6.3 vector in the plane

23
6.3 Vector in the Plane Magnitude Component form Unit Vector

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6.3 Vector in the Plane. Magnitude Component form Unit Vector. Vector is a Directed Line Segment. Terminal point Initial point Magnitude ( or Length): || PQ ||. Let P = (0,0) and Q = (3,4). To find the Magnitude || PQ || Direction (slope) is always important. Slope of . - PowerPoint PPT Presentation

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Page 1: 6.3 Vector in the Plane

6.3 Vector in the Plane

MagnitudeComponent form

Unit Vector

Page 2: 6.3 Vector in the Plane

Vector is a Directed Line Segment

Terminal point

Initial point

Magnitude ( or Length): || PQ ||

Q

P

Page 3: 6.3 Vector in the Plane

Let P = (0,0) and Q = (3,4)

To find the Magnitude || PQ ||

Direction (slope) is always important.Slope of

525169

0403 22

PQ

PQ

34

0304

PQ

Page 4: 6.3 Vector in the Plane

Vectors equality

If two vectors equal if they have the same magnitude and direction.

P

Q

R

VRVPQ

34

PQandRVofSlope

Page 5: 6.3 Vector in the Plane

is a vector in standard position

Vectors in Standard position have an initial point at the origin (0, 0).

PQ

Page 6: 6.3 Vector in the Plane

Component Form of a vector

P = (p1, p2 ); Q = (q1, q2 )

Which can be labeled by just a letter.

2211 , pqpqPQ

2211 , pqpqV

Page 7: 6.3 Vector in the Plane

Vector “V” can renamed

If || V || = 1, then V is a Unit Vector. || V || = 0 iff V is 0

2211 , pqpqV

222

1

21,

vvV

vvV

Page 8: 6.3 Vector in the Plane

Find the Component form and Magnitude

Let URS

2,5

4,86)2(413)5(8

2

1

uu

Page 9: 6.3 Vector in the Plane

Find the Component form and Magnitude

Let URS

2,5

4,86)2(413)5(8

2

1

uu

6,13U

Page 10: 6.3 Vector in the Plane

Find the Component form and Magnitude

Let URS

2,5

4,86)2(413)5(8

2

1

uu

6,13U

3.14205

613 22

U

U

Page 11: 6.3 Vector in the Plane

Vector Operations

Scalar Multiplication

Let

21, uKuKUK

30,20

65,45

56,4

KU

UK

KandU

Page 12: 6.3 Vector in the Plane

Vector Operations

Vector Addition

Let

2211 , vuvuVU

2,6

46,24

4,2

6,4

VU

V

U

Page 13: 6.3 Vector in the Plane

Parallelogram Law used in Addition of VectorsGraph the Vectors move the tail of one vector tothe head of the other vector.

6,4

4,2

Page 14: 6.3 Vector in the Plane

Parallelogram Law used in Addition of VectorsGraph the Vectors move the tailof one vectorto the head of the other vector.

6,4

4,2

2,6

Page 15: 6.3 Vector in the Plane

Properties of Vectors

U + V = V + U (Comm.)(U + V) + W = U + (V + W) (Asso.)U + 0 = U (Identity)U + (-U) = 0(Inverse)C(DU)=(CD)U (Comm.)(C + D)U = CU + DU (Dist.)1(U)=U; 0(U)=0|| cV|| =|c| x ||V||

Page 16: 6.3 Vector in the Plane

How to Find the Unit Vector

Let

657,

6547,4

651

74

7,4||||

7,4

22

vv

v

Page 17: 6.3 Vector in the Plane

Standard Unit Vector

Writing the Unit Vector as Standard Unit Vector.

i =j =

0,1

1,0

657,

654

||||

7,4

vvu

v

0,0 1

1

i

j

jiu657

654

Page 18: 6.3 Vector in the Plane

Direction Angle of a Unit Vector

What is the coordinate of the intersection of the vector and

unit circle?

Page 19: 6.3 Vector in the Plane

Direction Angle of a Unit Vector

What is the slope of the vector?What functionIs rise over run?

sin,cos

ab

bjai

Page 20: 6.3 Vector in the Plane

Direction Angle of a Unit Vector

What is the slope of the vector?What functionIs rise over run?

sin,cos

ab

bjai

ab

cossintan

ji sincos

Page 21: 6.3 Vector in the Plane

Direction of a Vector can be found if it is not a Unit Vector

jvivvv sincossin,cos

ab

vv

cossin

tan

Page 22: 6.3 Vector in the Plane

Homework

Page 436 – 437# 1, 7, 15, 19,

25, 31, 37, 43, 49, 55, 61, 67, 73

Page 23: 6.3 Vector in the Plane

Homework

Page 436 – 437# 32, 38, 54, 62,

70, 80