and v,
and the angle A between than, find l u + v l
(the resultant). Draw.
Vectors in a Coordinate System
1. l u l = 60, l v l = 40 with A = 40°
3. l u l = 1500, l v l = 2000 with A = 140°
5. l u l = 450, l v l = 250 with A = 85 °
7. l u l = 136, l v l = 119 with A = 114°
2. l u l = 260, l v l = 48 with A = 50°
4. l u l = 620, l v l = 140 with A = 54°
6. l u l = 6.8, l v l = 40.2 with A = 4°
8. l u l = a, l v l = b with 180°

110. Ex. C. Given the following points
A = (3, –2), B = (4, –3), C = (–2, 1), and D = (–3, 5)
Find the following vectors in the standard form. Draw.
Vectors in a Coordinate System
1. AB, BC and CD 2. AC, BD and CA
3. DA, DB and CA 4. AD, BA and DC
Ex. D. Given the vectors u = <4, –3>, v = <–2, 1>,
and w = <–3, 5> in the standard form,
find the following vectors and then magnitudes.
1. u + v, 2. u – w 3. w + v 4. u – v 5. 2u + 3v
7. 3w – 2v 8. 2u + 3v 9. –3w – v 10. 2u + 3v
11. w – 2v + w 12. u – v + 2w 13. 2w – v + 3u
14. 2u – 2v + w 15. 2(w – v) + u
16. u – 2(v + w) 17. –2(w – v) + 3u

111. Vectors in a Coordinate System
1. l u + v l = 94.2 3. l u + v l = 24.4
Ans. B
5. l u + v l = 530 7. l u + v l = 139
1. AB = <1, –1>, BC = <–6, 4> and CD = <–1, 4>
3. DA = <6, –7>, DB = <7, –8> and CA = <5, –3>
C
D.
1. u + v = <2, –2> 3. w + v = <–5, 6>
5. 2u + 3v = <2, –3> 7. 3w – 2v = <–5, 13>
9. –3w – v = <11, –16> 11. w – 2v + w = <–2, 8>
13. 2w – v + 3u = <8, 0>
15. 2(w – v) + u = <2, 5>
17. –2(w – v) + 3u = <14, –17>