Vectors in Two Dimensions

VECTOR REPRESENTATIONA vector represents those physical quantities such as velocity that have both a magnitude and a direction.Vectors can be represented as arrows whose length is proportional to the vector's magnitude and whose direction is the vector's direction.Manipulation of vector's can be greatly simplified when they are displayed on a Cartesian Coordinate System. This allows the vector to be manipulated in both Cartesian and Polar coordinates.One of the major simplifications comes from representing a vector in terms of its components along the x and y axes. The simplification comes from the fact that these components are scalar quantities that may be manipulated according to the rules of algebra. Of course we often want to take the components of a vector and represent the vector itself.Vectors may be added, subtracted, multiplied by a scalar, and multiplied by other vectors.

When representing a vector on a Cartesian Coordinate System, the x and y coordinates of the point of the arrow will be equal to the vector's x and y components. The direction of the vector will be the angle measured counterclockwise from the +x axes.

+x-x

+y

-y

origin

r V

V

θVx =V cosθ

VY =V sinθ

Vector Magnitude

Vector Direction

x-component

y-component

+x-x

+y

-y

Quadrant I0° < θ < 90°

V X > 0, V Y > 0

Quadrant III180° < θ < 270°V X < 0, V Y < 0

Quadrant IV270° < θ < 360°V X > 0, V Y < 0

Quadrant II90° < θ <180°V X < 0, V Y > 0

Determining a Vector from its Components

VY

VX

V

θ

V =V = V X2 + V Y

2

θ =tan−1 VYVX

⎛ ⎝ ⎜ ⎞

⎠ ⎟

r V

r V =V @ θ

For Vectors in Quadrants II, III, IV III

III IV

V =V = V X2 + V Y

2

θ =tan−1 VYVX

⎛ ⎝ ⎜ ⎞

⎠ ⎟

V =V = V X2 + V Y

2

V =V = V X2 + V Y

2 V =V = V X2 + V Y

2

θ =180° − tan−1 Vy

Vx

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

θ =180° + tan−1 Vy

Vx

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ θ =360° − tan−1 Vy

Vx

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Practice1. Find the components of the vector V= 15@130°

15130°

Vx =V cosθ

Vx =15⋅cos130°

Vx =−9.64

Vx

Vy =V sinθ

Vy =15⋅sin130°

Vy =11.5

Vy

2. Find the vector with the following x and y components:Vy =−12.8Vx =−8.1

Vx =−8.1

V y=−12.8V

θ

V = V x2 +V y

2

V = −8.1( )2 + −12.8( )2

V =15.1

θ =180° + tan−1 Vy

Vx

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

θ =180° + tan−1 − 12.8− 8.1

⎛ ⎝ ⎜ ⎞

⎠ ⎟

θ =180° + tan−1 1.58( )θ =180° + 57.7°

θ =237.7°

r V =15.1@ 237.7°

Operations with

Vectors

Adding and Subtracting VectorsThe sum of or difference between two or more vectors is a new vector.

r R =

r A +

r B +

r C

r R =R @ θ

R = R x2 +R y

2

θ

tan−1R yR x

⎛ ⎝ ⎜

⎞ ⎠ ⎟L Q : I

180 −tan−1R yR x

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟L Q : II

180 + tan−1R yR x

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟L Q : III

360 −tan−1R yR x

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟L Q : IV

⎧

⎨

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎧

⎨

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

Where :Rx =Ax +Bx +Cx

R y =Ay +By +Cy

Multiplying a Vector by a ScalarThe product of a vector and a positive scalar is a new vector in the same direction as the original vector but with a new magnitude.

r V =V @ θ

r R =s

r V

r R =sV@ θK s> 0

Rx =sVxR y =sVy

The product of a vector and a negative scalar is a new vector in the opposite direction as the original vector and with a new magnitude.

r R =sV@

θ +180°K θ <180°θ −180°K θ >180°

⎫ ⎬ ⎭s< 0

⎧ ⎨ ⎩

Multiplying Two VectorsThere are two ways of multiplying one vector by another.

1. Scalar (dot) product…yields a scalar. r A ⋅

r B =s

r A ⋅

r B =AxBx +AyBy

r A

r B

θ

r A ⋅

r B =ABcosθ

1. The scalar product is commutative : r A ⋅

r B =

r B ⋅

r A

2. If two vectors are pependicular(normal) their scalar product is zero (0) since θ = 90° and cos 90° = 0.

θ is the smaller of the two angles

between r A and

r B .

2. Vector (cross) product…yields a vector. r A ×

r B =

r C

r C =C=ABsinθ

The magnitude of the cross-product vector is given by:

The direction of the cross-product vector C is perpendicular to the plane generated by the vectors A and B and is given by the “right-hand-rule:

Point the fingers of the right hand in the direction of vector A so they could be curled in the direction of vector B. The thumb of the right hand will point in the direction of vector C.

θ is the smaller of the two angles

between r A and

r B .

r A

r B

θ

r C =

r A ×

r B

The vector product is not commutative :r A ×

r B =−

r B ×

r A

If two vectors are parallel their vector product is zero (0) since θ = 0° and sin 0° =0.

θ

r ′ C =

r B ×

r A

r B

r A

PracticeGiven the following two vectors:

Find:

r A +

r B

r A −

r B

r A ⋅

r B

r A ×

r B

r A =15@ 30°

Ax =15cos30° =13

Ay =15sin30° =7.5

r B =20@ 245°

Bx =20cos245° =−8.5

By =20sin245° =−18.1

r A +

r B =

r C

Cx =Ax +Bx =13 + −8.5( )=4.5

Cy =Ay+By =7.5 + −18.1( )=−10.6

r C is in Quadrant IV

θ =360° − tan−1 10.64.5( ) = 293°

r C =11.5@293°

C = Cx2 +Cy

2 = 4.5( )2 + −10.6( )2 =11.5

θ = 360° − tan−1 Cy

Cx

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

r B

245°

20

r A

30°15

r A +

r B =

r C

293°

11.5

r A +

r B =

r C

Geometrical Interpretation

r A −

r B =

r D

Dx =Ax −Bx =13− −8.5( )=21.5

Dy =Ay−By =7.5− −18.1( )=25.6

r D is in Quadrant I

θ =tan−1 25.621.5( ) = 50°

θ = tan−1 DyDx

⎛ ⎝ ⎜

⎞ ⎠ ⎟

r D = 33.4@50°

D = D x2 +D y

2 = 21.5( )2 + 25.6( )2 =33.4

r A −

r B =

r A + −1( )

r B

−1( )

r B = 20@(245° − 180°)

−1( )r B = 20@65°

r B

20

65°

r A

30°15

r A −

r B =

r D 33.4

50°

Geometrical Interpretation

r A ⋅

r B =s

s =AxBx +AyBy

s = 13( ) −8.5( )+ 7.5( ) −18.1( )

s =−246.25

r A

30°

15

r B

245°

20s =ABcosθ

s =−245.7

245°-30°=215°

360°-215°=145°s =(15)(20)cos(145)

What is the smallest angle between A and B?

r A ×

r B =

r E

E =ABsinθ

E =(15)(20)sin(145)

E =172.1

r A

r B

r E points "into" the

r A

r B plane.

r E

r A

r B

r B ×

r A =

r F

r F points "out of" the

r A

r B plane.

r F

F =ABsinθF=(15)(20)sin(145)F=172.1