v

Initial point

terminal point

A(x1 , y1)

B(x2 , y2)

V=AB, vector from the point A to the point B,

is a directed line segment between A and B.

v

A(x1 , y1)

B(x2 , y2)

MAGNITUDE OF VECTOR

║V║ = (x2−x1)2 + (y2−y1)

2

U

UNIT VECTOR

║U║ = 1

MAGNITUDE OF VECTOR IS 1

EXAMPLES

1- Vector v has the initial point A(3, 4) and the terminal point B( -2, 5).

Magnitude of vector v is (-2-3)2 + (5-4)2 = (-5)2 + 12 = 25+1 = 26

║V║= 26

* V is not a unit vector

2- Vector i has the initial point (0, 0) and the terminal point(1, 0).

Magnitude of the vector i is

= (1-0)2 + (0-0)2 = 12 + 02 = 1+ 0 = 1

3- Vector j has the initial point (0, 0) and the terminal point (0 ,1).

Magnitude of the vector j is

= (0-0)2 + (1-0)2 = 12 + 02 = 1+ 0 = 1

* i is a unit vector

* j is a unit vector

║ i ║

║j║

SCALAR MULTIPLICATION OF VECTOR

V

W−W

2W

0.5V

−2W

k

4k

VECTOR ADDITION

V

W

W+V

TRIANGLE METHOD

VECTOR ADDITION

V

W

W+V

PARALLELOGRAM METHOD

V

W

W+V

-W

V-W

VECTOR ADDITION

-V

W-V

VECTORS IN A COORDINATE PLANE

x

y

A(x1 , y1)

B(x2 , y2)

C(x2−x1 , y2−y1)

O

W=AB V

Vectors W and V are equivalent vectors. Since V starts from the origin we use a special notation for V

V = x2−x1 , y2−y1

VECTORS IN A COORDINATE PLANE

x

y

A(5 , 1)

B(1 , 4)C(−4 , 3)

O

W = AB

V

Vectors W and V are equivalent vectors.

V = −4 , 3

*−4 is the x-component ,and 3 is the y-component of the vector V

3

4

1

5-4 1

VECTORS IN A COORDINATE PLANE

x

y

A(1 , 4)

B(-3 , 1)

C(-4 , -3)

W = AB

V

Vectors W and V are

equivalent vectors. V = -4 , -3

- 3

4

1

- 3-4

1

VECTORS IN A COORDINATE PLANE

x

y

i

Vectors i and j are special unit vectors.

j = 0 , 1 1

1

j

i = 1 , 0

VECTORS IN A COORDINATE PLANE

Find a vector that has the initial point (3 , -1) and is equivalent to V = -2 , 3 .

x

A(3 , -1)

P(-2 , 3)

B( 1 ,2)

W = AB

Vectors W and V are equivalent vectors.

V

- 1

3

1- 2

2

3W

If ( x, y) is the terminal point of W, then

x−3 = −2 → x = 1 and

y−(−1) = 3 → y = 2

y

BASIC VECTOR OPERATIONS

V = a , b and W = c , d are two vectors and k is a real number.

1- ║V║ = a2 + b2

2- v+w = a , b + c , d = a+c , b+d

3- kV =k a , b = ka , kb

4- ║kV║ = k ║V║

BASIC VECTOR OPERATIONS

5V = 5 -2 , 3 = -10 , 15

║v║ = (-2)2 + 32 = 4 + 9 = 13

V +W = -2 , 3 + 4 , -1 = -2+4, 3-1 = 2 , 2

-3W = -3 4 , -1 = -12 , 3

V = -2 , 3 , W = 4 , -1

5V −3W = -10 , 15 + -12 , 3 = -10-12 , 15+3 = -22 , 18

ANY VECTOR CAN BE WRITTEN IN TERMS OF THE UNIT VECTORS i AND j

If V = a , b is any vector, then by using basic vector operations we get ;

V = a , b = a , 0 + 0 , b

= a 1 , 0 + b 0 , 1 = ai+ bj

V = a , b = ai + bj

i

j

4i

3j

4i+3j

P(4 , 3)

x

y

V = 4i+3j = 4 , 3

VECTORS WRITTEN IN TERMS OF THE UNIT VECTORS i AND j

VECTORS WRITTEN IN TERMS OF THE UNIT VECTORS i AND j

i j

4i

-3j

4i-3j

P(4 , -3)

x

y

V = 4i-3j = 4 , -3

DIRECTION ANGLE OF VECTORS

x

y

α direction angle of V β

V

W

V = x , y = V cosα , sinα =

P(x , y)

V cosα , V sinα

Q(a , b)

W = a , b = W cosβ , sinβ = W cosβ , W sinβ

Vcosα =

x

sinα = V

y

x

ytan

DIRECTION ANGLE OF VECTORS

If V = -2i + 3j , then find the direction angle of V.

tanα = = = −yx

3-2

V = -2i+3j = -2 , 3

α = tan-1[− ] in the second quadrant

32

32

DIRECTION ANGLE OF VECTORS

If V = , − , then find the direction angle of V.

tanα = = − = −yx

α = tan-1[− ] = −

3

12

32

3

123

2

3π

DIRECTION ANGLE OF VECTORS

If ║V║ = 6 and the direction angle of V i s , then

find the x and y-components of V. 67π

V = x , y =

V cosα , V sinα

V = = 6cos , 6sin6

7π6

7π−6 , −6

23 1

2

−3 3 , −3V=

UNIT VECTORS ON THE SAME DIRECTION WITH A GIVEN VECTOR

If V = x , y , then

U = , = V

x

V

y

is the unit vector on the same direction with V.

V = x , y = V

cosθ , sinθ

U

θ is the direction angle of V

UNIT VECTORS ON THE SAME DIRECTION WITH A GIVEN VECTOR

If V = -3 , 4 , then find the unit vector on the same direction.

U = , = V

x

V

y

v = (-3)2 + 42 = 9 + 16 = 25 = 5

5-3

54,

Now find the vector W on the same direction with magnitude 6.

W = 6U = 6 =

See the illustrations on the next slide

5-3

54,

5-18

524,

x

y

P(-3 , 4)

O

U

V

4

-3

W

V = 5U

W = 6U

X = -3U

X

║W║= 6

║V║ = 5

║X║=3

If V = −2i + 4j , then find the vector on the opposite direction

with magnitude 6.

First, find the unit vector on the direction of V.

U = = ,

Now, multiply the unit vector with −6. That will give you the answer.

W = −6U

The vector on the opposite direction with magnitude 6.

V

x

V

y −220

420

EXAMPLE

DOT PRODUCT OF VECTORS

V = a , b and W = c , d

V∙W = ac + bd

V∙V = a2 + b2 = ║V║2

DOT PRODUCT OF VECTORS

1− V∙W = W∙V dot product is commutative 2− U∙(V + W) = U∙V + U∙W distributive 3− a (V∙W) = (aV )∙W=V∙(aW) ,a is a scalar 4− V∙V = ║V║2

5− 0∙W = 0 zero vector

6− i∙i = j∙j = 1

7− i∙j = j∙i = 0

DOT PRODUCT OF VECTORS

║U+V║2 = (U+V).(U+V) = U.U+U.V+V.U+V.V

║U+V║2 = ║U║2 + 2U.V + ║V║2

SIMILARLY

║U−V║2 = (U−V).(U−V) = U.U−U.V−V.U+V.V

║U−V║2 = ║U║2 − 2U.V + ║V║2

ANGLE BETWEEN TWO VECTORS

VW

θ angle between V and W θ

V∙W = ║V║║W║cos θ

cos θ = __________║V║║W║

V∙W

EXAMPLE

V = −2 , 3 , W = 4 , − 1

V∙W = ac + bd = −2.4 + 3.(− 1) = − 8 − 3 = −11

V∙V = a2 + b2 = (−2)2 + 32 = 13 = ║V║2

cos θ = __________ =

_______║V║║W║

V∙W −1113 17

EXAMPLE

V = 3 , −6 , W = −1 , 2

V∙W = 3. (−1) − 6.2 = − 3−12 = −15

║V║= 45 , ║W║= 5

cos θ = __________ =

_______ =

______ =

_____ = −1║V║║W║

V∙W −1545 5

−15225

−1515

cos θ = −1, then θ = cos-1(−1) = π

EXAMPLE

V

W

If V∙W = 0 ,then V and W are perpendicular

θ is 90◦ , cos θ = 0

θ

EXAMPLE

If V = 4i-3j, then find a vector that is perpendicular to V

V∙W = 4x – 3y = 0

If W = xi + yj, then

Any choice of x and y that satisfies the equation above is an answer

Since 3 and 4 satisfy the equation

W = 3, 4 is one of the vectors

V

W

V and W are parallel

V

W

θ is 0◦, cos θ = 1

θ is 180◦, cos θ = -1

Same direction

Opposite direction

EXAMPLE

V = 3 , −6 , W = −1 , 2

cos θ = __________ =

_______ =

______ =

_____ = −1║V║║W║

V∙W −1545 5

−15225

−1515

cos θ = −1, then θ = cos-1(−1) = π

V and W are parallel with opposite direction

EXAMPLE

V = 3 , −6 , W = 1 , −2

cos θ = __________ =

_______ =

______ =

___ = 1║V║║W║

V∙W 1545 5

15225

15 15

cos θ = 1, then θ = cos-1(1) = 0

V and W are parallel with same direction

EXAMPLE

V = 3 , −6 , W = 4 , 2

cos θ = __________ =

_______ = 0║V║║W║

V∙W 045 20

cos θ = 0 , then θ = cos-1(0) = 90◦

V and W are perpendicular ( orthogonal ) vectors

EXAMPLE

If ║U + V║ = 7 , U is a unit vector and cos θ = ____ where θ is the

angle between U and V, then find the magnitude of V.

12

║U+V║2 = ║U║2 + 2U.V + ║V║2 = 7

1 + 2U.V + ║V║2 = 7 , ║V║2 + 2U.V − 6 = 0

cos θ = __________ =

___ , 2U.V = ║V║║U║║V║

U∙V 1 2

║V║2 + 2║V║ − 6 = 0 (║V║−2)(║V║+3) = 0

║V║= 2 or ║V║= −3, ║V║≥ 0 so ║V║= 2

SCALAR PROJECTION

VW

θ

projW

V = ║V║cos θ = _______

║W║

V∙W

EXAMPLE

If V = 2i + 2j , and W = −4i−2j , then find projW

V

projW

V = _______ = _____ = ______ = ____

║W║V∙W -12

20 -12 2 5

-6 5

EXAMPLE

If V = kW for any nonzero number k , then

projW

V = _______ = _________ = _________ = ________

║W║V∙W (kW)∙W

║W║k(W∙W)

║W║k║W║

2

║W║

projW

V = k║W║

How about projV

W ?.